5 mark questions X-STD MATHEMATICS Prove That A – (B C) = (A – B) (A – C) A A A A A BBB B B...
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Transcript of 5 mark questions X-STD MATHEMATICS Prove That A – (B C) = (A – B) (A – C) A A A A A BBB B B...
5 mark questions
X-STD
MATHEMATICS
PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21
Prove That A – (B C) = (A – B) (A – C)
AA A
A A
B B B
B B
C C C
C
C
(1)BC (2) A-(BC) (3) A – B
(4) A – C (5) (A – B) (A – C)
From the diagrams (2) and (5)
A – (B C) = (A – B) (A – C)
Prove that A – (B C) = (A – B) (A – C) using Venn diagram
A B
C
A A
A A
B B
B B
CC
C C
(1) BC (2) A – ( BC) (3) A – B
(4) A – C (5) (A – B)(A – C)
From the diagram (2)and(5) A – (BC) = (A – B)(A – C)
Prove that A (B C) = (A B) (A C) using Venn diagram.
A B
C
A A
A A
B B
BB
C C
C C
B C A (B C) A B
A C (A B)(A C)
From the figures 2 and 5
A(B C)=(AB)(AC)
Prove that A (B C) = (A B) (A C) using Venn diagram.
A B
C
A A
A A
B B
BB
C C
C C
B C A (B C) A B
A C (AB) (AC)
From the figures 2 and 5
A(BC)=(AB)(AC)
Prove that (A B)’ = A’ B’ using Venn diagram.
A BB
AA
AA B
B B
U U U
UU
1. AB 2. (AB)’ 3. A’
4. B’ 5. A’ B’
From the diagrams 2 and 5 (A B)’ = A’ B’
Prove that (A B)’ = A’ B’ using Venn diagram.
A BB
AA
AA B
B B
U U U
UU
1. A B 2. (A B)’ 3. A’
4. B’ 5. A’ B’
From the diagrams 2 and 5 (AB)’ = A’B’
Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}, show that (i) A (B C) = (A B) C. (ii) Verify (i) using Venn diagram.
Solution(i) B C = {3, 4, 5, 6} {5, 6, 7, 8}
= {3, 4, 5, 6, 7, 8}` A (B C) = {1, 2, 3, 4, 5} { 3, 4, 5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6, 7, 8}………… (1) A B = {1, 2, 3, 4, 5} {3, 4, 5, 6}
= {1,2,3,4,5,6}`(A B) C = {1,2,3,4,5,6} {5,6,7,8}
= {1, 2, 3, 4, 5, 6, 7, 8}……………. (2)
From (1) and (2), we have A (B C) = (A B) C.
Using Venn diagram, we have
A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}
A B
C
5 6
7 8
34
A B
C
5 6
7 8
341
2
(1) B C (2) A (B C)
Let A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.
(i) Show that A (B C) = (A B) C. (ii) Verify (i)
using Venn diagram.
Solution
Given A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.
B C = {a,c,e} {a,e} = {a,e}
A (B C) = {a,b,c,d} {a,e} = {a}…… (1)
A B = {a,b,c,d.} {a,c,e} = {a,c}.
(A B) C = {a,c} {a,e} = {a}…….. (2)
From (1) and (2) A (B C) = (A B) C.
A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.
A B
C
a e
(1) BC
A B
C
a
(2) A(BC)
A = {a,b,c,d}, B = {a,c,e} and C = {a,e}.
A B
C
a
c
(1) AB
A B
C
a
(2) (AB)C
from (2) and (4) , it is verified that A (B C) =(A B) C
Let A = {0,1,2,3,4}, B = {1, - 2, 3,4,5,6} and C = {2,4,6,7}.
(i) Show that A (B C) = (A B) (A C). (ii) Verify using Venn
diagram.
Solution
(i) B C = {1, - 2, 3, 4, 5, 6} {2, 4, 6, 7 } = {4, 6};
A (B C) = {0,1, 2, 3, 4} {4, 6} = {0,1,2,3,4,6}…….(1)
A B = {0,1,2,3,4} {1, - 2, 3,4,5,6} = {- 2, 0,1, 2, 3, 4, 5, 6},
A C = {0,1,2,3,4} {2,4,6,7} = {0,1, 2, 3, 4, 6, 7}.
(A B) (A C) = {- 2, 0,1, 2, 3, 4, 5, 6} {0,1, 2, 3, 4, 6, 7}
= {0,1, 2, 3, 4, 6}. …….(2)
From (1) and (2) ,we get A (B C) = (A B) (A C).
