Sslc Mathematics Workbook Old

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10TH STANDARD

MATHEMATICS

WORK BOOK

WITHPREVIOUS QUESTION PAPERS


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Guidance : Sri S.K.B. Prasad,Educational Officer, DDPI Office, ChitradurgaPh: 94486 94642

Concept: Sri K.P. Obaiah,Subject Inspector (Mathematics), DDPI Office, ChitradurgaPh: 94495 12518

Resource Persons :

(1) Sri K.S. Thippeswamy,Girisha Jr. College, HiriyurPh: 98866 07469

(2) Sri B.V. Guruprakash,Govt. Girls Jr. College, ChitradurgaPh: 99643 48058

(3) Sri M.J. Rudramuni,B.M.G.H.S., ChallakerePh: 98458 02338

(4) Sri B. Shivakumar,Govt. High School, Thimmappayyanahalli, Challakere Tq.Ph: 98459 05534

Over all Supervision :

Sri H. ManjunathDeputy Director for Public InstructionChitradurga

Sri N M RameshEducation OfficerDeputy Director for Public InstructionChitradurga

For copies :Contact the Resource Persons or Phone Numbers given above


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S.No. CHAPTERS PAGE No.

1. NUMBER SYSTEM 1 - 80

2. PERMUTATIONS AND COMBINATIONS 81 - 104

3. STATISTICS 105 - 122

4. FACTORS AND FACTORISATION 123 - 180

5. QUADRATIC EQUATIONS 181 - 222

6. MODULAR ARITHMETIC 223 - 232

7. PRACTICAL GEOMETRY 233 - 244

8. THEOREMS ON TRIANGLES AND CIRCLES 245 - 290

9. MENSURATION 291 - 316

10. POLYHEDRA AND NETWORKS 317 - 335

CONTENTS


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A B

SET THEORY• Set: A Set is a collection of well defined and well distinguished objects.

• Elements: The objects which constitute the set are said to be elements of the set.

• Representation of Set:

(a) Roster / Tabulation Method: In this method the set is represented by listing all itselements by commas and enclosing them in flower brackets.

Ex: A = {1, 2, 3, ................}

(b) Rule Method: Defining the set is called rule method.

P={Set of all natural numbers less than 6}

• Finite Set: A set is said to be finite if it has finite number of elements.

• Infinite Set: A set is said to be infinite if it has an infinite number of elemements.

• Order of a Set: Is the number of elements it contains.

• Empty Set: A set having no elements is said to be an empty set. It is also called null set or void set. It is denoted by {} or Φ. Order of null set A is n(A) = 0.

• Singleton Set: A set having only one element is said to be a singleton set.

• Subset: Let A and B be two sets them the set A is said to be a subset of the set B if each element of A is also an elements of B symbolically we write it as A ⊆ B.

• Equal Sets: Two sets are said to be equal if A ⊂ B and B ⊂ A. Two sets A and B are said to be equal if they have exactly same elements.

• Equivalent Sets: Two sets are said to be equivalent if they have same number of elements.

• Universal Sets: The main set under discussion or the set containing all possible values in the given frame of reference is said to be universal set and is denoted by U. It is the Union of all the sets.

• Operation on Sets: (a) Union of Sets: Let A and B be two sets. The Union of A and B is the set of al lthose elements which are either in A or in B or in Both. It is denoted symbolically by A ∪ B = {x : x A or x B}

NUMBER SYSTEM1

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(b) Intersection of Sets: Let A and B be two sets. Then intersection of A and B is the set of all those elements which are in both A and B. If it denoted symbolically by A ∩ B A ∩ B = {x : x A and x B}

• Disjoint Sets: Two sets are said to be disjoint if and only if they have no common elements.

A ∩ B = Φ n (A ∩ B) = 0

• Difference of Sets: Let A and B be two sets then (A - B) is the set of those elements of the Set A which are not in the set B

(A - B) = {x : x A and x B} (B - A) = {x : x B and x A}A}

(A - B) (B - A)

• Complement of Sets: Let U be the Universal set and A be any set then the complement of the set A is the set of all those elements of U which are not in the set A. This is denoted by A1 or (U - A).

A1 = {x : x U and x A}A}

A B

A B A B

U A

A B

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Fundamental Resluts

(1) Identity law : A ∪ Φ = AA ∩ Φ = Φ

(2) Commutative Law : i) (A ∪ B) = (Β ∪ Α)

ii) (A ∩ B) = (B ∩ A)

(3) Associative Law : i) A ∪ (B ∪ C) = (A ∪ B) ∪ C ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C

(4) Distributive Property : i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(5) DeMorgan’s Law : i) (A ∪ B)1 = A1 ∩ B1

ii) (A ∩ B)1 = A1 ∪ B1

(6) If A1 = Φ then A = U

(7) (A1)1 = A

(8) A ⊆ B then A1 ⊆ B1

(9) A ∪ A1 = U and A ∩ A1 = Φ

(10) If A ∩ B = Φ then A1 ∪ B1 = U

(11) A - (A - B) = A ∩ B

(12) If (A - B) = A then (A ∩ B) = Φ

(13) If B is subset of A then A ∩ B = B and A ∪ B = A

(14) If A, B, C are finite sets then

(a) n(A ∪ B) = n(A) + n(B) (If A and B are disjoint)

(b) n(A ∪ B) + n(A ∩ B) = n(A) + n(B)

(c) n(A ∩ B1) = n(A) - n(A - B)

(d) n(A ∩ A1) = n(B) - n(A ∩ B)

(e) n(A ∪ B) = n(A ∩ B1) + n(B ∩ A1) + n(A ∩ B)

(f) n(A − B) + n(A ∩ B) + n(B − A) = n(A ∪ B)

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I Fill the box with correct answer

II Write the given sets in Roster method.

(1) A = {x : x N and 1 ≤ x ≤ 5}A = { 1, 2, 3, 4, 5}

(2) B = {x : x is a perfect square less than 10}

(3) C = {x : x is a multiples of 3 less than 10}

(4) D = {x : x Z and -2 ≤ x ≤ 2}

(5) E = {x : x is even digit less than 10}

(6) F = {x : x is an odd numbers less than 10}

(7) G = {x : x is a prime numbers less than 15}

{x / x N & 1 ≤ x ≤ 5 }

A B A B A B

1 {a, b, c, d} {b, d, e, f} {a, b, c, d, e, f} {b, d}

2 {1, 2, 3, 4, 5} {1, 3, 5, 8, 9}

3 {c, a, t} {r, a, t}

4 {0, 1, 2}

5 {1, 3, 5, 7} {2, 4, 6, 8}

∪ ∩

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III Match the Following. (for Practice purpose)

(a) 1. De Morgan’s Law (a) (A∪B) = (B∪A) 1. ...................

2. Commutative Property (b) A∪(B∩C)=(A∪B)∩(A∪C) 2. ...................

3. Associative Property (c) (A∪B)1 = A1∩B1 3. ...................

4. Compliment of A (d) A∩(B∪C)=(A∩B)∪(A∩C) 4. ...................

5. Union is distributive over

intersection (e) U - A = A1 5. ...................

6. Intersection is Distributive

over Union (f) (A∪B)∪C=A∪(B∪C) 6. ...................

(b) 1. A ∪ B (a) Φ 1. .................................

2. A ∩ (B ∩ C) (b) U 2. .................................

3. A ∪ (B ∪ C) (c) B ∪ Α 3. .................................

4. (A ∩ B)1 (d) (A ∩ B) ∩ C 4. .................................

5. (A ∪ B)1 (e) A1 ∪ B1 5. .................................

6. (A1)1 (f) n(A∪B) + n(A∩B) 6. .................................

7. n(A) + n(B) (g) A 7. ..................................

8. A ∪ A1 (h) (A∪B)∪C 8. ..................................

9. A ∩ A1 (i) A1 ∩ B1 9. ..................................

IV Write the shaded portion represents in the given figure.

(1) (2) (3)

(A - B)

(4) (5) (6)

................................ ................................ ................................

................................ ................................ ................................

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(7) (8) (9)

V Draw a Venn Diagram for the follwoing sets and find A∪B and A∩B.

(1) If P = {x : x is an odd number less than 11} and Q = {x : x is prime number less than 15}

P = {1, 3, 5, 7, 9}Q = {2, 3, 5, 7, 11, 13}P ∪ Q = {1, 2, 3, 5, 7, 9, 11, 13}P ∩ Q = {3, 5, 7}

(2) If A = {0, 2, 4, 6, 8} and B = {x : x is even digit less than 5}

A =B =A-B =B-A =

................................ ................................ ................................

A B

C

A B

C

A B

C

P Q

357

21113

19

(3) If M = {x : x N and 5 < x < 15} and N = {odd natural numbers less than 10}

M =N =

(4) n(A) = 65, n(A∩B) = 25 and n(A∪B) = 75

n(B) =

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(5) In a school 60% of students invest in NDF and 70% of them invest in NSS.

n(NDF) =n(NSS) =n(NDF ∪ NSS) =n(NDF ∩ NSS) =

VI See the Venn’s diagram and write the following.

(a) U = {..........................................} (b) A = {...........................................}

(c) B = {..........................................} (d) A∪B = {......................................}

(e) A∩B = {......................................} (f) A-B = {.......................................}

(g) B-A={........................................} (h) B1 = {..........................................}

(i) A1 = {.........................................} (j) (A∪B)1 = {......................................}

(k) (A∩B)1 = {......................................}

A B

15 0

34

28

U

6 9

VII If U ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 7, 9} and B = {2, 4, 6, 7, 9} writedown the following.

A1 =

B1 =

A - B =

B - A =

A ∪ B =

A ∩ B =

(A ∪ B)1 =

(A ∩ B)1 =

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VIII Fill up the blanks with suitable answer.

(1) If n(A - B) = 5, n(B - A) = 4, n(A ∪ B) = 15 the n (A ∩ B) is .................................

(2) In the Venn Diagram unshaded portion represents ................................

(3) If A and B are two sets, (A ∩ B) = Φ then A and B are ..................................

(4) If A and B are sets, such that n(A) = 76, n(B) = 24, n(A ∩ B)=10 then n(UB) is

.....................................

(5) A, B and C are the sets. Intersection of sets is distributive over Union of Sets is

represented as ....................................

(6) If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (A ∪ B)1 = {1, 5, 7} then (A ∪ B) is ..................

(7) If A is subset of B and B is subset of A. Then the relation between A and B is ........

...................................

(8) If A and B are the subsets of U and A1 ∪ B1 = {2, 3, 5,} U = {1, 2, 3, 4, 5, 6} then

A ∩ B is ........................................

A

(9) A and B are the sets. If A-B = A then A ∩ B = ...................................

(10) If B ⊂ A then the value of A ∩ B is ........................................

(11) The value of A ∩ A is ..........................................

(12) Compliment of the Universal Set is ........................................

(13) In an Hotel 30 people speak only Hindi, 20 speak only English, 10 speak both

Hindi and English. The number of people in the Hotel is ....................................

(14) If P and Q are non empty sets and P - Q = P then P ∩ Q is ...................................

IX Choose the correct answer.

(1) In a distributive set P ∪ (Q ∩ R) = .................................(A) (P∪Q) ∪ (P∪R) (B) (P∪Q) ∩ (P∪R)(C) (P∩Q) ∪ (P∩R) (D) (P∩Q) ∩ (P∩R)

(M - 10)

(J - 06)

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(2) In a 1000 students 750 pupil plays cricket, 350 pupil plays voly ball and 150 plays

both number of pupils do not play any game is .................................(A) 200 (B) 500 (C) 250 (D) 50

(3) A = {a, b, c, d, e} B = {a, c, e, g, h} C = {c, f, g} then A∩(B∪C) is......................(A) {a,c,e} (B) {g,e} (C) {b,d} (D) {c}

(4) Union of the sets is distributive over intersection this can be represented by............(A) X∪Y = Y∪X (B) (X∪Y)∪Z = X∪(Y∪Z)(C) X∪(Y∪Z) = (X∪Y)∩(X∪Z) (D) X∩(Y∩Z) = (X∩Y)∪(X∩Z)

(5) If A = {1, 2, 3, 4} and B = {3, 4, 5} then A - B is ...........................(A) {3,4} (B) {1,5} (C) {1,2} (D) {0,1,2}

(6) Set A is subset of U, n(U) = 11 and n(A1) = 8 then the number of elements in A is

..............................(A) 19 (B) 11 (C) 8 (D) 3

(7) Set A = {a,b,c,d} B = {b,c,e} then n (A∩B) is .........................(A) 4 (B) 3 (C) 7 (D) 2

(J - 06)

(J - 06)

(M - 06)

(M - 06)

(M - 06)

(8) U = {0, 1, 2, 3, 4} A = {2, 3, 4} B = {0, 2, 3} then (A∩B)1 is .........................(A) {0,1,2,3,4} (B) {0,1,4} (C) {1,4} (D) { }

(9) In a 9 traveller 5 can speak Kannada, 2 can speak both Kannada and English then

the number of Traveller speak only English is .........................(A) 5 (B) 3 (C) 4 (D) 6

(10) In the set (A∪B)∪C = A∪(B∪C) this means .........................(A) Union of sets is commutative (B) Union of the sets is associative(C) Union of the sets is distributive over intersection(D) Intersection of the sets is distributive over union

(11) Universal set U = {2,3,5,6,10} subset A = {5,6} then A1 represents the diagram

..............................(A) (B) (C) (D) U

A5 6

2 310

U A5 6

2 310

U A5 6

2 310

U A5 6

2 310

(M - 07)

(M - 07)

(M - 07)

(J - 07)

(J - 07)

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(12) In a class of 60 students 22 member play volyball, 12 members play both volyball

and kho kho, 17 members can not play any game. The number of students play

kho-kho only is .........................(A) 32 (B) 28 (C) 33 (D) 21

(13) Set A = {x N / 1≤ x ≤ 4} and B = {3,4,5} then set A - B is equal to ......................(A) {5} (B) {1,2} (C) {3,4,5} (D) {1,2,3,4}

(14) (A∪B)1 = {2,4,6} then which set represents A1∩B1 ......................(A) {1,2,3,4,5,6} (B) {2,4,6} (C) {1,3,5} (D) { }

(15) n(A) + n(B) = n(A∪B) then n(A∩B) is equal to ..........................(A) 0 (B) Φ (C) 1 (D) 2

(16) Set A = {2,3,4,5} and B =- {4,5} which of the following is a Null Set ...................(A) A - B (B) B - A (C) A∪B (D) A∩B

(J - 07)

(M - 08)

(M - 08)

(M - 08)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(J - 08)

(17) P, Q and R are three sets then (P∪Q) ∩ (P∪R) is ............................(A) P∪(Q∪R) (B) P∩(Q∩R) (C) P∪(Q∩R) (D) P∩(Q∪R)

(18) Set A and B are the subsets of U, A1∪B1 = {2, 3, 5} and U = {1, 2, 3, 4, 5, 6} then

A ∩ B=..........................

(A) {2,3,5} (B) {1,4} (C) {1,2,3,4,5,6} (D) {1,4,6}

(19) In a class of 50 students every one should be the members of Science club or

Maths club. 29 students are in science club, 11 students are the members in both

the clubs. The number of students only in Maths club is .........................

(A) 21 (B) 18 (C) 11 (D) 10

(20) In a sets A and B, A - B = A then A ∩ B=..........................

(A) A (B) B (C) U (D) Φ

(21) If A = {1,2,4} B = {1,4,5} then A ∩ B is ..........................

(A) {4,5} (B) {1,5} (C) {1,4} (D) {2,5}

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(22) If U = {1,2,3,4,5,6} and A = {1,2,3} B = {2,4} then which diagram represents

A ∩ B ............................

(A) (B)

(C) (D)

(23) According to given Venn diagram n(A) is ............................

(A) 8 (B) 5

(B) 7 (D) 13

U A B13

56

2 4

U A B13

56

2 4

U A B13

56

2 4

U A B13

56

2 4

A B

8 5 7

(J - 08)

(J - 08)

(J - 08)

(J - 08)

(J - 09)

(J - 09)

(J - 09)

(J - 10)

(J - 10)

(24) In a given statement which one is DeMorgan’s rule ............................(A) n(A)+n(B) = n(A∪B)+n(A∩B) (B) n(A1) = n(U) - n(A)(B) A∪(B∩C) = (A∪B)∩(A∪C) (D) (A∩B)1 = A1∪B1

(25) A and B are the sets n(A) = 11, n(B) = 7, n(A∩B) = 3 then n(A∪B) is ...................(A) 21 (B) 15 (C) 8 (D) 10

(26) U = {0,1,2,3,4,5,6,7,8,9} and A = {0,1,3,5,7} then A1 is ...................(A) {0,2,3,4,6,8,9} (B) {0,2,4,6,8} (C) {2,4,6,8} (D) {4,6,8,9}

(27) Set A and B are have no common elements then n(A∩B) is ...................(A) 0 (B) Φ (C) {0} (D) {Φ}

(28) A = {1,2,3} B = {0,1,3,4} and C = {2,3,4} are the sets then A∪(B∩C) is ..............(A) {0,1,2,3} (B) {0,1,3,4} (C) {1,2,3,4} (D) {2,3,4}

(29) If P = {2,3,4} and Q = {3,5,7} then (P-Q) is .........................(A) {3,7} (B) {2,4} (C) {2,3,4,7} (D) {3}

(30) If A ⊂ B, B ⊂ C and C ⊂ A then which is correct .............................(A) A = B (B) B = C (C) A = C (D) One of them is null set

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(31) 5 boys eat apple only. 4 boys eat banana only, the number of boys eat apple out of

12 boys is .............................

(A) 5 (B) 4 (C) 8 (D) 9

(32) If A∩B = Φ then A1∪B1 is .............................(A) Null Set (B) Universal Set(C) Disjoint Set (D) Compliment Set

(33) If n(A-B) = 6, n(B-A) = 5, n(A∩B) = 3 then n(A∪B) is .............................(A) 8 (B) 9 (C) 11 (D) 14

(34) A = {a,b,c,d,e} B = {a,c,e,g,h} and c = {c,f,g} then A∩(B∩C) is .....................(A) {a,c,e} (B) {g,e} (C) {b,d} (D) {c}

(35) In a survey of 5000 persons it was found that 2800 read Indian Express and 2300

read Times of India, while 400 read both. Then the number of people who read

neither Indian Express nor Times of India .....................(A) 200 (B) 300 (C) 600 (D) 900

(J - 10)

(J - 10)

X Solve the following.

(1) If A = {3,4,5,6} and B = {4,5,6,7,8} find A∪B and A∩B and draw Venn Diagram

(i) A∪B = { } ∪ { }

= { }

(ii) A∩B = { } ∩ { }

= { }

A BU

A B

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(2) If A = {0,2,4,6,8} and B = {x/x is on even digit less than 5} verify(i) A ∪ B= B ∪ A and (ii) A ∩ B = B ∩ A

(i) A∪B = { } ∪ { }

= { }

B∪A = { } ∪ { }

= { }

(ii) A∩B = { } ∩ { }

= { }

B∩A = { } ∩ { }

= { }

(3) If P ={x/x is an odd numbers less than 11} Q = {x/x is prime number less than 15}verify (i) P∩Q = Q∩P and (ii) P∪Q = Q∪P.

(4) If A = {x/x N and 5 < x < 10} B = {odd natural numbers less than 8} andC = {even natural numbers less than 9}. Verify Union of sets is associative.

A = { } B∪C = { }

B = { } A∪B = { }

C = { }

A∪(B∪C) = { } ∪ { }

= { }

(A∪B)∪C = { } ∪ { }

= { }

∴ A∪(B∪C) = (A∪B)∪C

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(5) If A = {3,4,5,9} B = {4,5,6,8} and C = {5,7,8,9} verify that the intersection of setsis associative. (M - 07)

(J - 10)

(J - 06)

(6) If A = {5,6} B = {6,7,8} and C = {5,6,7} then find (A∪B)∩(A∪C)

(A∪B) = { } ∪ { }

= { }

(A∪C) = { } ∪ { }

= { }

(A∪B)∩(A∪C) = { }

(7) X = {x : x is a prime numbers less than 10} Y = {x : x is even digit less than 10}Z = {x : x is odd digit less than 10} verify that intersection of set is associative.

X = { } (X∩Y) =

Y = { } (Y∩Z) =

Z = { }

X∩(Y∩Z) = { } ∩ { }

= { }

(X∩Y)∩Z = { } ∩ { }

= { }

∴ X∩(Y∩Z) = (X∩Y)∩Z

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(8) A = {1,3,4,8,9,12} B = {1,4,9} and C = {2,4,8,10} verifyA∩(B∪C) = (A∩B) ∪ (A∩C)

B∪C = { } A∩(B∪C) = { } ∩ { }

A∩B = { } = { }

A∩C = { } (A∩B)∪(A∩C) = { } ∩ { }

= { }

∴ A∩(B∪C) = (A∩B) ∪ (A∩C)

(9) A = {2,4,6,8,10} B = {1,2,3,4,5,6} and C = {1,3,5,7,9,11,13} then verifyA∪(B∩C) = (A∪B) ∩ (A∪C)

B∩C = { } A∪(B∩C) = { } ∪ { }

A∪B = { } = { }

A∪C = { } (A∪B)∩(A∪C) = { }∩{ }

= { }

∴ A∪(B∩C) = (A∪B) ∩ (A∪C)

(10) If A={1,2} B = {2,3,5} C = {2,3,6,8} then show that A∪(B∩C) = (A∪B)∩(A∪C)(M - 07)

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(11) Set A={Positive even nos. less than 11} B={Positive odd nos. less than 11} thenshow that intersection of sets is distributive over union. (M - 08)

(12) U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {x / x is a perfect square less than 10}B = {x / x is a multiple of 3 less than 10} then verify(i) (A∪B)1 = A1∩B1 (ii) (A∩B)1 = A1∪B1

(i) A = { } (A∪B)1 = { }1

B = { } = { }

A1 = { } A1∩B1 = { }∩{ }

B1 = { } = { }

A ∪ B = { }

∴ (A∪B)1 = A1∩B1

(ii) (A∩B) = { } (A∩B)1 = { }1

= { }

A1∪B1 = { }∪{ }

= { }∴ (A∩B)1 = A1∪B1

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(13) If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1,2,4,6,8} and B = {1,3,5,6,9} verify thatthe compliment of union of sets is the intersection of their compliments.

A1 = { } (A∪B)1 = { }1

B1 = { } = { }

A∪B= { } A1∩B1 = { }∩{ }

= { }

∴ (A∪B)1 = A1∩B1

(14) U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {x / x is perfect square less than 10}B = {x / x is even digit less than 10} then show that (A∪B)1 = A1∩B1

(15) U = {0,1,2,3,4,5,6,7,8,9} A = {1,4,9} B = {2,4,6,8} then show that (A∪B)1 = A1∩B1

(M - 10)

(J - 09)

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(18) There are 60 students in a class. Every student have to learn at least one of thelanguages Kannada or English. 45 Students select Kannada and 30 English. Findhow many students select both the languages by using venn diagram.

Total No. of Students = n(K∪E) =

Kannada Selected students n(K) =

English Selected students n(E) =

No. of Students to select both n(K∩E) = ?

n(K) + n(E) = n(K∪E) + n(K∩E)

∴ n(K∪E) =

(19) In a group of 25 people 8 people drink only Tea, 7 can drink only Coffee and 4 candrink both Coffee and Tea. Some one can not drink either Coffee or Tea. Findthose number of person and draw Venn diagram.

No. of Persons who drink Coffee only = n(C) - n(C∩T) = 7No. of Persons drink both Coffe and Tea = n(C∩T) = 4No. of Persons who drink Tea only = n(T)-n(C∩T) = 8No. of Persons drink either Coffee or Tea = n(C∪T) = ?

n(C∪T) = 7 + 4 + 8 = 19

No. of Persons can not drink either Coffee or Tea is= U - n (C∪T)= 25 - 19 = 6

(16) In n(A-B) = 25 - x, n(B - A) = 24-x, n(A∩B) = 2x then find n(A∪B)[ n(A-B) + n(B-A) + n(A∩B) = n(A∪B)]∴

K E

A B

7 4 8

6

(J - 10)

(M - 10)

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(20) In a group of 50 persons each one should drink Coffee or Tea, 14 drink Tea butnot Coffee, 30 drink Tea. Find how many drink Coffee but not Tea?

(21) In a group of 110 girls 60 Pass in the First Test, 45 Pass in the Second Test, if 12Pass in both. How many girls failed in both the test (Ans = 17 Girls)

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(22) In a school 60% of students invest in NDF and 70% of them invest in NSS. Findthe percentage of students who invest in both the scheme? (Ans: 30%)

(23) There are 70 cycles in a cylce stand, 50 cycles have carriers, 35 have bells andcarriers, while some have bell but no carriers. Find the number of cycles whichhave bell but no carrier (Ans: 55)

(24) In a college of 800 students if 500 select History, 250 select both History andGeography. Find the number of students who select Geography only and Historyonly? (Ans: 300, 250)

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(25) 62 factories in a country manufacture pens only and 48 manufacture pencils andpens. Some of the factories manufacture pencils only. Find their number if the totalnumber of factories in the country manufacturing pens or pencils is 132. (Ans: 22)

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SEQUENCE

• Sequence: It is an ordered arrangement of numbers according to a given rule. The term of a sequence in successive order is denoted by Tn. The nth term Tn is called general term of the sequence

• Finite Sequence: A Sequence having a finite number of terms.T1, T2, T3 ................... Tn is a Sequence of nth term and Tn is the last term.

• Infinite Sequence: A Sequence having infinite number of terms.T1, T2, T3 ........................

• Series: Sum of the terms of the Sequence is called a Series.T1 + T2 + T3 + ...................... + Tn Finite SeriesT1 + T2 + T3 + ...................... Infinite SeriesSn = T1 + T2 + T3 + ...................... + Tn

• Arithmetic Progression (AP): It is a sequence in which defference between a term and its preceding term is a constant. The constant term is called common difference and it is denoted by ‘d’.

• General form of an A.P.: ‘a’ is the first term, ‘d’ is common difference and Tn is nth term then the General form is

a, a+d, a+2d, ................... a + (n-1)dand nth term is Tn = a + (n-1)d

Tn+1 = Tn + d and Tn-1 = Tn - dTo find the common difference ‘d’

d = and Tp = Tq + (p - q) d

If Sequence is in ascending order then the ‘d’ is +ve and if it is in descending orderthen ‘d’ is -ve.

• Sum of n terms of an AP: (1) If First term (a) last term ‘l’ and number of terms ‘n’ are given

Sn = (a + l)

The middle term = (average)

(2) a, d and n are given

Sn = [2a + (n-1)d]

Tp - Tq

P - Q

n2

a + l 2

n2

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(3) Sum of all natural numbers Σn =

(4) Sum of first n odd natural numbers Sn = n2

(5) Sum to first n even natural numbers Sn = n(n+1)

• If each term of an AP is increased or decreased or multiplied or divided by the same quantity except zero the resulting series is also in AP.

I Is it a Sequence or not

(1) 1, -1, 1, -1, 1, -1, ....................... It is a sequence

(2) 3, 7, 11, 15, ...........................

(3) 3, 7, 15, 23, ......................... It is not a sequence

(4) 5, 10, 15, 20, 25, .........................

(5) 7, 4, 1, -2, -5, ...........................

(6) 1, 4, 9, 16, 25, ...........................

(7) 2, 6, 8, 54, ................................

(8) , , , .....................

(9) 1, , , ....................

(10) , , , .....................

II Find the nth term of the sequence.

n(n+1) 2

12

23

34

45

110

1100

11000

14

34

74

94

(1) 10, 100, 1000, ............ 10, 102, 103, ....... Tn = 10n

(2)

(3) 5, 9, 13, 17, .................. 5+0, 5+4, 5+8, 5+12 .... 5+4(n-1) = 4n+1

(4) 2, 5, 8, 11, .................... 3x1-1, 3x2-1, 3x3-1,.....

(5) 2, 5, 10, 17, ................. 12+1, 22+1, 32+1, ......

, , , ........... , , ............12

23

34

45

11+1

22+1

33+1

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III (A) Find the first 4 terms of the given sequence.

III (B) Find number of terms in a given sequence.

IV Calculate and fill in given box.

Tn T1 T2 T3 T4

(1) 4n + 3 7

(2) n2+2 6

(3)

(4) (-1)n 1

(5) 4n2-1 3

n-1n+1

24

Tn Tn+1 Tn-1 Tn+2 Tn-2

(1) 5n+3 5(n+1)+35n+8

5(n-1)+35n-5+3

5n-2

5(n+2)+35n+10+3

5n+13

5(n-2)+35n-10+3

5n-7

(2)

(3)

(4)

3n2 - 2

n2 + 2

nn+1

(1) Tn = 2n2+1 Tn = 73

73 = 2n2+12n2 = 72n2 = 36n = 6

(2) Tn = n2+1 Tn = 65

(3) Tn = 2n2+5 Tn = 2553

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V Write ‘a’ and ‘d’ for the following AP.

VI In an AP, Tn = a+(n-1)d then calculate unknown one and fill in the right box.

AP a d AP a d

(1) 4, 7, 10, 13, ....... 4 7-4=3 (5)

(2) 10, 20, 30, 40, ..... (6)

(3) 7, 3, -1, -5, ...... (7)

(4) 8, 6, 4, 2, ..... (8)

23

43

53

, 1, , ........

, 1, , 4 ........12

32

, 2, , 3 ........32

52

12

72

, , 3 ........

a d n Tn

(1) 2 3 8 ...............

(2) 5 ............... 10 23

(3) 3 8 ..............

(4) ............ 4 15 59

(5) 15 -2 ............... 5

(6) ............ 3 12 31

(7) ............... 15

(8) 8 -4 ............... 28

(9) 4 -5 ............... -106

(10) 5 ............... 7 -7

12

12

152

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VII Match the following.A B

(1) nth term of an AP (a) n(n+1) (1).............................

(2) Sum of n terms of an AP (b) (2)..............................

(3) Common difference d (c) Tn (3)..............................

(4) Arithmetic mean (d) n2 (4)..............................

(5) Sn - Sn-1 (e) [2a+(n-1)d] (5)..............................

(6) Sum of natural numbers (f) a+(n-1)d (6)..............................

(7) Sum of odd numbers (g) (7)..............................

(8) Sum of even numbers (h) (8)..............................

VIII (A) Find the sum of the series.

n(n+1)2

n2

a+b2

Tp-Tq

p-q

n2182

905

Series a l n d Sn / Tn

(1) 1+6+11+…+86 1 86 18 5

Tn = a+(n-1)d 86 = 1+(n-1)5 86 = 1+5n-5 5n = 86+4=90 n = = 18 Sn = (a + l) = (1 + 86) = 9 x 87 = 782

(2) 3+7+11+….+123

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(B) Find the number of terms of the following Arithmetic Series.

Series a l n d Sn / Tn

(3) 1+4+7+…+103

27

Series a d Sn n

(1) 1+2+3+…= 1275 1 1 1275

Sn = [2a + (n-1)d] 1275 = [2 x 1 + (n–1) 1] 1275 = [2 + n – 1] 1275 = (n + 1) n2 + n – 2550 = 0 n2 + 51n – 50n – 2550 = 0 n(n+51)-50(n+51) = 0 (n+51) (n-50) = 0 n = -51 n = 50

(2) 1+3+5+….= 625

n2

n2

n2

n2


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(C) In an AP find Sn.

Series a d Sn n

(3) 1+4+7+….= 590

152

n2152152

Term T1=a T2 d=T2-T1 Sn

(1) Tn = 5 – 2n 3 1 -2 S15

Sn = [2a + (n-1) d] = [2 x 3 + (15-1) -2] = [6 – 28] = x -22 = -165

(2) Tn = 4n – 3 S15

(3) Tn = 5n + 2 S20

(J - 06)

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(D) Find the sum of the arithmetic series which contains n term and middle term.

Term T1=a T2 d=T2-T1 Sn

(4) Tn = 10-3n S40

(M - 08)

(M - 07)

n2

No. of Terms Middle Term Sum of Series

(1) 19 26

Sn = (a + l) is a middle terms Sn = 19 x 26

= 498

(2) 25 20

(3) 21 17

a + l2

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(E) If the three numbers are in AP. Sum of the numbers and product of thenumbers are given. Find the numbers.

153

(2) 30 120

Sum ofthe

numbers

Productof the

numbersThe numbers are a, a+d and a-d

(1) 15 105

Sum: a + a + d + a - d = 15

3a = 15

a = = 5

Product: a (a+d) (a-d) = 105

5 (52-d2) = 105

25 - d 2 = 21

d 2 = 25 - 21 = 41

d = 2

the numbers area = 5a+d = 5+2 = 7a-d = 5-2 = 3They are (3, 5, 7)

1053 = 21

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Sum ofthe

numbers

Productof the

numbersThe numbers are a, a+d and a-d

(2) 36 1620

(3) -24 288

(3)

(4)

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(F) Find ‘x’ if the arithmatic mean between a and b is A.

a A b A =

(1) 2x+3 x-1 2-x

x-1=

x-1 =

2x-2 = x+5

x = 7

(2) 5 x 19

(3) 6-a x 6+a

(4) 2 (x-1) 4

a+b2

2x+3+2-x2

x+52

(J - 10)

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33

(5) 3x+1 5x-1 5x+1

(6) 2x x+10 3x+2

(7) 2x+3 x-1 2-x

IX (A) If Σn = find the value of

(1) Σn = (2) Σn =

= 465

(3) Σn - Σn (4) Σn + Σn

n(n+1)2

30 x 312

30

1

25

1

60

1

30

1

40

1

30

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34

(B) Find the value of n

(1) Σn = 36 (2) Σn = 210

(3) Σn = 820 (4) 3Σn = 630

(5) Σn-1 = 78 (6) Σn-1 = 820

X (1) Find the sum of all naturals between 50 and 150 ( Σ149 - Σ50) (Notice the use of words betwen, from and to)

(2) Find the sum of all naturals between 100 and 200 which are divisible by 5.

105 + 110 + 115 + ..................... + 195

= 5(21 + 22 + 23 + ................... + 39)

= 5(Σ39 - Σ20)

∴


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(3) Find the sum of all naturals from 101 to 200 which are not divisible by 4.(Hint: Find sum of 101 to 200 and sum of divisible nos. and subtract)

(4) Find the sum of all odd natural number from 1 to 100.

(5) Find sum of all even natural numbers from 1 to 100

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36

XI Find an AP if the ratio between the numbers are given.

Ratio

(1) 3rd 9th 1 : 3 T18 = 90=

=

3a + 6d = a+8d2a = 2d

a = d AP is 5, 10, 15, 20......

T18 = 90

a + 17d = 90

18d = 90

d = = 5

a = 5

(2) 4th 8th 1 : 2 T10 = 30

(3) 5th 10th 1 : 2 T12 = 36

T3

T9

13

a+2da+8d

13 90

18

(J - 09)


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XII (1) In an AP if T10 = 2T6 then prove that T4 = 2T3.

(2) 7th term of an AP is -15 and 16th term is 30 find the sum of the 1st and 19th

terms.

d = Tn = a + (n-1)d T19 = ?

(3) In a sequence T1 = 3, Tn = 3Tn-1 + 2 for all n > 1. Find first 4 terms of asequence

Tp - Tq P - Q

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(4) In an AP the sum of first 10 terms is 175 and the sum of next ten terms is 175find APS10 = 175, S20 = 350

Sn= [2a + (n-1) d]

S10 = 5 (2a + 9d) = 1752a + 9d = 35 -- (1)|||ly 2a + 19d = 70 -- (2)equate (1) and (2)

(5) The sum of the 6 terms of an AP is 345 and difference between first and lastterm is 55 find the terms of an AP

Sn = (a + l)

345 = (a + l)

a + l = 115 -- (1)l - a = 55 -- (2)equate (1) and (2)

(6) In an AP of 21 terms sum of the middle 3 terms is 129 and sum of the last 3terms is 237 find the sequence.T10 + T11 + T12 = 129a + 9d + a + 10d + a + 11d = 129

a + 10d = 43 -- (1)T19 + T20 + T21 = 237a + 19d = 79 -- (2)equate (1) and (2)

(7) 8th term in an AP is double the 13th term, show that 2nd term is double the10th term.T8 = 2T13 T2 = 2T10

n2

n2

62

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(8) Three numbers are in AP whose sum is 18 if 2, 4, 11 are added to themrespectively the resulting numbers are in GP find those numbers.

(9) In an AP of 21 terms the first and last terms are 4 and 64 respectively. Findthe sum of the series.

(10) The angles of a quadrilateral are in AP. If the smallest angle is 150. Find theangles of the quagrilateral.(Assume angles are a-d, a+d, a+3d, a-3d)

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(11) Find the sum of the series 3 + 7 + 11 + ............. up to 42 terms.

(12) 3 nos are in the ratio 2 : 5 : 7. If 7 is substracted from 2nd no. the resultingnumbers are in AP. Find the numbers.(The numbers are 2x, 5x, 7x and T2 = 5x - 72x, 5x-7, 7x are in AP)

(13) Three angles of a triangle are in AP. The smallest angle is 400 then find theangles of the triangle.[Assume angles are (a-d), a, (a+d)]

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(14) A person deposits Rs. 1000 in the first month. Then every month he increasesthe monthly deposit by Rs. 60. Calulate his total investment at the end of2 years.