A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}.
A B
C
4 6
(1) BC
A B
C
4
(2) A(BC)
6
3
01
2
A B
C
4
3
(3) AB
A B
C
0
(4) A C
from (2) and (4) , it is verified that A (B C) =(A B) C
A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}.
1
2
4
3
01
26
7
-2
6
5
A B
C
4
(5) (AB)(AC)
6
3
01
2
from (2) and (5) , it is verified that
A (B C) = (A B) (A C)
Given that U = {a,b,c,d,e, f,g,h}, A = {a, b, f, g}, and B = {a, b, c}, verify De Morgan’s laws of complementation.
U = {a, b, c, d, e, f, g, h} A = {a, b, f, g} B = {a, b, c}.De Morgan’s laws(AB)’ = A’ B’(AB)’ = A’ B’VERIFICATION A B = {a, b, c, f, g}(A B)’ = {d, e, h}……..(1)A’ = {c, d, e, h}B’ = {d, e, f, g, h}A’B’ = {d, e, h}……….(2)From (1) and (2) (A B)’ = A’B’
A B = {a, b}(A B)’ = {c, d, e, f, g, h}....(1)A’ = {c, d, e, h}B’ = {d, e, f, g, h}A’ B’ = {c, d, e, f, g, h}….(2)From (1) and (2)
(A B)’ = A’ B’
Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}..A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7}
C = {3,9,10,12,13}..
De Morgan’s laws
(1) A\(BC) = (A\B)(A\C) (2) A\(BC) = A\B) (A\C)
VERIFICATION
B C = {1, 2, 3, 5, 7, 9, 10, 12, 13}
A\(BC) = {11, 15}……..(1)
A\B = {3, 9, 11, 13, 15}
A\C = {1, 5, 7, 11, 15}
(A\B)(A\C) = {11, 15}……….(2)
From (1) and (2) A\(BC) = (A\B)(A\C)
Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}..
A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7}
C = {3,9,10,12,13}..
BC = { }
A\(B C) = {1, 3, 5, 7, 9,11,13,15}……..(1)
A\B = {3, 9, 11, 13, 15}
A\C = {1, 5, 7, 11, 15}
(A\B) (A\C) = {1, 3, 5, 7, 9,11,13,15}……….(2)
From (1) and (2) A\(B C) = (A\B) (A\C)
Let A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5,10,15, 20, 30}and C = {7, 8,15,20,35,45, 48}. Verify A\(B C) = (A\B) (A\C)
A = {10,15, 20, 25, 30, 35, 40, 45, 50},
B = {1, 5, 10, 15, 20, 30}
C = {7, 8, 15, 20, 35, 45, 48}
BC = {15, 20}
A\(B C) = {10, 25, 30, 35, 40, 45, 50}……..(1)
A\B = {25, 30, 35, 40, 45, 50}
A\C = {10, 25, 30, 40, 50}
(A\B) (A\C) = {10, 25, 30, 35, 40, 45, 50}……….(2)
From (1) and (2) A\(B C) = (A\B) (A\C)
13. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and f : A "
B be defined by f(x) = . Represent f by (i) an arrow
diagram (ii) a set of ordered pairs (iii) a table (iv) a graph .3
3x
3
3)(
xxf
13
3
3
36)6(
f
23
6
3
39)9(
f
43
12
3
315)15(
f
53
15
3
318)18(
f
63
18
3
321)21(
f
6
9
15
18
21
1
2
4
5
6
A B
ARROW DIAGRAM
SET OF ORDERED PAIR
f = {(6, 1), (9, 2), (15, 4), (18, 5), (21, 6)}
X 6 9 15 18 21Y 1 2 4 5 6
Table
Graph
| | | | | | | | |
3 6 9 12 15 18 21 24 27
– 7
– 6
– 5
– 4
– 3
– 2
– 1 (6, 1)
(9, 2)
(15, 4)
(18, 5)
(21, 6)
Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let f : A B be a function given by f (x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph.
Given A= {0, 1, 2, 3},
B = {1, 3, 5, 7, 9}
f(x) = 2x + 1
f(0) = 2(0) + 1 = 0 + 1 = 1
f(1) = 2(1) + 1 = 2 + 1 = 3
f(2) = 2(2) + 1 = 4 + 1 = 5
f(3) = 2(3) + 1 = 6 + 1 = 7
(i)Set of ordered pairs
{(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) Table
x
f(x)
0
1
1
3
2
5
3
7
4
6
8
10
3
4
5
6
7
ABARROW
DIAGRAM
Graph
| | | | | | | |
0 2 4 6 8 10 12 14 16
– 7
– 6
– 5
– 4
– 3
– 2
– 1
(4, 3)
(6, 4)
(8, 5)
(10, 6)
A function f: [1, 6)R is defined as follows
f(x) =
6 x 4 103
4 x 2 12
2 x 1 1
2x
x
x Find the value of (i) f(5), (ii) f(3), (iii) f(1), (iv) f(2) – f(4) (v) 2f(5) – 3f(1)
Since 5 lies between 4 and 6
f(x) = 3x2 – 10
f(5) = 3(5)2 – 10
= 75 – 10
= 65
Since 3 lies between 2 and 4
f(x) = 2x – 1
= 2(3) – 1 = 6 – 1
= 5
Since 1 lies in the interval 1 x < 2
f(x) = 1 + x
f(1) = 1 + 1 = 2
Since 2 lies in the interval 2 x < 4
f(x) = 2x – 1
= 2(2) – 1 = 4 – 1 = 3
Since 4 lies in the interval 4 x < 6
f(x) = 3x2 – 10
f(4) = 3(4)2 – 10
= 48 – 10 = 38
f(2) – f(4) = 3 – 38 = – 34
2f(5) – 3 f(1) = 2 (65) – 3(2)
= 130 – 6
= 124
A function f: [-3, 7)R is defined as follows
6 x 4 ; 32
4 x 2 ; 23
2 x 3- 1;4
)(
2
x
x
x
xf
Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv)
(i) Since 5 lies between 4 and 6
f(x) = 3x – 2
f(5) = 3(5) – 2
= 15 – 2 = 13
Since 6 lies in the interval 4 < x 6
f(x) = 2x – 3
f(6) = 2(6) – 3 = 12 – 3
= 9
f(5) + f(6) = 13 + 9 = 22
)1()6(2
)1()3(
ff
ff
A function f: [-3, 7)R is defined as follows
f(x) =
6 x 4 ; 32
4 x 2 ; 23
2 x 3- 1;4 2
x
x
x Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4)(iv)
(ii) Since 1 lies in the interval -3 x < 2
f(x) = 4x2 – 1
f(1) = 4(1)2 – 1
= 4 – 1 = 3
Since –3 lies in the interval-3 x < 2
f(x) = 4x2 – 1
= 4(-3)2 – 1
= 36 – 1 = 35
f(1) – f(-3) = 3 – 35 = – 32
)1()6(2
)1()3(
ff
ff
A function f: [-3, 7)R is defined as follows
f(x) =
6 x 4 ; 32
4 x 2 ; 23
2 x 3- 1;4 2
x
x
x Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv)
(iii) Since -2 lies in the interval -3 x < 2
f(x) = 4x2 – 1
f(1) = 4(-2)2 – 1
= 16 – 1 = 15
Since 4 lies in the interval 2 x 4 f(x) = 3x – 2
= 3(4) – 2
= 12 – 2 = 10
f(-2) – f(4) = 15 – 10 = 5
)1()6(2
)1()3(
ff
ff
A function f: [-3, 7)R is defined as follows
f(x) =
6 x 4 ; 32
4 x 2 ; 23
2 x 3- 1;4 2
x
x
x Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv)
(iv) Since 3 lies in the interval 2 x 4 f(x) = 3x – 2
f(3) = 3(3) – 2
= 9 – 2 = 7
Since -1 lies in the interval -3 x < 2
f(x) = 4x2 – 1
= 4(-1)2 – 1
= 4 – 1 = 3
f(6) = 9, f(1) = 3
)1()6(2
)1()3(
ff
ff
3)9(2
37
)1()6(2
)1()3(
ff
ff
318
10
15
10
3
2
A function f: [-7, 6)R is defined as follows
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;12
)(
2
x
x
xx
xf
Find the value of
(i) 2f(-4) + 3f(2)
(ii) f(-7) – f(– 3),
(iii) (i) Since -4 lies in the interval -5 x 2 f(x) = x + 5
f(-4) = – 4 + 5
= 1
Since 2 lies in the interval -5 x 2 f(x) = x + 5
f(2) = 2 + 5
= 7
)1(3)6(
)4(2)3(4
ff
ff
2f(-4) + 3f(2)
= 2(1) + 3(7)
= 2 + 21
= 23
A function f: [-7, 6)R is defined as follows
f(x) =
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;122
x
x
xx Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