(15) In an AP seven times the 7th term is equal to the eleven times the 11th term.Then show that the 18th term is equal to zero.( 7T7 = 11T11)

(16) Check whether 301 is a term in the AP 5, 11, 17, 23..................

(M - 09)

(M - 10)

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(17) The sum of four numbers in AP is 16 their product is 105. Find them.(Assume the numbers as (a-3d) (a+d) (a-d) (a+3d) and c-d is 2d)

(18) Sum of 3 numbers in AP is 27 the sum of their sequence is 293. Find thosenumbers (Let the numbers be (a-d) (a) (a+d))

(19) Sides of the right angle triangle are in AP. Show that they are in the ratio3 : 4 : 5 [Assume the sides are a + d, a, a - d, Apply Pythagorous Theorem)

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XIII Fill up the blanks with suitable answer

(1) The sum of first 13 terms of an AP of which 7th term is 40 is ....................

(2) The sum of the first twenty positive multpiles of 7 is ...............

(3) S9 = 81, T9 = 27 Value of S8 is ...........................

(4) If Tn = (-1)n the sum of 4 terms is ...........................

(5) If a is constant then a + 2a + 3a + ....... + na = .....................

(6) Sum of the first twenty odd numbers is ........................

(7) Sum of first twenty even natural numbers is .......................

(8) If first term of an AP is ‘a’ and its common defference is d. Then the formula to

find the nth term of AP is ......................

XIV Choose the correct answer.

(1) The common difference of an AP is d then the relation between 5th and 12th term is

..........................

(A) T5 = T12-6d (B) T12=T5+6d

(C) T5 = T12-7d (D) T12 = T5+8d

(2) In an AP the common difference is ..........................

(A) d = Tn+Tn+1 (B) d = Tn-1- Tn+1

(C) Tn+1 - Tn (D) Tn+1 + Tn

(3) Sn - Sn-1 is ......................

(A) Sn (B) Tn (C) d (D) Sn+1

(4) In an AP 20th and 30th terms are 201 and 301 respectively the common differnece

is .............................

(A) 10 (B) 100 (C) 5 (D) 2

(J - 10)

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(5) Σn - 1 is given by the formula is ........................

(A) (B) (C) (D)

(6) Sum of 1+3+5+ .............. + n is .............................

(A) n2 (B) n + 1 (C) n(n+1) (D) 2n(n+1)

(7) Which of the following is the term of the sequence 3, 6, 9, 15, ....... is .................

(A) 25 (B) 38 (C) 45 (D) 42

(8) The value of Σ (-1)9 is ......................

(A) 0 (B) 1 (C) -1 (D) 10

(9) Pth term of an AP is q and Qth term is p then (P+Q)th term is ......................

(A) 0 (B) (P + Q) (C) (P - Q) (D) -(P + Q)

(10) Three numbers are in AP the sum of first and third terms is 14, the middle term is

.........................

(A) 6 (B) 8 (C) 18 (D) 7

n(n+1) 2

n(n-1) 2

n(n+2) 2

n(n-2) 2

10

n=1

(11) 2 + 4 + 6 + 8 + ............... + 2n can be written as ....................

(A) 2Σn (B) Σ (C) 3Σn (D) Σn

(12) If the terms of an AP are in ascending the common difference is ....................

(A) Positive (B) Negative

(C) Zero (D) May be +ve or -ve

(13) If (5a - x), 6a, (7a + x) are in AP then the 5th term is ....................

(A) 3(3a+x) (B) 9(a+3x) (C) 9(a+x) (D) 3(3a-x)

(14) n(n2+1) is nth term of a sequence the sequence is in ....................

(A) AP (B) GP

(C) HP (D) None of these

(15) In an AP S5 = 35 and S4 = 22 then, 5th term is ....................

(A) 35 (B) 10 (C) 13 (D) 22

n2

(M - 06)

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(16) 3, 7, 11, 15, ........... the nth term is ....................

(A) 4n - 1 (B) 4n + 1 (C) 4n + 3 (D) 3n + 4

(17) Ramu put a dot in 1st square, 2 in 2nd, 3 in 3rd and so on. Total number of squares

he required to put 55 dot is ............................

(A) 55 (B) 11 (C) 9 (D) 10

(18) In a sequence Tn = n2 - 1 and Tn = 35 then value of n is ....................

(A) 6 (B) 36 (C) 34 (D) -6

(19) Geetha climbed 15 steps of a building in the first minute after that she climbed 3

steps less than what she had climbed in the previous minute. If she climbed the

building in 5 minute the number of steps she climbed is ....................

(A) 75 (B) 105 (C) 45 (D) 50

(20) In a sequence Tn = 2n-1 the 4th term is ....................

(A) 23 (B) 9 (C) 5 (D) 7

(21) The value of Σn is ........................

(A) 10 (B) 11 (C) 55 (D) 110

(22) A person put 3 marbles in first box, 5 in second box, 7 in third box and so on.

Number of marbles can be put in 16th box is ....................

(A) 66 (B) 33 (C) 31 (D) 35

(23) In an AP which relation is true ....................................

(A) Tn-5 = Tn-4 + d (B) Tn-5 = Tn-6 + d (C) Tn-5 = Tn + d (D) Tn-5= Tn-d

(24) Sum of 15 terms of an AP is 180 then 8th term is ....................................

(A) 8 (B) 12 (C) 15 (D) 18

(25) 2x+1, 4x, 13-x are in AP the value of x is ....................................

(A) 2 (B) 3 (C) 4 (D) 5

10

1

(M - 06)

(M - 06)

(J - 06)

(J - 06)

(M - 07)

(M - 07)

(J - 07)

(M - 08)

(M - 08)

(M - 08)

45


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(26) In an AP Tn = 3n-1 the common difference is ....................................

(A) 1 (B) 2 (C) 3 (D) 4

(27) In an AP Tn+5 = 35 and Tn+1 = 23 then common difference is ......................

(A) 3 (B) 2 (C) 3n (D) 2n

(28) In a sequence Tn = 2n2 + 1 then S2 is ....................................

(A) 9 (B) 12 (C) 10 (D) 11

(29) In an AP common difference is 3, first term is 1 then the 10th term is ...........

(A) 27 (B) 29 (C) 30 (D) 28

(30) If Tn = (-1)n the correct relation between the terms is .................................

(A) S1 = S2 (B) S2 = S3 (C) S3 = S4 (D) S2 = S4

(31) In an arithmatic sequence if T4=8 and a=2, then its common difference is ..........

(A) 6 (B) 4 (C) 2 (D) 10

(M - 09)

(M - 09)

(J - 09)

(J - 09)

(M - 10)

(J - 10)

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GEOMETRIC PROGRESSION (GP)

• A Sequence in which the ratio of a term and its preceding term is a constant

Example: 2, 6, 18, 54, ..............................

T1, T2, T3, T4 .............................

= = .............................

The constant ratio is called the ‘common ratio’ and is denoted by ‘r’ and the first term is ‘a’

• The General Form of a G.P. ‘a’ is the first term, ‘r’ is the common ratio and ‘Tn’ is the ‘n’th term, then the general form is

a, ar, ar2, ar3 .................. arn-1

The nth term is Tn = arn-1

• If ‘Tn’ and ‘r’ are given, then

The succeding term Tn+1 = Tn x r

The preceding term Tn-1 =

• To find common ratio r =

also we have = rp-q

Note: If each term of a GP is multiplied or divided by the same quantity then the resultingterms are also in GP.

( Verify this with a suitable example)

T2

T1

T3

T2

T4

T3

Tn

r

T2

T1

Tp

Tq

47


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• Sum of the terms of a GP is called Geometric Series

Example: 2, 4, 8, 16, 32, ....................... is a GP

2 + 4 + 8 + 16 + 32 + ............... is a Gemetric Series.

• Sum of ‘n’ terms of a Geometric Series

(i) If r > 1 then Sn = a

(ii) If r < 1 then Sn = a

(iii) If r = 1 then Sn = na

• Sum of infinite (∞) terms of a geometric series is

S∞ =

• The geometric mean (G) begween a and b is

G = ab or G2 = ab

• Sn : S2n = 1 : rn + 1

S2n : Sn = rn + 1

rn - 1r - 1

1 - rn

1 - r

a1 - r

48


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I Write any 5 Geometric Progressions.

(1)

(2)

(3)

(4)

(5)

II Write ‘a’ and ‘r’ of the following Geometric Progressions.

(1) 2, 6, 18, 54 ..........................

a = ...................... r = ..........................

(2) 5, 10, 20, 40 ........................

(3) 1, -1, 1, -1, .......................

(4) 1, , , ............................

(5) 1, , , ............................

III Write the Geometric Progressions for the given ‘a’ and ‘r’.

(1) a = 3, r = 2

(2) a = 1, r =

(3) a = , r =

12

14

18

13

19

127

13

12

12

49


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IV Which of the following sequences are in GP. Justify your answer.

(1) 1, 2, 4, 8.............................

(2) 3, 6, 9, 12...........................

(3) 8, 4, 2, 1 ............................

(4) 4, 12, 36, 108 .....................

(5) 12, 6, 3, ...........................

V Find the nth terms of the following GP.

(1) 3, 12, 48, ......................

(2) 1, 3, 9, 27 ......................

(3) 3, 6, 12, 24, ......................

(4) 1, 2, 4, 8, ......................

VI Find the first 4 terms of the follwoing GP’s.

32

Tn T1 T2 T3 T4

2 x 5n-1

2 x 3n-1

3n-1

2n-1

50

1-n


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VII Fil up the boxes in the following table with suitable answer.

VIII Four alternatives are given to each questions choose the right answer andwrite it in the space provided for that.

(1) In a GP if T4 = and r = , then T5 is ....................

(A) (B) 8 (C) (D)

Ans

(2) The (n + 1)th term of a GP is ........................

(A) Tn+1 = Tnrn (B) Tn+1 = Tnr

n-1 (C) Tn+1 = Tn rn+1 (D) Tn+1= Tnx r

Ans

(3) If x, 1, are in GP then the common ratio is ........................

(A) x (B) 1 (C) (D)

Ans

(4) In a GP, which of the following is correct ........................

(A) Tn = r x Tn-1 (B) Tn-1 = r x Tn (C) Tn-1 x Tn = r (D) Tnxr = Tn-1

Ans

14

12

14

12

18

1x

1x

1x2

51

S.No. T1 T2 T3 T4

1 1 2 256

2 2 8 32

3 6

4 3 2 6

13

1729

.......................

.......................

.......................

.......................


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(5) In a GP, if S∞ = 3, r = then the first term is ....................

(A) 2 (B) 3 (C) (D)

Ans

(6) In a GP the first term is 3, and the common ratio is 1, then the sum of ten terms is

........................

(A) 3 x 110 (B) 30 (C) 3 x 19 (D) 10

Ans

(7) The nth term of the GP, 10, 100, 1000 .......... is .........................

(A) 10n (B) n10 (C) 10n (D)

Ans

(8) The 20th term of the GP Tn = xn-1 is ........................(A) x20 (B) 20x (C) 20x (D) x19

Ans

(9) The missing term in the GP 2, 4, ........., 16, 32 ........ is .........................(A) 6 (B) 8 (C) 10 (D) 12

Ans

(10) The nth term of the GP 1, 3, 9, 27, .......... is ........................

(A) 3 x 3n-2 (B) 3 x 3n-1 (C) 3n-1 (D) 3n + 1

Ans

(11) The common ratio of the GP Tn = 2 x 5n-1 is ........................

(A) 2 (B) (C) 5 (D) 2.5

Ans

13

23

32

10n

32

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(12) In a GP, if T8 : T5 = 8 : 1 then the common ratio is ....................

(A) -2 (B) 2 (C) ± (D) x5

Ans

(13) The 9th term of the GP Tn = 2n-1 is ........................(A) 256 (B) 64 (C) 32 (D) 128

Ans

(14) If 16, x, 25 are in GP, the value of x is .........................

(A) 20 (B) 200 (C) 20 (D) 400

Ans

(15) IF ‘b’ is the GM between ‘a’ and ‘c’ then = ........................

(A) (B) (C) (D)

Ans

(16) ABCD is a square EFGH is the another square obtained by joining the mid points

of ABCD, |||ly PQRS is the another square obtained by joining the midpoints of

EFGH and so on. Then the areas of these squares are in .........................(A) AP (B) HP (C) GP (D) None of these

Ans

(17) The AM and GM of two numbers are 5 & 4 respectively, then the numbers are ......

(A) 2, 8 (B) 4, 16 (C) 3, 7 (D) 1, 4

Ans

(18) The first term of a GP is ‘a’ and common ratio is ‘r’ then the 4th term is ...............

(A) ar3 (B) ar4 (C) ar2 (D) ar

Ans

ba

ab

ac

bc

53

ab


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(19) In a GP which of the following is not correct ?

(A) Tn+1 = Tn x r (B) Tn-1 = (C) r = (D) Tn+1 = Tn+r

Ans

IX Four alternatives are given to each questions, choose the right answer andfill in the blanks. (Previous Question Paper Questions)

(1) The 6th term of the sequence 1, , , ....... is ...................................

(A) (B) (C) - (D) -

(2) As ‘n’ approaches ‘∞’ the value of S∞ is ...................................

(A) (B) arn-1 (C) (D) ar0

(3) The GM between 4 and 16 is ...................................

(A) 4 (B) 16 (C) 8 (D) 12

(4) The formula to find the sum of ‘n’ terms a geometric series is ................................

(A) (B) a (C) (n2+1) (D)

(5) In a GP, if T5 : T2 = 1 : 8, then the common ratio is ................................

(A) 3 (B) (C) 2 (D)

(6) In a GP if Tn = 2 x 3n-1, then T5 is ................................

(A) 486 (B) 162 (C) 243 (D) 8

(7) The GM of three numbers is 4, then their product is ................................

(A) 16 Clue: G = ab a, G, b(B) 64 G2 = ab ab x G(C) 128 42 = ab 16 x 4 = 64(D) 256 ab = 16

Tn

rT4

T3

-13

19

-127

181

1243

181

1243

a1-r

1-ra

n(n+1)2

1 - rn

1 - rn2

rn-1a(r-1)

13

12

(J - 06)

(M - 07)

(M - 07)

(M - 07)

(M - 08)

(M - 08)

(M - 09)

54


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(8) In a GP, if Tn is the nth term, and common ratio is r, then Tn-1 term can be deter

mined by ................................

(A) Tn x r (B) (C) (D) Tn+1 x r

(9) The 4th term of the sequence 3 , 3, 3 3 is ................................

(A) 9 (B) 27 3 (C) 21 (D) 9 3

(10) p, 8, q are in GP then pq = .................................

(A) 64 (B) 2 (C) 2 2 (D) 64

X Solve the following problems. (Previous Question Paper Questions)

(1) In a GP if a = 3, r = 2 and Tn = 96 then ST Sn = 189.

Solution: Tn = 96arn-1 = 96 Sn = a

= ..........................................

(2) Find S3 : S6 of the series 4 + 12 + 36 + ................

Solution: r = Clue:Sn : S2n = 1 : rn + 1S3 : S8 = 1 : r3 + 1

(3) In a GP S8 : S4 = 97 : 81 find r

Solution:

Tn-2

rTn

r

rn - 1r - 1

...................................

...................................

T2

T1

(J - 09)

(J - 10)

(J - 09)

(M - 06)

(J - 06)

(M - 07)

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(4) In a GP if T6 = 32 and r = 2 find a

Solution: Use Tn = arn-1

(5) The 10th term of a GP is 8 times of its 13th term if the first term is 3, then find thesum of infinite terms.

Solution: T10 = 8 x T13 a = 3 r = .................

= 8 S∞ = .................................

= .................................................. = 8

= ...............................................................

= ...................................r = ..........................

= ...................................

(6) Find the sum of the series 2 + 4 + 8 + ............. + 256

Solution: a = .............. Tn = arn-1 Sn = a

r = ................... = .................. =

Tn = ................. = ................. =

n = ? n = ................. =

=

T10

T13

rn - 1r - 1

(M - 09)

(J - 09)

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(7) The first term of a GP is 64 and the common ratio is ‘r’. If the average of 1st and4th term is 140 find the value of ‘r’.

Solution: a = 64r = ?T1 = 64T4 = ar3

(8) Three Nos. are in GP, whose sum is 26 and their product is 216, find the numebrs.

Solution: Let the nos. be , a, ar

Product: x a x ar = ..............

........................... = ......................... .................................

a3 = ................. ..................................

a = .................. ..................................

r = .......................

The numbers are

.................., ...................., ...................

= 140

64 + ar3 = 140 x 2

ar3 = .............. - ..............

64 r3 = ....................

r3 =

r =

a + ar3

2

64

ar

ar

Sum : + a + ar = 26ar

.................................

(A - 10)

(J - 10)

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(9) Three numebrs are in GP, whose sum is 28 and their product is 512. find the Nos.

Solution:

(10) For what value of K, in K, K-2, K+1 makes a GP.

Solution: Clue K, K-2, K+1a G b

G = ab

...........................................

...........................................

...........................................

(11) If a, b, c are in AP and x, y, z are in GP, then show that x x = 1

Solution: a, b, c are in AP & x, y, z are in GP

b = y = xz

LHS x x

= x x

= .............................................................

= .............................................................

= .............................................................

xb

xc

yc

ya

za

zb

a+c2

xxc

a+c2 ( xz )c

( xz )a

za

za+c

2

x x xxc

a2

c2 ( x )c x ( z )c

( x )c x ( z )c

za

z x za 2

c 2

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(12) The sum of 2 terms of a GP is 6 and first term is 2. Find the common ratio.

Solution: a = T1 = ....................

S2 = T1 + T2 r =

6 = ........... + T2 .........................................

................................... .........................................

T2 = .......................... .........................................

(13) S8 : S4 = 97 : 81, T15 = 16, write the GP.Solution:

(14) A person invests Rs. 5 as initial amount, then after he deposite Rs. 10 at the end of1st month and Rs.20 at the end of 2nd month and soon, if he deposite Rs.3,27,680 atthe end of 16th month, find the total amount that he invested.Solution: 5 + 10 + 20 + ............................+ 3,27,680

T1 T2 T17

a =r =n = 17Sn = ?

T2

T1

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(15) Sn is the sum of ‘n’ terms of a geometric series, a = 3 and r = 2, then find S8 - S3.

Solution:

(16) How many terms of the series 2+4+8+....... should be added to get a sum of 1022.

Solution: a = ......................

r = .......................

Sn = 1022

n = ?

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(17) Find the number of terms of the following series 3 + 9 + 27 + ........... = 1092

Solution: Sn = a Cluea = 3r =Sn = 1092n = ?

(18) In a GP show that =

Solution: Hint: Tn-1 =

Tn+1 = Tn x r

(19) Three numbers whose sum is 18 are in AP, if 2, 4 & 11 are added to themrespectively, the resulting numbers are in GP. Find the numbers?

Solution: Let the numbers bea-d, a, a+d in APSum is 18

....................................

....................................

a = 6

If 2, 4, 11 are added to a-d, a, a+d respectivelya-d+2, a+4, a+d+11 are in GP T1 T2 T3

=

.....................................................

rn - 1r - 1

Tn-1

Tn+1

1r2

Tn

r

T2

T1

T3

T2

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(20) Find the GM between (x + y) and x3 + 3xy (x + y) + y3

Solution: a = x + yb = x3 + 3xy (x + y) + y3

G = ab

G = .................................................

G = .................................................

G = .................................................

G = .................................................

G = x2 + y2 + 2xy

(21) If 8, x+1, T2 are in GP find x.

Solution: HintG = ab

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HORMONIC PROGRESSION (HP)

• A Hormonic Progression (HP) is a Sequence in which the reciprocal of terms of it form an arithmatic progression.

• If Tn is the nth term of a HP then will be the nth term of the AP.

• nth term of a HP is Tn =

• General term of HP is , , , , ...................

• is in AP then is in HP

• If a, H, b are in HP, then Hormonic mean between a and b is H =

• If A, G, H are the AM, GM and HM of two positive numbers a and b then A . G . H are in GP (G = AH )

• A > G > H

I State which of the Sequences are in HP.

(a) 1, , , .......... (b) 6, 4, 3 ..............

Reciprocal of the given Sequence

1, 2, 3, 4 ..........

T2 - T1 = 2 - 1 = 1

T3 - T2 = 3 - 2 = 1

it is in AP hence the reciprocal of the

Sequence is in HP.

(c) , , , .......... (d) 1, , , ..........

63

1Tn

1a + (n-1) d

1a

1a + d

1a + 2d

1a + 3d

1a + (n-1) d

23

32

2aba + b

12

13

14

12

15

110

117

23

12

25


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II (a) , , is in HP then find Tn and 2, 5, 8 is in AP

2 = 2 + 0 = 2 + 3 x 05 = 2 + 3 x 1 = 2 + 3 x 18 = 2 + 6 = 2 + 3 x 2

Tn = ................................................

(b) In , , , Find 12th Term and 8, 13, 18, 23 ...... is in AP

a = d = n = 12

Tn = a + (n-1) d

=

(c) In , , 1, -1 ................ find 10th term.

III (1) In a HP T4 = and T10 = find T19.

If Tn =

T4 = =

a + 3d = 12 -- (1)a + 9d = 42a + 3d = 12 6d = 30

d = = 5

12

15

18

12

15

18

18

15

13

112

142

1a + (n-1) d

1a + ( ) d

112

- - -

306

T12 = =

a + 9d = 42 -- (2)a + 3d = 42a + 3 x 5 = 12 a = 12 - 15

a = - 3

1a + ( ) d

142

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T19 = =

T19 =

(2) In a HP T5 = and T1 = find T7 and T12.

(3) In a HP T10 = and T3 = then find T11.

(4) In a HP T7 = and T13 = find a.

1 + ( )

187

619

29

619

29

(M - 08)

(M - 10)120

138

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IV (1) Find the HM between 6 and 3.

(2) If A, G, H are AM, GM and HM of the number 9 and 16 then verify A, G, H are in GP.

a = 9 A = G = ab H = b = 16

∴ a = A x H

=

(3) If A, G, H are AM, GM and HM of the number 2 and 8 then show that A > G > H.

a = 2 A = G = ab H =b = 8

66

a + b2

2aba + b

a + b2

2aba + b


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(3) If a, b, c are in HP then find the value of

, , are in AP

T2 - T1 = T3 - T2

(5) If A, G, H are the AM, GM and HM of two positive numbers a and b then, show that A, G and H are in GP

(6) If , and are in HP then show that a, b, c are also in HP.

(By adding or dividing a constant term to an each term of an HP there is no change in the given HP)

67

a - bb - c

1a

1b

1c

ab+c

bc+a

ca+b


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V Previous Exam paper questions.

(1) If x, y, z are AM, GM and HM between 2 numbers respectively, then which of

following relation is true.

(a) x < y < z (b) x > y > z

(c) x = y = z (d) x < y < z

(2) The 6th term of the sequence , , ......... is ..........................

(a) (b) (c) (d)

(3) In a HP T4 = , T10 = then d is ..........................

(a) 6 (b) 5 (c) 3 (d) 4

(4) The common difference of the sequence , , ....... is ..........................

(a) 2 (b) -2 (c) (d) 1

(5) If 11, 13, 15, 17, 19 ............ is an AP, then the terms in HP are

(a) 1, 2, 3, 4, ............. (b) 1, 3, 5, 7, .............

(c) , , , , ...... (d) , , .........

(6) If 3, x, 7 are in HP, then the value of x is

(a) (b) 5 (c) 21 (d)

(7) The AM and GM between two distinct numbers are 5 and 4 respectively, then their

HM is ........................................

(a) 4 (b) 4.5 (c) 5 (d) 3.2

(8) The HM between 1 and 4 is ........................................

(a) (b) (c) (d)

68

(M - 06)

(J - 06)

(J - 06)

(M - 07)

(M - 09)

(M - 07)

(M - 07)

12

15

18

111

114

117

120

112

142

12

14

16

12

111

113

115

117

119

110

112

114

215

521

75

85

65

74


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(9) The Harmonic mean of P and Q is

(a) (b)

(b) (d)

(10) If x, y, z are in HP, then the harmonic mean is .............................

(a) (b) (c) (d)

(11) If A, G, H are AM, GM and HM of a and b then .............................

(a) A, G, H are in AP (b) A, G, H are in GP

(c) A, G, H are in HP (c) A, G, H are not in AP, GP & HP

(12) The HM between 1 and 2 is .............................

(a) 1 (b) 1 (c) 1 (d) 1

(13) An example for HP among the following is

(a) 1, , , ........... (b) 1, , , ...........

(c) 1, , , ........... (d) 1, , , ...........

(14) The reciprocal of the term of a HP are in .............................

(15) The GM of two numbers is 4 3 and HM is 6, then its AM is .............................

69

(M - 09)

(J - 09)

(J - 09)

(M - 10)

(M - 10)

(J - 10)

(J - 11)

2 (P + Q)PQ

2 (P + Q)P - Q

2 PQP + Q

2 P + QPQ

2xzx + z

2xyx + y

2yzy + z

2xzx + y

12

14

13

23

12

23

34

13

16

19

23

12

25

14

17

19


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MATRICES• Matrix:

It is a rectangular arrangement of numbers written in the form of rows and columnsis called Matrix. The arrangement of numbers are enclosed by brackets like [ ] or ( )or || ||. If a matrix has ‘m’ rows and ‘n’ columns then it is called a matrix of orderm x n and it has ‘mn’ elements.

• Types of Matrix:(a) Zero Matrix: The matrix in which all its elements are zero.

(b) Row Matrix: A matrix having only one row is called a row matrix and it’sorder is 1 x n.

(c) Column Matrix: A matrix having only one column is called a columnmatrix and its order is n x 1

(d) Rectangular Matrix: A matrix in which the number of rows is not equal tonumber of columns and its order is m x n.

(e) Square Matrix: A matrix in which the number of rows equal to the numberof columns and its order is n x n

(f) Diagonal Matrix: In a square matrix except the principle diagonal all otherelements are zero.

1 0 00 2 00 0 3

(g) Scalar Matrix: A diagonal matrix in which the principle diagonal elemntsmust be same.

3 0 00 3 00 0 3

(h) Unit Matrix (Identity Matrix): A diagonal Matrix in which principlediagonal element must be one. It is denoted by I. Identity Matrix is also aScalar Matrix.

1 0 00 1 0

0 0 1

3 x 3

3 x 3

3 x 3

71


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(i) Symmetric Matrix: If a Square Matrix is folded along its principle diagonal theelements which coinside are said to be symmetric with respect to the principlediagonal of the matrix.

5 1 31 0 43 4 2

(j) Skew Symmetric Matrix: A Square Matrix is called a Skew Symmetric Matrix ifthe elements which are symmetric with the principle diagonal are equal but oppistein sign while diagonal elements are zero.

0 2 1-2 0 -3-1 3 0

• Equal Matrix : Two matrix are said to be equal if their order are same and thecorresponding elements are equal.

• Addition of two Matrix: Let A and B be two matrix of order m x n. The sum A + Bof the matrix A and B is a matrix obtained by adding their corresponding elementsorder of A + B is also m x n.

• Scalar Multiplication of the Matrix: Let A be a matrix of order m x n and K be anyscalar. The scalar multiplication of the matrix A by a scalar K is matrix denoted byKA and its elements are scalar multiple of the elements of A. The order of K is alsom x n.

• Multiplication of Matrices:

If A = and B = then AB is

AB = =

If the number of columns of matrix A and the number of rows of matrix B are equalthen only AB exists. If A is of order m x n and B is of order n x p. The product AB is of orderm x p.

• Transpose of a Matrix: Let A be a matrix of order m x n. The matrix obtained fromA by interchanging the rows in to columns is called the transpose of the matrix A andit is denoted by A1 and its order is n x m.

a b e fc d g h

a b e f ae + bg af + bhc d g h ce + dg cf + dh

3 x 3

3 x 3

72


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• If A and B be any two matrices of same order then,

(a) (A + B)1 = A1 + B1

(b) (A1)1 = A

(c) If A and B are two matrices such that (A•B) is defined them (A•B)1 = B1•A1

(d) If A = A1 then the matrix A is a Symmetric Matrix.

(e) If A = -A1 then the matrix A is a Skew Symmetric Matrix.

(f) If A and B are two matrices A - B = B - A and IA = AI = A

(g) A + A1 is a Symmetric Matrix and A - A1 is Skew Symmetric Matrix.

(h) AB = BA.

I. Matrix A Type Order Transpose A1 -2A

(1) 1 0 0 0 1 0 0 0 1

(2) 1 2 3 0 1 -3- 1 2 0

(3) 2 3 1 0 1 2

(4) 1 2 3 2 0 4 3 4 2

(5) -3 0 0 0 -3 0 0 0 -3

(6) [ 2 -1 0 ]

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II. Match the Following:

(1) Null Matrix (a) 0 3 (1) ................................................... -3 0

(2) Diagonal Matrix (b) 2 8 0 (2) ...................................................1 5 2

(3) Scalar Matrix (c) 0 0 (3) ...................................................0 0

(4) Unit Matrix (d) 3 6 (4) ...................................................6 1

(5) Rectangle Matrix (e) 1 0 (5) ...................................................0 1

(6) Row Matrix (f) 3 0 (6) ...................................................0 3

(7) Column Matrix (g) 2 0 (7) ...................................................0 1

(8) Symmetric Marix (h) [ 2 1 3 ] (8) ...................................................

(9) Skew Symmetric Matrix (i) 3 (9) ...................................................01

III. Find the value of x for the given:

Symmetric Skew Symmetric Scalar Identity

(1) 1 x-2 0 x-2 2x-1 0 2x+1 0 2x-1 -1 2x-1 0 0 2+x 0 1 2x-1 = x-2 2x-1 = -(x-2) 2x-1 = 2+x 2x+1 = 1 2x-x = -2+1 2x-1 = -x+2 2x-x = 2+1 2x = 1-1 x = -1 3x = 3 x = 1 x = 3 x = 0

(2) 3 2+x 0 5 2x 0 6+x 0 2x-5 5 x+6 0 0 x-3 0 1

....................... ....................... ....................... .......................

....................... ....................... ....................... .......................

....................... ....................... ....................... .......................

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(3) 2 x-3 0 2x 3x+4 0 x+5 0 3x-1 -2 6+x 0 0 2x+5 0 1

....................... ....................... ....................... .......................

....................... ....................... ....................... .......................

....................... ....................... ....................... .......................

IV. Fill up the blanks with suitable answer:

(1) The number of elements of the matrix 3 x 4 is .......................

(2) The number of columns in the matrix given the total number of elements and the

number of rows 30 and 10 is .......................

(3) The principle diagonal elements of the given matrix1 2 32 3 4 is ...........................5 6 7

1 -1(4) If A = -1 2 then A1 is ...........................

2 0

(5) If the order of the given matrix A is (m x n) then the order of it’s transpose A1 is

...............................

(6) If A is a square matrix and I is a Unit Matrix of the same order then AI = ...........

(7) A is a matrix then [{(A1)1}1]1 is ......................

(8) Matrices A, B and C are of order 3 x 3, 3 x 2 and 3 x 2 respectively then the order

of A (B + C) is ...............................

(9) If A is a Square Matrix of order 2 x 2 then A + A1 is ...............................

(10) If A is a Square Matrix of order 2 x 2 then A - A1 is ...............................

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V. Choose correct answer and fill in the blanks:

(1) If A = A1 then the matrix A is .......................(A) Scalar Matrix (B) Square Matrix(C) Symmetric Matrix (D) Skew Symmetric Matrix

(2) If A = -A1 then the matrix A is .......................(A) Scalar Matrix (B) Square Matrix(C) Symmetric Matrix (D) Skew Symmetric Matrix

(3) The value of x, y and z is in 3+x 0 1+y = 1 is ....................... 3-z 2

(A) (-3, 0, 2) (B) (-3, 0, 1) (C) (-3, 0, -1) (D) (1, 3, -2)

(4) A is a matrix of order 3 x 4, B is a matrix of order 3 x 2 and C is a matrix of order

2 x 4. Which of the following has the order 3 x 3 = ..........................

(A) ABC (B) BCA (C) BCA1 (D) CAB

(5) Which of the following is a Scalar Matrix ..........................

(A) 2 0 (B) 0 2 (C) 2 2 (D) 2 00 2 2 0 2 2 2 0

(6) Which of the following is a Symmetric Matrix ..........................

(A) 0 -3 (B) 2 0 (C) 0 5 (D) 2 33 0 0 5 3 0 3 0

(7) Transpose of a row matrix is ....................................(A) Column Matrix (B) UnitMatrix(C) Square Matrix (D) Null Matrix

(8) Which of the follwoing relation is true ....................................(A) A - B = B - A (B) A + B = A1 + B1

(C) A + B = B + A (D) AB = BA

(J - 07)

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4 5 12 15 6 7 18 21

(A) 1 (B) 2 (C) 3 (D) 4

1 2 8 9 3 4 6 7

(A) 7 7 (B) 9 11 (C) 3 3 (D) 11 113 3 9 11 7 7 9 9

(11) Which one is a Skew Symmetric Matrix .......................

(A) 2 1 (B) -3 0 (C) 0 5 (D) 0 -5 -2 -1 0 -3 -5 0 -5 0

1 x-2 2x-1 -1

(A) 1 (B) 2 (C) 3 (D) -1

1 2 0 2 40 3 1 2 5

(A) A + B (B) A - B (C) AB (D) BA

3 2 1 0 1 4 0 1

(A) 4 2 (B) 3 2 (C) 3 1 (D) 2 31 5 1 4 2 4 4 1

(15) 3x 1 5 2 8 3 then x is equal to ............................. 5 4 1 0 6 4(A) 3 (B) 0 (C) -3 (D) 1

(16) 0 5 is a Skew Symmetric Matrix then the value of x is .......................x+6 0

(A) -6 (B) 0 (C) 5 (D) 6

+ =

(9) If A = and KA = then the value of K is .......................

(10) If A = and A+B = then the matrix B is ..........................

(12) If A = is a Skew Symmetric Matrix then the valyue of x is ..............

(13) A = B = then which one is possible......................

(14) A = I = then IA is ..........................

(M - 06)

(J - 06)

(M - 07)

(M - 07)

(M - 07)

(J - 07)

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1 2 3

(A) 1 x 3 (B) 3 x 1 (C) 1 x 1 (D) 3 x 3

(18) 0 x 0 6 then the value of x is ............................. 1 5 1 5

(A) -6 (B) 0 (C) 5 (D) 6

(19) 2x 0 is a Scalar Matrix then the value of x is ............................. 0 x-3

(A) 0 (B) -1 (C) -2 (D) -3

2 3 4 5 1 0 5 6

(A) AB (B) A1B (C) A1B1 (D) BA

(21) 2x y-3 6 3 then value of x and y are ............................. 1 2 1 2

(A) (0.0) (B) (1.2) (C) (2.3) (D) (3.6)

1 3 5 0 5 6

(A) 23 (B) 23 30 (C) 5 18 (D) 2030 5 0 0 30 30

(23) 2 2x-6 7is a diagonal matrix then x is equal to ............................. 1 3

(A) 0 (B) 1 (C) 2 (D) 3

2 3 5 6

(A) 2 5 (B) 2 3 (C) 2 3 (D) 5 63 6 6 5 5 6 2 3

=

=

(17) If A = [ 1 2 3 ] B = then the order of the matrix A x B is .....................

(20) A = and B = then which matrix is possible ..........................

(22) P = and Q = then PQ = .............................................

(24) (AB)1 = then B1A1 = .............................................

(J - 07)

(J - 07)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(M - 09)

(M - 09)

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(25) If A is a Skew Symmetric MAtrix then which one of the following is true ......................

(A) A = A1 (B) A = -A1 (C) A1=(A1)1 (D) A= -A1

6+x 0 0 1

(A) -5 (B) -6 (C) 5 (D) 1

1 2 0 1

(A) 1 2 (B) 5 2 (C) 5 2 (D) 0 20 1 2 1 0 2 1 2

(28) 3x 2 1 4 10 6 then the value of x is ......................... 1 3 2 1 3 4

(A) 2 (B) 3 (C) 4 (D) 5

(29) Order of the matrix A is m x n and order of the matrix B is n x p then order of the

matrix AB is .......................................

1 x+2 2x+1 3(A) 2 (B) -1 (C) -2 (D) 1

< (31) If the order of a given matrix is p x q then the order of its transpose matrix is

.......................................

VI. Solve the following problems:

1 6 3 2 5 1 4 5 -4 5 0 3

(i) A + B (ii) 2B + C (iii) B1 + 2C

(1) If A = B = and C = then find the follwoing:

(26) I = is an Identity Matrix then value of x is ...........................

(27) A = then AA1 is .............................................

(30) If A = is a Symmetric Matrix then the value of x is .................

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(M - 10)

(J - 10)

(J - 10)

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(2) Prove that A + A1 is a Symmetric for any Matrix

A = A1 =

A + A1 = + =

(3) Prove that A-A1 is a Skew Symmetric for any Matrix.

(4) Prive that A + B = B + A for any Matrix

A = B =

A+B = + B+A = +

= =

(5) Prove that A - B = B - A for any Matrix.

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4 -13 2

(6) If A = show that 1/2 (A+A1) is Symmetric

2 5 3 2-1 0 -2 4

A + 1/2 X = B ∴ X = 2 -1/2 X = (B - A) X = 2 (B - A)

= 2

=

1 2 2 1 3 4 1 3

(7) If A = B = and A + 1/2 X = B then find X.

(8) If A = and B = then verify (A+B)1 = A1+B1.

A1 = B1 =

A+B = + A1+B1 = +

= =

(A+B)1 =

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3x 2 1 4 7 6 1 3 2 1 3 4

(9) Solve for x given that + =

1+x 0 3-x 0 -1 2 -1 2

(10) If = find x.

2x 1 3 2 9 3 0 4 2 1 2 5

x2 3 x 1 0 1 4 5 -1 0 6 5

(11) If + = find x.

(12) If - 2 = find x.

(J - 10)

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(13) If = find a, b, c and d.

(14) If = find x and y

\

(15) If = Solve for x and y.

(16) Solve for x and y =

a+1 b-2 0 0c+3 d-1 0 0

x+y 2 4 3 -1 x-y -1 8

1 5 x 2-2 3 y 0

x 2 -1 2 y 3 4 -1

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(17) Find x and y if =

(18) If A = then find A2 and AA1.

A2 = A x A = AA1 =

= =

(19) If A = and B = then verify (AB)1 = B1A1

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

1 3 2 4

1 2 5 -2

x y

-2 3 0 1

2 -1

2 1 -3 0 2 6

1 3 0 -1 2 4

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(20) Prove that A(B+C) = AB + AC for any matrix.

(21) Prove that Multiplication of two matrices is not commulative (AB = BA)

(22) If A = then find A2 2 1 -2 -1

(M - 06)

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(23) If A = then find AA1

(24) If A = then find AA1

(25) A = then find AA1

(26) A = then find A2-2A

1 0 2 3

2 3 5 1

1 2 3 0

1 2 0 3

(J - 06)

(M - 09)

(M - 10)

(M - 07)

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(27) If A = then show that A2-A-3I = 0

(28) If A = then show that A2-8A+13I = 0

(29) If A = then show that A2-5A+7I = 0

(30) If A = and B = then verify A2-B2=(A+B)(A-B)

1 3 1 0

3 1 2 5

3 1 -1 2

1 3 5 7

4 5 7 8

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(31) Find x, if y = and 2x + y =

(32) Find X and Y, if X + Y = and X - Y =

(33) If + 2 = find x and y.

**************

3 2 1 4

1 0 -3 2

7 0 2 5

3 0 0 3

3 -2 -1 4

2x 1

-4 5

8 4y

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PERMUTATIONS AND COMBINATIONS2

93

• Points to remember

n! = n x (n - 1) x (n - 2) ................ x 3 x 2 x 1

n! = n(n - 1)!

0! = 1

nPr = n(n - 1) .......................... (n - r + 1)

nPn = n(n - 1) .......................... 3 x 2 x 1 = n!

nPr = r ≤ n

nCr =

nCr =

nCr = r ≤ n

nCr = nCn-r or nCX = nCY => X = Y or X + Y = n

nC0 = nCn = 1

nC1 = nCn-1 = n

nCr = x n-1Cr-1

nCr + nCr-1 = n+1Cr

n!(n-r)!

nPr

r!

n(n - 1) (n - 2) ......... (n - r + 1)r!

n!r! (n-r)!

nr


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90

FACTORIALI Convert the following products into factorial:

Model Problems

(1) 1. 2. 3. 4. 5 = 5!

(2) 6. 7. 8. 9. 10 = =

(3) 2. 4. 6. 8. 10 = 25 (1. 2. 3. 4. 5) = 25. 5!

II Calculate the following

(1) = =

(2) =

10!5!

10!8!

1. 2. 3. 4. 5. 6. 7. 8. 9. 101. 2. 3. 4. 5

10 x .......... x ............8!

10!6! 4!

10 x .......... x ............ x .......... x ............

6! x .......... x ............ x .......... x ............

III Convert the following product into factorials.

(1) 5. 6. 7. 8. 9. 10

(2) 3. 6. 9. 12. 15. 18

(3) (n+1) (n+2) (n+3) ..................... (2n)

(4) 1, 3, 5, 7, 9............... (2n-1)

IV Find ‘n’ from the following.

(1) (n + 1)! = 12 x (n-1)!

= 12

= 12

............ x ............ = 12

............ x ............ = 3 x 4 ∴ n = ....................

(n+1)!

...........

.......... x ............ x ..........

(n-1)!


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91

(2) (n + 2)! = 60 (n-1)!

= 60

= 60

............ x ............ x ............ = 5 x 4 x 3

∴ n = ....................

(3) If (n + 3)! = 56 (n+1)! find n.

(n+2)!(n-1)!

.......... x ............ x .......... x (n-1)!

(n-1)!


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92

FUNDAMENTAL COUNTING PRINCIPLE

If one event occurs in ‘m’ different ways and another event occurs independently in ‘n’different ways then the two events together can be done in (m x n) different ways.

An Illustration:

A coin is tossed, afterwards, a die is thrown. Let us find the total number of outcomes ofthis experiment.

Total number of outcomes Number of outcomes Number of outcomes

of the experiments of tossing the coin of throwing the die

[h, t] [1, 2, 3, 4, 5, 6]

12 = 2 x 6

Examples:

(1) A and B are connected by five routes and B and C by four routes. Then the

numbber of ways of travelling from A to C enroute B is .......... x ........... = ............

(2) I throw a die twice. The total number of outcomes is ........... x ........... = ...............

(3) The total number of 2 digit numbers in which the digits are non repeating is

............. x .............. = ...............

(4) The total number of two digit numbers is ............. x .............. = ...............

(5) The total number of three digit numbers is ............. x .............. = ...............

(6) The total number of 3 digit numbers with no digit repeated

................ x ................. x .............. = ....................

(7) In a class there are 14 boys and 11 girls. The number of ways one boy and one girl

pair can be formed is ............. x .............. = ...............

(8) In a restaurant, 5 types of sweets, 4 types of snacks and 6 types of drinks areavailable. The number of ways a customer can order for a sweet, a snack and a

drink is ............. x .............. x .............. = ...............

(9) There are 20 buses playing, between Bangalore and Hyderabad. Number of ways

can a person go from Bangalore to Hyderabad is ............. x .............. = ...............

= x

x


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93

PERMUTATION

Permutation is the arrangement of objects which can be made by taking some or allobjects from a set of given objects.

4P3 = 4 x 3 x 2

Note:

(1) On the LHS 4 is at the top of P and first factor on the RHS is 4.

(2) On the LHS 3 is the bottom of P and the number of factors on the RHS is 3.

(3) Successive factors starting with 4 (i.e. the first factor) are got by declaring eachfactor by 1 generalisation

(4) The last factor in (1) on the RHS is got as follows n, (n-1), (n-2) .......... from anAP with first term n, CD = -1.

Tn = a + (n-1) d rth term = n + (r - 1) (-1) = n - r + 1

I Fill up the blanks.

(1) 7P4 = 7 x ............... x ............... x ...............

(2) ................. = 10 x (10 - 1)

(3) ................. = 6 x (6 - 1) x (6 - 2)

(4) 8P5 = ............... x ............... x ............... x ............... x ...............

II Find the value of the following.

(1) 5P2 = ............... x ............... =

(2) 10P3 = ............... x ............... x ............... =

(3) 6P6 = ...............

(4) 4P4 + 4P0 + 4P3 = ............... + ............... + ............... =

(5) nPn = ...............


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94

(6) nP0 = ...............

(7) 0! = ...............

III Write the following in the form of nPr.

(1) 8(8 - 1) =

(2) 8(8 - 1) (8 - 2) =

(3) 8(8 - 1) (8 - 2) (8 - 3) =

(4) 8(8 - 1) (8 - 2) (8 - 3) (8 - 4) =

(5) 8(8 - 1) (8 - 2) (8 - 3) (8 - 4) (8 - 5) =

IV Write the following in the form of nPr.

(1) n(n-1) =

(2) n(n-1) (n-2) =

(3) n(n-1) (n-2) (n-3) =

(4) (n+1) n(n-1) =

(5) (n+3) (n+2) (n+1) n =

(6) (n-1) (n-2) (n-3) =


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95

Teaching Tips:

(1) The following method (of expressing the value as the product of consecutivenatural numbers) is easier

(2) It is always possible to express the value of nPr as the product of consecutivenatural numbers

Their being done, the number of factors in the product is r, while n is the biggest ofthe factors.

V Finding unknown ‘n’ or ‘r’.

A (1) If nPn = 5040 find n.

nPn = n!

n! = 5040

n! = 1 x 2 x ........... x ............. x .............

n! = ......... !

∴ n = ...........

Strategy: Express the given value as product of consecutive natural numbers.

(2) If nPn = 120 find n.

n! = 120

n! = 1 x 2 x .................................

n! = ........... !

n = ..............

(3) n! = 720

n! = ........... x ............. x ..................................

n! = ............!

n = ..............

1 5040

2 5040

3 2520

4 840

5 210

6 42

7

1 120

2 120

3


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96

(B) (1) If 11Pr = 990 find r

11Pr = 11 x 10 x 9

11Pr = 11P3

∴r = 3

(2) 8Pr = 336

8Pr = 8 x .......... x ...............

8Pr = 8P.....

∴r = ....................

(3) 13Pr = 156

13Pr = ............. x .............

13Pr = ...... P.....

∴r = ....................

(C) (1) nP2 = 30

n(n-1) = 6 x 5

n(n-1) = 6 x (6-1)

∴n = 6

(2) nP3 = 210

n(n-1) (n-2) = ............ x ............. x .............

n(n-1) (n-2) = ............ x (...... - 1) x (...... - 2)

∴n = ..........

11 990

10 90

9

8 336

7

13 156


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97

(3) n+1P2 = 6

(n+1) x .............. = 3 x 2

(n+1) x ........... = (2+1) x 2

∴n = ..........

(4) (n+3)P2 = 20

................ x .............. = 5 x 4

................ x .............. = (2+3) x (2+2)

∴n = ..........

(D) (1) If nP4 = 12 nP2 then find n.

n x (n-1) x ............. x ................ = 12 x n x ................

................. x ................ = 12

n2 - 3n - ......... + 6 = 12

n2 - .......... + 6 = 12

n2 - .......... + 6 - 12 = 0

n2 - 5n - 6 = 0

n2 - .......n - ........n - 6 = 0

n (n- ........) - 2 (n-3) = 0

(.............) (.............) = 0

.......... - .......... = 0 or ......... - ......... =0

∴n = .......... or n = ..........


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98

(2) 9(n-1) P3 = nP4 find n.

9 ( ) x ( ) x ( ) = n x ................ x ................ x ................

∴ .................... = ........................

∴ n = ........................

(E) (1) Find the number of permutations of the letters of the word “MILK”.

The word “MILK” consists of .................. different letters

Number of permutaions = 4P4

= 4 x ................ x ................ x 1\

= ....................

(2) Find the number of permutations of the letters of the word “WORLD”.

The word “WORLD” consists of .................. different letters

Number of permutaions = ...................................

= ................ ................ ................

= ....................

(F) (1) In how many ways can 5 students be seated on a bench?

Ist Method:

5 Students are arranged in a row

∴ Total number of arrangements = 5P5

= 5!

= 120


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99

IInd Method:

1st 2nd 3rd 4th 5th ---> Places

Let us take 5 places for five students.

The first place can be filled with 5 students. After allocating one, the 2nd can be filled with4 students, 3rd with 3, 4th and 5th with 2 and 1 respectively.

\ Total number of ways = 5 x 4 x 3 x 2 x 1

= 120 Ways.

(2) Seven athletes are participating in a race. In how many ways can first three prizesbe won?

1st 2nd 3rd

The required number of ways = ................ x .................. x ..................

= ..........................

(3) In how many ways 4 people out of 6 people can be seated in a row for aphotograph.

1st 2nd 3rd 4th

The required number of ways = ............... x ................. x ................. x ................

= ..........................

5 4 3 2 1


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100

Model Problems

(1) How many different numbers of 3 digits can be formed with digit 1, 2, 3, 4, 5(a) Without repeating (b) With repeatation

(a) Without repeatation: Ist Method

100th 10th Unit

The hundered place can be filled in 5 ways.

The tenth place can be filled in 4 ways

The unit place can be filled in 23 ways

∴ The total number of 3 digits = 5 x 4 x 3 = 60

IInd Method

This is nothing but number of permutations of five digits, taking three at a time.

5P3 = 5 x 4 x 3 = 60

(b) With repeatation:

100th 10th Unit

The number of 3 digit numbers that can be formed = 5 x 5 x 5 = 125

(2) A 3-digit numbers can be formed using 0, 1, 2, 3 without repeating any digit?

100th 10th Unit

Any digit except 0 can occupy the 100th place. Therefore, the first place can be

filled in 3 ways, after this second place can be filled in ................. ways, the third

place can be filled in ................ ways.

∴ The total number of ways of following the

places = ................ x ................ x ................ = .................. ways

∴ The total number of 3 digit numbers = ....................

5 4 3

5 5 5


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101

(3) How many numbers between 400 and 1000 can be made with digits 2, 3, 4, 5, 6and 0.

Solution: H T U

100th Place can be filled only with digits 4, 5, 6 = 3 ways.

10th Place can be filled in ................ ways

and Units place can be filled in ...................... ways

∴ The total number of number formation = ............. x ............. x ............. s

= ....................

(4) How many even numbers can be formed using all of 1, 2, 3, 5, 6 without repetitionof the digit in a number.

Solution: 10,000th 1,000th 100th 10th U

The unit place can be filled only in 2 ways (2 or 6)

The remaining box can be filled in ............! ways

∴ The total number of ways = ...........! x 2

= ............. x .............

= ....................

(5) How many seven digit numbers can be formed with digits 1, 2, 3, 4, 5, 6, 7 suchthat they start with 2 and end with 6 (without repetition).

Solution: In this case the first and last position are fixed.

Therefore the question is about arranging 5 digits in the rest of 5

places which is .............! ways.

= ...................... ways

2 6


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102

(6) Find the number of arrangements of word “ARUN” if all the word begin withletter A.

The first place can be filled with ............. and remaining in ............... ways.

∴ Number of arrangements = .....................

= ...............! = .................

(7) How many permutations of all the letters of the word CHEMISTRY

(a) begin with ‘M’ (b) begin with M and end with ‘T’?

Solution: (a)

Number of permutation = ....................

= ....................

(b)

Number of permutation = ....................

= ....................

(8) In how many ways five friends can sit in a row? In how many of there two friendsA and B are side by side?

Solution: (a) Number of arrangements = ......................

= .................... = ................... ways

(b) Here we take the two friends (A and B) as one (as a strategy). Then the

number of friends is 4 and hence can be permutated in =........................ ways

=........................ ways

A and B may be permutated among themselves in = ............. ways

∴ Therefore the total number of arrangements when A and B are side by side

= ....................... = ..................... ways


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103

COMBINATION

Combination is mere a selectio of different objects.

Tips to teach: When n is big and r is close to n, you can avoid unnecessary calculators, using the property nCr = nCn-r.

Case 1: nCx = nCy => x = y

Case 2: nCx = nCn-y => x = n - y => x + y = n

A (1) If nC6 = nC4 find the value of n.

Solution: nC6 = nC4

nC6 = nCn-4

6 = n - 4

n = 10

(2) If nC6 = nC16 find ‘n’

Solution: nC6 = nC16

................. = .................

................. = .................

n = .................

(3) Find r, if (a) 10Cr = 10C12-2r

Solution: r = 12 - 2r or r + 12 - 2r = 10

3r = 12 12 - r = 10

r = 4 r = 2

(4) 10Cr = 10C12-5r

Solution: r = 12 - 5r or r + 12 - 5r = ................

6r = ................ ............... = ...............

r = ................... r = .................


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(5) If 24Cp = 24C2p+3 find P

Solution: ................................ or .................................

................................ ................................

................................ ................................

B (1) Find n if nC2 = 15

Solution: nC2 = 15 nCr =

= 15

= 15

n x ........... = 15 x 2

n x ........... = 30

n x ........... = 6 x 5

∴ n = .................

(2) Find n if nC2 = 10

Solution: nC2 = 10

= 10

= 10

........... x ........... = 10 x 2

........... x ........... = 20

............ x ........... = 5 x 4

∴ n = .................

nP2

2!

n x .......2 x 1

nPr

r!

nP2

2!

....... x .......2 x 1


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(3) If 6Pr = 360 and 6Cr = 15, find r

Solution: nPr = nCr x r!

r! =

r! = ................

r! = .............!

∴ r = .................

(4) If nPr = 720 and nCr = 120, find r

Solution: nPr = nCr x r!

720 = 120 x r!

r! = ................

r! = ................!

r = ..............

(5) If 36nCr = 3nPr find the value of r.

Solution: 36nCr = 3nPr

36 = 3nPr

= 3

................ = ................

r = ..............

........

........

nPr

r!

36

.........


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(6) n+1C3 = 4 x nC2

Solution: = 4 x

= 4 x

=

.............. = ..............

n = ..............

(7) Find the value of (a) 100C2 (b) 100C99

(a) 100C2 = (b) 100C99nCr =

nCn-r

=nC1 = n

= 4950

(c) 10C10 = 1 nCn = 1

(d) 52C0 = 1 nC0 = 1

C (1) There are 10 non-collinear points. How many

(a) Straight Lines (b) Triangles can be drawn by joining these points.

Solution: (a) A line is a 2 combination of points.

Ten points are given, 2 points joined give a line

∴ The number of straight lines formed = 10C2

= ...................

= ...................

= ...................

(n+1)P3

3!

nP2

2!

(n+1) x ........... x .........

.......... x ........... x .........

........... x .........

........... x .........

..............

..............

..............

..............

100P2

2!

100 x 992 x 1

= 100C100-99

= 100C1

= 100


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Solution: (b) A triangle is formed by joining any three non-collinear points in pairs

∴ There are 10 non collinear points.

∴ The number of triangles formed = ...................

= ...................

= ...................

= ...................

(2) How many diagonals can be drawn in a pentagon?

I Method

Consider, a Pentagon, vertices are 5 non collinear points.

No. of straight lines can be drawn

by using 5 non collinear points = 5C2

=

= ...................

= ...................

No. of straight lines = No. of side + No. of diagonals.

∴ No. of diagonals = -

= 10 - 5

= 5

II Method

If a polygon has ‘n’ sides then No. of diagonals = nC2 - n

In a pentagon no. of diagonals = 5C2 - 5

= - 5

= - ..............

= ....................

5P2

.........

No. of straight lines No. ofcan be drawn diagonals

............

.........

............

.........


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III Method

No. of diagonals =

In a pentagon no. of diagonals =

=

= 5

(3) A polygon has 90 diagonals. How many vertices are there for the polygon.

Solution:

Key Idea: When all straight lines are got by joining the vertices of the

polygon, the diagonals are those other than the sides.

No. of diagonals = = 90

n (n - 3) = ............. x ............

n (n - 3) = 15 x ...........

n (n - 3) = 15 x (15 - ........)

∴ n = ...............

n(n-3)2

5(5-3)2

5 x 22

n(n-3)2

(4) A polygon has 44 diagonals. Find the number of sides.

Solution: No. of diagonals =

44 = ..............................

n (n-3) = ............... x ...............

n (n-3) = ............... x (.......... - ..........)

n = .......................

n(n-3)2


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D (1) From a group of 20 cricket players, a team of 11 players is to be chosen. In howmany ways can this be done?

Solution: The total number of forming

the team of eleven

(2) How many groups of 6 persons can be formed from 8 men and 7 women?

Solution: Required number of ways = 15C6

= 20C11

= .........................

= .........................

= .........................

= .........................

= .........................

= .........................

(3) Find the number of combination of the word ‘CAKE’

Solution: The word ‘CAKE’ has 4 different letters.

Select all 4 letters at a time

∴ The number of combinations = ..............................

= ..............................

(4) Find the number of combination of the word ‘SWETHA’

Solution: Required number of ways = ..............................

= ..............................

= ..............................


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(5) A team of cricket eleven has to be formed 20 players Ajit is one. In how manyways the team can be formed. In how many ways the team can be formed so as to

(a) Include Ajit (b) Exclude Ajit

Solution: (a) If the team is to contain Ajit, the choice has to made from theother 19 players to select 10 players.

This can be done in = ........................ ways

= ........................

= ........................

∴ The total number of ways of forming the team including Ajit = ........

(b) If the team does not contain Ajit, such a team is to be formedfrom 19 players (excluding Ajit) selecting 11 of the team.

∴ The total number of ways of forming the team excluding Ajit = ........

= ........................

= ........................

= ........................

Note: 19C10 + 19C11 = 20C11 because any team formed either includes Ajit ordoes not include Ajit.

E (1) In a school of 16 teachers there are 10 men and rest women. A committee of 5teacher to be formed to represent the class. In how many ways this group can bechosen such that at least has 2 women.

Solution: The following possiblities are possible.

No. of Men No. of Women No. of ways

3 2 10C3 x 6C2 =......................

2 3 ............. x ............. = .................

1 4 ............. x ............. = .................

0 5 ............. x ............. = .................

Total number of ways =................ + .................. + ................. + ..............

= ..................


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111

(2) There are 5 bowlers and 10 batsmen in a cricket club. In how many ways can ateam of 11 be selected so that the team contain exactly 3 bowlers.

Solution: No. of teams selected = 5C3 x 10C8

= ............. x .............

= ............. x .............

= ...............

(3) There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a teamof 11 be selected so that the team contain at least 4 bowlers

Solution: The following possiblities are possible.

Bowlers (6) Batsmen (9) No. of ways

4 7 6C4 x 9C7 = ......................

5 6 ............. x ............. = .................

..................... ..................... ............. x ............. = .................

Total number of ways =................ + .................. + .................

= ..................

F (1) A purse contain 3 coins out of which 6 are golden and 4 are silver. 3 coins areremoved from the purse. Find the number of combinations to remove 2 gold coins.

(2) From 7 English men and 4 Americans a committee as 6 is to be formed. In howmany ways can this be done

(i) When the committee contain exactly 2 Americans.

(ii) At least 2 Americans.

Solution 1:

The number of ways in which the American can be chosen: ....................

The number of ways in which the Englishmen can be chosen: ....................

The required number of ways: ............. x ...............


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112

Solution 2:

Americans (4) Englishmen (7) No. of ways

2 4 ...................................

3 3 ...................................

4 2 ...................................

∴ Total number of ways: ............. x ............... x ...............

= ......................

(3) Out of 7 consonants and 4 vowels, how many words can be made each containing3 consonants and 2 vowels.

The number of combined groups, each containing

3 consonants and 2 vowels

Further, each of these groups contian 5 letters, which may be arranged among

themselves = ................. ways

= .............................

= .............................

****************

= .............. x ...............


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113

STANDARD DEVIATION FOR AN UNGROUPED DATA

POINTS TO REMEMBERX = MeanX = Score

• Arithmatic Mean ( X ) = N = Number of scores

Σ = Sigma

Σx = Sum of all Scores

Ex: Find the mean of scores 20, 23, 26, 27, 29

Σx = 20 + 23 + 26 + 27 + 29

= 125

N = 5

∴ X = = = 25

I Fill in the blanks

• Deviation : Deviation = Score - Mean

D = x - X

Ex: 1, 2, 3, 4, 5

X = = = 3

ΣxN

ΣxN

125 5

STATISTICS3

N X

130 10

240 20

14 20

8 12

360 18

480 16

Σx

1 + 2 + 3 + 4 + 55

15 5


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114

D = x - x= 1 - 3 = -2 2 - 3 = -1 3 - 3 = 0 4 - 3 = 1 5 - 3 = 2

Find the deviation for the following scores:5, 8, 12, 13, 17

X = ΣX = ...................

N = ..................

To find Variance and standard deviation of ungroupded data:

• Deviation : 2 = 2 = Variance N = Number of Scores D = Deviations

Standard Deviation ( ) = Variance

=

Ex: Find the variance and standard deviation for the follwoing scores:16, 18, 23, 25, 28

ΣX = 110

N = 5

X =

X = = 22

Scorex x - X = D

58121317

ΣD2

N

ΣD2

N

ΣXN

1105


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115

Scorex x - X = d D2

1618232528

16 - 22 = -618 - 22 = -423 - 22 = 125 - 22 = 328 - 22 = 6

3616010936

• Variance 2 =

= = 19.60

• Standard Deviation

= Variance =

= 19.60

= 4.3

II (1) Find the variance and standard deviations for the following scores: 20, 25, 35, 30, 45, 43

Σx = ............................

N = ..............................

X = = ...............................

• Variance ( 2) =

= .............

• Standard Deviation= Variance

=

= .......................

ΣD2

N985

ΣD2

N

ΣxN

Scorex x - X = D D2

202535304543

ΣD2 =

ΣD2

N

ΣD2

N

ΣD2 = 98

D


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116

(2) Find variance and standard deviation for the following scores: 55, 50, 35, 45, 40

Σx = ............................

N = ..............................

X = = ...............................

• Variance 2 =

=.............

• Standard Deviation= Variance

= = .......................

= ..........................

Ex: If Variance = 1.69 then SD = 1.69 = 1.3

If SD = 3 then Variance = (SD)2

= 32

= 9III Fill up the blanks

ΣxN

ΣD2

N

ΣD2

N

Scorex x - X = D D2

5550354540

ΣD2 =

Variance( 2) SD ( )

1.44

64

0.01

2.5

0.18

7

0.02


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117

STANDARD DEVIATION FOR GROUPED DATA

POINTS TO REMEMBERX = Score

• Mean of Grouped Data ( X ) = f = Frequency

N = Number Scores

X = Mean

Ex: Find the mean from the following table

Score(X) 12 14 16 18 20

frequency (f) 6 8 4 7 5

Mean X =

= =

= 15.8

Ex: Find the variance and standard deviation from the given data.

ΣfXN

47430

15810

ΣfXN

ScoreX

frequencyf fx

1214161820

68475

7211264126100

N = 30 474

Score X 20 30 40 50 60

frequency f 5 4 2 4 5

X f fx X - X = D D2 fD2

2030405060

54245

10012080200300

20 - 40 = -2030 - 40 = -1040 - 40 = 050 - 40 = 1060 - 40 = 20

4001000

100400

20004000

4002000

N = 20 ΣfX=800 ΣfD2 = 4800


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118

• X = = = 40

• Variance ( 2) =

= = 240

• Standard Deviation ( ) =

= 240

= 15.48

IV (1) Find the variance and standard deviation for the following data.

• X = = ..........................

= ..........................

ΣfXN

80020

ΣfD2

N480020

ΣfD2

N 1 240.0000 1 1 25 140 5 125 304 1500 4 12163088 28400 24704 3696

15.48

X 15 20 25 30 35 40

f 8 5 3 2 5 7


15

20

25

30

35

40

8

5

3

2

5

7

ΣfX= ΣfD2 =N=

ΣfXN


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119

• Variance( 2) = ................................

= .............................

= ................................

• ..................................... = Variance

= ..........................................

= ..........................................

= ..........................................

(2) Find the variance and standard deviation for the following data.

X 20 23 25 30 32 35

f 1 2 5 5 2 5


20

23

25

30

32

35

1

2

5

5

2

5

ΣfX= ΣfD2 =N=

• X = ..........................

= ..........................

= ..........................

• Variance( 2) = ................................

= .............................

= ................................


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120

• SD ( ) =

= ..........................................

= ..........................................

(3) Find the variance and standard deviation for the following data.

X 32 35 38 39 40 42

f 2 4 10 6 7 1


32

35

38

39

40

42

2

4

10

6

7

1

ΣfX= ΣfD2 =N=

• ............ =

= ..................................................

= ..........................

• ............ =

= .............................

= ................................

• ............ = Variance

= .............................

= ..............................

= ..............................

ΣfXN

ΣfD2

N


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121

STANDARD DEVIATION FOR GROUPED SCORES WITHEQUAL CLASS INTERVALS

POINTS TO REMEMBERX = Mid point of the class interval

• X = f = frequency

N = Number of frequencies

Ex: Find the variance and standard deviation for the following data:

C I 1 - 5 6 - 10 11 - 15 16 - 20

f 1 2 3 4

• X =

= = 13

• Variance ( 2) =

=

= 25

• SD ( ) = Variance

=

= 25

= 5

ΣfXN

CI f Mid PtX fX X - X = D D2 fD2

1 - 5

6 - 10

11 - 15

16 - 20

1

2

3

4

3

8

13

18

3

16

39

72

3 - 13 = -10

8 - 13 = -5

13 - 13 = 0

18 - 13 = 5

100

25

0

25

100

50

0

100

N = 10 130 ΣfD2 = 250Σfx =130

ΣfXN

13010

ΣfD2

N25010

ΣfD2

N


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122

V (1) Find the variance and standard deviation for the following data:

\

• Mean X = ..............................

= ..............................

= ...............................

• Variance ( 2) =

= ..............................

= ..............................

• SD ( ) =

= ..............................

= ..............................

= .................................

C I 5 - 9 10 - 14 15 - 19 20 - 24 25 - 29

f 1 2 4 3 5

CI f Mid PtX fX X - X = D D2 fD2

5 - 9

10 - 14

15 - 19

20 - 24

25 - 29

1

2

4

3

5

N = ΣfD2 =ΣfX =

ΣfD2

N


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123

(2) Find the variance and standard deviation for the following data:

\

• ................ =

X = ..............................

= ...............................

• Variance ( 2) = ..............................

= ..............................

= ..............................

• .............................. = Variance

= ..............................

= ..............................

= .................................

C I 30 - 34 25 - 29 20 - 24 15 - 19 10 - 14 5 - 9

f 1 4 6 5 3 1

CI f X fX X - X = D D2 fD2

30 - 34

25 - 29

20 - 24

15 - 19

10 - 14

5 - 9

1

4

6

5

3

1

N = ΣfD2 =ΣfX =

ΣfXN


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124

(3) Find the variance and standard deviation for the following grouped data:

\

• X = ..............................

= ..............................

= ...............................

• Variance ( 2) = ..............................

= ..............................

= ..............................

• SD ( ) = ...............................

= ................................

= ..................................

= .................................

C I 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44

f 3 4 6 8 7 2


15 - 19

20 - 24

25 - 29

30 - 34

35 - 39

40 - 44

3

4

6

8

7

2

N = ΣfD2 =ΣfX =


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125

(4) Find the variance and standard deviation for the following data:

\

• X = ..............................

= ..............................

= ...............................

• Variance ( 2) = ..............................

= ..............................

= ..............................

• SD ( ) = ...............................

= ................................

= ..................................

= .................................

C I 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60

f 1 5 6 5 3

CI f X fX X - X1 = D D2 fD2

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

N = ΣfD2 =ΣfX =


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126

CO-EFFICIENT OF VARIATION (C.V.)

POINTS TO REMEMBER∴ CV = Coefficient of Variation

• CV = X 100 = Standard Deviation

X = Mean

Coefficient of Variation determines Consistency or Variability.

• The measure of despersion having less co-efficient of variation is more Consist or less Variation.• The measure of despersion having more CV is less Consistant or more Variation

Ex: The performance of two teams in an examination is given below. Which team is more Consistant.

Team Mean Score( X )

SD( )

A 13 0.2

B 15 0.3

CV = X 100

CV of A Team = X 100

=

= 1.53

CV of A Team = X 100

=

= 2

∴ CV of A team is less than CV of B team. A team more Consistant.

X

0.213

X

2013

0.3153015


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127

VI (1) The performance of Rama and Ravi in Mathematics is given below. Which one is more Consistant?

CV = ................................

CV of Rama = ................................

= ................................

= ................................

X

Rama 60 0.3

Ravi 75 0.5

0.5(........)

CV of Ravi = X 100

= ................................

= ................................

CV of ............................. is less than CV of ..............................

∴ ................................ is more Consistant

(2) The mean and standard devation of heights and weights of 20 persons are given below

In which carecteristic are they more variable?

CV = ................................

CV of Height = ................................

= ................................

= ................................

CV of ............................. is more than the CV of ..............................

∴ ................................ is more Variable.

X

Height in cm 160 0.4

Wight in K gs 50 0.3

CV of Weight = ................................

= ................................

= ................................


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128

(3) The performance the yeild of mango trees Ramappa, Seenappa and Rajappa are

given below. The yeild of whom tree is more Consistant.

CV = x 100

CV of Ramappa = ................................

= ................................

= ................................

CV of Seenappa = x 100

= ................................

= ................................

CV of Seenappa = x 100

= ................................

= ................................

On comparing CV of ............................. is least.

Hence performance of .............................. is more Consistant

Trees of X

Ramappa 50 0.6

Seenappa 75 0.3

Rajappa 60 0.5

X

0.3......

........60


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129

(4) Find the co-efficient of variation for the follwoing data.

\ • X = ..............................

= ..............................

= ...............................

• SD ( ) = Variance

= ..............................

= ..............................

= ..............................

• CV = x 100

= ................................

= ..................................

= .................................

C I 5 - 9 10 - 14 15 - 19 20 - 24

f 1 2 3 4


5 - 9

10 - 14

15 - 19

20 - 24

1

2

3

4

N = ΣfD2 =ΣfX =

X


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130

VII

(1) The mid point of the class interval 25 - 29 is .......................

(2) The mean of 4 scores is 20 and three scores among those are 16, 19, 21 then the

4th score is ..............................

(3) The mid point of the class interval below the class interval 40 - 44 is .......................

(4) Formula to find the mean of individual scores is ...........................

(5) Formula to find the man of grouped score is ............................

(6) Formula used to find the standard deviation of individual scores is .........................

(7) Formula used find the standard deviation of grouped scrores is ...........................

(8) Formula used to find the CV is ...........................

(9) The measure compared to find the conisistancy is ...........................

(10) The measure compared to understand the variability is ...........................

(11) The variance of the scores is 0.0625. Then the standard deviation is .......................

(12) The SD of scores is 1.6 then variance is ...........................

(13) The CV is 30 and the mean is 10, then SD is ............................

(14) The CV = 1.5, = 0.3 then X = ............................

(15) If = 0.6, X = 30 then CV = ............................

(16) The CV of two players Rama and Ramesh are 3/5 and 5/7. The more consistant

player is ...........................

(17) The CV of two teams A and B are 2/3, 4/5 then the team which varies more is

.............................

**************


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FACTORS AND FACTORISATION4

POINTS TO REMEMBER

• a2 + 2ab + b2 = (a+b)2

• a2 - 2ab + b2 = (a-b)2

• a2 - b2 = (a+b) (a-b)

• a3 + b3 = (a+b) (a2 - ab + b2)

• a3 - b3 = (a-b) (a2 + ab + b2)

• a4 + a2b2 + b4 = (a2 + ab + b2) (a2 - ab + b2)

H.C.F. and L.C.M. by factorisation method:

Expression Factors HCF LCM

1

5x – 10

5x2 - 20

5x – 10

= 5(x-2)

5x2 – 20

= 5(x2 – 4)

= 5(x2 – 22)

= 5(x+2)(x-2)

5(x-2)

5(x-2)(x+2)

2

3x – 9

5x2 - 45

3x – 9

= 3( )

5x2 – 20

= 5( )

= 5( )

= 5( )( )

................

.......................

3

a2+7a+12

a2+8a+15

a2+7a+12

= ...................

= a( ) +3( )

= (a+3) ( )

5x2 – 20

= 5( )

= 5( )

= 5( )( )

................

.......................

4

a2b2(p2-16)

ab(p3-64)

a2b2 (p2-16)

= a2b2 ( )

= a2b2 ( )( )

ab(p3 – 64)

= ab(p3 - 43)

= ab(p - 4)( )

= ab(p - 4)( )

= a3 – b3

................

.......................

Factors

131


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5

a2 – b2

a3– b3

a2 – b2

a3 – b3

................

.......................

6

p2 – 4

p2– 5p + 6

p2 – 4

= p2 – 22

= ( )( )

p2– 5p + 6

= p2– 3p – 2p + 6

= p( )-2( )

= (p-2) ( )

................

.......................

7

8x2 – 1

4x2+ 2x + 1

8x3 – 1

4x2+ 2x + 1

................

.......................

8

a – b

b – a

a – b

b – a

................

.......................

9

m2 – n2

(m + n)2

m2 – n2

(m + n)2

................

.......................

10

18x3y4z9

15x4y7z9

18x3y4z9

15x4y7z9

................

.......................

Factors

3

132


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11

a + b

a – b

a + b

a – b

................

.......................

12

(x + y)2

(x – y)2

x2 – y2

................

.......................

13

ab, bc, ca

................

.......................

14

a – 2

a3 – 8

a – 2

a3 – 8

................

.......................

15

a2 – 9

a2 + 6a + 9

a2 – 9

a2 + 6a + 9

................

.......................

16

7(m+n)2(m-n)2

14(m2 - n2)

7(m+n)2(m-n)2

14 (m2 – n2)

................

.......................