(ii) Since -7 lies in the interval -7 x < -5
f(x) = x2 + 2x + 1
f(-7) = (–7)2 + 2(–7) + 1
= 49 – 14 + 1 = 36
Since -3 lies in the interval -5 x 2 f(x) = x + 5
f(2) = -3 + 5
= 2
)1(3)6(
)4(2)3(4
ff
ff
f(-7) – f(–3)
= 36 – 2
= 34
A function f: [-7, 6)R is defined as follows
f(x) =
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;122
x
x
xx Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
(iii) Since -3 lies in the interval -5 x 2 f(x) = x + 5 f(-3) = -3 + 5 = 2
Since -6 lies in the interval -7 x < -5
f(x) = x2 + 2x + 1
f(-6) = (–6)2 + 2(–6) + 1
= 36 – 12 + 1 = 25
)1(3)6(
)4(2)3(4
ff
ff
4 lies in the interval 2 < x < 6
f(x) = x – 1
f(4) = 4 – 1
= 3
Since 1 lies in the interval -5 x 2 f(x) = x + 5
f(1) = 1 + 5
= 6
A function f: [-7, 6)R is defined as follows
f(x) =
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;122
x
x
xx Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
(iii) f(-3) = 2
f(-6) = 25
f(4) = 3
f(1) = 6
)1(3)6(
)4(2)3(4
ff
ff
)6(325
)3(2)2(4
)1(3)6(
)4(2)3(4
ff
ff
1825
68
27
14
If verify that (AB)T = BTAT .
11
12 ,
37
25BA
LHS = RHS (AB)T = BTAT .
11
12 ,
37
25BA
11
12
37
25AB
13)1(7)1(327
12)1(5)1(225
37314
25210
411
38
LHSAB T ........43
118)(
11
12 ,
32
75 TT BA
32
75
11
12TT AB
317)1(215)1(
3)1(722)1(52
3725
314210
RHSAB TT ........43
118
If verify that (AB)T = BTAT . 631 ,
5
4
2
BA
LHSAB T ........
302412
15126
542
)(
631 ,
5
4
2
BA
631
5
4
2
BA
)6(53515
)6(43414
)6(23212
BA
30155
24124
1262
BA
If verify that (AB)T = BTAT . 631 ,
5
4
2
BA
LHS = RHS (AB)T = BTAT .
LHSAB T ........
302412
15126
542
)(
6
3
1
,542 TT BA
542
6
3
1
TT AB
5)6(46)2(6
5343)2(3
5141)2(1TT AB
RHSAB TT ....
302412
15126
542
If then show that A2 – 4A + 5I2 = O
32
11A
32
11A
32
11
32
112 AAA
33)1(22312
3)1()1(12)1(11
9262
3121
78
41
128
44
32
1144A
50
05
10
0155 2I
50
05
128
44
78
4154 2
2 IAA
5127088
044541
00
00
A2 – 4A + 5I2 = O
7. Find X and Y if 2X + 3Y =
)........(104
3232
YX
)2(..........51
222Y3X
)3......(04
322642)1(
YX
)4.......(51
2236Y9X3(2)
51
223
04
3225X)4((3)
153
66
08
64
15038
6664
1511
1225 X
1511
122
5
1 X
51
222Y3X and
04
32
)........(104
3232
YX
)2(..........51
222Y3X
)3......(04
323963)1(
YX
)4.......(51
222Y4X62(2)
51
222
04
3235Y)4((3)
102
44
012
96
100212
4946
1014
1325 Y
1014
132
5
1 Y
11. An electronic company records each type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets.
TV DVD VIDEOGAMES CD players
WEEK I
Store I 30 15 12 10
Store II 40 20 15 15
Store III 25 18 10 12
WEEK II
Store I 25 12 8 6
Store II 32 10 10 12
Store III 22 15 8 10
Find the sum of the items sold out in two weeks using matrix addition.
1081522
12101032
681225
,
12101825
15152040
10121530
Let BA
1081522
12101032
681225
12101825
15152040
10121530
BA
101281015182225
1215101510203240
61081212152530
22183347
27253072
16202755
If verify (AB)C = A(BC) .
12 and
2
1
0
,321
121
CBA
LHS = RHS (AB)C =A(BC)
2
1
0
321
121AB
231201
211201AB
620
220
8
4
128
4)(
CAB
1828
1424LHS.....
816
48
12
2
1
0
BC
1222
1121
1020
24
12
00
24
12
00
321
121)(BCA
231201432201
211201412201
6201240
220440RHS.....
816
48
If then prove that
(A + B)2 A2 + 2AB + B2.