Factors

133


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134

H.C.F. by Division Method:Note: (1) Arrange the terms of the expressions in decreasing order of the powers in two columns

(2) If the last remainder is a constant and not zero, then the HCF of two expressions is 1

(1) Find the HCF of 2m2 + 2m + m3 + 1 and 2m + 1 + m2

[Hint: Arrange the terms of the expressions in decreasing order of the powers]

m m3 + 2m2 + 2m + 1 m2 + 2m + 1 m

m3 + ...........+............. ............+ m

+ m + 1 m + 1

HCF = (m + 1)

(2) x3 - 86x + 35 - 6x2 x3 + 40 - 5x2 - 99x

(3) x3 + 2x2 + 2x + 1 x3 - x2 - x - 2

(4) Find the HCF of expressions x3 - 7x2 + 14x - 8 and x3 - 6x2 + 11x - 6

x3 - 7x2 + 14x - 8 x3 - 6x2 + 11x - 6


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(5) Find the HCF of expressions x3 - 9x2 + 26x - 24 and x3 - 6x2 + 11x - 6

x3 - 9x2 + 26x - 24 x3 - 6x2 + 11x - 6

(6) Find the HCF of expressions x4 + 3x3 - x - 3 and x3 + x2 - 5x + 3

x4 + 3x3 - x - 3 x3 + x2 - 5x + 3

(7) Find the HCF of 3x4 + 6x3 - 12x2 - 24x and 4x4 + 14x3 + 8x2 - 8x

3x4 + 6x3 - 12x2 - 24x

3x(x3 + 2x2 - 4x - 8) [Taking 3x as C.F.]

4x4 + 14x3 + 8x2 - 8x

2x(2x3 + 7x2 + 4x - 4) [Taking 2x as C.F.]

x3 + 2x2 - 4x - 8 2x3 + 7x2 + 4x - 4

HCF =

HCF of 3x and 2x is ............................

HCF of the expressions = ( ) ( )


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I Relation Between Two Expressions and their HCF and LCM

The product of the HCF and LCM of two expressions is equal to the product of the

expressions.

A x B = H x L

(1) If A and B are two expressions and their HCF is H, then their LCM can be calculated

using the formula is ......................................

(A) L = (B) L =

(C) L = (D) L =

(2) If H and L are the HCF and LCM of the algebraic expression, A and B respectively.

Then which of the following is correct? ...................................

(A) A x H = B x L (B) =

(C) = (D) =

(3) HCF and LCM of two algebriac expressions are ax and 12ax2b3y respectively. If one

of the expressions is 4axy, then the other is ........................................

(A) 3xa2b2 (B) 3x2a (C) 3x2ab3 (D) 3xab2

(4) If H = (a-3), L = (a-3) (a-2) (a-1) and A = a2-5a+6 then the other experssion ‘B’ is

.......................................

(A) a2+3a+3 (B) a2-4a+3 (C) a2+2a-3 (D) a2-3a-1

(5) The product of HCF and LCM of two expressions is 6a3b4c2. If one expression is

2a3b3c2, then the other is .......................................

(A) 3abc (B) 6bc (C) 3b (D) 3bc

(6) HCF and LCM of two expressions are 5x2y2 and 10x3y3 respectively. If one of the

expressions is 5x2y3 then the other is .......................................

(A) 10x3y3 (B) 10x2y2 (C) 10x3y2 (D) 5x3y2

H x AB

A x BH

BA x B

AH x B

AH

LB

AL

BH

AB

HL


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LCM by Division Method

II (A) (1) Find the LCM of x3 + 2x2 + 2x + 1 and x3 - x2 - x - 2

L = x B

A = x3 + 2x2 + 2x + 1 B = x3 - x2 - x - 2

First find the HCF of two expressions A and B

1 x3 + 2x2 + 2x + 1 x3 - x2 - x - 2

x3

HCF =

∴LCM of the two given expressions can be find by using the relation

L = x B

L = x x3 - x2 - x - 2

............................

.................. x3 + 2x2 + 2x + 1

∴L = .................................... x (x3 - x2 - x - 2)

AH

AH

x3 + 2x2 + 2x + 1

......................


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(2) Find the LCM of x3 - 7x2 + 14x - 8 and x3 - 6x2 + 11x - 6

(3) Find the LCM of m4 + 3m3 - m - 3 and m3 + m2 - 5m + 3


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(4) x3 - 3x2y - 18xy2 + 40y3 and x3 - 4x2y - 11xy2 +30y3

B (1) HCF and LCM of two expressions are (x-3) and x3 - 5x2 - 2x + 24 respectively. Ifone of the expression is (x2 - 7x + 12). Find the other.

B = x H

= x x - 3

..............................

x2 - 7x + 12 x3 - 5x2 - 2x + 24

B = ................................. x (x - 3)

LA

x3 - 5x2 - 2x + 24

x2 - 7x + 12

H = x - 3L = x3 - 5x2 - 2x + 24A = x2 - 7x + 12B = ?


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(2) The HCF and LCM of two algebraic expressions are (a-3) and a3 - 5a2 - 2a + 24respectively. If one of the expressions is a2 - 7a + 12. Find the other.

B = x H

..............................

a2 - 7a + 12 a3 - 5a2 - 2a + 24

B = ................................. x (a - 3)

(3) The product of two expressions is a4 - 9a2 + 4a + 12 and their HCF is (a-2). Findtheir LCM.

L =

=

..............................

a-2 a4 - 9a2 + 4a + 12

L = .................................

LA

A x BH

a4 - 9a2 + 4a + 12

a-2


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III (1) The HCF and LCM of two expressions are a-3 and a3 + a2 - 17a + 15 respectively.Find the two expressions.

a2 + 4a - 5

a - 3 a3 + a2 - 17a + 15

-a3 - 3a2

4a2 - 17a

-4a2 - 12a

- 5a + 15

- 5a + 15

0 0

H H

A = (a - 3) (a + 5) B = (a - 3) (a - 1)

= ................................... = ...................................

= ................................... = ...................................

(2) The HCF and LCM of two expressions are (m-3) and m3 + m2 - 17m + 15

respectively. Find the two expressions.

....................................

m - 3 m3 + m2 - 17m + 15

A = (m-3) x ................................... (Factor 1)

B = (m-3) x ................................... (Factor 2)

+

+

+ -

a2 + 4a - 5

= a2 + 5a - a - 5

= a (a+5) - 1 (a+5)

= (a+5) (a-1)

F1 F2

F1 F2


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IV (1) M and N are two prime numbers, then their HCF is .........................................

(2) Factors of a2 - b2 is .........................................

(3) If one of the factors of a3 - b3 is (a-b), then the other one is ..................................

(4) (a+b) and a2 + b2 - ab are the factors of ..................................

(A) a3 + b3 (B) a3 - b3 (C) (a + b)3 (D) (a - b)3

(5) Algebraic expression which is having the factors (x+1) and (x+2) is ......................

(A) x2 + 2x + 2 (B) x2 + 3x + 2 (C) x2 + x + 2 (D) x2 + 3x + 2D)

(6) If one of the factor of a4+a2b2+b4 is a2+b2+ab then the other factor is ......................

(A) a2 + b2 (B) a2 - b2 (C) a2 -ab + b2 (D) a4+b4-ab

D)

(7) The expression obtained by s ubtracting x3 - 7x2 + 14x - 8 from x3 - 6x2 + 11x - 6

is ........................................

(A) x2 - 3x + 2 (B) x2 + 3x - 2 (C) x2 + 3x + 2 (D) x2 - 3x + 2


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CYCLIC SYMMETRY

An expresion in three variables a, b and c is said to posses “Cyclic Symmetry” if we get backthe original expression by changing a to b, b to c and c to a in order. Then the expression iscalled a cyclically symmetric expression in a, b, c

Ex: (1) Show that x2 (y - z) + y2 (z - x) + z2 (x - y) is cyclically symmetrical expression.

x2 (y - z) + y2 (z - x) + z2 (x - y) --- (1)

Changing the variables x to y, y to z and z to x in cyclic order.

y2 (z - x) + z2 (x - y) + x2 (y - z) --- (2)

Compare equation (1) and (2) both are same

∴x2 (y - z) + y2 (z - x) + z2 (x - y) is cyclically symmetrical expression.

I Show That the following expressions are cyclically symmetrical expression.

(1) a (b - c) + b (c - a) + c (a - b) --- (1)

b (c - a) + ............ + a (.........) --- (2)

(2) ab (a + b) + bc (b + c) + ca (c + a) --- (1)

................+ ...............+ ab(a + b) --- (2)

(3) a2b (a - b) + b2c (b - c) + c2a (c - a) --- (1)

............... + c2a (c - a) + ............... --- (2)

(4) xy2 (x - y) + yz2 (y - z) + zx2 (z - x) --- (1)

yz2 (y - z) + ............... + ............... --- (2)


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II Write the following using ‘Σ’ Notation

Ex: (1) a + b + c

= Σ a

(2) a2 + b2 + c2 - ab - bc - ca

= a2 + b2 + c2 - (ab + bc +ca)

= Σ a2 - Σ ab

(3) ab (a + b) + bc (b + c) + ca (c + a)

(4) x (y2 - z2) + y (z2 - x2) + z (x2 - y2)

(5) ab2 - ac2 - a2b + bc2 - cb2 +ca2

= ab2 + bc2 + ca2 - a2b - cb2 - ac2

= ab2 + bc2 + ca2 - a2b - b2c - c2a

= ab2 + bc2 + ca2 - [a2b + b2c + c2a]

III (1) The missing term of cyclically symmetrical experssion

ab2(a - b) + .............. + ca2 (c - a) is

(a) ba2 (b - a) (b) bc2 (b - c) (c) bc2 (a - b) (d) bc2 (c - b)

(2) Which of the following is a cyclically symmetrical? .......................................

(a) (a+b) (b+c) (c+a) (b) xy (x-y) + yz (y-z) + zx (z-x)

(c) (a-b) (b-c) (c-a) (d) 3 (a2+b2) + 2ab + a + b

abc

abc abc


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(3) Which of the following is cyclically symmetrical? ........................................

(a) (a-b) + (b-c) + (c+a) (b) a2 (b-c) + b2 (c+a) + c2 (c-b)

(c) ab(a-b) + bc(b-c) + ca(c+a) (d) a2 (b+c) + b2 (c+a) + c2 (c+b)

(4) Which of the following is not cyclically symmetrical? ........................................

(a) a+b+c (b) a2 + b2 + c2 + 2ab + 2bc + 2ca

(c) a(a-b) + b(b-c) - c(c-a) (d) ab2 (a-b) + bc2 (b-c) + ca2 (c-a)

(5) Which of the following has no cyclic symmetry? ..........................................

(a) x + y + z (b) xy + yz + zx

(c) xy - yz - zx (d) -x2 - y2 - z2

(6) Which of the following is be added to ab2+a2c to make it cyclic symmetry.......................

(a) a2b (b) bc2 (c) b2c (d) ac2

(7) Which of the following has to be added to a3 + b3 + c3 + 3ab (a + b) to make it

cyclically symmetric? ....................................

(a) bc ( b + c) + ca (c + a) (b) 3bc (b + c) + ca (c + a)

(c) 3 [bc (b+c) + ca (c+a)] (d) 3abc (bc+c) + (c+a)

(8) Which of the following expression is in cyclic symmetry if a, b and c are

variables.................................

(a) a + b + c) (b) a2 - b2 - c2 (c) a - a2 - a2 (d) a - b - c

(9) Which of the following is in cyclic with x, y, z variables?

(a) x2 + xy (b) x2 + z2 + xz + yz

(c) x2 + y2 + z2 (d) x2 + y2 + xy + yz


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IV (1) a + b + c in Σ notation is ...................................................

(a) Σ a (b) Σab (c) Σabc (d) Σ ac

(2) a2 + b2 + c2 - ab - bc - ca is represented in Σ notation is ...........................................

(a) Σ a2 (b) Σa2+ab (c) Σa2-Σab (d) Σa2-ab

(3) ab + bc + ca can be written by using Σ as ...........................................

(a) Σab (b) Σabc (c) Σa (d) Σa+b+c

(4) Write xy2 - xz2 + yz2 - yx2 + zx2 - zy using Σ notation by ...........................................

(a) Σ x(y2-z2) (b) Σx (y2+z2) (c) Σ xy(z-x) (d) Σ xy(z+x)

(5) ab + ac + bc + ab + ca + bc can be written by using Σ notation as .....................

(a) 2Σb(a+b) (b) 2Σc(a+c) (c) 2Σa(b + c) (d) 2Σa(a+b)

(6) The one which is equal to ab(a + b) + bc(b + c) +ca(c + a) is.....................

(a) Σab(a+b) (b) Σc2(a+b) (c) Σa2(b+c) (d) Σb2(a+b)

(7) ab2 - ac2 - a2b + bc2 - cb2 + ca2 can be written as follows.....................

(a) Σab (a+b) (b) Σab (b-a) (c) Σa (b2-c2) (d) Σb(a-b)

(8) a2 + b2 + c2 + 2ab + 2bc + 2ca can be expressed using Σ notation as ....................

(a) Σa2 - Σ2ab (b) Σa(a-2b) (c) (Σa)2 (d) Σ(a+b+c)2

(9) Σab (b-c) is same as .....................

(a) Σab(a-b) (b) Σbc(c-a) (c) Σca(a-c) (d) Σab2+Σabc

(10) Which of the following cannot be written using Σ notation.

(a) a2+b2+c2 (b) ab2+bc2+ca2+3abc

(c) a3-b3-c3 (d) (a-b) + (b-c) + (c-a)

abc abc abc abc

abc abc abc abc

abc abc abc abc

abc abc abc abc

abc abc abc abc

abc abc abc abc

abc abc abc abc

abc abc abc abc

abc abc abc abc abc

abc

abc


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(11) When Σ notation is used, the expression x2 + y2 + z2 - x - y - z becomes .....................

(a) Σ (x2+x) (b) Σ(x-x2) (c) Σx2+Σx (d) Σ(x2+x)

(12) c + + b + + c + can ne expressed using Σ notation is .................

(a) Σ a2 (b) Σa2+ab (c) Σa2-Σab (d) Σa2-ab

V (1) Σx = 0, then 3xyz is equal to ...........................................

(a) Σxy (b) Σx2 (c) Σx3 (d) Σxyz

(2) Σa2 + Σ2ab is equal to ...........................................

(a) (a+b)2 (b) (a+b)3 (c) (a2+b2+c2) (d) (a+b+c)2

(3) If a, b, c are variables, then Σa2 + b is .....................................

(a) a2 + b + b2 + c + c2 + a (b) a2 - b + b2 - c + c2 - a

(c) a2 + b - c + b2 + c - a (d) a2 + b2 + c2

(4) Σx(y-z) is expanded and simplified to obtain .....................................

(a) xy + xz (b) 0 (c) xz - xy (d) y(z-x)

(5) The expression of Σp2 is ....................................

(a) p2 + q2 + r2 (b) p2 (c) q2 (d) pqr

(6) The expanded form of Σ a(a-b) is ...........................................

(a) (a-b) + (b-c) + (c-a) (b) a(b-c) + b(c-a) + c(a-b)

(c) a - b - c (d) a(a-b) + b(b-c) + c(c-a)

(7) Σa2 (b2 - c2) when expanded and simpliefied ...........................................

(a) a2b2 - a2c2 (b) a2b2 - a2c2 + b2c2

(c) a2b2 + b2c2 + c2a2 (d) 0

1b2

1c2

1a2

xyz xyz xyz xyz

xyz xyz xyz xyz

xyz

abc abc abc abc

abc abc

abc

xyz

abc

pqr

abc


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(8) Σa2 - Σb2 is ............................

(a) 1 (b) 0

(c) a2 + b2 + c2 (d) 2a2 + 2b2 + 2c2

(9) Σ (a+b) - Σ (c+a) is ..............................

(a) 0 (b) a+b+c

(c) 2a+2b+2c (d) None of the above

(10) Σa2 (b2-c2) when expanded and simplified ...........................................

(a) a2b2 - a2c2 (b) a2b2 - a2c2 + b2c2

(c) a2b2 + b2c2 + c2a2 (d) 0

(11) When Σ (x+y) is expanded, we get ...........................................

(a) x + y + z (b) 2x + 2y + 2z (c) 3x + 3y + 3z (d) 3xyz

(12) The value of Σ (a-b) is equal to .....................................

(a) a - b - c (b) a + b + c (c) 1 (d) 0

(13) When Σx(y - z) expanded and simplified, its value is .....................................

(a) 0 (b) xy - yz - zx

(c) xy - xz (d) xy + yz + zx

(14) Σa - Σb is ....................................

(a) a + b + c (b) 2a + 2b + 2c (c) 0 (d) a - b

VI (1) If the sum of three numbers is 0 and the sum of the cubes of the same numbers is 99,

then the product of these numbers is ...........................................

(a) 9 (b) 33 (c) 24 (d) 30

abc abc

abc abc

abc

abc

abc abc

xyz

xyz


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(2) If a + b + c = 0 the Σ is ............................

(a) 3 (b) 0

(c) -3 (d) None of the above

(3) The sum and the product of three numbers are 0 and 30 respectively. The sum of

their cubes is ..............................

(a) 0 (b) 90 (c) 160 (d) 900

a+b3abc


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CONDITIONAL IDENTITY

Points to be remember

(1) (a+b)2 = a2 + b2 + 2ab

(2) (a-b)2 = a2 + b2 - 2ab

(3) a2 - b2 = (a+b) (a-b)

(4) (x + a) (x + b) = x2 + x (a+b) + ab

(5) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(6) (a + b)3 = a3 + b3 + 3ab (a+b)

(7) (a - b)3 = a3 - b3 - 3ab (a-b)

(8) a3 + b3 = (a+b) (a2 + b2 - ab)

(9) a3 - b3 = (a-b) (a2 + b2 + ab)

(10) a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

(11) a4 + a2b2 + b4 = (a2 + b2 + ab) (a2 + b2 - ab)

(12) (x + a) (x + b) (x + c) = x3 + x2 (a+b+c) + x (ab + bc + ca) + abc

************

I (1) What is an Identity? Give an example.

(2) What is meant by conditional identity? Give an example.


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II (1) If a + b + c = 0, prove that a3 + b3 + c3 = 3abc.

a + b + c = 0

a + b = ..................

Cubing both sides

(a + b)3 = ...................................

a3 + .......... + 3ab (a + b) = ..............................

a3 + .......... + 3ab (- c) = ..............................

a3 + b3 + c3 = ...........................

(2) If a + b + c = 0, show that (b + c) (b + a) = ac

a + b + c = 0

b + c = ..................

b + a = ..................

LHS = (b + c) (b + a)

= .................. x ..................

= ........................................

(3) If a + b + c = 0, prove that (b + c) (b - c) + a (a + 2b) = 0

LHS = (b + c) (b - c) + a (a + 2b)

= b2 .......... ( ................... ) + a2 + ..........

= a2 + b2 + 2ab - ( ................... )

= (a + b)2 .......... ( ....................)

= (a + b + c) (......................)

= 0 (...............................)

LHS = 0

LHS = RHS


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(4) If a + b + c = 0, prove that a (a2 - bc) + b (b2 - ca) + c (c2 - ab) = 0

LHS = a (a2 - bc) + b (b2 - ca) + c (c2 - ab)

= a3 ......... (.................) + b3 ......... (.................) + c3 ......... (.................)

= a3 + b3 + c3 - .............................................

= 3abc - ..............................

LHS = ..............................

LHS = ..............................

(5) If a + b + c = 0, prove that a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = -3abc

LHS = a (b2 + c2) + b (c2 + a2) + c (a2 + b2)

= ab2 + ac2 + bc2 + ba2 + ca2 + cb2

= ab2 + ba2 + bc2 + cb2 + ac2 + ca2

= ab (.................) + bc (.................) + ac (.................)

= ..................... + ..................... + ......................

LHS = ........................................

LHS = RHS

(6) If a + b + c = 0, prove that + + = 3

LHS = + +

=

=

=

= ....................

a2

bcb2

cac2

ab

a2

bcb2

cac2

ab

.......... + ............ + ...........abc

.......................abc

.......................abc


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III (1) If a + b + c = 0, then the value of (b + c) (b + a) = .....................................

(2) If a + 2b + 3c = 0, then the value of a3 + 8b3 + 27c3 = .....................................

(3) If x + y + z = 0, then + + = ..........................

(4) If a + b + c = 0, then which is equal to (b + c) (c + a)?

x2

yzy2

xzz2

xy


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(5) If x2 + y2 = 5 and xy = 2 then x + y = .....................................

(6) x + - x - is equal to .....................................

(7) If a + b + c = 0, then (a + b) (b + c) (c + a) = .....................................

(8) If a - 2b - 3c = 0 then a3 - 8b3 - 27c3 = .....................................

1x

2 1x

2


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(9) If a + b + c = 0 then, + + + +

is equal to .....................................

(10) If + + = 0, then + + = ........................

(11) If x2 - 3x + 1 = 0 then x + = ........................

x2 - 3x + 1 = 0

÷x - + = 0

∴ x - 3 + = 0

x + = .............

a+bc[ b+c

ac+ab ] c

a+b[ ab+c

bc+a ]

ab

bc

ca

a3

b3b3

c3c3

a3

1x

x2

x3xx

1x

1x

1x


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(12) If p + = 2 then p2 + is ................................

p + = 2

Squaring both sides

p + = 22

.....................................

.....................................

.....................................

(13) If a + b + c = 0, then the value of a + b - 2c is .....................................

(14) If x + y = 7 and xy = 10 then the value of is .....................................

(x - y)2 = (x + y)2 - 4xy

= ..................................... =

= .....................................

= .....................................

1p

1p2

1p

1p

2

x + yx2 - y2

x + yx2 - y2

x + y(x + y) (x - y)

= .....................................

= .....................................


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(15) The value of (a + b)2 - (a - b)2 is .................................

(16) If a + b + c = 0, show that b2 - 4ac is a perfect square.

a + b + c = 0

a + c = -b

Squaring both sides

(a + c)2 = (-b)2

............................. = .............................

Subtract 4ac both sides

............................. - 4ac = ............................. - 4ac

............................. = .............................

∴b2 - 4ac = .............................

∴b2 - 4ac is a perfect square

(17) x + = 3, then prove that x3 + 2x + + = 30

LHS = x3 + 2x + + = 30

= x3 + + 2 x +

= ..................... + 2 (.............)

= .........................................

1x

2x

1x3

2x

1x3

1x3

1x

x + = 31x

Cubing both sides

x + = 33

x3 + + 3 x x x = 27

x3 + = 27 - (...........)

x3 + = .....................

1x

3

1x

1x3

1x3

1x3


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(18) If a + b + c = 0, show that + + + = 0

LHS = + + + = 0

= +

= +

= + [Substitute a3+b3+c3 = 3abc]

= = ...........................

= =

= ....................................

(19) If a + b + c = 0, then prove that a2 - bc = b2 - ca = c2 - ab = - (ab+bc+ca)

a2 - bc a + b + c = 0

= a x a - bc b = -b - c

= a (-b - c) - bc

= -ab - ca - bc

= .............................. = ..............................

(20) If a + b + c = 0, then prove that Σa (a2 - bc) = 0

Σa (a2 - bc)

= a (a2 - bc) + ....................... + .......................

= a3 - abc + ....................... + .......................

= a3 + ............. + ................. - 3abc

= .............................................

= .............................................

a2 + b2 + c2

a3 + b3 + c3

23

1a

1b

1c

a2 + b2 + c2

a3 + b3 + c323

1a

1b

1c

a2 + b2 + c2

a3 + b3 + c323

bc+ca+ababc

a2 + b2 + c2

a3 + b3 + c32ab + 2bc + 2ca

3abc

a2 + b2 + c2

a3 + b3 + c32ab + 2bc + 2ca

3abc

.......................................

.......................................

........................

........................

........................

........................

abc

abc


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(21) If 3x - 4y = 5 then P.T. 27x3 - 64y3 - 180xy = 125

LHS = 27x3 - 64y3 - 180xy

= (3x)3 ......... (.............)3 ......... 3 x (.........) x (.........) x (...............)

= (..............) ......... (..............) ......... (....................)

= ...........................................................

= ...........................................................

(22) If x + y + z = 0, P.T. (xy + yz + zx)2 = x2y2 + y2z2 + z2x2

LHS = (xy + yz + zx)2 Hint: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca

= (x + y + z) = 0

(23) If a + b + c = 0 then P.T. + + + + = 9

LHS = + + + + = 9

Hint: a+b+c = 0

a+b = -c

b+c = -a

c+a = -b

ab+c[ b

c+ac

a+b ] b+ca[ c+a

ba+b

c ]a

b+c[ bc+a

ca+b ] b+c

a[ c+ab

a+bc ]


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(24) If ab + bc + ca = 1, then S.T. =

LHS =

= [Substitute 1 = ab + bc + ca]

(25) If 2(a2 + b2) = (a+b)2 show that a = b

(26) If x + y = 5 and xy = 10, P.T. = 3

LHS = Hint: (x-y)2 = (x+y)2 - 4xy

= = .........................

= ................................................

= ................................................

= ................................................

a+b1-ab

1c

a+b1-ab

a+bab+bc+ca-ab

x3 - y3

x2 - y2

x3 - y3

x2 - y2

(x-y)3 + 3xy (x-y)(x+y) (x-y)

= .........................

= .........................


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(27) If x + y = 8 and xy = 12, Find the value of

+ ÷ + 1 + 1

(28) If b2 = c2 + a2, P.T. (a + b + c) (a - b + c) = 2ac

LHS = (a + b + c) (a - b + c)

= [(a+c)+b] [(a+c)-b]

= ..................................................

= ..................................................

= ..................................................

(29) If xy (x + y) = 1, P.T. - x3 - y3 = 3

LHS = - x3 - y3 Hint: xy (x + y) = 1

= - [x3 + y3]

= ............................

= ............................

x2

y[ ]y2

xxy[ ] y

x[ ]

(x + y) =

Cubing Both Sides

(x + y)3 =

............................

............................

1x3y3

1x3y3

1xy

31xy

1xy

3


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(30) If x + y = a, xy = b, S.T. (1 + x2) (1 + y2) = a2 + (1-b)2

RHS = a2 + (1-b)2

= (x + y)2 + (1 - xy)2

(31) If x + = 4, S.T. x3 + 4x2 + - = 148

(32) If x2 + = 7, P.T. x3 + = 18

1x

4x2

1x3

1x2

1x3


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(33) If x2 + = 27, then P.T. x + = ± 291x2

1x

VI If a + b + c = 2S, then

(a) a + b - c = a + b - c + c - c [Add and Subtract c] = a + b + c - 2c = 2S - 2c = 2 (S - c)

(b) a + c - b =

(c) b + c - a =

(d) a - b - c =


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(e) a2 + b2 - c2 + 2ab

= a2 + b2 + 2ab - c2

= (.................................)2 - c2

= (a + b + c) (a + b - c)

= 2S ( ......................... ) = .......................................

(f) b2 + c2 - a2 + 2bc =

(g) c2 + a2 - b2 + 2ca =

(h) a2 - b2 - c2 + 2bc = a2 - (b2 + c2 - 2bc)

= a2 - (b - c)2

= [a+ (b-c)] [a-(b-c)]

(i) b2 - c2 - a2 + 2ac =

(j) c2 - a2 - b2 + 2ab =

V If a + b + c = 2S then prove the following identities.

(1) S2 + (S - a)2 + (S - b)2 + (S - c)2 = a2 + b2 + c2

LHS = S2 + (S - a)2 + (S - b)2 + (S - c)2

= S2 + S2 + a2 - 2as + ......... + ......... - ......... + ......... + ......... - ..........3

= 4S2 + a2 + b2 + c2 - (................) ......... (................) ......... (................)

= 4S2 + a2 + b2 + c2 - 2a (......................................)

= 4S2 + a2 + b2 + c2 - 2a (.............)

= 4S2 + a2 + b2 + c2 - (.............)

LHS = a2 + b2 + c2

∴ LHS = RHS


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(2) a2 + b2 - c2 + 2ab = 4S (S - C)

LHS = a2 + b2 - c2 + 2ab

= a2 + b2 + 2ab - c2

= (..............................) - c2 a2 - b2 = (a+b) (a-b)

= (..............................) (..............................)

= (a + b + c) (a + b - c + C - C) Add and subtract ‘C’

= (a + b + c) (a + b + c -2C)

= 2S (2S - 2C)

= 2S x 2(S - C)

= 4S (S - C)

(3) b2 + c2 - a2 + 2bc = 4S (S - a)

LHS = b2 + c2 - a2 + 2bc

= .............................. - a2

= (..............................) x (..............................)

= (..............................) x (..............................)

= ..............................

= ..............................

= ..............................

(4) c2 + a2 - b2 + 2ca = 4S (S - b)


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(5) a2 - b2 - c2 + 2bc = 4(S - b) (S - c)

LHS = a2 - b2 - c2 + 2bc

= a2 - [b2 + c2 - 2bc]

= a2 - (b - c)2 a2 - b2 = (a + b) (a - b)

= [a + (b - c)] [a - (b - c)]

= [a + b - c] [a - b + c]

= [a + b + c - 2c] [a + b + c - 2b]

= (2S - 2C) (2S - 2b)

LHS = 4 (S - b) (S - C)

LHS = RHS

(6) b2 - c2 - a2 + 2ca = 4(S - C) (S - a)

LHS = b2 - c2 - a2 + 2ca

= b2 - [..........................]

= b2 - (.................)2

= [...........................] [..........................]

= [...........................] [..........................]

= [...........................] [..........................]

= (...................) (...................)

LHS = (...................) (...................)

LHS = RHS


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(7) (2bc + a2 - b2 - c2) (2bc - a2 + b2 + c2) = 16S (S-a) (S-b) (S-c)

LHS = (2bc + a2 - b2 - c2) (2bc - a2 + b2 + c2)

= [a2 - b2 - c2 + 2bc] [b2 + c2 + 2bc - a2]

= [a2 - (b2 + c2 - 2bc)] [b2 + c2 + 2bc - a2]

= [a2 - (....................)2] [(....................)2 - a2]

= (..................)(..................) (..................)(..................)

= .............................................................

= .............................................................

= .............................................................

= .............................................................

= .............................................................

= .............................................................

(8) (2ab + c2 - a2 - b2) (2a - c2 + a2 + b2) = 16S (S-a) (S-b) (S-c)


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(9) [2ca + b2 - a2 - c2] [2ca - b2 + a2 + c2] = 16S (S-a) (S-b) (S-c)

(10) =

LHS =

=

=

a2 + b2 - c2 + 2aba2 - b2 + c2 + 2ac

S - cS - b

a2 + b2 - c2 + 2aba2 - b2 + c2 + 2ac

a2 + b2 + 2ab - c2

a2 + b2 + 2ac - b2


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(11) =b2 + c2 - a2 + 2bcb2 - c2 + a2 + 2ab

S - aS - b

(12) =c2 + a2 - b2 + 2cac2 - a2 + b2 + 2bc

S - bS - a


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(13) If a + b + c = 2S, then show that =

LHS =

=

=

=

=

=

a2 + b2 - c2 + 2aba2 - b2 - c2 - 2bc

SS - b

a2 + b2 - c2 + 2aba2 - b2 - c2 - 2bc

a2 + b2 + 2ab - c2

a2 - [b2 + c2 + 2bc]

(..................)2 - c2

a2 - (..................)

(a+b+c) (a+b-c)(a+b+c) [a-(b+c)]

a + b + c - 2Ca - b - c

(14) If a + b + c = 2S, prove that, S2 + (S - a)2 + (S - b)2 + (S - c)2 = a2 + b2 + c2

LHS = S2 + (S - a)2 + (S - b)2 + (S - c)2

= S2 + S2 + a2 - 2as + (..........)+(.........)-(.........)+(.........)+(.........)-(..........)

= 4S2 + a2 + b2 +c2 - 2aS - 2bS - 2cS

= 4S2 + a2 + b2 +c2 - 2S (a + b + c)

= ........................................................

= ........................................................

= ........................................................


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(15) c2 - a2 - b2 + 2ab = 4 (s - a) (s - b)


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SURDS

• The irrational root of a rational number is called a Surd.

Example: 2 , 3 , 5

• The Standard form of a Surd is a

• a ---> ‘a’ radicand---> ‘n’ order of the surd

• Pure Surd: The surds having 1 as their rational co-efficients are called Pure Surds.

Ex: 2 , 4 , 7 etc.

• Mixed Surd: The surds having their rational co-efficients other than 1 are called Mixed Surds.

Ex: 2 3 , -7 3 , 8 3 etc.

• Like Surd: A Group of Surds of the same order, having the same radicand in their reduced form are caled Like Surds.

Ex: (1) 2 , 2 2 , 4 2

(2) { 8 , 18 , 50 , 32 } ---> { 2 2 , 3 2 , 5 2 , 4 2 }

• Unlike Surd: A Group of surds of different order, diffferent radicand or both in their reduced form are called Unlike Surds.

Ex: (1) 2 , 3 , 4

(2) 3 , 5 , 7

(3) 5 , 5 , 5 etc

• Binomial Surd: The algebraic sum of two distinct monomial surds or the algebraic sum of a rational number and a monomial surd is called Binomial Surd.

Ex: (1) 2 + 3 , 2 - 7 , p + 2 q etc

Think !

4 , 9 , 8 are

not Surds, why

3 5

n

n

3

3

3 33

3 33

3

3

4

4


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• Only like surds can be added or subtracted.

• Multiplication and Division of Surds having same order.

(1) Rule: a x b = a x b

Ex: 3 x 2 = 3 x 2 = 6

(2) Rule: =

Ex: = = 2

Rationalisation of Surds and Rationalising Factors

The process of multiplying a surd, by another surd to get a rational number is calledRationalisation.

Then the each surd is the rationalising factor of the other.

3 x 3 = 9 = 3

Here 3 is the R.F. of 3

n n n

ab

ab

nn

n

147

147

3

3

3

3


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I Write the respective orders and radicands of the surds in the following table.

II Write the surds having the following order and radicand.

3

5

4

8

4

III Express the following surds in the index form.

IV Express the following in Surd form.

S.No. Order Radicand Surd S.No. Order Radicand Surd

Ex 2 8 8 3 2 14

1 3 6 4 3 7

2 4 18 5 6 15

S.No. Surd Order Radicand S.No. Surd Order Radicand

1 Ex 3 2 3 5 8 17

2 5 6 19

3 50 7 5 19

4 7

Surd Index Form Surd Index Form

Ex: 3 3½ 5 19

9 142

7 5 6

17 7

3

4

8

4

3

Index Form Surd Form Index Form Surd Form

Ex: 201/3 20 5(2)1/3

5½ 3(27)1/8

81/5 7(x2)1/3

3


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Reduction of Surds

Ex: 75 can be simplified as

75 = 25 x 3 = 52 x 3 = 5 3

Here 75 is a pure surd

5 3 is a mixed surd

V Reduce the following surds.

VI Conversion of mixed surd into pure surd

Ex: 11 2 ----> 112 x 2 = 121 x 2 = 242 Mixed form

Express the following surds into their pure form

Pure Surd Simplification Mixed Surd

48

32

200

405

300

3

4

Mixed Surd Pure form

8 4

3 2

3 5

10 3

5 6

3

4


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Classification of Like Surds

Ex: 5 3 , 4 2 , 7 3 , 10 2 , 11 3 , 8 2

Set A: 5 3 , 7 3 , 11 3

Set B: 4 2 , 10 2 , 8 2

VII Classify the following surds into set of like surds (reduce them wherever necessary)

(1) 5 2 , 6 5 , 3 2 , 8 2 , 10 5 (2) 18 , 3 2 , 4 3 , 32 , 2 24

(3) 125 , 45 , 12 , 20 , 27, 300 (4) a3, 3 a , 4 2a , 6 16a , 10 4a

3 3 3

3 3 3

3 3 3 3

Addition and Subtraction of Surds

(Only like surds can be added or subtracted)

Ex: (1) Addition of 5 a and 2 a

= 5 a + 2 a

= (5+2) a

= 7 a

Index form = 7(a)1/2

Sum or difference of thereduced surds can beobtained by adding orsubtracting the co-efficient.


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VIII Simplify:

(1) 10 x - 8 x

(2) Find the sum of 3 a , 5 a and 4 a and express in the index form.

Ex: Simplify: 18 + 5 2 + 200

9 x 2 + 5 2 + 100 x 2

3 2 + 5 2 + 10 2

(3+5+10) 2

= 18 2

IX (1) Simplify: 7 2 + 16 - 54 (2) Simplify: 4 3 - 3 12 + 2 75

3 3

3 3 3

(3) Simplify: 45 - 3 20 + 4 5 (4) Simplify: 2 2a + 4 8a - 3 2a


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Addition of Binomial Surds

Ex 1: Find the sum of Ex 2: Subtract (5 a + 3 b ) from

(2 x + 3 y ) and (5 x - y ) (8 a + 5 b )

Soln: 2 x + 3 y + 5 x - y Soln: (8 a + 5 b ) - (5 a + 3 b )

2 x + 5 x + 3 y - y 8 a + 5 b - 5 a - 3 b

(2+5) x + (3-1) y 8 a - 5 a + 5 b - 3 b

7 x + 2 y (8-5) a + (5-3) b

3 a + 2 b

X (1) Find the sum of (2) Add : 2 + 3 + 5 3 - 2 +

8 + 12 , 3 2 + 4 3 , 27 - 8 2 2 - 4 3

Multiplication of Surds (having same order)

Rule : a x b = ab

Ex : 2 x 3 = 2 x 3 = 6

Muliply the following surds:

(1) 6 x 3 =

(2) 2 7 x 2 =

(3) 5 a x 3 b =

(4) a ( b + c ) =

(5) ( 3 + 2 ) ( 5 - 7 ) =

n n n


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Multiplication of Surds (having different order)

Ex : Multiply 3 x 2 The orders are 2 & 3, LCM of 2 & 3 is 6

3 = (3)1/2 = 31/2 x 6/6 = 33/6 = (33)1/6 = 33 = 27

2 = (2)1/3 = 21/3 x 6/6 = 22/6 = (22)1/6 = 22 = 4

∴ 3 x 2

26 x 4 (Now they have reduced to same order)

= 27 x 4

= 108

XI (1) Multiply 5 and 2

6 6

6 63

3

3

6 6

6

6

3 4

(2) Multiply 2 and 3

(3) Find the product of 3 x 2

(4) Simplify 5 x 3

3

3 4

4


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(5) Find the product of 3 and 6

(6) Multiply 3 x 5

3

3 4

Rationalisation of Surds and Rationalising Factor

Ex: 5 is a surd. When 5 is multiplied by 5

5 x 5 = 25 = 5 is a rational number.

∴ 5 is the Rationalising factor of 5

Write the rationalising factors of the following surds.

S.No. Surd RF

1 3 a

2 x + y

3 m

4 5 x

5 7 y

6 4 p+q

7 6 a-b


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The RF of a binomial surd (Conjugate)

Ex: ( 3 + 2 ) is a binomial surd, if this is multiplied by ( 3 - 2 )

( 3 + 2 ) ( 3 - 2 ) ---> [(a+b)(a-b) = a2 - b2 form]

( 3 )2 - ( 2 )2

= 3 - 2

= 1

∴ 3 - 2 is the RF of 3 + 2

3 + 2 is the RF of 3 - 2

Write the RF of the following binomial Surds

Simplification by rationalising the denominator.

Ex (1) Rationalise the denominator and simplify.

x (Multiply and divide by RF of the denominator)

=

S.No. Binomial Surd RF (Conjugate)

1 5 a + 3 b

2 m + n

3 4 x - 2 y

4 10 a - b

5 3 + b

6 7 - 3

3x

3x

xx

3 x( x )2

3 xx


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XII Rationalise the denominators and simplify the following.

(1) (2)

(3) (4)

(5) (6)

25

ab

7x

53

3a5

32x

Ex (2) : Rationalise the denominator and simplify

The RF of denominator is 3 - 2

x [(a+b)(a-b) = a2 - b2]

=

= = 5 ( 3 - 2 )

53 + 2

53 + 2

3 - 23 - 2

5 ( 3 - 2 ) ( 3 )2 - ( 2 )2

5 ( 3 - 2 ) 1


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XIII Rationalise the denominator and simplify the following.

(1) (2)2

x - y10

7 + 2

(3) (4)

(5) (6)

XIV Rationalise the denominators and simplify the following.

(1)

3 26 + 3

7 310 - 3

mm + n

63 + 2

5 + 35 - 3


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(2) 2 3 + 2 3 2 + 5

(3) +

(4) -

(5)

23 - 2

3 3 + 2

32 + 3

2 2 - 3

3 - 23 + 2


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(6) +6

3 + 2

3 26 + 3

(7) -

(8) +

Think

(1) The RF of x2 is x why ?

(2) S.T. the RF of x1/3 + y1/3 is x2/3 - (xy)1/3 + y2/3

(Clue: Use a3 + b3 formula)

7 310 - 3

2 56 + 2

45 - 3

2 5 + 3

3 3


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(9) Find a and b if = a + b 3

(10) Find a and b if = a - b

(11) Simplify: +

186

3 - 13 + 1

4 + 22 + 2

4 + 54 - 5

4 - 54 + 5


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XV EXAM QUESTIONS (PREVIOUS)

(1) The answer obtained by the subtraction of (5 a - 3 b ) from (8 a + 5 b )

is ..........................

(2) The RF of x1/2 - y1/2 is ..........................

(3) The product of 2 and 3 is ..........................

(4) Rationalise the denominator and simplify

(5) The sum of 3 a , - 13 a , 6 a is ..........................

(6) The value of (2 2 + 3 3 )2 is ..........................

(7) The conjugate of x-y is ..........................

(8) Simplify: + by rationalising the denominator.

(9) The pure surd of 2 5 is ..........................

(10) The RF of a b + c is ..........................

(11) The answer obtained by rationalising the denominator of is ....................


(13) Which of the following is a pair of like surds

(a) 2 , 3 (b) 2 , 2 (c) 2 , 8 (d) 4 3 , 3 4

(14) If 3 - 2 is subtracted from 2 - 3 , then the result is ..........................

(15) The product of 3 5 and ( 3 - 2) is ..........................

(16) The RF of p q - q p is ..........................

(17) The index form of a b is ..........................

(18) The product of 2 and 5 is ..........................

(19) When 2 - 8 + 50 is simplified we get ..........................

(20) The area of a rectangle with length (2 5 + 3 ) unit and breadth (2 5 - 3 )units is ..........................

(21) Simplify by rationalising the denominator.

3

5 + 35 - 3

2 3 - 2

3 3 + 2

3

3

3

3

3

35

7 310 - 3


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(22) The sum of 9a and 25a is ..........................

(23) When is simplified by rationalising the denominator, the result is ............

(24) When 10 - 2 is multiplied by its RF, the result is ...................

(25) 32 + 50 = ..........................

(26) The RF of 5 p - q is ..........................

(27) When 2 x - y is subtracted from 5 x + 2 y , the result is ....................

(28) The order and radicand’s of a x are .................. and ..................

(29) The sum of 50 + 6 2 + 128 is ..................

(30) The product of 2 and 5 is ..................


+

(32) Conjugate of a x + b y is ..........................



(35) If 9 x = 12 + 147 , the value of x is ..........................

(36) The order and radicands of 5 3m are ..........................


(38) If 10 + 84 = a + 2 b then the value of b ..........................

(a) 84 (b) 21 (c) 10 (d) 2

(39) If x = 7 - 4 3 then ST 7x + = 4.

Clue ( 7 - 4 3 = 4 - 3 )

**************

n

2 3 - 2

3 3 + 2

34

3 2 + 2 3 3 2 - 2 3

2 3 3 + 2

1x

82


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QUADRATIC EQUATIONS5

• An equation involving a variable whose highest degree is one is a linear equation, linear equation has only one root.

General form is mx + c = 0

• An equation involving a variable whose highest degree is two is a quadratic equation. The quadratic equation has only two roots.

• Quadratic equation involving a variable only in second degree is a pure quadratic equation.

General form is ax2 + c = 0

• Quadratic equation involving a variable in second degree as well as in first degree is an adfected quadratic equation.

Standard form of quadratic equation where a, b and c are real numbers and a ≠ 0 is

ax2 + bx + c = 0

• A quadratic equation can be solved by factors method or by using the formula

x =

• The nature of the roots of ax2 + bx + c = 0 depends on the value of (b2 - 4ac) which is called the discriminant and denoted by Δ (delta)

• If b2 - 4ac = 0 the roots are real and equal.

• If b2 - 4ac > 0 the roots are real and distinct.

• If b2 - 4ac < 0 the roots are complex or imaginary.

• If b = 0 the roots are equal but opposite in sign.

• If a = c the roots are reciprocal to each other.

• If c = 0 one root is zero.

• If a and c have the same sign but ‘b’ opposite both roots are positive.

• If a, b and c have the same sign both roots aer negative.

• If a and c have opposite signs the roots have opposite sign.

• The sum of the roots of ax2 + bx + c = 0 is given by

• The product of the roots of ax2 + bx + c = 0 is given by

• When the roots are given we can form the quadratic equation by

x2 - (sum) x + product = 0

-ba

ca

- b ± b2 - 4ac2a


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190

95

95

12

12

Vhπ

πVh

πhV

Vπh

- b ‐ b2 + 4ac2a

- b ± b2 + 4ac2a

+ b ± b2 - 4ac2a

- b ± b2 - 4ac2a

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(M - 06)

I (1) In a perticular quadratic equation a stright line cuts the parabola at (-2, 4) and (1, 1)

then the roots of the quadratic equation are ..................................

(a) (-2, 4) (b) (1, 1) (c) (-2, 1) (d) (4, 1)

(2) Select the pure quadratic equation in the following..................................

(a) 2x2 + x + 5 = 0 (b) 3x2 + 1 = 28 (c) x2 - x - 7 = 0 (d) x - x2 = 0

(3) Nature of the roots of the equation x2 - 6x + 9 = 0 is ..................................

(a) Real and Rational (b) Real and Irrational

(c) Equal to each other (d) Imaginary

(4) Sum of the roots of x2 - 5x = 9 is ..................................

(a) 5 (b) -5 (c) (d) -

(5) The quadratic equation whose roots are 3 + 2 and 3 - 2 is .........................

(a) x2 + 6x - 7 = 0 (b) x2 - 6x + 7 = 0

(c) x2 - 6x - 7 = 0 (d) x2 + 6x + 7 = 0

(6) If the product of the roots of the equation x2 + 3x + q = 0 is zero then the value of q

is ................................

(7) In which of the following equation the parabola obtained? ................................

(a) Equation of Straight Line (b) Cubic Equation

(b) Quadratic Equation (d) Linear Equation

(8) Positive root of the equation (2x - 1) (x - 3) = 0 is ................................

(a) 3 (b) -3 (c) - (d)

(9) If V = πr2h then value of r is ................................

(a) ± (b) ± (c) ± (d) ±

(10) The value of x in the equation ax2 + bx + c = 0 is ................................

(a) (b)

(c) (d)


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(11) The roots of the equation x2 - 3x = 0 are ................................

(a) (0, 3) (b) (0, -3) (c) -3 (d) 2

(12) The sum of a number and twice its square is 105. The equation form is ......................

(a) x2 + 2x = 15 (b) 2x2 - x = 105

(c) 2x2 + x = 105 (d) 2x2 + x + 105 =0

(13) If 3a2 - 27 = 0 then the value of a is ......................

(a) ±9 (b) ±3 (c) ±27 (d) ±1

(14) If (a+8)2 - 5 = 31 then the value of a is ......................

(a) 2, 14 (b) +2, -14 (c) -2, +14 (d) -2, -14

(15) Roots of the equation x2 - x = 6 ......................

(a) +1, -6 (b) +3, -2 (c) -3, -2 (d) -1, +6

(16) Sum of the number and its square is 20 then the number is ......................

(a) 5 (b) 10 (c) 4 (d) 2

(17) m and n are the roots of the equation x2 - 6x + 2 = 0 then the value of +

is ......................

(a) 6 (b) 1.5 (c) 3 (d) 2

(18) General form of pure quadratic equation is ......................

(a) (bx + c)2 = 0 (b) ax2 + bx + c = 0 (c) ax2 + c = 0 (d) ax2 + bx = 0

(19) The roots of the quadratic equation depends on ......................

(a) b2 - 4ac (b) b2 - ac (c) b2 + 4ac (d) b2 + ac

(20) Roots of the equation x2 - 2x + 1 = 0 is ......................

(a) Not a real no.s (b) Distinct (c) Equal (d) Zero

(21) The quadratic equation whose roots are 2 + 3 and 2 - 3 is ......................

(a) x2 + 4x + 1 = 0 (b) x2 - 4x + 1 = 0

(c) x2 + 2x + 3 = 0 (d) x2 - 2x - 3 = 0

1m

1n

(M - 06)

(M - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)


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(22) Product of the roots of the equation 6k2 - 3k = 0 is ................................

(a) 2 (b) - (c) (d) 0

(23) The value of m for which the equation x2 + mx + 4 = 0 has equal roots is ......................

(a) ±4 (b) ± 2 (c) 0 (d) ±1

(24) If we solve the equation 4a = the value of a is ......................

(a) ±9 (b) + 3 (c) ‐ 3 (d) ± 3

(25) Standard form of the equation 2x = 5 - x2 is ......................

(a) 2x - 5 + x2 = 0 (b) x2 + 2x - 5 = 0

(c) x2 - 2x + 5 = 0 (d) 2x - 5 - x2 = 0

(26) The quadratic equation whose roots are 3 ± 5 is ......................

(a) x2 - 6x + 4 = 0 (b) x2 - 3x + 5 = 0

(c) x2 + 3x - 5 = 0 (d) x2 + 6x + 4 = 00

(27) If the roots of the equation are real and distinct which one is correct in the given

below ......................

(a) Δ > 0 (b) Δ < 0 (c) Δ = 0 (d) Δ ≤ 0

(28) Sum of the roots of the equation 2x2 - 5x + 6 = 0 is ......................

(a) (b) 3 (c) (d)

(29) The value of m for which the equation mx2 + 6x + 1 = 0 has equal roots is ....................

(a) 6 (b) 1 (c) 9 (d) 5

(30) One root of the equation x2 - 5x = 0 is ‘0’ then other one is ......................

(a) 0 (b) - 5 (c) + 5 (d) ± 5

(31) The quadratic equation whose roots are 5 and -6 is .....................

(a) x2 - 30x - 1 = 0 (b) x2 - x - 30 = 0

(c) x2 + x - 30 = 0 (d) x2 - x + 30 = 0

(32) The product of the roots of the equation 3x2 - 7x + 9 = 0 is ......................

(a) (b) 3 (c) (d) -3

12

12

36a

-52

52

25

73

13

(J - 06)

(J - 06)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(J - 07)

(J - 07)


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(33) If S = gt2 after solve the equation t is ................................

(a) t = ± (b) t = (c) t = (d) t = ±

(34) Product of the roots of the equation x2 - 5x + 6 = 0 is ......................

(a) Real and Disticnt (b) Real and Equal (c) Imaginary (d) Equal

(35) A stright line graph Y = 3 is ......................

(a) Passes through the origin

(b) Perpendicular to X axis

(c) Parallel to X axis and passes through Y axis at 3

(d) Parallel to Y axis and passes through X axis at 3

(36) If b = 0 in the equation ax2 + bx + c = 0 then the equation is ......................

(a) Pure quadratic equation (b) Adfected quadratic equation

(c) Equation of the straight line (d) Linear equation

(37) The value of m for which the equation x2 - mx + 16 = 0 has equal roots is ......................

(a) ±4 (b) ±16 (c) ±2 (d) ±80

(38) ‘m’ and ‘n’ are the roots of the equation x2 - 6x + 2 = 0 ......................

(a) 12 (b) 6 (c) 2 (d) 3

(39) If graphically solving the equation the roots are ..........................

(40) Which one of the following is pure quadratic equation ....................

(a) 2x + 5 = 13 (b) x2 + 5 = 26x (c) x2 = 5x (d) x2 + 2x = 3

(41) The quadratic equation whose roots are +2 and -2 is ......................

(a) Adfected quadratic equation (b) Linear equation

(c) Simple linear equation (d) Pure quadratic equation

12

2Sg

2gS

2Sg

2gS

O

Y

Y1

X1 X0 1 2-1-1

-2

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(M - 08)

(M - 08)


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(42) One of the root of the equation 2x (x + 3) = 0 is ................................

(a) 0 (b) (c) +3 (d) 5

(43) The base of a triangle is 4cm longer than its altitude. If its area is 30 Sqcm. Which of

the equation is represented it? ......................

(a) x(x+4) = 30 (b) 2x(x+4) = 40 (c) x(x+4)=5 (d) x(x+4) = 60

(44) Product of the number and twice of it is 200 then the number is ...............

(a) ±10 (b) ±15 (c) ±20 (d) ±25

(45) Discriminent of the equation 2x2 = 5x is ......................

(a) 27 (b) 25 (c) 23 (d) 10

(46) The value of m for which the equation x2 + mx + 36 = 0 has equal roots is ......................

(a) ±6 (b) ±8 (c) ±12 (d) ±180

(47) Sum and product of the roots of the equation 2x2 = 3x is ......................

(a) and 0 (b) 0 and (c) and 0 (d) 0 and

(48) The quadratic equation whose roots are 3 + 2 2 and 3 - 2 2 is ..........................

(a) x2 + 6x + 1 = 0 (b) x2 + 6x - 1 = 0 (c) x2 - 6x + 1 = 0 (d) x2- 6x - 1 =0

(49) If the roots of the equation mx2 + 6x + 1 = 0 are equal then the value of ‘m’ is ............

(a) 6 (b) 1 (c) 9 (d) 5

(50) If a2 = b2 + c2, then c is equal to ...................................

(a) ± b2 + a2 (b) ± a2 + b2 (c) ± a - b (d) ± a2 - b2

(51) IF A = 4πr2, then r is ...................................

(a) ± (b) (c) 4Aπ (d)

(52) “Twice of the square of a number is added to thrice of that number, the sum is 65” the

equation for this statement is ...................................

(a) 3x2 + 2x = 65 (b) 2x2 + 3x = 65

(c) 2x2 - 3x = 65 (d) 3x2 - 2x = 65

23

32

32

152

-152

A4π

4πA

A4π

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(J - 08)

(J - 08)

(J - 08)

(J - 08)


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(53) In a quadratic equation b2 - 4ac = -7, then nature of the roots is ................................

(a) real and equal (b) real and distict (c) imaginary (d) -ve number

(54) The descriminant of the equation ax2 + bx + c = 0 is .................................

(a) (b) b2 - 4ac

(c) (d)

(55) The roots of the equation x2 + 4x + c = 0 are equal then the value of c is ......................

(a) 3 (b) 4 (c) 5 (d) 12

(56) The sum and product of the equation 4x2 + 1 = 0 are ......................

(a) 1 and 4 (b) 0 and 1 (c) 0 and (d) 0 and

(57) In a right angled triangle the hypotenous is 13cm one of the remaining sides is 5 cm

more than the other, then the relation between them is ......................

(a) x + (x + 5) = 13 (b) x2 + (x2 + 5) = 13

(c) x2 + (x + 5)2 = 132 (d) x2 + (5 - x)2 = 132

0

(58) The roots of a quadratic equation are 0 and , then the equation is ......................

(a) 2x2 + x = 0 (b) x2 + = 0 (c) 2x2 + 1 = 0 (d) 2x2 - x = 0

(59) Which one of the following is pure quadratic equation ..........................

(a) 2x2 - x = 0 (b) 5x = 3 (c) 4x = 9x2 (d) 2x2 = 16

(60) The graph of a parabola is by ...............................

(a) a linear equation (b) Simultaneous equation

(c) a quadratic equation (d) a polynomial equation

(61) A equation has only one root those equation is ...............................

(a) Quadratic equation (b) Equation of Stright line

(c) Cubic Equation (d) Linear Equation

(62) If F = then v is ...................................

(a) (b) (c) (d)

+ b ± b2 - 4ac2a

ca

-ba

-14

14

-12

12

mv2

rFmr

mrF

Frm

Frm

(J - 08)

(J - 08)

(J - 08)

(J - 08)

(J - 08)

(J - 08)

(M - 09)

(J - 08)

(M - 09)

(M - 09)


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(63) A positive root of the equation (2x - 7) (3x - 5) = 0 is ................................

(a) (b) (c) (d)

(64) The value of x in the equation px2 + qx + r = 0 is .................................

(a) (b)

(c) (d)

(65) Length of the rectangle is 4 cm longer than its breadth. If its area is 60 Sq.cm. The

equation represented for it is ......................

(a) x + (x + 4) = 60 (b) x + (x + 4) - 60 = 0

(c) (x + 4) x + 60 = 0 (d) (x + 4) x - 60 = 0

(66) The nature of the roots of the equation depends upon the value of ......................

(a) b2 - 4ac (b) b2 + 4ac (c) b - 4ac (d) b + 4ac

(67) Product of the roots of the equation 2x2 = 3x is ......................

(a) (b) (c) 0 (d)

(68) The value of m for which the equation x2 - mx + 25 = 0 has equal roots is ......................

(a) 20 (b) 10 (c) 15 (d) 5

(69) Sum and product of the roots of the equation are -5 and 4 then the equation is ..........

(a) x2 + 5x + 4 = 0 (b) x2 - 5x + 4 = 0

(c) x2 + x - 20 = 0 (d) x2 - x - 20 = 0

(70) a and b are the roots of the equation x2 - 5x + 7 = 0 then ab (a+b) ...............................

(a) 5 (b) 25 (c) 35 (d) 49

(71) Product of the roots of the equation x2 + 5x + (K+4) = 0 is zero then K = .......................

(a) -5 (b) -4 (c) 4 (d) 5

(72) If a = 0 in the quadratic equation ax2 + bx + c = 0 then the equation is .............................

(a) Pure quadratic equation (b) Adfected quadratic equation

(c) Linear equation (d) Simultaneous equation

72

27

35

57

- p ± p2 - 4pq2p

- p ± r2 - 4pq2r

- q ± q2 - 4pr2p

- p ± p2 - 4pq2q

-23

32

12

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(M - 09)

(J - 09)


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(73) Roots of the equation 3x2 - 3x = 0 are ................................

(a) 0 and 1 (b) 0 and 3 (c) 1 and 3 (d) 0 and -3

(74) If we solve the equation 7y = then the value of y is ...............................

(a) ± 7 (b) ± 3 (c) ± (d) ± 5

(75) Sum of a number and its square is 42 it can be represented as ......................

(a) x2 + x + 42 = 0 (b) x2 + x - 42 = 0 (c) 2x2 + x - 42 = 0 (d) x2-x- 42 = 0

(76) The standard form of 2m2 = 2 - m is ......................

(a) 2m2 + m - 2 = 0 (b) 2m2 - m - 2 = 0

(c) 2m2 - m + 2 = 0 (d) 2m2 + m + 2 = 0

(77) The co-efficients a, b, c of the equation 2k2 - 2k - 5 = 0 are substituted the roots

obtained are ......................

(a) k = (b) k =

(c) k = (d) k =

(78) In a quadratic equation b2 = 4ac then the roots are ......................

(a) Real and equal (b) Real and distinct

(c) Imaginary (d) Imaginary and equal

(79) ‘m’ and ‘n’ are the roots of the equation then the general form of the quadratic

equation is ...............................

(a) x2 + (m + n) x + mn = 0 (b) x2 - (m + n) x - mn = 0

(c) x2 + (m - n) x + mn = 0 (d) x2 - (m + n) x + mn = 0

(80) If m and n are the roots of the equation 2x2 - 6x + 1 = 0 then m2n + mn2 is ...................

(a) (b) (c) (d)

(81) The graphs y = x2 and y = 2 - x are intersects at the points (1, 1) and (-2, 4) then the

roots of the expected equation are ...............................

(a) 2 and 2 (b) 1 and -2 (c) 0 and -2 (d) 0 and 4

35y

57

- (-2) ± (-2)2 - 4(2)(5)2

- 2 ± (-2)2 - 4(2) (-5)2 (2)

2 ± (-2)2 - 4(2) (-5)2 (2)

- 2 ± (-2)2 - 4(2) (5)2 (2)

32

23

-32

12

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)


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(82) If v2 = u2 + 2aS then the value of u is ................................

(a) v2 = 2aS (b) ± v2 + 2aS (c) ± v2 - 2aS (d) 2aS - v2

(83) The quadratic equation whose roots are 1 and -1 is .................................

(a) ax2 - x - 1 = 0 (b) ax2 - 1 = 0 (c) x2 = 1 (d) x2 + 1 = 0

(84) The pure quadratic equation in the following is ......................

(a) 4x = (b) x + = 5 (c) (x + 2)2 = 3x (d) 5 - x2 = x

(85) If l2 = r2 + h2 then value of h is ......................

(a) ± l2 - r2 (b) ± r2 - l2 (c) ± l2 + r2 (d) ± l - r

(86) If b = 0 in the equation ax2 + bx + c = 0 the equation becomes ......................

(a) Adfected quadratic equation (b) Pure quadratic equation

(c) Linear equation (d) Simultaneous equation

(87) Sum of the roots of the equation 2x2 = 6x - 5 is ......................

(a) (b) (c) - (d) 3

(88) Sum of the number and twice of its square is 78. It can be represented as ........................

(a) x + 2x2 = 78 (b) x + (2x)2 = 78

(c) x2 + 2x = 78 (d) x2 + (2x)2 = 78

(89) x + = 2 is same as ...................

(a) x + = 22 (b) x2 + 2x + 1 = 0

(c) x2 - 2x + 1 = 0 (d) x2 + 2x = 0

(90) If in ax2 + bx + c = 0, a = c the roots are ...............................

(a) Additive inverses (b) Multiplicative inverses

(c) Equal (d) Zero

(91) One root of 4x2 - 8mx - 9 = 0 is the negative of the other root. So the value of m is

...............................

(a) 0 (b) (c) - (d) -

81x

1x

13

52

52

1x

1x

2

98

94

98

(M - 10)

(M - 10)

(M - 10)

(J - 10)

(J - 10)

(J - 10)

(J - 10)


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(92) The equation 5x2 - 26x + p = 0 is has reciprocal roots, so the value of p is

................................

(a) 25 (b) 1 (c) 5 (d)

(93) The graph of x2 - 6x + 9 = 0 is .........................................

(a) Cuts the X - axis at two points (b) Cuts the Y - axis at two points

(c) Touches the X - axis at one point (d) Has no contact with the X - axis

(94) The equation (b-c)x2 + (c-a)x + (a-b) = 0 has ......................

(a) equal roots (b) irrational roots

(c) rational roots (d) none

(95) If the roots of the equation ax2 + b = 0 are real and distinct then ......................

(a) ab > 0 (b) a = 0 (c) ab < 0 (d) a>0, b>0

(96) If the product of the roots of the equation m2 + 6x + (2m - 1) = 0 is -1 then the value

of m is ......................

(a) 1 (b) -1 (c) (d) -

(97) If α, β are the roots of x2 - 2x + 2 = 0 then α2 + β2 = ......................

(a) 2 (b) 0 (c) 1 (d) 4

(98) If the roots of the equation 12x2 + mx + 5 = 0 are in the ratio 3 : 2 then m is .....................

(a) (b) (c) 5 10 (d) 10

(99) In a quadratic equation with a = 1, a student reads the co-efficient 16 of x wrongly as

19 and obtains the roots as -15 and -4 the correct roots are ..........................

(a) 6, 10 (b) -6, -10 (c) 8, 8 (d) -8, -8

(100) If (x - 4) (2x + 5) = 0 the incorrect statement is ...............................

(a) x = 4 (b) x = (c) x = (d) 2x = -5

(101) If α and β are the roots of x2 - px + 36 = 0 and α2 + β2 = 9 then p = ...............................

(a) ± 3 (b) ± 6 (c) ± 8 (d) ± 9

15

13

13

112

512

512

52

-52


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200

(102) The nature of the graph of y = x2 + 5x - 24 is ...............................

(a) (b)

(c) (d)

II Which of these are quadratic equations

III Classify as pure and adfected quadratic equation.

1) x2 – 6x + 4 5) x + = 2

2) x2 + 2x = 4 6) 5(x-2)(x+3) = 0

3) x2 + = 2 7) x + 3x = 5

4) = x 8) k = mv2

1x2

1x

x+52

12

1x

35y

1) x2 + x = 5 5) x2 + 2x = 5

2) p (p – 3) = 1 6) k2 – k = 0

3) x + = 2 7) 7y =

4) 2m2 = 32 8) x2 + 2 = 6

Y

Y1

X X1

Y

Y1

X X1

Y

Y1

X X1

Y

Y1

X X1


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201

IV (A) Solve the pure quadratic equation

B (1) If A = 2πr2 then solve for r and find the value of r, when A = 77 and π =

(2) If B = then solve for ‘a’ and find the value of a when B = 16 3

(3) If V = πr2h then solve for r and find the value of r, when v = 176 and h = 14

81a

(x - 4)2

1829

x2

234

14

1) 5x2 = 125

2) m2 – 1 = 143

3) 4a =

4) (2m – 5)2 = 81

5) =

6) - = 7

227

3 a2

4

x2 =

x2 = 25

x = ± 25

x = ±5

1255

(M - 07)


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202

(4) If r2 = l2 + d2 then solve for d and find the value of d, when r = 5 and l = 4.

(5) If k = mv2 then solve for v and find the value of v, when k = 100 and m = 2.

(6) If v2 = u2 + 2aS solve for v, if u = 0, a = 2 and S = 100, find the value of v.

V Find the roots of the following equations.

12

1) x (x – 3) = 0

2) (2m+1)(3m-2) = 0

3) m2 – 4m = 0

4) (y+6)(y+9) = 0

5) (b-3)(b-5) = 0

6) (5z-2)(7z+3) = 0


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203

VI Solve the quadratic equations. (By factorisation method)

(1) x2 + 15x + 50 = 0 (2) a2 + 5a + 6 = 0

(3) y2 = y + 2 (4) 6 - p2 = p

(5) 13m = 6 (m2 + 1) (6) 0.2t2 - 0.04t = 0.03

(Multiply by 100)

(7) 7 x2 - 6x - 13 7 = 0


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204

VII Solve the following equation by using formula.

(1) x2 - 7x + 12 = 0 (2) a2 = 4a + 6

(3) x2 - 8x + 1 = 0 (4) 2x2 - 2x = 5

(5) 3x2 = 48 (6) x2 + 1 = 8x

(7) x2 - 2x - 4 = 0 (8) 2y2 + 6y = 0

(M - 07)

(J - 09)

(J - 10)

(M - 09)

(M - 08)


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VIII Solve the following equations..

(1) (x+4)(x-4) = 6x (2) 2(a2-1) = a(1-a)

(3) (n-3)2 + (n+1)2 = 16 (4) 3(b-5) (b-7) = 4(b+3)

(5) 8(S-1) (S+1) + 2 (S+3) = 1 (6) 11(m+1) (m+2) = 38 (m+1) + 9m

IX (A) Solve (we first reduce to Standard form and solve by any method)

(1) = (2) =3x - 8x-2

5x - 2x+5

11m+39m+13

12m+110m+11

(J - 07)


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(3) + = (4) + =

(5) + = (6) + =

B (1) The product of two consecutive integers is 182. Find them.

(Let the integers be x and (x+1))

(2) The product of two consectutive even numbers is 48. Find them

( Let the number be x and x+2)

yy+1

y+1y

2512

1x-2

2x-1

2x

n+1n-1

n+2n-2

2n+13n+1

2m+2

52(m+4)

6m+6

∴


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(3) The sum of the squares of three consecutive integers is 194. Find them.

(Let the integers are x2, (x+1)2 and (x+2)2)

(4) The perimeter of a rectangular plot is 32m and its area is 60 Sq.m. Find the dimensions. ( l+b = 16, lb = 60)

(5) Sum of the two numbers is 18 and sum of the squares of those number is 290 find the numbers. ( x + y = 18 and x2 + y2 = 290)

(6) The length of a rectangular field is three times its breadth. The area of the field is 147 Sq.m. Find the length of the field.

∴

∴

(M - 10)

(J - 07)


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(7) The hypotenuse of a right tirangle is 20m. If the other two sides one is 4m longer then the other. Find the length of those two sides.

(8) The perimeter of a right triangle is 30cm. Its hypotenuse is 13cm. Find the other two sides.

(9) The base of a triangle is 4cm longer than its altitude. If its area is 48 Sq.cm. Find the base and the altitude of the triangle.

(10) An aeroplane takes one hour less for a journey of 1200 km if its speed is increased by 60km ph. Find its usual speed.

(M - 07)


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(11) A train required one hour less to return from a city 180km away, by travelling 6 km/h faster. How fast did it travel each way?

(12) Sailor Raju takes one hour forty minutes to go 8km downstream and return. If the speed of the current is 2 km/h. Find the speed of the boat in still water.

(13) Yashu bought some books for Rs. 60. Had he bought 5 books more for the same amount each book would have cost him one rupee less. Find the number of books bought by Yashu.

(J - 06)

(M - 06)


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(14) A dealer sells an article for Rs. 24 and gains as much percent as the cost of the article in rupees. Find the cost of the article.

(15) A man bought a certain number of glass vessels for Rs. 600. Two of them were broken. He sold the remaining at a profit of Rs. 10 each gaining Rs. 50 on the whole. How many glass vessels did he buy?

(16) Varsha takes 6 days less than Usha does, to do a piece of work. Together they can do it in 4 days. Find the number of days each will take to do it separately.


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(17) Two pipes together can fill a tank of water in 6 hours one of them takes 5 hours less than the other to fill it working alone. Find the time each will take to fill the tank.

(18) What number increased by its reciprocal equals:

(19) In an auditorium the number of seats in each row is 8 fewer than the number of rows. How many seats are in each row if the auditorium seats 609?

658


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(20) The head master of a school distributed Rs. 1500 equally among the rank holder of X standard class. If 5 more of them have secured the rank each would have got Rs. 25 less. Find the number of rank holders.

(21) The altitude of a triangle is 5 cm smaller than its base, if its area is 150 Sq.cm., find the base of the triangle. (M - 09)


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X Discuss the nature of the roots.

Q.E. Δ = b2 – 4ac Conclusion

1

3d2 – 2d + 1 = 0

a = 3 Δ = (-2)2 – 4(3)(1)

b = -2 = 4-12

c = 1 = -8 < 0

Roots are

Imaginary

2

2n2 – 9n + 8 = 0

3

x2 – 2x + 1 = 0

4

x2 – 2x – 5 = 0

5

2x2 + 5x – 1 = 0

6

x2 + 7x + 12 = 0


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XI For what positive value of m or p are the roots of the equations are equal, distinctand imaginary.

Quadratic Equation Δ = b2 – 4ac

Roots are Imaginary

Equal Distinct 1

x2 – mx + 9 = 0

a = 1 = (-m)2 – 4(1)(9)

b = -m = m2 - 36

c = 9

m2-36=0

m2 = 36

m = 6

m > 6

m < 6

2

mk2 – 3k + 1 =0

3

px2 + 2(p-4)x + 2= 0

4

x2 – (m+1)x + 9 = 0

5

3x2 + 2mx + 4 = 0

Imaginary

Roots are

±6


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XII Find the sum and product of the roots of the following equations.

-ba

-(-4)3

43

ca

53

Quadratic Equation a b c Sum m + n Product mn

1

x2 – mx + 9 = 0

3

-4

5

m + n =

=

=

mn = =

2

x2 – 5 = 4x

3

6x2 – 5x = 0

4

3x2 + 5 = 0

5

2ax2 + bx + ab = 0

6

x2 + mnx + m + n=0

7

2(a2+ b2)x2 +

2(a+b)x + 1 = 0

8

3a2x2 +8abx +4b2=0

3x2 - 4x + 5 = 0


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XIII Find the quadratic equations whose roots are given.

53

35

53

35

53

35

Roots m+n mn Equation

1

and

+

= =

x

=

x2-(m+n)x + mn = 0

x2- x + 1 = 0

15x2- 34x + 15 = 0

2

5 and -3

3

and

4

3 and

5

3 + 5 and 3 - 5

6

2 + 1 and 2 - 1

7

-2 ± 3 5

25+915

3415

13415

35

-23

25


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XIV (1) x = 2 is one root of 3x2 - 5x + c = 0. Find the other root.

(2) x = 7 is one root of x2 - bx - 28 = 0. Find the other root.

(3) Find the value of k, so that the equation x2 + 4x + k + 2 = 0 has one root equal to zero.

(4) The equation 4x2 - 8mx + 9 = 0 has one root as the negative of the other. Find the value of m.


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(5) The equation 5x2 - 26x + p = 0 has reciprocal roots. Find p.

(6) Find the value of q so that the equation 2x2 - 3qx + 5q = 0 has one root which is twice the other.

(7) Find the value of ‘p’ so that the equation 4x2 - 8px + 9 = 0 has roots whose difference is 4.


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(8) If one root of the equation x2 + px + q = 0 is twice the other prove 3p2 = 16q

(9) If one root of the equation px2 + 3x + 2 = 0 is reciprocal of the other then find the vlaue of p.

(10) If the roots of the equation (b-c) x2 + (c-a) x + (a-b) = 0 are equal. Then prove that 2b = a + c

(M - 10)

(J - 10)


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XV (1) If ‘a’ and ‘b’ are the roots of 3m2 = 6m + 5 find

(a) + (b) + (c) + (d) +

(2) If p and q are the roots of 2x2 + 3x + 6 = 0 find the value of

(a) p2 + q2 (b) (p+2q) (2p+q)

ab

ba

1a

1b

1a2

1b2

1a3

1b3


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(3) If m and n are the roots of 3x2 - 6x + 4 = 0 find the value of

(a) m2n + mn2 (b) + (c) + + 2 + +3mn

(4) If p and q are the roots of the equation 2x2 + 3x + 6 = 0, find the value of

(a) p2 + q2 (b) p3 + q3 (c) p2q + pq2

1m3

1n3

mn

nm

1n

1m


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(5) If m and n are the roots of the equation x2 + 6x + 9 = 0 find the value of

(a) m2n + mn2 (b) +

(c) + (d) + + 2 + + 3mn

(6) From the equation whose roots are the squares of the roots of the equation x2- 2x+ 4 = 0

(7) Find the value of m for which the equation 2x2 + 3x + m = 0 has equal roots.

1m2

1n2

m2

nn2

m1m

1n

1m3

1n3

(J - 07)

(J - 10)


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XVI (1) Draw the graph of y = x2 and find the value of 7

(2) Draw the graph of y = 2x2 and verify that the value of 4 = ±2

(J - 06)


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(3) Draw the graph of y = x2 and find the value of 3

(4) Draw the graph of y = x2 and y = x + 2 and hence solve x2 - x - 2 = 0. (M - 08)


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(5) Draw the graph of y = x2 and y = 2x + 3 and hence solve the equation x2 - 2x - 3 = 0

(6) Solve graphically x2 + x - 2 = 0.

(J - 09)

(M - 07)


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(7) Solve graphically 2x2 + 3x - 5 = 0

(8) Draw the graph of x212


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(9) Draw the graph of x2

(10) Draw the graph of y = x2

32

(M - 06)


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(11) Draw the graph of y = 2x2

(12) Draw the graph of y = x2 and y = 6 - x and hence solve the equation x2 + x - 6 = 0

(M - 09)

(M - 10)


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(13) Draw the graph of y = x2 and y = 2 - x and hence solve the equation x2 + x - 2 = 0

(14) Solve Graphically: 2x2 - 3x - 5 = 0

(J - 10)


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(15) Solve Graphically: 2x2 + 3x - 5 = 0

***************


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MODULAR ARITHMETIC6

• The face of the clock shows only 12 hours.

Ex (1) : 13th hour of the day is equivalent to 1st hour. The relation is expressed as13 congruent 1 (mod 12)

or 13 ≡ 1 (mod 12)

|||ly 17 ≡ 5 (mod 12)

Ex (2) : In an English calendar, if the 1st day is monday then 8th day, 15th day,22nd day these are also monday. This is an example of modulo 7.

i.e. 8 ≡ 1 (mod 7)

15 ≡ 1 (mod 7)

In general

a ≡ b (mod m) --> a - b ≡ 0 (mod m) that means (a-b) is a multiple of m

or

‘m’ is a divisor of (a-b)

** If a ≡ b (mod m) then (a-b) is always a multiple of ‘m’.

Set of residues: If any positive integer is divided by ‘m’ then the reminder will be

{0, 1, 2, 3, ............ (m-1)}

i.e. Zm = {0, 1, 2, 3, 4, ............ (m-1)}

Complete the following

(1) Z3 = { }

(2) Z4 = { }

(3) Z5 = { }

(4) Z6 = { }


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Modular Addition

When ‘a’ and ‘b’ are any two integers and ‘m’ is a positive integer then

a ⊕m b = r means,

‘r’ is the remainder obtained when (a+b) is divided by ‘m’.

Ex: 2 ⊕4 3 = 1

i.e. 2+3 = 5, when 5 is divided by 4 then reminder is 1.

Modular Multiplication

‘a’ and ‘b’ are any 2 integers, m is a positive integer then

a ⊗m b = r means,

‘r’ is th remainder obtained when (a x b) is divided by ‘m’.

Ex: 2 ⊗4 3 = 2

i.e. 2 x 3 = 6, if 6 is divided by 4, then reminder is 2.

Check whether the following relations are true or false

Ex: 17 ≡ 5 (mod 3)

Solution: 17 - 5 = 12

12 is a multiple of 3

∴ relation (congruence) is true.

(1) 75 ≡ 10 (mod 15)

(2) 121 ≡ (-8) (mod 17)

(3) 12 ≡ (-4) (mod 4)


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(4) 9 ≡ 23 (mod 12)

(5) 15 ≡ 29 (mod 7)

(6) 175 ≡ 5 (mod 17)

(7) 12 ≡ -8 (mod 6)

Find the sum of the following

Ex: 3 ⊕4 2

Solution 3 + 2 = 5

When 5 is divided by 4, the reminder is 1

∴ 3 ⊕4 2 = 1

(1) 8 ⊕10 9 (2) 5 ⊕7 4

(3) 4 ⊕5 3 (4) (3 ⊕7 6) ⊕7 4


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(5) (10 ⊕12 2) ⊕12 3

(6) 5 ⊕7 (6 ⊕7 8)

Find the value of ‘x’

(1) 3 ⊕5 4 = x (2) 5 ⊕6 x = 5

(3) 3 ⊕4 x = 1 (4) 2 ⊕3 x = 0

Find the product of the following

Ex: 5 ⊗11 10

Solution: 5 x 10 = 50

dividing 50, by 11, the reminder is 6

∴ 5 ⊗11 10 = 6

(1) 5 ⊗6 5 (2) 4 ⊗5 3


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(3) 5 ⊗7 8 (4) (4 ⊗11 3) ⊗11 7

(5) (5 ⊗12 2) ⊗12 6 (6) (3 ⊗4 5) ⊗4 6

Choose the value of ‘x’ or ‘y’ from the following.

Ex: x + 2 ≡ 4 (mod 5)

(a) 4 Solution: x + 2 - 4 = 5

(b) 7 x - 2 = 5

(c) 5 x = 5 + 2

(d) 3 x = 7

(1) 2x + 1 ≡ 1 (mod 10)

(a) 2

(b) 9

(c) 1

(d) 5

(2) 4 ⊕ y ≡ 0 (mod 5)

(a) 0

(b) 1

(c) 4

(d) 5


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(3) 3 ⊕ y ≡ 2 (mod 6)

(a) 4

(b) 2

(c) 5

(d) 6

(4) 2y ≡ 1 (mod 5)

(a) 2

(b) 5

(c) 6

(d) 3

(5) y ⊗ y ≡ 1 (mod 8)

(a) 6

(b) 3

(c) 8

(d) 4

CAYLEY’S TABLE

Cayley’s table represents the modular arithmetic system.

Ex (1) : Construction of Cayley’s table under addition modulo 3

i.e. (Z3, +) or ⊕3

(1) Z3 = {0, 1, 2}

0 ⊕30 = 0 0 ⊕31 = 1 0 ⊕32 = 2

1 ⊕30 = 1 1 ⊕31 = 2 1 ⊕32 = 0

2 ⊕30 = 2 2 ⊕31 = 0 2 ⊕32 = 1

⊕3 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1


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(2) Similarly ⊗3 [Multiplication table of Z3]

0 ⊗30 = 0 0 ⊗31 = 1 0 ⊗32 = 2

1 ⊗30 = 1 1 ⊗31 = 2 1 ⊗32 = 0

0 ⊗30 = 2 2 ⊗31 = 0 2 ⊗32 = 1

Construct the Cayley’s table for the following.

(1) Z5 under modulo 5 addition.

(2) ⊗6 , Z6

(3) Z4, ⊕4

⊗3 0 1 2

0 0 0 0

1 0 1 2

2 0 2 1


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(4) Z4, ⊗4

(5) Q = {0, 2, 4, 6, 8} under ⊕ mod 10.

(6) S = {1, 5, 7, 11} under ⊗ mod 12

Previous Exam Questions

(1) Which one of the following is true?

(a) 2 ≡ 3 (mod 3) (b) 7 ≡ 13 (mod 5)

(c) 3 ≡ 5 (mod 5) (d) 7 ≡ 12 (mod 5)

(2) If a ≡ b (mod m) then (a-b) is always

(a) greater than ‘m’ (b) multiple of ‘m’

(c) lesser than ‘m’ (d) m = 0

(M - 06)

(M - 06)


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(3) The value of 4 ⊗7 8 is ............................

(4) If 2y ≡ 1 (mod 5) the value of y is ............................

(5) The value of (3 ⊕7 6) ⊕7 4 is ............................

(6) The value of a is a ⊗ a ≡ 2 (mod 14) is ............................

(7) The value of (4 ⊗11 5) ⊗11 7 is ............................

(8) Modulo sustem which satisfies 15 ≡ 1 is

(a) mod 7 (b) mod 5 (c) mod 3 (d) mod 6

(9) If 2y ≡ 1 (mod 5) then y is ............................

(a) 2 (b) 5 (c) 6 (d) 3

(10) 17th hour of a day is equivalent to 5th hour this relation can be written as

(a) 17 ≡ 5 (mod 12) (b) 12 ≡ 5 (mod 17)

(c) 17 ≡ 12 (mod 5) (d) 17 ≡ 5 (mod 24)

(11) If y ⊗ y ≡ 1 (mod 8) the value of y is ............................

(12) The set of residues of modulo 4 is ............................

(13) The departure time of a bus from Mangalore to Bangalore is 21 : 00 hours, this isequal to

(a) 9:00 AM (b) 9:00 PM (c) 10:00 AM (d) 10:00 PM

(14) If 3y ≡ 2 (mod 2) the value of y is equal to

(a) 1 (b) 3 (c) 4 (d) 5

(15) Which one of the following is a correct relation

(a) 8 ≡ -7 (mod 3) (b) 7 ≡ (-2) (mod 4)

(c) 10 ≡ 6 (mod 3) (d) 18 ≡ 10 (mod 5)

(16) The value of x in 3x ≡ 2 (mod 4) is ............................

(a) 1 (b) 2 (c) 3 (d) 4

(M - 06)

(M - 06)

(J - 06)

(J - 06)\

(J - 06)

(J - 06)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(M - 08)

(M - 08)

(J - 08)

(J - 08)


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(17) The highest residue obtained by the division of a positive integer by (m+1) is

(a) 0 (b) 1 (c) m-1 (d) m

(18) If x ⊕10 x ≡ 2, the value of x is ............................

(19) If y ⊗ y ≡ 1 (mod 3), the value of y is ............................

(a) 1 (b) 3 (c) 6 (d) 2

(20) The sum of (3 ⊕7 6) ⊕7 5 is ............................

(21) Construct Cayley’s table under ⊗10 on S = {2, 4, 6, 8}

(22) Construct Cayley’s table under addition mod 3, on Z3.

**************

(M - 09)

( M - 0 9 )

( J - 0 9 )

( J - 0 9 )

( A - 1 0 )

( J - 1 0 )


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PRACTICAL GEOMETRY7

I (1) Two circles having radius 4cm and 3cm touching externally. Then the distance

between their centres is ..............................................

(2) Two circles having radius 4cm and 3cm touching internally. Then the distance

between their centres is ..............................................

(3) In figure, AB = CD = 8cm and OX = 3cm

then OC is ..............................................

(4) In the given figure, the transverse common

tangent is ................................................

(5) If two circles are intersecting each other, then the number of common tangents drawn

is .................................................

(6) In the figure common tangent is ................................

(7) In the figure common tangent is ................................

(8) Two circles having the radius 3cm and 4.5 cm, the distance between their centers is

5cm then the cirlces are ................................

(9) In the figure, the segment which makes right angle

in a semi circle is ................................

A

C

B

D

x

y

O

XYP

O1O2

A Q B

. .

A

B P

O

C

A CB

E

D F.

D

AE

F G

OB

C


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(10) Number of common tangents to be drawin to two concentric circles is ...........................

(11) If two circles having radius 4cm and 2cm, the distance between their centres is 7cms.

Then the circles are ................................

(12) AB and CD are two equal and parallel chords in a circle. If the distance from the

centre of the circle to the chord AB = 2x units, then the distance between the chords

is .......................................

(13) ABC is an angle in a major arc. Then ABC is ...................................

(14) Radii of two circles are 5cm and 3cm respectively and the distance between the

centres is 6cm. Then they are ...................................

(15) ‘O’ is the center of a circle. AB is a chord.

From the figure ACB is ...................................

(16) Two circles when touch externally, then number of transverse common tangents that

can be drawn is ...................................

(17) The angle in a semi circle is ...................................

(18) If two circles are touching internally, the distance between their centres is equal to

difference of their ...................................

(19) As the length of the chord increases, the perpendicular distance ...................................

(20) As the length of the chord decreases, the perpendicular distance ...................................

(21) The biggest chord in a circle is ...................................

(22) The perpendicular distance between diameter and center of the circle is .......................

(23) Minor arc subtends ...................................

(24) Major arc subtends ...................................

C

A B

O.


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(25) What are concentric circles?

(26) What are congruent circles?

(27) What is a secant?

(28) What is a tangent?

(29) How many common tangents can be drawn to the externally touching circles?

(30) In the figure POB and PBO are complementary

angle then PB is a .......................................

(31) In the figure OP ⊥ AB and OQ ⊥ AC.

If OP = OQ then the relation between

AB and AC is .......................................

(32) In the figure AB is a tangen ‘A’ is a point

of contact. I OBA = 450 which of the

following is true .......................................

(A) OA > AB (B) AB > OA

(C) AB > OB (D) OB = AB

O

P

B

A

B C

P QO

O

A B


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(33) No common tangents can be drawn to the following circles ...................................

(A) Externally touching circles (B) Internally touching circles

(C) Intersecting circles (D) Concentric Circles

(34) The tangents drawn to a circle at the ends of a diameter are ...................................

(35) The length of the transverse common tangent drawn to two circles is calculated

using the formula...................................

(36) The length of the direct common tangent drawn to two circles is caculated using the

forumla ...................................

(37) The number of tangents that can be drawn to internally touching circles at their point

of contact is ...................................

(38) In the figure AB = 4cm OP = OQ then CQ is ............................

II (1) Construct a tangent to a circle of radius 3cm at any point ‘p’ on it.

C

A

D

B

Q

P

O


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(2) Given PQ = 4cm with PQ as diameter draw a circle. Draw two tangents to the circleat ‘P’ and ‘Q’

(3) In a circle of radius 3cm, draw two radii such that the angle between them is 600.Draw two tangents at the ends of the radii.

(4) In a circle of radius 3cm, draw two tangents such that the angle between them is 1100.


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(5) Construct two tangents to a circle of diameter 4cm from a point 6cm away from thecentre.

(6) Construct two tangents to a circle of radius 3cm from a point 2.5cm away from thecircle.


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(7) In a circle of radius 2.5 cm, draw a chord of length 3.5cm. Draw two tangents at theends of chord.


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III (1) What is meant by a direct common tangent?

(2) What is meant by a transverse common tangent?

Common Tangents

S.No. Figure No. of Common tangents No. of DCT No. of TCT

1

2

3

4

5

A

C

B

D


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Direct Common Tangent

(1) Draw a direct common tangent to two congruent circles of radii 3cm, whose centresare 8cm apart.

(2) Construct a direct common tangent to two congruent circles of radii 2.5 cm each andwhose centres are 5cm apart.


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(3) Two circles of radii 3cm and 2cm have their centres 7cm apart. Draw direct commontangent to the circles.

(4) Construct a direct common tangent to two circles of radii 3.5 cm and 2cm and whosecentres are 1.5 cm apart.

(5) Draw two direct common tangents to two circles of radii 4.5cm and 2.5cm and theircentres are 7cm apart.


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(6) Two circles of radii 3cm and 2cm have their centres 3.5cm apart. Draw direct common tangents to the circles.


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Transverse Common Tangent

(1) Draw two congruent circles of radii 3cm having their centres 8cm apart. Draw atransverse common tangent.

(2) Construct two circles of radii 2cm and 3cm where centres are 8cm apart. Constructa transverse common tangents and measurethe length of trasverse common tangentand verify by calculation.

***************


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253

THEOREMS ON TRIANGLES AND CIRCLES8

• Two polygons having the same number of sides are Similar if and only if

(i) The angles of one triangle are equal to the corresponding angles of theother and

(ii) The sides of one triangle are proportional to the corresponding sides of theother.

• Congruent triangles and squares always similar.

• The symbol ‘|||’ means similar to

• Basic proportionality theorem (Thales Theorem):

A stright line drawn parallel to a side of a triangle divides the other two sides proportionately.

• Conversely if a line divides two sides of a triangle proportion the line is parallel to the third side of the triangle.

• Corollary of BPT: If a line is drawn parallel to a side of a triangle then the sides of new triangle formed are proportional to the sides of the given triangle.

THEOREM - 1• If two triangles are equiangular then their corresponding sides are proportional.

• Converse of Theorem 1: If the corresponding sides of two triangle are proportional then the triangles are equiangular.

• (i) If A = D, B = E and C = F then

Δ ABC ||| Δ DEF

(ii) If XY || BC then

Δ AXY ||| Δ ABC

(a) = =

(b) =

(c) = =

ABDE

ACDF

BCEF

AXAB

AYAC

XYBC

AXBX

AYCY

^ ^ ^ ^ ^ ^A

B C

D

E F

X Y^^


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THEOREM - 2

• Areas of similar triangles are proportional to the square on the corresponding sides.

• Areas of similar triangles are proportional to the square on the corresponding medians.

• Areas of similar triangles are proportional to the square on the corresponding altitude.

• Areas of similar triangles are proportional to the square of the radii of their circum circles.

THEOREM - 3

• In a right angle triangle the square on the hypotenuse is equal to the sum of the square on the other two sides.

• Conversely if the sum of the squares on any two sides is equal to the square on the third side those two sides include a right angle.

• If the square on the longest side of a triangle is greater than the sum of the squares on the other two sides the triangle is obtuse angled.

• If the square on the longest side of a triangle is less than the sum of the squares on the other two sides the triangle is Acute angled.

• If a set of three numbers the square of one equal the sum of the squares of the other two the three numbers form a pythagorean triplet.

• According to Boudhayana Shulba Sutra the diagonal of a rectangle produces both area which its length and breadth produce separately.

• Bhaskaracharya (1114 AD) has given a proof for the property of the right angle triangle.

I (1) In the figure MN || PR, if QN : NR = 2 : 3 and PQ = 11.5 cm then QM is ........................

(A) 10 cm (B) 2.3 cm

(C) 4 cm (D) 4.6 cm

(2) Sides of a triangle are of lengths 2cm, 3cm and 4cm respectively. Which of the sets

of numbers are sides of a triangle similar to the above triangle. ...............................

(A) 4, 5, 6 (B) 5, 6, 7 (C) 12, 13, 14 (D) 6, 9, 12

(3) The area of ΔABC = 144 Sq.cm and area of ΔPQR = 25 Sq.cm. Altitude of

ΔABC=6cm. If ΔABC ||| ΔPQR then the correspoinding altitude of ΔPQR is .................

(A) 2.5cm (B) 5cm (C) 12cm (D) 6cm

P

M

Q N R

. .

(M - 06)

(M - 06)

(M - 06)


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(4) In the figure ABCD is a trapezium AB || DC, which of the

following is equal to ? ...............................

(A) (B) (C) (D)

(5) In the figure MN || QR, is equal to .......................

(A) (B)

(C) (D)

(6) In the figure APQ = ACB and AQP = ABC then AP . AB is equal to ...............................

(A) AQ . AC (B) AP . AB

(C) AC . BC (D) BC . AB

(7) The corresponding sides of two equiangular triangles are ............................

(A) equal (B) parallel (C) proportional (D) unequal

(8) A vertical pole of 10m casts a shadow of 8m at certain time of the day. The length of

the shadow cast by a tower standing next to the pole of height 110m is

...............................

(A) 80m (B) 88m (C) 100m (D) 18m

(9) In a parallelogram ABCD, P is point on BC. In ΔDCP and ΔBLP, DP : PL ......................

(A) DC : BL (B) DC : BP

(C) PC : BL (D) PC : PL

ADBC

ABCD

BCAD

AOAB

OBOD

A

D

B

C

O

^^

Area of ΔPQRArea of ΔPMN

QRMN

PQ2

PM2

PQ2

PR2MN2

QR2

P

Q R

M N

^ ^ ^ ^

A

B C

P Qx

x

^

D C

A B

P

L

^^̂

(10) The name of the mathematician who proposed the basic proportionality theorem is

......................................

(A) Euclid (B) Thales (C) Pythagoras (D) Shreedhar

(M - 06)

(M - 06)

(M - 06)

(M - 06)(J - 07)

(M - 06)

(J - 06)

(J - 06)


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(11) In the figure XY || AB, AX = 9cm, XC = 7cm, BC = 20cm,

BY = ......................................

(A) 11.25 cm (B) 10.25 cm

(C) 10 cm (D) 15 cm

(12) If the two given triangles are similar then the ratio between their sides is ........................

(A) = = (B) = =

(C) = = (D) = =

(13) In the trapezium ABCD, AB || CD and the diagonal intersect

at O, then is equal to ...........................

(A) (B) (C) (D)

(14) In the given figure value of PQ is ...........................

(A) 10 m (B) 7.5 m

(C) 9.5 m (D) 3.5 m

(15) Select the set of numbers in the following which can form similar triangles .....................

(A) 9, 12, 18 and 3, 4, 6 (B) 3, 4, 6 and 9, 10, 12

(C) 8, 6, 12 and 2, 6, 3 (D) 3, 4, 5 and 2, 4, 10

(16) Two similar triangles have areas 120 Sq.cm and 480 Sq.cm respectively. Then the

ratio of any pair of corresponding sides is ...........................

(A) 1 : 4 (B) 1 : 2 (C) 4 : 1 (D) 2 : 3

^^

C

A B

X Y

ABPQ

ACQR

BCPR

ABQR

BCRP

ACPQ

ABPR

BCQR

ACPQ

ABQR

BCPQ

ACRP

•

•

A

P

B C RQx x

D

A

C

B

O

^^ OB

OAABCD

ACDB

OCOD

P

Q RX

Y

1.5m

8m 2m

ODOC

(J - 06)

(J - 06)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(M - 07)

(17) In the given figure ABC = AYX then the ratio of the

corresponding sides is ......................................

(A) = = (B) = =

(C) = = (D) = =

^ ^

^^

A

B C

X Y

AXAC

ABAY

CBXY

ABAX

ACAY

BCXY

ABAY

BCXY

AXAC

AXAC

AYAB

XYCB


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(18) A ladder 13 m long rests against a wall at a height 12m from the ground. Then the

distance of the foot of the ladder from the wall is ......................................

(A) 1m (B) 25m (C) 5m (D) 12.5m

(19) In the following figure DE || AB. If AD = 7cm, CD = 5cm, CE = 10cm then the length

of BE is ......................................

(A) 17cm (B) 14cm

(C) 12cm (D) 20cm

(20) In the trapezium ABCD, AB || DC which of the following is a

correct statement ......................................

(A) AO . OD = OB . OC (B) AO . OB = OC . OD

(C) AB . DC = OB . OD (D) AO . AB = OC . DC

(21) In a right angled triangle ABC, CAB = 900. If AD ⊥ BC then

the angle equal to ACD is ......................................

(A) ABD (B) DAB

(C) CAD (D) ADB

(22) In the figure OA = AB = BC = CD = 1 unit. The unit of OD is

.............................

(A) 1 (B) 2

(C) 3 (D) 4

^^

C

A B

D E

A

D

B

C

O

^^

^

^

C

A B

D

^

^

^

^

DC

B900

900

900

O A

(M - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(23) In a Rhambus ABCD diagonals intersect at O. The sum of

AC2 + BD2 is ......................................

(A) 4AB2 (B) 4AC2

(C) 4BD2 (D) 4AO2

A

C

B DO


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(24) In a right angled triangle ABC if CAB = 900 which of the following is correct.................

(A) BC2 = AC2 + AB2 (B) AC2 = AB2 + BC2

(C) AB2 = BC2 + AC2 (D) BC2 = AB2 - AC2

(25) Which of the following are the sides of the right angled triangle? .............................

(A) 6, 7, 8 (B) 20, 30, 10 (C) 24, 26, 10 (D) 16, 17, 18

(26) Which of the following polygons are always similar?.............................

(A) Isosceles triangles (B) Right angled triangle

(C) Rhombuses (D) Squares

(27) In the figure ΔABC, BE ⊥ AC and CF ⊥ AB. Then which of

the followig relation is correct?.............................

(A) AE . EC = AF . AC (B) AE . FC = AF . EB

(C) AB . BC = AC . EB (D) AE . BC = AB . CF

(28) In ΔXYZ if XY2 - YZ2 = XZ2 then the hypotenuse and right angled vertex are.................

(A) XY and X (B) XY and Z (C) YZ and X (D) YZ and Y

(29) Perimeter of a square is 20 cm. Then the length of the diagonal is .............................

(A) 10 2 cm (B) 10 cm (C) 5 2 cm (D) 5 cm

(30) In the figure B = C = 900. If AB = 5cm, BC = 6cm and

CD = 3cm then AD is equal to ............................

(A) 8 cm (B) 9 cm

(C) 10 cm (D) 12 cm

A

B C

F E

^ ^ ^ ^

D

A

B C

(J - 07)

(J - 07)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(31) The dimensions of a rectangula plot is 12m and 16m. The length of the longest line

that can be drawn in it is .............................

(A) 16m (B) 20m (C) 24m (D) 28m


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(32) In the figure DE || BC, AD : AB = 1 : 2, BC = 6cm then DE is

................................

(A) 1cm (B) 2cm

(C) 3cm (D) 4cm

(33) In ΔABC and ΔDEF if = = then the correct pair of corresponding

equal angles is .................................

(A) A and E (B) C and F (C) B and D (D) A and F

(34) The two corresponding sides of similar triangles are 3cm and 4cm. The area of

largest triangle is 48 Sq.cm. The area of smallest triangle is .....................................

(A) 80 Sq.cm (B) 64 Sq.cm (C) 36 Sq.cm (D) 27 Sq.cm

(35) Which of the following is a correct statement...........................................

(A) All the rectangles are similar

(B) All the rhombuses are similar

(C) All the right angled triangles are similar

(D) All the equilateral triangles are similar

(36) In ΔABC, PQ || AB the correct relation is ...........................

(A) = (B) =

(C) = (D) =

(37) In the figure ABC = AQP = 900 then = ................

^^

A

B C

D E

ABDE

BCEF

ACDF

^ ^ ^ ^ ^ ^ ^ ^

A

B CQ

P BQBA

CPCA

PQBQ

ABBC

APPC

PQQC

ABAP

(A) (B)

(C) (D)

BCPQ

QPBC

ACPQ

APAB

A

B C

P

Q

BQQC

AQAB

(J - 08)

(J - 08)

(J - 08)

(M -09)

(M -09)

(M -09)

(M - 09)

(38) If the perimeter of two similar triangles aer in the ratio of 4 : 1 then the ratio between

their areas will be .............................

(A) 16 : 1 (B) 4 : 1 (C) 2 : 1 (D) 2 : 1


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(39) ΔABC ||| ΔDEF the area of ΔABC is 45cm2 and the area of ΔDEF is 20 cm2 one side

of ΔABC is 3.6 cm then the length of corresponding side of ΔDEF is .............................

(A) 3.4 cm (B) 2.4 cm (C) 1.4 cm (D) 4.4 cm

(40) “If the square on one side of a triangle is equal to the sum of the squares on the other

two sides then those two sides contain a right angle” this statement refers to

..........................................

(A) Pythagoras Theorem (B) Thales Theorem

(C) Converse of Thales Thorem (D) Converse of Pythagorus Theorem

(41) The length of a diagonal of a square of side 5 cm is .............................

(A) 5 2 cm (B) 2 5 cm (C) 10 cm (D) 10 2 cm

(42) From the figure the length of AD is ...........................

(A) 12 cm (B) 14 cm

(C) 11 cm (D) 13 cm

(43) Which one of the following is Pythagorian Triplet?.............................

(A) 8, 15, 16 (B) 8, 15, 18 (C) 8, 15, 17 (D) 8, 15, 19

(44) The corresponding sides of two similar triangles are in the ratio 4 : 9. The ratio

between their areas is .............................

(A) 2 : 3 (B) 16 : 81 (C) 81 : 16 (D) 14 : 19

(45) The diagonal of a square is 10 2 cm then the length of its sides is .............................

(A) 2cm (B) 10cm (C) 8cm (D) 20cm

(46) The Mathematician who proposed Basic Proportionality Theorem is .......................

^^

^̂ ^̂

A

E

B C

D11cm12cm

12cm

6cm

(47) In ΔABC and ΔDEF, A = D, C = F, we can conclude .............................

(A) ΔABC is similar to ΔDEF (B) B is not equal to E

(C) ΔABC is not similar to ΔDEF (D) ΔABC is congruent to ΔDEF

^ ^ ^ ^

^ ^

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(A - 10)

(A - 10)

(A - 10)


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(48) ΔABC has sides of 5, 6 and 7 units while ΔPQR has a perimeter of 360 units.

ΔABC ||| ΔPQR when the sides of Δ PQR are .............................

(A) 2 times the sides of ΔABC (B) 10 times the sides of ΔABC

(C) 20 times the sides of ΔABC (D) 18 times the sides of ΔABC

(49) In the figure A = B, AD = BE the incorrect statement

is ..........................................

(A) AC = BC (B) CD = CE

(C) DE || AB (D) DE = AB/2

(50) The areas of similar triangles are not proportional to the square on the corresponding

..........................................

(A) sides (B) medians (C) altitudes (D) angular bisectors

(51) In ΔABC, A = 900 AD ⊥ BC so we can write .............................

(A) AD2 = BD . CD (B) AD2 = BC . CD

(C) AD2 = BD + DC (D) AD2 = BC . BD

(52) The relation 392 = 362 + 152 is mentioned as back as in 2000 BC in .............................

(A) Taittiriya Samhita (B) Bhaskara’s Work

(C) Apasthamba Sutras (D) Kathyayana Sutras

(53) An example for a pythagorean triplet is .............................

(A) 9, 10, 14 (B) 7, 25, 24 (C) 5, 5, 5 (D) 6, 6, 8

(54) The sides of a triangle are (a+b), (a-b) and 2 ab units it is .............................

(A) A right triangle with hypotenuse (a+b)

(B) A right triangle with altitude (a+b)

(C) A right triangle with hypotenuse 2 ab

(D) A right triangle with hypotenuse (a-b)

C

A B

D E

^ ^

^C

A B

D

(55) ABCD is a rhombus. AC2 + BD2 is equal to .............................

(A) 2AB2 (B) 3AD2 (C) 4BC2 (D) CD2


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(56) In ΔABC, A = 900, AD ⊥ BC which of the following is

correct .............................

(A) ΔABC ||| ΔADC (B) ΔABC ||| ΔADB

(C) ΔADC ||| ΔADB (D) Both A, B and C

(57) In figure = ..........................................

(A) (B)

(C) (D)

(58) In the figure ABD = BDC, CD = 4AB then

BD = ..........................................

(A) BE (B) 4BE

(C) BD = AC (D) 5BE

(59) In ΔABC, BAC = 900 AD ⊥ BC then the value of x is

.............................

(A) 5 (B) 2 5

(C) 3 5 (D) 5 5

(60) In a trapizium ABCD, AB || DC and AB = 2CD, if the

diagonals intersect at ‘O’ the triangle AOB is

.............................

(A) 4 ΔAOD (B) 4 ΔCOB

(C) 4 ΔCOD (D) 4 ΔADC

(61) D, E, F are the midpoints of a triangle ΔABC then ΔDEF is equal to .............................

(A) ΔABC (B) ΔABC (C) ΔABC (D) ΔABC

A

D C

a + ba

aa + b

a + ca

b + db

c + dc

B

D

C

A B

E

b

a c

d

^ ^

A

B

D

C

E

D

A C

B

x

4

5

O

^

D

A

^

B

C

12

14

18

(62) In an equilateral triangle ABC, AN ⊥ BC then AN2 = .............................

(A) 3BN2 (B) BN2 (C) 2BN2 (D) 4BN2


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(63) The length of the diagonal of a square of side x unit is .............................

(A) x x (B) 2x (C) x 2 (D)

(64) The ratio of the length of a side of an equilateral tiangle and its height is ......................

(A) 2 : 1 (B) 1 : 2 (C) 2 : 3 (D) 3 : 2

(65) In Δ ABC, AB = AC and BD ⊥ AC then BD2 + CD2 = ..........................................

(A) 2 AC . CD (B) AC . CD (C) (D) 2 (AC-CD)

(66) If ABC and DEF are similar triangles in which A = 470 and E = 830 then C is I

.............................

(A) 600 (B) 700 (C) 500 (D) 400

x2

ACCD

^ ^ ^


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II (1) In figure PQ || BC fill the blanks in the following

(1) = (2) =

(3) = (4) =

(2) In the figure XY || QR write the ratios in terms of a, b and c

(1) = ===>

(2) = ===>

(3) = = ===>

(3) In the figure XY || BC and XZ || AC, show that = =

(4) In ΔABC, XY || BC, AB = 4.5 cm, AC = 3.5 cm and AX = 7.2 cm. Find XY.

(5) In the adjoining figure AD = 11.2 cm, AB = 16.8 cm, AE = 7.6 cm and AC = 11.4 cm.Show that BC || DE

ABAP

QCAQ

ABPB

AQAP

P

A

B C

Q

X

P

Q R

Y

b

a c

dx

y

^^

PXXQ

PYYR

PXPQ

PYPR

PXPQ

PYPR

XYQR

AXAB

AYAC

XYBC

X

A

B C

Yd

^

^

Z

P

A

B C

Q

D

C

A

E

B


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(6) In ΔABC, DE || BC, AD = 4x-3, AE = 8x-7, BD = 3x-1 and CE = 5x-3. Find the valueof x

(7) Prove that any two medians of a triangle divide each in the ratio 2 : 1

(8) In the figure XY || BC, AX = a-2 and BX = 2a-3 and = find a.

(9) ABCD is a parallelogram. P is any point on BC . DE meets AB produced at L. Provethat DP : PL = DC : BL.

E

A

B C

D

G

AYCY

25

X

A

B C

Y

^^

D C

A B

P

L


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(10) Show that in a trapezium the line joining the mid points of non parallel sides isparallel to the parallel sides.

(11) In the figure ABD = BDC, CD = 4AB. Show that BD = 5BE

(12) In ΔABC, AB = AC, D is a point on AC. Such that BC2 = AC x CD.

(13) In the figure A = B, AD = BE, show that DE || AB.

X

P

A B

Y

^^D C

^ ^

A

B

D

C

E

^ ^

D

C

A B

E


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(14) Fill with suitable words.

If two triangles are equiangular then their corresponding sides are .................................

Data: In ΔMJR and ΔAPN

JMR = PAN

MJR = .....................

MRJ = .....................

To Prove: = =

Construction:Mark V on MJ and Y on MR such that MV=AP and MY=...........and join VY

Proof: In ΔMVY and ΔAPN

M = ................. (Data)

MV = AP (...........................................)

MY = AN (Construction)

∴ Δ ............. ≅ Δ ............. (S.A.S).

∴ VY = PN and MVY = APN

MYN = APN = MJR

∴ MVY = MJR (Corresponding Angles)

∴ VY || .................

∴ = = (BPT and Corollary)

∴ = =

V

M

J R

YA

P N

^ ^

^

^

MJAP

JRPN

.........

.........

^ ^

^ ^ ^

^ ^

MJMV

JR........

MR.........

MJ..........

JR........

MR.........

^


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(15) A man whose height is 1.5 m standing 8 m from a lamp post observes that hisshadow cast by the light is 2 m in length. How high is the lamp above the ground.

(16) If one diagonal of a trapezium divides the other in the ratio 2 : 1 then prove that oneof the parallel sides is twice the other. (M - 09)


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(17) Prove that the areas of similar triangles are proportional to the squares on thecorresponding sides.

Data: Let PQR and XYZ are similar triangles in which QR and YZ are corresponding sides.

To Prove: =

Construction: Draw PA ⊥ QR and XB ⊥YZ

Proof: =

= x (1)

In ΔPAQ and ΔXBY

PQA = ...........................

PAQ = ................ = 90

QPA = YXB

∴ ΔPAQ ||| ........................ (Equiangular)

= (Corresponding Sides)

= =

= (2)

= = (Substitute (2) in (1))

∴ =

P

A RQ B ZY

X

Area of a ΔPQRArea of ΔXYZ

..................

..................


1/2 x ..................

1/2 x ..................

QRYZ

.............

.............

^

^

^ ^

PAXB

PQXY

PA........

PQ........

QR........

PAXB

QRYZ


QRYZ

QRYZ

Area of a ΔPQR............................

............YZ2


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(18) Two corresponding sides in two similar triangles are 3.6 cm and 2.4 cm respectively.If the area of the bigger triangle is 45 cm2 find the area of the smaller one.

(19) Two similar triangle have areas 392 cm2 and 200 cm2 respectively. Find the ratio ofany pair of corresponding sides.

(20) The area of ΔABC is 25.6 cm2 XY || BC.It divides AB in the ratio 5 : 3. Find the areaof ΔAXY.


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(21) In ΔABC, BE ⊥ AC, CF ⊥ AB, BE and CF meet O. Show that =

(22) XY is drawn parallel to base BC of ΔABC. If ΔAXY : trapezium XBCY = 4 : 5, showthat AX : XB = 2 : 1

(23) Prove that the areas of similar triangles have the same ratio as the squares of thecorresponding altitudes.

ΔBOFΔCOE

BF2

CE2


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(24) Prove that the areas of similar triangles have the same ratio as the squares of thecorresponding medians.

(25) Prove that the areas of similar triangles have the same ratio as the square on theircircum radii.


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(26) ΔABC is right angled at A, AD ⊥ BC, Show that =

(27) In ΔABC, A = 900, AD ⊥ BC. Show that AD2 = BD . CD

(28) D, E and F are the mid point of ΔABC, show that ΔDEF = ΔABC

ΔABDΔACD

AB2

AC2

^

14


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274

(29) Prove that in a right angled triangle the square on the hypothenuse is equal to thesum of the squares on the other two sides.

Data: In ΔMJR, JMR = 900

To Prove: JR2 = ..................+ ..................

Construction: Draw MP ⊥ JR

Proof: In ΔMJR and ΔPJM

JMP = .................. = 900 (Data and Construction)

MJR = MJP (Common)

∴ ΔMJR ||| ........................ (Equiangular)

∴ = (Raio of corresponding sides)

∴ JP . JR = ....................... (1)

In ΔMJR and ΔPMR

JMR = MPR = 900 (Data and Construction)

ΔMRJ = ............... (Common)

∴ ΔMJR ||| Δ................ (Equiangular)

=

PR x JR = ....................... (2)

JP x JR + PR x JR = .......................... (Adding i and ii)

JR (JP + PR) = MJ2 + MR2

∴ JR . JR = MJ2 + MR2 ∴ .............................. = ..............................

M

P RJ

^

^ ^

MJJP

JR..........

^ ^

^

JRMR

MR.........


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(30) Test whether the following are pythagoren triplet or not.

(31) Find the length of the diagonal of a squares of side

(a) 12 cm (b) 5 unit

i) 5, 3, 4 52 = 42 + 32 25 = 16 + 9 25 = 25 They form pythagoren triplet

ii) 8, 15, 17 iii) 9, 10, 14

iv) 7, 24, 25

v) 12, 13, 5 vi) (a+b), (a-b), 2 ab

vii) 2.5, 3.5, 4.5

viii) 2.5, 6, 6.5 ix) 3, 3, 5


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(32) The diagonal of square is given find each side of the squaer and perimeter

(a) 10 cm (b) 8 cm

(33) Calculate the altitude of an equilateral triangle of side

(a) 8 cm (b) 5 cm

(34) Two poles 10m and 18m high stand vertically 15 m away from each other. Find thedistance between their top


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(35) The man walks 8km due north then 5 km due east and from there 4km to north. Howfar is he from the starting point.

(36) The base of a right circular cone has a diameter of 15 cm and its slant height is 8.5cmfind its vertical height.

(37) A ladder is placed in such a way that its foot is at a distance of 5m from the wall andits top reaches a window 12m above the ground. What is the length of the ladder.


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(38) ABCD is a trapezium in which AB || CD and BC ⊥ AB, AB = 7.5 cm, AD = 13cm andCD = 12.5 cm, find BC.

(39) ABC is a triangle right angled at A, AD ⊥ BC. Find the sides of the triangle ABC ifBD = 8cm and CD = 2cm.

(40) In ΔABC, AD ⊥ BC, prove that AB2 - BD2 = AC2 - CD2.


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(41) In an equilateral ΔABC, AN ⊥ BC, show that AN2 = 3BN2.

(42) In ΔABC, AB = AC, BD ⊥ AC, Prove that BD2 + CD2 = 2AC . CD

( 4 3 ) I n ΔABC, AD ⊥ BC, DB : CD = 3 : 1, Prove that BC2 = 2 (AB2 - AC2)


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(44) ABCD is a rhombus, prove that AC2 + BD2 = 4AB2.

(45) ΔABC is right angled at C, P is a point on AC, Q is a point on BC prove that

AQ2 + BP2 = AB2 + PQ2

(46) P is a point of trisection of the base BC of an equilateral ΔABC, prove that 9AP2=7BC2

(47) In ΔABC, B = 900 P is the midpoint of BC, PN ⊥ AC, prove that AN2 - NC2 = AB2.


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(48) A ladder of length 2.6m is leaned against a wall when it is at a distance of 2.4 m fromthe foot of the wall. The top of the ladder touches the bottom edge of the window inthe wall. If the foot of the ladder is moved 1.4 m towards the wall is touches the topedge of the window. Find the height of the window. (A - 10)


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TOUCHING CIRCLES

If two circles touch each other the point of contact and the centre of the circles arecollinear.

If the two circles touch externally then the distance between their centre is d = R + r

If the two circles touch internally then the distance between their centre is d = R - r

Three circles of radii r1, r2 and r3 touches externally then perimeter of the triangleformed by joining the centres of the circles is 2(r1 + r2 + r3)

The tangents drawn to a circle formed an external point are equal.

If the sides of a quadrilateral are tangents to a circle then the sum of opposite sidesequal to the sum of the other pair of sides.

Nature Direct

Total Common

DCT Tangents

TCT Circles are

1 d > R + r 4 2 2 Separated

2 d = R + r 3 2 1 External Touch

3 d < R + r 2 2 0 Intersect

4 d = R – r 1 1 0 Internal Touch

5 d < R – r 0 0 0 One within other

6 d = 0 0 0 0 Concentric

TotalCommonTangent

Common Tangents

A B

A B

A B

A B

A B

A


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Mention the kinds of circles in each case.

I Multiple Choice Questions

(1) Two circles of radii 8cm and 5cm touch internally. Then the distance between the

centres is ..............................

(A) 18 cm (B) 3 cm (C) 5cm (D) 6cm

(2) A Tangent drawn to a circle of radius 8cm from a point which is at a distance of 10cm

from the centre of the circle. Then the length of the tangent is ..............................

(A) 8 cm (B) 18 cm (C) 2cm (D) 6cm

(3) AC, CE, EH are tangents drawn to the circle at B, D and F respectively. If

CE = 10cm and DE = 3.5 cm then EF is equal to ..............................

(A) 6.5cm (B) 3.5cm

(C) 10cm (D) 5cm

(4) A Tangent of length 16cm is drawn to a circle at a distance of 20cm away from the

centre of the cirlce. The radius of the circle is ..............................

(A) 12 cm (B) 16 cm (C) 20 cm (D) 8 cm

C

S.No. d R r Kind of Circles

1 15 8 5

2 3 7 4

3 0 4 2

4 1 5 3

5 7 5 3

6 9 6 3

E

FB

A H

D

(M - 07)

(M - 07)

(M - 06)

(M - 06)

.


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(5) AP is a tangent to a circle with centre O as shown in the figure. If P = 450 and radius

of circle is 5cm then OP is equal to ..............................

(A) 5cm (B) 10cm

(C) 9cm (D) 5 2 cm

(6) Two circles of radii 4cm and 3cm touch each other then the distance between their

centres will be ..............................

(A) 7 cm (B) 1 cm

(C) Either 7cm or 1cm (D) 0cm

(7) Three circles of radii 4cm, 3cm and 2cm touch each other externally. The perimeter

of the triangle formed by joining their centres is ..............................

(A) 9cm (B) 15cm (C) 18cm (D)12cm

(8) APB is a tangent at P to the circle with centre O if QPB = 600 then POQ is .....................

(A) 1200 (B) 900

(C) 1000 (D) 600

(9) PA and PB are the tangents to a circle with centre O as shwon in the figure. If

AOB = 1400 then the measure of APB is ..................................

(A) 400 (B) 200

(C) 900 (D) 1400

(10) In a circle of radius 5cm the distance of a chord of length 7cm from the centre is

..................................

(A) 4cm (B) 13cm (C) 2.5cm (D) 3cm

(11) Two circles of radii 6.9cm and 2.8cm touch each other externally. Then the distance

between their centre is ................................

(A) 3.45cm (B) 1.4cm (C) 4.1cm (D) 9.7cm

O P

A

^ ^

^ ^

A

O

B

P1400

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(J - 06)

(J - 06)

(J - 06)

A BP

QO

600


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(12) Three circles with centres A, B and C touch each other as shown in the figure. If the

radii of these circles are 8cm, 3cm and 2cm respectively then the perimeter of ΔABC

is ..............................

(A) 13cm (B) 16cm

(C) 3cm (D) 26cm

(13) For a circles of radius 5cm, two tangents PA and PB are drawn from a point P. If

PA = 12cm and PAB = 600 then the length of AB is ..............................

(A) 10cm (B) 12cm (C) 2.5cm (D) 6cm

(14) Two circular discs of radii 4.5cm and 2cm are fixed to a string of length 10cm as

shown. Then the diameter of another disc which touches in circular disc at P and Q

is ............................

(A) 6.5cm (B) 2.5cm

(C) 1.75cm (D) 3.5cm

(15) From the figure AP = 3cm and PC = 8cm then the length of the tangent CD is ................

(A) 11cm (B) 5cm

(C) 7cm (D) 8cm

(16) In the figure PA and PB are the tangents and AOB = 1400 then the measure of

APO is .......................................

(A) 900 (B) 400

(C) 200 (D) 1800

B

A C

^

A BP Q

D

A

O

PB

C.

^

^

A

O 1400

B

P

(J - 06)

(J - 06)

(M - 07)

(M - 07)

(M - 07)


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(17) If two circles of radii 9cm and 4cm are touching internally then the distance between

their centres in cm is ..............................

(A) 13 (B) 36 (C) 8 (D) 5

(18) O is the centre of a circle PA and PB are tangents at A and B respectively. If P = 650

then the measurement of AOB is ..............................

(A) 1300 (B) 650

(C) 1150 (D) 150

(19) In two concentric circles of radii 6cm and 10cm with centre O, OP is the radius of the

smaller circle OP ⊥ AB, AB cuts the outer circle at A and B then length of AB is

..........................................

(20) In the figure the length of OP is ..........................................

(A) 5cm (B) 4cm

(C) 3cm (D) 25cm

(21) In the figure AB, AC, DC are tangents. If AB = 3cm and CD = 2cm then length of AC

is ...................................

(A) 6cm (B) 10cm

(C) 5cm (D) 1cm

(22) In the figure if PA and PB are tangents and AB = AP then APB is ..........................

(A) 300 (B) 900

(B) 450 (D) 600

^

A

O

B

P

A

3 4

O?

P

DB

CA

^

A

B

P

^

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)


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(23) Two circes of radii 3cm and 4.5cm are drawn with their centres 5cm apart. They are

..............................

(A) Concentric Circles (B) Intersecting Circles

(C) Congruent Circles (D) Non-intersecting Circles

(24) In the figure AB, AC and PQ are tangents. The perimeter of the equilateral Δ APQ is

15cm then AB is equal to ..............................

(A) 5cm (B) 6cm

(C) 6.5cm (D) 7.5cm

(25) Radii of two concentric cirlces are 8cm and 10cm respectively. The length of the

greatest chord which is a tangent to the inner circle is ..............................

(A) 6cm (B) 8cm (C) 12cm (D) 20cm

(26) The common tangent in the figure is ..............................

(A) AC (B) AD

(C) CE (D) CF

(27) Two circles of radii 4cm and 2cm have their centres 7cm apart. These circles

...............................................

(A) Touch each other externally (B) Touch each other internally

(C) Do not touch each other (D) Intersect each other

(28) Two circles touch each other internally. The distance between their centres is 1.5cm.

If the radius of one circle is 3.5cm then the radius of the other circle is ...........................

(A) 5cm (B) 4cm (C) 3cm (D) 2cm

B

C

P

Q

A

E

FD

A B C

(M - 08)

(M - 08)

(M - 08)

(M - 08)

(J - 08)

(J - 08)


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(29) In the given figure O is the centre of the circle. AC and BC are the tangents. If

BOC = 650 then ACO is ..............................

(A) 250 (B) 350

(C) 650 (D) 1150

(30) In the given figure O is the centre of the circle XY is a tangent. If PQY = 550

OPQ is .......................

(A) 1250 (B) 1200

(C) 1100 (D) 350

(31) In the figure AB is tangent to the circle with centre O. If AOB = 300 then A and B

respectively are .......................

(A) 750, 750 (B) 1000, 500

(C) 800, 700 (D) 900, 600

(32) Radii of two circles are 5cm and 3cm respectively and the distance between their

centre is 6cm then they are .......................

(A) Touching Externally (B) Intersecting Circles

(C) Touching Internally (D) Concentric Circles

(33) In the figure A and B are the centres of two circles with radii 6cm and 2cm

respectively. CD is the diameter then MD is .......................

(A) 8cm (B) 6cm

(C) 4cm (D) 2cm

^ ^

A

O C

B

450

^

^

O

X YQ

550

P

^

A

O B

^ ^

CM

A B. ..

(J - 08)

(J - 08)

(M - 09)

(M - 09)

(M - 09)


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(34) In the figure AB, AC and BD are the tangents as shown in the figure. If AB = Xcm

and BD = Ycm then AC is ..............................

(A) x cm (B) y cm

(C) (x-y) cm (D) (x+y) cm

(35) If two circles of radii 4.5cm and 3.5cm are touching externally then distance between

their centre is ..............................

(A) 8.0 cm (B) 1.0 cm (C) 7.0 cm (d) 7.5

(36) PA and PB are tangents drawn from point P to a circle of centre ‘O’ of APO = 300

then AOB is ..............................

(A) 600 (B) 1200

(C) 1100 (D) 1000

(37) In a circle of radius 10cm, O is the centre OP ⊥ AB. If OP = 6cm then the length of

chord AB is ..............................

(A) 8cm (B) 12cm

(C) 20cm (D) 16cm

(38) In the figure XP, XQ and XR are tangents to the circle. If the length of XQ = 9cm

then the length of tangent XR is ..............................

(A) 18cm (B) 10cm

(C) 9cm (D) 12cm

(39) In the given figure if PAO = 300 the measure of POQ is ............................

(A) 600 (B) 1200

(C) 900 (D) 300

^

^

C .O

D

BA

O

B

300 P

A

A B

O

P

Q

XR

P

AB

..

P

O

Q

A

^ ^

(M - 09)

(J - 09)

(J - 09)

(J - 09)

(J - 09)

(A - 10)


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(40) In the given figure O is the centre of the cirlce AB, BC and CA touch the circle at L,

M and N respectively. If B = 700 and C = 600 LON = ..............................

(A) 500 (B) 600

(C) 700 (D) 1300

(41) In the figure shown PO and PR are the tangents to the circle. If PQR = 600 and

PQ = 9cm then the length of QR is ..............................

(A) 9cm (B) 7cm

(C) 5cm (D) 3cm

(42) Two concentric circles are of radii 13cm and 15cm. The length of the chord of the

outer circle which touches the inner circle is ..............................

(A) 27.8cm (B) 24cm (C) 16.9cm (D) 12cm

(43) AT and BT are tangents to a circle. Another tangent PQ is drawn such that TP = TQ

(A) Congruent (B) Similar (C) Scalene (D) Non equiangular

(44) In the figure XY and PC are common tangents to two touching circles then XPY is

..............................

(A) 900 (B) 600

(C) 450 (D) 300

(45) AB, AP and AC are tangents to the circle then ..............................

(A) AB = BP (B) BP = CP

(C) AB = AC (D) AC = BP

^ ^ ^

^

ΔTAB and ΔTPQ are ..............................

A

L N

MB C

O700 600

O

O .

Q

R

P600

^

P

X C Y

B

A

C

P.

.


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II (1) Three circles touch each other externally, their centres are joined to form a triangle.If the sides of the triangle formed are 9cm, 8cm and 7cm respectively find the radii ofeach circle.

(2) In the given figure O is the centre of the circle, AB = 2cm, AC = 3cm, CE = 6cm.

Find DE.

(3) Three circles of radii 3.5cm, 4.5cm and 5cm touch each other externally as shown infigure. Find the sides of ΔABC and perimeter of ΔABC given that A, B and C are thecentres of the circle.

A

CB

EA

BC

.

D

O

PA B

C

Q R

(M - 08)

(J - 08)

(J - 08)


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(4) TP and TQ are the tangents drawn to a circle with centre O. Show that PTQ = 2OPQ

(5) The sides of the quadrilateral ABCD are tangents to the circle with centre O. Thenshow that AB + CD = AD + BC.

(6) In the figure PQ and PR are tangents to the circle with centre ‘O’ if QPR = 900. Showthat PQOR is a square.

^ ^

^

(M - 09)

(J - 09)

Q

O

R

P


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(7) In two concentric circles of radii 6cm and 10cm with centre O. OP is the radius of thesmaller circle OP ⊥ AB which cuts the outer circle at A and B. Find the length of AB.

(8) AB = 10cm, with centre A, a circle of radius 4.5cm and with centre B, a circle ofradius 2cm are drawn. These circles pass AB at P and Q respectively. What is theradius of another circle which touches these circles at P and Q?

(9) Two circles of radius 8cm and 5cm with centres A and B have external contact. PQ isdirect common tangent to them. Calculate the length of PQ.


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(10) AB = 8cm, M is the mid point of AB. Semi circles are drawn on the same side of ABwith AM, BM and AB diameter. A circle with centre O touches all these semi circles.Prove that the radius of this cirlce is AB/6.

(11) Two circles touch each other externally at P. Two diameters AB and CD are drawn,one in each circle, parallel to each other. Show that BPC and APD are straight lines.

(12) In figure AP = 3cm, PC = 8cm find CD.

MA C D B

O

P Q

D

A B

CO.

P


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(13) PQ and PR are tangents. OQ = 9cm, PQR = 600. Find the length of chord QR.

(14) AT and BT are tangents to a circle. Another tangent PQ is drawn such that TP = TQ.Show that ΔTAB ||| ΔTPQ.

(15) In ΔABC, AB = AC. The sides of the triangle touch a circle at P, Q and R as given.Prove that Q is the midpoint of BC.

^

A

P

B Q

R

C

.


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(16) The sides of a quadrilateral ABCD are tangents to a circle. Show that AB+CD=AD+BC

(17) The sides of quadrilateral ABCD are tangents to a circle with centre O. Show thatAOB + COD = 1800.

(18) AB is a tangent to a circle with centre O and A is the point of contact. If OBA = 450.Prove AB = OA.

^ ^

^

N

MD C

BA

L

K


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(19) Prove that the tangents drawn at the ends of the diameter to a circle are parallel.

(20) Tangents to circle with centre O are drawn at the ends of the diameter AB. Anothertangent is drawn to intersect these at C and D. Prove that CD = AC + BD.

B

A

O

D

C

R


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(21) Two concentric circles are of radii 13cm and 5cm. Find the length of the chord of theouter circle with touches the inner circle.

**************


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MENSURATION9

• Mensuration is a branch of Mathematics which deals with the measurements of lengths of lines, areas of surfaces and volumes of solids.

• Two parts of mensuration

(a) Plane Mensuration

(b) Solid Mensuration

• The formula to find the area of a circle is πr2

• The length of the circumference of a circle is 2πr

• Area of a triangle is x base x height

• Area of a rectangel is length x breadth

Points to be remember

(1) Circumference of a circle C = 2πr

(2) Diameter d = 2 x r

(3) The relation between l, r and h is l2 = r2 + h2

(4) In case of Hollow Cylinder, opened at both ends Surface Area = 2πrh

(5) Cylinder opened at one end, the Surface Area = 2πrh + πr2 (or)

= πr(2h + r)

(6) Area is expressed in Square Units.

(7) Volume is expressed in Cubic Units.

12

Solid Structure Area of

Flat Surface

Lateral Surface Area

(LSA)

Total Surface Area

(TSA)

Volume (V)

Cylinder 2πr2 2πrh 2πr(r+h) πr2h

Cone πr2 πrl πr(r+l) πr2h

Hemisphere πr2 2πr2 3πr2 πr3

Sphere ‐ 4πr2 πr3

13

23

43


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I (A) (1) What is a Cylinder?

(2) Name two types of Cylinders. Write their diagrams

(3) Give any 4 examples for a solid cylinder.

(4) Give any four examples for a hollow cylinder.

(5) A rectangle of length 5cm and breadth 3.5cm is rotated on its length. Name thesolid formed and write its radius and height.

(6) Write the formula to find the lateral surface area of a cylinder.

(7) Write the formula to find the total surface area of a cylinder.

(M - 08)(M - 10)

(M - 09)(J - 09)


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(8) Which formula you can use to find the surface area of a circular tin opened at oneend.

Hint:

area of Lateral Surface +

Area of Circular Base

(9) Find the curved surface area of a cylinder whose circumference is 44 cm andheight 10cm.

(10) Fill up the following table, with suitable answer. (Use extra sheets of paper ifnecessary)

(11) The circumference of a thin hallow cylindrical pipe is 44 cm and length is 20 m.Find the surface area of the pipe.

Solution: C = 2πr Since the cylinder is hallow

44 = 2πr Surface area means LSA

.......................... LSA = ...................................

.......................... ..................................

r = ................. cm ...................................

Length of the pipe = 20m ...................................

h = 20 x 100 cm

S.No. Radius

r Height

h LSA of the Cylinder

TSA of the Cylinder

1 14 cm 10 cm

2 10 cm 18 cm

3 14 cm 1056 Sqcm

4 10 cm 1760 Sqcm

5 7 cm 12 cm

8 70 cm 1.3 mtr


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(12) The circumference of a thin hollow cylindrical pipe is 88cm and radius is 12cm.Find the surface aera of the pipe.

(13) A mansion has 12 cylindrical pillers each having the circumference 50 cm andheight 3.5 m. Find the cost of painting the lateral surface of the pillers at Rs. 25/-per square meter.

Clue: LSA of 1 Pillar = 2πrh

LSA of 12 Pillars = 12 x 2πrh

(14) The diameter of a thin cylindrical vessel opened at one end is 3.5 cm and height is5 cm. Calculate the surface aera of the vessel

d = 3.5 cm Clue: Surface area means

r = ...................... = LSA + aera of the base

= 2πrh + πr2

=


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(15) A roller having radius 35cm and length 1m takes 200 complete revolutions tomove once on a playground. What is the area of the playground

r = 35cm LSA = 2πrh

l = h = 1m = 100cm =..........................................

= ..................... x 200 Sq.cm

= ..................... Sq.cm


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VOLUME OF A CYLINDER

Volume of a Cylinder V = Area of the base x height

V = πr2 x h Cubic Units

V = πr2h

B (1) Fill up the following table with suitable answer. (Use extra sheets of paper)

(2) Area of the base of a right circular cylinder is 154 Sq.cm and height is 10 cm.Calculate the volume of the Cylinder.

Solution: B = πr2 = 154 Sq.cm V = πr2h

h = 10 cm

V = ?

S. No.

Radius ‘r’

Diameter ‘d’

Circumference of the base

‘c’

Height ‘h’ V

1

7cm

120m

2

7cm 120m 462 cm3

3

88cm 10cm

4

5cm 28cm

5 20cm 3080cc

6

20cm 4400cc

(J - 07)


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(3) A cylindrical vessel of height 35 cm contains 11 ltr of Juice. Find the diamter of thevessel. (Clue: 1 ltr = 1000 cc)

Solution: V = 11 ltr = 11000 cc

h = 35 cm

r = ?

d = ?

(4) The height of the water level in a circular well is 7m, and its diameter is 10m.Calculate the volume of water stored in well in ltrs.

(5) A thin cylindrical tin can hold only one ltr of paint. What is the height of the tin ifthe diameter of the tin is 14cm?


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CONE

II (A) (1) What is a cone?

(2) Write the diagram of a right cone?

(3) Write the formula to find the lateral surface area of the cone.

(4) Write the formula to find the total surface area of the cone.

(5) Fill up the following table with suitable answer (Use extra sheets for working)

Clue: l2 = h2 + r2

S. No.

Radius ‘r’

Diameter of the base

‘d’

Slant Height

‘l’

Height‘h’

Circumference of the base

‘c’

LSA or

CSA TSA

1

10cm

28cm

2

3.5cm

10cm

3

10cm 12cm

4

12cm 66cm

5

10cm 440

Sq.cm

6

14cm

9cm

7

12cm

8cm

(J - 09)

(J - 07)


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(6) The curved surface area of a conical tomb is 528 Sqm and radius is 8m. Find theheight of the tomb?

CSA = 528 Sqm r = 8 m

(7) The height of the cone is 5.6 cm and diameter of the base is 8.4 cm. Find the areaof the curved surface.

(8) The height of a conical tent is 28m and the diamter of the base is 42m. Find thecost of canvas used at Rs. 20 per Sqm. (Find LSA)

Solution: h = 28md = 42m

r = .............. LSA = ...................................

= ...................................

= ...................................

= ................................... Sqm

Cost / Sqm = ..................... Rs.

Total Cost = .............. x 20 Rs.

= ........................ Rs.


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VOLUME OF A CONE

Volume of a Cone = x Area of the base x height

V = πr2h Cubic Units

Volume of a cone is ‘one third’ that of a cylinder having same radius and height

B (1) Fill up the blanks in the following table with suitable answer.

(2) A meter long metal rod (cylndrical) of radius 3.5cm is melted and recast to formcones of radius 1cm and height 2.1 cm. Find the no. of cones so formed?

(Hint: Even the shape changes, volume doesnot alter)

Solution: Volume of Cylinder V = ...................................

r = ................ cm = ...................................

l = h = 1m = 100 cm = ...................................

= ........................... cc

13

13

S. No.

Radius of the base ‘r’

Slant Height

‘l’

Height ‘h’

Diameter‘d’

Circumference ‘C’

Area of the base

Volume V

1

12cm 154

Sqm

2

7cm

18cm

3

15cm 300

Sqm

4

5cm

550cm3

5

10cm

21cm

6

6cm 22cm

7

21cm

3080cm3


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r = 1cm Volume of 1 Cone = ................................

h = 2.1 cm = ...................................

= ...................................

= ...................................

= ...................................

No. of Cones =

= ...................................

= ...................................

(3) A right angled triangle of sides 21 cm, 28 cm and 35 cm is revolved on the side 28 cmName the solid formed and find its volume.

13

Volume of rodVolume of 1 cone

28 35

21


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SPHERE

III (A) (1) What is a Sphere?

(2) What is a hemisphere?

(3) Name the solid obtained when a sphere is cut into two halves.

(4) Write the diagram of a hemisphere.

(5) Write the formula to find the surface area of a sphere.

(6) Write the formula to find the flat surface area of a hemisphere.

(7) Write the formula to find the curved surface area of a hemisphere.

(8) Write the formula to find the total surface area of a hemisphere.

(J - 09)(J - 08)

(M - 10)


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(9) Complete the following table with proper answer.

(10) Find the total surface area of a hemisphere of radius 14cm.

(11) The circumference of a hemispherical dome is 44cm. Calculate the cost of paintingat Rs. 20 per Sq.mtr

C = 44cm Surface aera of a

C = 2πr hemispherical dome = ........................................

44 = ................. = ........................................

r = .................. = ........................................

= ........................................

= ........................................

Cost per Sq.mtr = 20 Rs.

Total cost = ............... x 20 Rs.

= ...................... Rs.

S. No. Radius ‘r’

Diameter ‘d’

Circumference ‘C’

Surface area of a Sphere

1

14cm

2

616 Sq.cm

3

21cm

4 88cm

5 154 Sq.cm

6

7

(J - 09)


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VOLUME OF A SPHERE

(B) (1) Write the formula to find the volume of a sphere.

(2) Write the formula to find the volume of a hemisphere.

(3) Find the volume of a sphere whose radius is 21cm.

(4) Find the volume of a sphere whose diameter is 6cm.

(5) The diameter of a shot-put is 9cm. Calculate the volume of the shot-put.

(6) Total volume of 21 steel balls in a bearing is 88 cm3. Find the diameter of each ball

Solution: Volume of 21 balls is = 88 cm3

Volume of each ball is = cc

V = πr3

= πr3

r3 = .............................

r3 = .............................

d = 2 x r d = .................... cm

8821

43

8821

43

(M - 07)(J - 08)

(J - 10)


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(7) A hemispherical bowl has radius 14 cm. How many litres of water does it hold?(Hint: 1 ltr = 1000 cc)

r = 14 cm V = πr3

= ........................ ltrs

(8) The depth of a hemispherical water tank is 2.1 m at the centre. Find the capacity ofwater tank in ltrs.

Solution: Depth = Radius = 2.1 m V = = 210 cm

V = ................. cc

V = ltr

V= ................. ltr

(9) 21 leads marbles of even size are recast to form a big shpere. Find the volume ofthe new sphere when the radius of each marble is 2 cm.

Solution: r = 2 cm Volume of 1 marble = πr3

Volume of 21 marbles = 21 x ..................................

= ..................................

= ..................................

23

cc1000

43

(M - 08)

(J - 06)


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IV Provious Question Paper Questions

(1) The total surface area of a cylinder whose radius ‘r’ and height ‘h’ is .....................(A) 2πrh (B) 2πr(r+h) (C) 2π(r+h) (D) 2πh(r+h)

(2) The relation between the slant height, altitude and radius of a right cone is ............

(3) The area of the base of a cone 300 Sq.cm and height is 15 cm, then its volume is..................................

(4) The circumference of a sphere is 88cm, then its surface area is ..............................

(A) 2464cm2 (B) 352cm2 (C) 616cm2 (D) 1032cm2

(5) The diameter of a sphere is 14cm, then its surface area is ..............................

(A) 2464cm2 (B) 154cm2 (C) 88cm2 (D) 616cm2

(6) The height of the water stored in a circular well of diameter 10 mtr is 7m, thevolume of water in the well is ..................................

(A) 550m3 (B) 70m3 (C) 35m3 (D) 110m3

(7) The circumference of a hollow cylinder is 14cm and height is 20cm, then itssurface area is ..................................

(A) 280cm2 (B) 1760cm2 (C) 880cm2 (D) 140cm2

(8) Slant height of a cone is 15cm and radius is 9cm, then its vertical height is .............

(A) 6cm (B) 3cm (C) 5cm (D) 12cm

(9) The circumference of the base of a cone is 66cm and slant height is 12cm, then itsLSA is ..................................

(A) 396cm2 (B) 792cm2 (C) 78cm2 (D) 54cm2

(10) The factor that doesnot alter when a plastic sphere is melted and recast into a cubeis ..................................

(A) Length (B) Breadth (C) Surface Area (D) Volume

(11) A hemispherical bowl has radius 21cm, the volume of the hemisphere is .................

(A) π(21)2 cm2 (B) π(21)3 cc

(C) π(21)2 cm2 (D) π(21)3 cc

43

23

43

23

(M - 06)

(M - 06)

(M - 06)

(M - 06)

(J - 06)

(J - 06)

(J - 06)

(J - 06)

(M - 07)

(M - 07)

(J - 07)


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(12) The volumes of a cone and a cylinder with same height are equal, then the ratiobetween their radii is ..................................

(A) 3 : 1 (B) 2 : 1 (C) 3 : 1 (D) 2 : 1

(13) The volume of a cone is 90 cc, then the volume of a cylinder, having the sameradius as that of cone is ..................................

(A) 30 cc (B) 45 cc (C) 90 cc (D) 270 cc

(14) A = 2πr (r+h) by using this formula, we can find..................................

(A) LSA of a cylinder (B) TSA of a cylinder

(C) Volume of a cylinder (D) Area of a sphere

(15) Volume of a cone is 60 cc and area of its base is 20cm, then its height is ...............

(A) 6cm (B) 9cm (C) 12cm (D) 18cm

(16) The ratio between the radii of two solid spheres is 2 : 3, then the ratio betweentheir volume is ..................................

(A) 8 : 27 (B) 4 : 9 (C) 2 : 3 (D) 2 : 3

(17) A metal sheet of length 2mtr and breadth 44cm is rolled into a hollow pipe, thenthe radius of the pipe is ..................................

(A) 44cm (B) 22cm (C) 11cm (D) 7cm

(18) A sheet of aluminium foil having a shape as shown in the figure is rolled to make acone. If AB = 25cm and Arc BC = 44cm, find the volume of the obtained cone.

(19) The radii and heights of a cylinder and a cone area equal, if the volume of acylinder is 27cc, then the volume of the cone is ....................

(A) 99cc (B) 27cc (C) 81cc (D) 3cc

(20) Name the solid having two circular flat surfaces and a curved surface ..................

A

B C

(M - 08)

(J - 08)

(J - 08)

(M - 09)

(M - 09)

(M - 09)

(M - 10)

(J - 10)


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SCALE DRAWING

Scale drawing is the representation of irregular shaped field into known geometricalrectilinear figures, taking the measurements to the scale.

Area of a triangle = x base x height = bh

Area of a rectangle = length x breadth = l x b

Area of trapezium = x height [ Sum of Parallel sides ] = h (a + b)

Steps of Scale Drawing

(1) Irregular shaped field is divided into known geometrical shaped fragments.

(2) Measurements are recorded and a sketch is drawn to the scale.

(3) Measurements are recorded in the Surveyor’s field book.

(4) Total area of the land is the sum of the area of all the right angled triangle and thetrapezium etc.

Note: Area of the land is expressed in hectares

1 Hectare = 10,000 Sq.mtr

Example: Planout and find find the area of the field from the following notes from the field book

12

12

12

12

To ‘D’ (in mtr)

150

(R) 100 70 to C

To ‘E’ 80 80 (Q)

(P) 30 40 to B

From A


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Solution: Take a suitable scale

Let the scale be 20m = 1 cm

Mark the points on base line

Construction (Steps)

(1) Draw Base line AD = 7.5 cm ( )

(2) Mark R on base line such that AR = 5cm ( )

(3) Mark Q on base line such that AQ = 4cm ( )

(4) Mark P on base line such that AP = 1.5cm ( )

(5) Draw PB ⊥ AP (PB = 2cm) ( )

(6) Draw QE ⊥ AQ (QE = 4cm) ( )

(7) Draw RC ⊥ DR (RC = 3.5cm) ( )

(8) Join AB, BC, CD, DE and EA

(Observe the arrow marks)

Calculation of Base line Calculation of Perpendicular from base line

(i) From A to D = = 7.5 cm ( ) (1) From P to B = = 2 cm ( )

(ii) From A to R = = 5cm ( ) (2) From Q to E = = 2 cm ( )

(iii) From A to Q = = 4cm ( ) (3) From R to C = = 3.5 cm ( )

(iv) From A to P = = 1.5cm ( )

15020

10020

8020

3020

4020

8020

7020

D

E

A

CR

Q

P

(3)(4)

(2)(5)

(1) B30

40

50

80

70

50

20


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I Calculation of Area

(1) Area of Δle APB = x AP x PB = x 30 x 40 = 600 Sqm

(2) Area of Δle AQE = x AQ x QE = x 80 x 80 = 3200 Sqm

(3) Area of Δle DQE = x DQ x QE = x 70 x 80 = 2800 Sqm

(4) Area of Δle DRC = x DR x RC = x 50 x 70 = 1750 Sqm

(5) Area of trapezium PBCR

= PR [PB + RC]

= x 70 (40 + 70) = x 70 (110) = 3850 Sqm

Total area = 600 Sqm + 3200 Sqm + 2800 Sqm + 1750 Sqm + 3850 Sqm

= 12200 Sqm

= 1.22 Hectare 1 hectare = 10000 Sqm

II (1) Draw a plan and calculate the area of the level ground using the information givenbelow.

Scale: 40mtr = 1cm

12121212

12

12

12

To ‘C’ (in mtr) 220

To ‘D’ 120 210 120 200 to B

To ‘E’ 180 80 From A


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(2)

Scale: 20mtr = 1cm

To ‘D’ (in mtr) 200 140 50 to C

To ‘E’ 60 120 40 30 to B From A


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Previous Exam Question

(3) Calculate the are aof the field given below (in mtr)

(4) Draw a plan and calculate the area of the field.

C

D E A

B

Q

7055

50

4065

60P

To ‘D’ (in mtr) 300 200 150 to C 150 100 to B From A

(M - 06)

(M - 07)


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(5) From the data given below draw a plan no need to calculate area (Scale 20m = 1cm)

(6) From the data given below draw a plan (Scale 10m = 1cm)

To ‘D’ (in mtr) 200 120 70 to C

To ‘F’ 60 80 50 100 to B From A

To ‘C’ (in mtr) 100

To D 30 70 To E 30 50 20 to B

30 From A

(M - 08)

(J - 08)


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(7) Draw a plan using the data given below.

(8) Draw a plan from the following data.

To ‘D’ (in mtr) 200 160 60 to C

To E 80 100 40 50 to B From A in mtrs

To ‘D’ (in mtr) 250 150 100 to C

To E 100 100 50 50 to B From A

(M - 09)

(J - 10)

(J - 09)


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(8) Draw a plan from the following data (Scale 20cm = 1cm)

(9) Draw a plan of the following Data (Scale 20m = 1cm)

To ‘D’ (in mtr) 140 120 60 to C

To E 80 100 50 40 to B From A

(M - 10)

To ‘D’ (in mtr) 250 150 100 to C

To E 100 100 50 50 to B From A

20014012040

50

3060


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(10) Planout and find the area of the field from the following notes from the field book.

**************

To ‘D’ (in mtr) 140 120 60 to C

To E 80 100 50 40 to B From A

1501008030

70

4080


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POLYHEDRA AND NETWORKS10

Polygon: “A closed figure bounded by straight line segments” is a Polygon.

Polygonal Region: The region bounded by a polygon is Polygonal Region.

Regular Polygon: A polygon having equal sides and equal angles.

Polyhedron: A closed figure in the space bounded by polygonal faces is Polyhedron.

Note: A Polyhedron divides the space into 2 regions

Polyhedral Solid: A solid bounded by a polyhedron is a polyhedral solid.

For a polyhedron the no. of faces are F, no. of edges are E and number of vertices are V.

Regular Polyhedra: is a polyhedra, having the faces which are regular polygons.

Note: There are only 5 types of regular polyhdera they are.

(1) Tetra hedron (Fire)

(2) Hexa hedron (Earth)

(3) Octa hedron (Air) Platonic Solids

(4) Dodeca hedron (Universe)

(5) Icosa hedron (Water)

F + V = E + 2

There are FIVE Platonic solids

F + V = E + 2

This formula gives the relation between number of faces, vertices and edges of agiven polyhderal solid.

This formula is known as EULER’s FORMULA for polyhedral solids

Name of the Solid Tetrahedron Hexahedron Octahedron Decahedron Icosahedron

Shape of each face

Equilateral Triangle Square Equilateral

Triangle Regular

Pentagon Equilateral Triangle


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I Tabulate the number of faces, vertices and edges of the following polyhedral solids and

verify Euler’s Formula.

II Find the number of faces, vertices and edges in each of the following polyhedral solidsand verify Euler’s Formula.

(1) F = F + V = E + 2

V =

E =

(2) F = F + V = E + 2

V =

E =

(3) F = F + V = E + 2

V =

E =

S.No. Name of the

Polyhedra F V E F + V E + 2

1 Tetrahedron

2 Hexahedron

3 Octahedron

4 Decahedron

5 Icosahedron

6 Δler Prism

7 Square Based Prism

8 Triangular Pyramid

(J - 10)

(M - 10)


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(4) F = F + V = E + 2

V =

E =

(5) F = F + V = E + 2

V =

E =

III Find F, V and E for the following prisms and verify Euler’s formula.

Clue: If ‘n’ be the number of sides of the base of a prism, then

F = n+2 for example = for a triangular based prism

V = 2n n = 3

E = 3n then F = 3 + 2 = 5

V = 3 x 2 = 6

E = 3 x 3 = 9

(1) Pentagonal based prism.

F = F + V = F + V = E + 2

V = E + 2 =

E

(2) Hexagonal based prism

(3) Octagonal based prism

F = F + V = F + V = E + 2

V = E + 2 =

E

A

B D

C

(J - 10)


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Note: If ‘m’ be the number of sides of the base of a pyramid then

F = m+1 For example

V = m+1 In a Triangular based pyramid

E = 2 x m m = 3

F = 3 + 1 = 4

V = 3 + 1 = 4

E = 2 x 3 = 6

IV Find F, V and E for the following pyramids.

(1) Triangular based pyramid

F =

V =

E =

(2) Square based pyramid

(3) Pentagonal based pyramid

(4) Hexagonal based pyramid

V Answer the following

(1) What is a polygon?


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(2) What is a regular polygon?

(3) What is a polyhedron?

(4) What is a regular polyhedra?

(5) Write the Euler’s formula for polyhedral solids.

(6) Name the each face of a tetrahedron

(7) Name the each face of a hexahedron

(8) Name the each face of a octahedron.

(9) Name the each face of a dodecahedron.

(10) Name of the each face of a icosa hedron

(11) Write the names of platonic solids


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VI Previous Exams Multiple Choice Questions

(1) Each face of a regular polyhedral solid is a regular pentagon, then the solid is

.........................................

(A) Tetra hedron (B) Hexa hedron

(C) Octa hedron (D) Dodeca hedron

(2) The number of edges in a octagon are ................................

(A) 8 (B) 6 (C) 12 (D) 14

(3) Each face of a hexagon is .............................................

(A) Equilateral Triangle (B) Regular Pentagon

(C) Square (D) Rectangle

(4) In the list of platonic solids a cube is called as ................................


(C) Octa hedron (D) Icosa hedron

(5) The number of faces and edges of a solid are 8 and 18 respectively, then the number

of vertices is ................................

(A) 10 (B) 12 (C) 14 (D) 16

(6) The solid obtained by the given network is ................................

(A) Icosa hedron (B) Dodeca hedron

(C) Hexa hedron (D) Tetra hedron

(7) The number of vertices and edges of a tetra hedron are respectively........................

(A) 4, 6 (B) 6, 4 (C) 8, 6 (D) 6, 8

(8) The one which is not a platonic solid is ................................


(C) Square based pyramid (D) Octa hedron

(9) The number of edges of an octahedron is ........................

(A) 12 (B) 14 (C) 8 (D) 6

(M - 06)

(J - 06)

(M - 06)

(M - 08)

(M - 08)

(J - 08)

(M - 09)

(J - 09)

(J - 09)


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Graph: A set of points together with line segments joining the points in pairs.

Nodes in a graph: A point is a node if there is atleast one path starting from it or reaching it.

Arcs in a graph: The line segment joining two nodes is an arc.

Region: An area bounded by Arcs is called a region. (Considering outside is also a region)

Note: Number of nodes, arcs and regions are denoted by N, A and R respectively.

N + R = A + 2 is the Euler’s formula for networks

Order of a node: The number of paths starting from a node (or reaching it) is called order ofthat node.

The order of a node with a loop: A loop can be traced in the clockwise as well as in theanticlock wise directions. Therefore the order of a node with a loop is 2.

Even Node: If the order of a node is an even number then it is an even node.

Odd Node: If the order of a node is an odd number, then it is an odd node.

Traversibility of a network: A graph is said to be traversible if it can be drawn in one sweepwithout lifting the pencil from the paper and tracing the same arc twice.

Euler’s solution for traversibility of a network:

A graph is traversible if

(a) It has only even nodes.

(b) It has only two odd nodes.

**********

VII Define the following.

(1) Graph or Network

(2) Nodes in a graph


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(3) Arcs in a graph

(4) Region in a graph

(5) Traversible network

(6) The order of a node

(7) The order of a node with a loop

(8) Even node

(9) Odd node

(10) Euler’s conditions for traversibility of a network.


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VIII Find the number of Nodes (N), Arcs (A) and regions (R) in the following networks andverify Euler’s formula for networks.

Network No. of Nodes

N

No. of Arcs

A

No. of Region

R N + R A + 2

A E

B C D

A B C D

P

Q

R

A C

B

P

S

R

Q

A B

D C

A

B C


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IX For each of the given networks find the order of each node and verify the traversibility.

Graph Order of

each node

No. of even

Nodes

No. of Odd

Nodes Traversibility Reason

A – 3 B – 3 C – 3 D – 3

4 0 Not traversable

More than 2 odd nodes

C

A BD

F E D

A B C

A

B

C D

E

P

R

Q S

D

A

A

BC

B

C

A

B

C

E

D

A

BE

C D

A

CF

D

F

B

E

0 4


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X Which of the following networks are traversible and give reasons.

(a) The graph is not traversable because it has more than 2 odd nodes.

(b)

(c)

(d)

(e)

(f)

AB

C

BA

E

C

D

A

B

D

C

NK

Q

L MP

A DE

B C

F G

E F

D C

A B


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Matrix of a Graph

The information regarding the number of arcs, connecting the nodes can be displayed by amatrix

Ex: Note:

If a node is not connected to itself or to anyother node then the result is indicated as zero.

0 3 0 This is the matrix of the given graph or network.

3 0 2

0 2 0

Note: Number of arcs of the graph is 5, sum of the orders of the nodes is 10.

XI Construct the matrix for each of the following networks and complete the given table.

A B

A B C

A 0 3 0

B 3 0 2

C 0 2 0

C

Network Matrix Number of arcs

Sum of the order of the nodes

B

A

C

Q

P

R

X Y

Z

A to A ---- 0A to B ---- 3A to C ---- 0

B to A ---- 3B to B ---- 0B to C ---- 2

Number of Arcs = Sum of the orders of nodes2


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Network Matrix Number of arcs

Sum of the order of the nodes

DB

C

A

P

S

Q

R

X Y

Z

AB

D

C

A C

B

A

BCD

A

C

F

D

B E


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To draw the graph from a given matrix.

A B C

Ex: Draw the graph of the matrix A 0 1 2

B 1 0 3

C 2 3 0

XII Draw the graph for the following matrices.

Note: (1) The sum of the orders of nodes is an even number.

(2) The matrix of a given network is always a symmetric matrix.

(3) In case of a loop, the order is 2, but no. of arc is considered as 1.

A

C

B

Matrix Network / Graph

0 2 2

2 0 1

2 1 0

0 2

2 0

0 1 0

1 0 1

0 1 0

0 1 2

1 0 1

2 1 0

2 1 0

1 4 1

0 1 2

A

B C


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XIII Previous Exam Question

(1) The correct matrix of the network is

(A) 0 1 2 (B) 2 1 0 1 0 1 1 1 0 2 1 0 2 1 0

(C) 1 5 5 (D) 0 1 3 3 2 0 1 0 1 1 2 7 3 1 0 .............................

(2) In this graph, the order of the node A is .....................

(A) 2 (B) 4 (C) 1 (D) 6

(3) Veryify Euler’s formula for the network.

(4) This network is not traversible because .........................................

(A) All are even nodes (B) All are odd nodes

(C) Has only 2 odd nodes (D) Has more than 2 odd nodes

(5) From the matrix of the given network, the number of arcs are ..........................

1 5 5 (A) 12 (B) 6 3 2 0 1 2 7 (C) 3 (D) 5

(6) The number of regions in the given network are ..........................

(A) 3 (B) 2 (C) 5 (D) 4

B

A

C

C

BA

A B

D C

F E

P

RT

S

V

Q

U

(M - 06)

(M - 06)

(M - 06)

(J - 06)

(J - 06)

(M - 07)

E

A

FC B

D


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(7) Sum of the order of the nodes in the network .........................

(A) 3 (B) 4

(C) 5 (D) 10

(8) In the graph, the order of Q is............................

(A) 2 (B) 3

(C) 4 (D) 5

(9) The matrix of the network is ............................

(A) 0 3 (B) 3 0

3 0 0 3

(C) 3 3 (D) 2 3

3 3 3 2

(10) The network is not traversible because .............................................

(A) Network has 4 nodes

(B) Contains more than 2 odd nodes

(C) No even nodes

(D) Euler’s formula cannot be verified

(11) Verification of Euler’s formula N + R = A + 2 for the network is .......................

(A) 3 + 8 = 9 + 2 (B) 4 + 7 = 3 + 8

(C) 9 + 2 = 5 + 2 (D) 4 + 7 = 9 + 2

(12) In the given graph network the even nodes are .......................

(A) C & B (B) B & A

(C) C & D (D) A & B

X

Y

Z

R

P Q

S

N

Q

R

S

P

A

BD

C

C

AB D

(M - 07)

(J - 07)

(J - 07)

(J - 07)

(J - 07)

(M - 08)


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(13) The no. of nodes with loop in the graph of the given matrix .........................

(A) 1 (B) 2

(C) 3 (D) 4

(14) The traversible network among the following is ...........................

(A) (B)

(C) (D)

(15) Construct the matrix for the given network.

(16) In the given network the number of regions and nodes respectively are .................

(A) 3, 2 (B) 3, 3

(C) 4, 2 (D) 4, 5

(17) Construct the matrix for the network and write the sum of elements of the matrix and

relate the order of the nodes with the sum of the elements.

0 1 2

1 0 1 2 1 0

P

Q

R

STP R

Q

SP

QP

S

Q

R

T

C

AB

A B

A

B C

(M - 08)

(J - 08)

(J - 08)

(M - 09)

(M - 09)


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(18) Euler discovered that a graph is not traversible if it has .........................

(A) All even nodes (B) More than 2 odd nodes

(C) Two odd and 2 even nodes (D) Only 2 nodes

(19) Draw the graph for the following matrix

(20) Verify the Euler’s formula for the solid given below

(21) The matrix of the network is ............................

(A) 2 0 (B) 2 1

0 2 1 2

(C) 2 1 (D) 1 1

2 1 1 1

(22) A non - traversable network among the following is ............................

(A) (B)

(C) (D)

0 2 2 2 0 1 2 1 0

A

B C

D

E F

P Q

(J - 09)

(A - 10)

(A - 10)

(A - 10)

(A - 10)


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343

(23) The Euler’s formula for Network is .........................

(A) R + A = N + 2 (B) N + R = A + 2

(C) N + R = A - 2 (D) A + N = R + 2

(24) Construct the network for the given matrix

*************

0 2 2 2 0 1 2 1 0

(J - 10)

(J - 10)


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1 SET THEORYI (1) ...........

(2) A ∪ B = {1, 2, 3, 4, 5, 8, 9} A ∩ B = {1, 3, 5}

(3) A ∪ B = {a, c, r, t} A ∩ B = {a, t}

(4) A ∪ B = {0, 1, 2, 3, 4, 5} A ∩ B = {1, 2}

(5) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} A ∩ B = { }

II (1) {1, 2, 3, 4, 5} (5) {0, 2, 4, 6, 8}

(2) {1, 4, 9} (6) {1, 3, 5, 7, 9}

(3) {3, 6, 9} (7) {2, 3, 5, 7, 11, 13}

(4) {-2, -1, 0, 1, 2}

III (a) (1) c (b) (1) c

(2) a (2) d

(3) f (3) h

(4) e (4) e

(5) b (5) i

(6) d (6) f

(7) g

(8) b

(9) a

IV (1) A - B (2) A ∩ B (3) A ∪ B

(4) B - A (5) (A ∪ B)1 (6) A1

(7) A ∩ (B ∩ C) (8) A ∪ (B ∪ C) (9) (A ∩ B) ∪ C

V (1) ........

(2) A ∪ B = {0, 2, 4, 6, 8} A ∩ B = {0, 2, 4}

(3) M ∪ N = {1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} M ∩ N = {7, 9}

(4) n (B) = 35

(5) 30

ANSWERS

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VIII (1) 6 (2) A1 (3) Disjoint Sets

(4) 90 (5) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(6) {0, 2, 3, 4, 6, 8, 9} (7) A = B (8) {1, 4, 6}

(9) φ (10) B (11) A

(12) φ (13) 60 (14) φ

IX (1) B (2) D (3) A

(4) C (5) C (6) D

(7) D (8) B (9) C

(10) B (11) B (12) D

(13) B (14) B (15) A

(16) B (17) C (18) D

(19) A (20) D (21) C

(22) B (23) D (24) D

(25) B (26) C (27) A

(28) C (29) B (30) A

(31) C (32) B (33) D

(34) D (35) B

X (1) (i) {3, 4, 5, 6, 7, 8} (ii) {0, 2, 4}

(2) (i) {0, 2, 4, 6, 8} (ii) {0, 2, 4}

(3) (i) {3, 5, 7} (ii) {1, 3, 5, 7, 9, 11, 13}

(6) {5, 6, 7} (16) 40 (18) 15

(19) 6 (20) 20 (21) 17

(22) 30% (23) 20 (24) 300

(25) 22

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SEQUENCES & API (1) It is a Sequence (2) Is a Sequence

(3) Is not a Sequence (4) Is a Sequence

(5) Is a Sequence (6) Is a Sequence

(7) Is not a Sequence (8) Is a Sequence

(9) Is a Sequence (10) Is not a Sequence

II (1) 10n (2) (3) 4n+1

(4) 3n-1 (5) n2+1

III (A) (1) T1 = 7 T2 = 11 T3 = 15 T4 = 20

(2) T1 = 3 T2 = 6 T3 = 11 T4 = 18

(3) T1 = 0 T2 = 1/3 T3 = 1/2 T4 = 3/5

(4) T1 = -1 T2 = 1 T3 = -1 T4 = 1

(5) T1 = 3 T2 = 15 T3 = 35 T4 = 63

(B) (1) 6 (2) 8 (3) 5

IV

V

nn+1

n+1n+2

n - 1n

n+2n+3

n - 2n - 1

S.No. Tn+1 Tn-1 Tn+2 Tn-2

1 5n + 8 5n – 2 5n + 13 5n – 7

2 3n2 + 6n + 1 3n2 - 6n + 1 3n2 + 12n + 10 3n2 - 12n + 10

3 n2 + 2n + 3 n2 - 2n + 3 n2 + 4n + 6 n2 - 4n + 6

4

S.No. a b S.No. a b

1 4 3 5 2/3 1/3

2 10 10 6 1/2 1/2

3 7 -4 7 3/2 1/2

4 8 -2 8 1/2 3

346

d d


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VI (1) 23 (2) 3 (3) -1/2 (4) 3

(5) -4 (6) -2 (7) 1/2 (8) 6

(9) 25 (10) -2

VII (1) f (2) e (3) h (4) g

(5) c (6) b (7) d (8) a

VIII (A) (1) 1, 86, 18, 5 782

(2) 3, 123, 31, 4 1953

(3) 1, 103, 37, 3 1820

(B) (1) 1, 1, 1275 n = 50

(2) 1, 2, 625 n = 25

(3) 1, 3, 590 n = 20

(C) (1) 3, 1, -2 -165

(2) 1, 5, 4 435

(3) 7, 12, 5 1090

(4) 8, 6, -2 -1240

(D) (1) 498 (2) 500 (3) 357

(E) (1) 3, 5, 7 (2) 8, 10, 12 (3) 9, 12, 15

(4) -18, -8, 2

(F) (1) 7 (2) 12 (3) 6

(4) 4 (5) 2 (6) 6

(7) 7

IX (A) (1) 465 (2) 325 (3) 1365

(4) 1285

(B) (1) 8 (2) 20 (3) 40

(4) 20 (5) 13 (6) 41

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X (1) 9900

(2) 2850

(3) 11250

(4) 2500

(5) 2450

XI (1) 5, 10, 15, 20, .........................

(2) 3, 6, 9, 12, .........................

(3) 3, 6, 9, 12, .........................

XII (2) 0

(3) 3, 11, 35, 107

(4) 4, 7, 10, ...............

(5) 30, 41, 52, .................

(8) 3, 6, 9

(9) 714

(10) 150, 650, 1150, 1650

(11) 3570

(12) 28, 78, 98

(13) 400, 600, 800

(14) 40560

(15) 0

(16) 75th Term

(17) 1, 3, 5, 7

(18) 4, 9, 14

XIII (1) 520 (2) 1470 (3) 54

(4) 0 (5) aΣn (6) 400

(7) 420 (8) a+(n-1)d

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XIV (1) C (11) A (21) C

(2) C (12) A (22) B

(3) B (13) A (23) B

(4) A (14) D (24) B

(5) B (15) C (25) A

(6) A (16) A (26) C

(7) C (17) D (27) A

(8) C (18) A (28) B

(9) D (19) C (29) D

(10) D (20) D (30) D (31) C

GEOMETRIC PROGRESSIONII (1) 2, 3 (2) 5, 2 (3) 1, -1

(4) 1, 1/2 (5) 1, 1/3

III (1) 3, 6, 12, .......................

(2) 1, 1/3, 1/9, .......................

(3) 1/2, 1/4, 1/8, .......................

IV (1) GP (2) Not GP (3) GP

(4) GP (5) GP

V (1) 3 x 4n-1 (2) 1 x 3n-1 (3) 3 x 2n-1

(4) 2n-1

VI 2, 10, 50, 250, ..............................

2, 6, 18, 54, ..............................

1, 3, 9, 27, ..............................

1, 2, 4, 8, ..............................

VII (1) 9 (2) 1/4 (3) 3

(4) 96

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VIII (1) A (8) D (15) D(2) D (9) B (16) C(3) C (10) C (17) A(4) C (11) C (18) A(5) A (12) B (19) D(6) B (13) A(7) C (14) A

IX (1) D (4) B (7) B(2) A (5) C (8) B

(3) C (6) B (9) A (10) D

X (1) n=6, Sn=189 (8) 2, 6, 18 (15) 744

(2) 1 : 28 (9) 4, 8, 16 (16) 9

(3) 2/3 (10) 4/5 (17) 6

(4) 1 (11) 1 (18) ......

(5) 6 (12) 2 (19) 3, 6, 9

(6) 510 (13) 81, 54, 36, ..... (20) ......

(7) 3/2 (14) 655355 (21) 23

HORMONIC PROGRESSIONI (a) HP (b) HP (c) Not a HP

(d) HP

II (a) 3n - 1 (b) 1/63 (c) -1/13

III (2) 2/5 & 6/5 (3) 3/13 (4) 2

IV (1) 4 (3) a/c

V (1) B (6) A (11) B

(2) C (7) D (12) C

(3) B (8) B (13) C

(4) A (9) B (14) AP

(5) C (10) A (15) B

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MATRICESI

II (1) c (2) g (3) f

(4) e (5) b (6) h

(7) i (8) d (9) a

III Symmetric Skew Symmetric Scalar Identity

(1) x = -1 x = 1 x = 3 x = 0

(2) x = 7 x = -11 x = -3 x = -5

(3) x = -1 x = -2 x = 1 x = -4

S.No. Matrix A Type Order Transpose A1 -2A

1

1 0 0 0 1 0 0 0 1

Unit Matrix 3 x 3

1 0 0 0 1 0 0 0 1

-2 0 0 0 -2 0 0 0 -2

2

1 2 3 0 0 -3 1 2 0

Square Matrix 3 x 3

1 0 1 2 1 2 3 -3 0

-2 -4 -6 0 -2 6 -2 -2 0

3

2 3 1 0 1 2

Rectangular Matrix 2 x 3

2 0 3 1 1 2

-4 -6 -2 0 -2 -4

4

1 2 3 2 0 4 3 4 2

Symmetric Matrix 3 x 3

1 2 3 2 0 4 3 4 2

-2 -4 -6 -4 0 -8 -6 -8 -4

5

-3 0 0 0 -3 0 0 0 -3

Scalar Matrix 3 x 3

-3 0 0 0 -3 0 0 0 -3

6 0 0 0 6 0 0 0 6

6 [ 2 -1 0 ] Row Matrix 1 x 3

2 -1 0

[ -4 2 0 ]

[ ] [ ][ ][ ] [ ]

[ ] [ ][ ] [ ]

[]

[ ]

[ ] [ ]

[ ][ ] [ ]

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IV (1) 12 (2) 3 (3) 1, 3, 7

(4) 1 -1 2 (5) n x m (6) A

-1 2 0

(7) A (8) 3 x 3 (9) Symmetric Matrix

(10) Skew Symmetric Matrix

V (1) C (11) C (21) B

(2) D (12) D (22) D

(3) B (13) C (23) C

(4) (14) B (24) C

(5) A (15) D (25) B

(6) D (16) (26) A

(7) A (17) C (27)

(8) C (18) D (28) B

(9) C (19) D (29)

(10) A (20) B (30) D (31)

VI (1) (i) 4 8 (ii) 11 5 (iii) 13 -2

0 10 -8 13 2 11

(7) -2 6 (9) x = 2 (10) x = 1

-6 8

(11) x = 3 (12) x = 0 (13) a=-1, b=2, c=-3, d=1

(14) x = 6, y = -2 (15) x = 6/13 y = 4/13

(16) x = 5, y = 2 (17) x = -1, y = 2

(18) 7 10 5 11 (22) 2 1

15 22 11 26 -2 -1

(23) 1 2 (24) 13 13 (25) 5 3

2 13 13 28 3 9

(26) -1 4

0 3

[ ]

2B + C = B1 + 2C =

[ ]

[ ]

[ ]x =

[ ]

A2 = AA1 = A2 =

[ ]

[ ]

[ ]

[ ]

[ ]

[ ] AA1 = AA1= AA1=

[ ]A2 - 2A =

352


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2 PERMUTATION AND COMBINATIONII (1) 90 (2) 210

III (1) 10!/4! = 10P6 (2) 3(6!) (3) 2n!

(4) (2n-1)!

IV (1) 3 (2) 3 (4) 5

PERMUTATION

I (1) 7 x 6 x 5 x 4 (2) 10P2 (3) 6P3

(4) 8 x 7 x 6 x 5 x 4

II (1) 5 x 4 (2) 10 x 9 x 8 (3) 6 x 5 x 4 x 3 x 2 x 1

(4) 4! + 1 + 4 = 29 (5) n! (6) 1 (7) !

III (1) 8P2 (2) 8P3 (3) 8P4

(4) 8P5 (5) 8P6

IV (1) nP2 (2) nP3 (3) nP4

(4) n+1P3 (5) n+3P4 (6) n-1P3

V (A) (1) n = 7 (2) n = 5 (3) n = 6

(B) (1) r = 3 (2) r = 3 (3) r = 2

(C) (1) 6 (2) n = 7 (3) n = 2

(4) n = 2

(D) (1) 3 (2) 9

(E) (1) 24 (2) 120

(F) (1) 120 (2) 7 x 6 x 5 = 210

(3) 6 x 5 x 4 x 3 = 360

MODEL PROBLEMS

(2) 3 x 3 x 2 = 18 (3) 3 x 4 x 3 = 36 (4) 1 x 2 x 3 x 4 x 2 = 48

(5) 5P5 = 5! = 120 (6) 3P3 = 6 (7) (a) 8P8 (b) 6P6

(8) (a) 5P5 (b) 4P4 x 2P2

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COMBINATIONS

(A) (2) n = 20 (3) r = 4 (4) r = 2

(B) (1) 6 (2) 5 (3) 4

(4) 3 (5) (6) 11

(7) (a) 4950 (b) 100 (c) 1 (d) 1

(C) (1) (a) 45 (b) 120 (2) 5C2 - 5 = 5

(3) n = 15 (4) 11

(D) (1) 503880 (2) 5005 (3) 1

(4) 1 (5) (a) 92378 (b) 75582

(E) (1) 2856 (2) 180 (3) 1170

(F) (1) 60 (2) (i) 210 (ii) 371 (3) 210

3 STATISTICSI Σx N X

(1) 130 10 13

(2) 240 20 12

(3) 280 14 20

(4) 96 8 12

(5) 360 20 18

(6) 480 30 16

II (1) 9.03 (2) 7.07

III Variance SD Variance SD

(1) 1.44 1.2 (5) 0.0324 0.18

(2) 64 8 (6) 49 7

(3) 0.01 0.1 (7) 0.0004 0.2

(4) 6.25 2.5

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IV (1) 9.88 (2) 4.66 (3) 2.29

V (1) 6.27 (2) 6.20 (3) 6.36 (4) 11.22

CO-EFFICIENT OF VARIATION (C.V.)VI (1) Rama (2) Weight is more variable

(3) Seenappa (4) 29.41

VII (1) 27 (7) (13) 3

(2) 24 (8) CV = x 100 (14) 20

(3) 37 (9) CV (15) 2

(4) (10) CV (16) Rama

(5) (11) 0.25 (17) B

(6) (12) 2.56

NΣx

NΣfx

NΣD2

NΣfD2

σX

4 FACTORS AND FACTORISATIONI

(1) D (3) C (5) C

(2) C (4) B (6) C

IV (1) 1 (4) a3 + b3 (7) x2 - 3x + 2

(2) (a+b) (a-b) (5) a2 - ab + b2

(3) a2 + ab + b2 (6) a2 - ab + b2

CYCLIC SYMMETRY

III (1) b (4) c (7) c

(2) b (5) c (8) a

(3) d (6) b (9) c

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IV (1) a (4) a (7) b (10) c

(2) c (5) c (8) d (11) c

(3) a (6) a (9) b (12) b

V (1) c (5) a (9) a (13) a

(2) d (6) d (10) d (14) c

(3) a (7) d (11) b

(4) b (8) b (12) d

VI (1) b (2) b (3) b

CONDITIONAL IDENTITY

IV (a) ac (d) ab (g) 18abc (j) 3

(b) 18abc (e) 4 (h) 9

(c) 3 (f) -abc (i) -3

SURDS

VIII (1) 2(x)1/3 (2) 12(a)1/2

IX (1) 5 2 (2) 8 3 (3) 5 (4) 7 2a

X (1) 3 2 + 9 3 (2) 2 3 + 2 2

XI (1) 5000 (2) 72 (3) 648 (4) 75

(5) 972 (6) 10125

XII (1) (2) (3) (4)

(5) (6)

3

12 6 12 4

6 12

105

abb

7 xx

5 33

6x2x

15a5

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XIII (1) (2) 2 7 - 2 2 (3)

(4) 3 2 - 2 3 (5) 2 3 - 6 (6) 30 + 3

XIV (1) 4 + 15 (2)

(3) 5 (4) 2 - 5 3 (5) 5 - 2 6

(6) 3 2 - 6 (7) 3 + 2 5 (8) 3 5 + 3

XV (1) 3 a + 8 b (2) x1/2 + y1/2 (3) 72

(4) 4 + 15 (5) -4 a (6) 35 + 12 6

(7) x - y (8) 6 (9) 40

(10) a b - c (11) (12) 18

2 x + 2 yx - y

m - mnm - n

6 6 - 2 15 + 6 - 1013

6

3

3 55

3

(13) 2 , 8 (14) 2 ( 2 - 3 ) (15) 3 ( 15 - 2 5 )

(16) P Q + Q P (17) a (b)1/3 (18) 200

(19) 4 2 (20) 23 + 4 15 (21) 3 + 30

(22) 8 a (23) 4 2 (24) 6

(25) 9 2 (26) P - Q (27) 3 x + 3 y

(28) n, x (29) 5 (30) a x - b y

(31) 5184 (32) 5 + 2 6 (33) 3

(34) 2, 3m (35) 2 (3 - 6 ) (36) 21

6

12

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5 QUADRATIC EQUATIONI (1) (c) (2) (b) (3) (c) (4) (a)

(5) (b) (6) 0 (7) (b) (8) (d)

(9) (c) (10) (b) (11) (a) (12) (c)

(13) (b) (14) (d) (15) (b) (16) (c)

(17) (c) (18) (c) (19) (a) (20) (c)

(21) (b) (22) (d) (23) (a) (24) (d)

(25) (b) (26) (a) (27) (a) (28) (c)

(29) (c) (30) (c) (31) (b) (32) (b)

(33) (a) (34) (b) (35) (c) (36) (a)

(37) (d) (38) (39) x = -1, +2 (40) (a)

(41) (d) (42) (a) (43) (d) (44) (a)

(45) (b) (46) (c) (47) (a) (48) (c)

(49) (c) (50) (d) (51) (a) (52) (b)

(53) (c) (54) (b) (55) (b) (56) (d)

(57) (c) (58) (a) (59) (d) (60) (c)

(61) (d) (62) (c) (63) (b) (64) (b)

(65) (d) (66) (a) (67) (c) (68) (b)

(69) (a) (70) (c) (71) (b) (72) (c)

(73) (a) (74) (d) (75) (b) (76) (a)

(77) (c) (78) (a) (79) (d) (80) (a)

(81) (b) (82) (c) (83) (d) (84) (a)

(85) (a) (86) (b) (87) (d) (88) (a)

(89) (c) (90) (b) (91) (a) (92) (d)

(93) (c) (94) (c) (95) (c) (96) (c)

(97) (b) (98) (c) (99) (b) (100) (b)

(101) (d) (102) (b)

IV (A) (1) +5 (2) +12 (3) +9/2 (4) 7, -2

(5) 6, 2 (6) +4

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B (1) +3/2 (2) +8 (3) + 14 (4) +3

(5) +10 (6) +20

V (1) 0, 3 (2) -1/2, 2/3 (3) 0, 4 (4) -6, -9

(5) 3, 5 (6) 2/5, -3/7

VI (1) x = -5, -10 (2) a = -2, -3 (3) y = 2, -1

(4) p = 2, -3 (5) m = 2/3, 3/2 (6) t = 1/2, -3/10

(7) x = - 7 ,

VII (1) x = 4, 3 (2) a = 2 + 10 (3) x = 4 + 15

(4) x = (5) x = +4 (6) x = 4 + 15

(7) x = 1 + 5 (8) x = 0, -3

VIII (1) x = 8, -2 (2) a = 1, -2/3 (3) n = 3, -1

(4) b = 31/3, 3 (5) S = -3/4, 1/2 (6) m = 2, -8/11

IX (A) (1) x = 11/2, 4 (2) m = 5, 2 (3) y = -4, 3

(4) x = (5) n = 5, 6/5 (6) m = 6, m = -10/3

B (1) 13, 14 or -14, -13 (2) 6, 8 or -8, -6

(3) 7, 8, 9 (4) 10m, 6m (5) 17, 1

(6) 21m (7) 12m, 16m (8) 12cm, 5cm

(9) 12cm, 8cm (10) 240 km/h (11) 30km/h (arrival)

36km/h (return)

(12) 10km/h (13) 15pen (14) 20 Rs

(15) 15 (16) 6 days (17) 1 Pipe = 15h, 2 pipe = 10 h

(18) 8 or 1/8 (19) 21 (20) 15 (21) 20

XIV (1) -1/3 (2) -4 (3) k = 2 (4) m = 0

(5) P = 5 (6) Q = 5 (7) Q = +5/2 (9) P = 2

XV (1) (a) -22/5 (b) -6/5 (c) 66/25 (d) -486/125

137

1 + 112

-1 + 172

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(2) (a) -15/4 (b) 15/2

(3) (a) 8/3 (b) 0 (c) 8

(4) (a) -15/4 (b) 81/8 (c) -9/2

(5) (a) 54 (b) 6/27 (c) 2/27 (d) 103/3

(6) x2 + 4x + 16 = 0 (7) m = 9/8

8 THEOREMS ON TRIANGLESI (1) D (2) D (3) A (4) D

(5) D (6) A (7) C (8) B

(9) A (10) B (11) B (12) A

(13) A (14) B (15) A (16) B

(17) D (18) C (19) B (20) A

(21) A (22) B (23) C (24) A

(25) C (26) D (27) B (28) B

(29) C (30) C (31) B (32) C

(33) B (34) D (35) D (36) B

(37) C (38) A (39) B (40) D

(41) A (42) D (43) C (44) B

(45) B (46) Thales Thorem

(47) ΔABC is similar to ΔDEF (48) C (49) d

(50) D (51) A (52) A (53) B

(54) A (55) C (56) D (57) D

(58) D (59) B (60) C (61) B

(62) A (63) C (64) D (65) A

(66) C

II

1 (1) (2) (3) (4)

2 (1) = (2) = (3) = =

ACAQ

ACQC

PBAP

QCPB

ab

cd

aa+b

cc+d

aa+b

cc+d

xy

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(4) AY = 5.6cm (6) x = 1 or x = 1/2 (8) a = 4

(15) 7.5cm (18) 20cm2 (19) 7 : 5

(20) 10cm2 (30) (ii), (iv), (v), (vi), (viii) are triplets and other are not

(31) (a) 12 2 cm (b) 5 2 Unit

(32) (a) 5 2 cm (b) 4 2 Unit

(33) (a) 4 3 cm (b) 5/2 3 Unit

(34) 17m (35) 13km (36) 4cm (37) 13m

(38) 12cm (39) AB = 4 3 cm, AC = 2 5 cm, BC = 10cm

(48) 1.4m

TOUCHING CIRCLES

I (1) B (2) D (3) A (4) A

(5) D (6) A (7) C (8) A

(9) A (10) A (11) D (12) B

(13) B (14) D (15) B (16) C

(17) D (18) C (19) 16cm (20) A

(21) C (22) D (23) B (24) B

(25) C (26) A (27) C (28) D

(29) A (30) D (31) D (32) B

(33) A (34) C (35) A (36) B

(37) D (38) C (39) B (40) D

(41) A (42) C (43) B (44) A

(45) C

II

(1) 4cm, 3cm, 5cm (2) 4cm

(3) AB = 8cm, BC = 9.8cm, AC = 8.5cm, Perimeter = 26cm

(7) AB = 16cm (8) 1.75cm (9) 4 10 cm

(12) CD = 5cm (13) QR = 9cm

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9 MENSURATIONI (A) CYLINDER

(1) A Right Circular Cylinder is a solid described by rotation of a rectangle aboutone of it’s sides which remains fixed.

(2) Hollow Cylinder, Solid Cylinder (3) A Garden Roller

(4) Pipe (5) Cylinder, r = 0.79cm, h = 3.5cm

(6) A = 2πrh (7) A = 2πr (r + h) (9) A = 440 Sq.cm

(11) A = 8.80 Sq.m (12) A = 1056 Sq.cm (13) 525 Rs

(14) A = 64.62 Sq.cm (15) A = 440 Sq.cm

(B) Volume of a Cylinder

(2) v = 15040 cm3 (3) d = 20cm (4) 550 cm3

(5) h = 6.49 cm

II (A) CONE

(1) A right circular cone is a solid generated by the revolution of a right angletriangle about one of the sides containing the right angle.

(3) A = πr1 (4) A = πr (r+1) (6) h = 19.4cm

(7) A = 92.4 Sq.cm (8) 46200 Rs.

(B) Volume of a Cone

(2) 1750 (3) v = 12936 cm3

III (A) SPHERE

(1) A Sphere is a solid described by the rotation of a semi circle about a fixeddiameter.

(2) A plane through the centre of the sphere divides into two equal parts each iscalled Hemisphere.

(3) Hemisphere (5) A = 4πr2 (6) A = 2πr2

(7) A = 2πr2 (8) A = 3πr2 (10) A =1848 Sq.cm

(11) 616 Rs.

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(B) Volume of a Sphere

(1) v = 4/3πr3 (2) v = 2/3πr3 (3) v = 38808 cm3

(4) v = 113.14 cm3 (5) v = 381.85 cm3 (6) d = 2cm

(7) 5.75 litre (8) 19404 litre (9) v = 704cm3

IV Previous Question Paper Questions(1) B (2) l = + r2+h2 (3) 1500cm3 (4) A

(5) D (6) A (7) A (8) D

(9) A (10) D (11) D (12) A

(13) D (14) B (15) B (16) A

(17) D (18) V = 1232 cm3 (19) A (20) Cylinder

10 POLYHEDRA AND NETWORKS

VI (1) D (2) C (3) C

(4) Hexohedren (5) B (6) D (7) A

(8) C (9) A

XIII (1) A (2) B (4) D (5) C

(6) C (7) D (8) D (9) A

(10) B (11) D (12) C (13) A

(14) C (15) (16) C

(17) Sum of the order of the nodes = (18) B

Sum of the elements of the matrix

(19) (21) B (22) A

(23) B (24)

[ ] 2 3 03 0 20 2 0

[ ] 2 3 03 0 20 2 0

C

A

B

C

A

B

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Sslc Mathematics Workbook Old

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