23
61 ,
32
41BA
23
61 ,
32
41BA
23
61
32
41BA
11
20
2332
6411
11
20
11
20)( 2BA
11211101
12201200
1210
2020LHS...
31
22
32
41
32
412A
33)4()2()2(31)2(
3)4()4(1)2()4(11
9862
12481
178
169
23
61
23
612B
)2()2(633)2()1(3
)2(66)1(36)1()1(
41863
126181
229
1819
If then prove that
(A + B)2 A2 + 2AB + B2.
23
61 ,
32
41BA
23
61 ,
32
41BA
23
61
32
41AB
)2(36)2(33)1()2(
)2()4(613)4()1(1
61292
86121
1811
1413
229
1819
1811
14132
178
1692 22 BABA
229
1819
3622
2826
178
169
2236179228
18281619269
36391722
34282628
RHS....35
62
LHS RHS
(A + B)2 A2 + 2AB + B2.
If find (A + B)C and
AC + BC. Is (A + B)C = AC + BC ?
64
32 and
90
78 ,
67
33CBA
64
32 and
90
78 ,
67
33CBA
90
78
67
33
BA
157
1011
9607
7383
64
32
157
1011)(
CBA
615)3(741527
610)3(11410211
90216014
60334022
6974
2762
64
32
67
33AC
66)3(74627
63)3(34323
36212414
189126
1538
918
64
32
90
78BC
69)3(04920
67)3(84728
540360
42242816
5436
1844
If find (A + B)C and
AC + BC. Is (A + B)C = AC + BC ?
64
32 and
90
78 ,
67
33CBA
5436
1844
1538
918BCAC
54153638
1894418
6974
2762
(A+B)C = AC + BC
The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term.
Given t10 = 41
t18 = 73
tn = a + (n – 1)d
t10 = 41
a + (10 – 1)d = 41
a + 9d = 41…………(1)
t18 = a + (18 – 1)d = 73
a + 17d = 73 …….(2)
(2) a + 17d = 73
(1) a + 9d = 41
(2) – (1) 8d = 32
d = 32/8
d = 4
Sub d = 4 in (1)
a + 9(4) = 41
a + 36 = 41
a = 41 – 36
a = 5
t27 = a + (27 – 1)d
= 5 + (26)4
= 5 + 104 = 109(-) (-) (-)
The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers.
Let a – d, a, a + d be three consecutive terms in an A.P.
Sum = 6
a – d + a + a + d = 6
3a = 6
a = 6/3 = 2
Product = – 120
(a – d) a (a + d) = – 120
(2 – d) 2 (2 + d) = – 120
(22 – d2) 2 = – 120
4 – d2 = – 120/2 = – 60
– d2 = – 60 – 4
– d2 = – 64
d2 = 64
d = 8The given numbers are
2 – 8, 2, 2 + 8
– 6, 2, 10 (or)
10, 2, –6
A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the 15th year.
Given t7 = 1000
t10 = 1450 tn = a + (n – 1)d
t7 = 1000a + (7 – 1)d = 1000 a + 6d = 1000…………(1)
t10 = a + (10 – 1)d = 1450 a + 9d = 1450 …….(2)(2) a + 9d = 1450(1) a + 6d = 1000
(2) – (1) 3d = 450
d = 450/3
d = 150
Sub d = 150 in (1)
a + 6(150) = 1000
a + 900 = 1000
a = 1000 – 900
a = 100
t15 = a + (15 – 1)d
= 100 + (15)150
= 100 + 2250 = 2350
Find the sum of the first 40 terms of the series 12 – 22 + 32 – 42 + ……. .
12 – 22 + 32 – 42 + ……. . = 1 - 4 + 9 - 16 + 25 ….. to 40 terms
= (1 – 4) + (9 – 16) + (25 – 36) + …… to 20 terms.
= (-3) + (-7) + (-11) +………20 terms
It is an AP with a = –3, d = –4, n = 20
dn )1(2a2
nSn
)4)(120(2(-3)2
20S20
)4(19)62
20
7662
20
822
20
10
= 10(-82)
= – 820
Find the sum of all 3 digit natural numbers, which are divisible by 9.
First number = 100 + (9 – 1) = 108
Last number = 999
The AP is
108 + 117 +……+ 999
a = 108, l = 999, d = 9 1d
a-ln
19
108-999n
110919
981
n = 110
)al(2
nsn
)1107(2
110
)999108(2
110s110
= 55 1107
= 60885
55
1 1
9 1 0 0
9
1 0
9
1
1 1 1
9 9 9 9
9
9
9
9
9
0
The three digit numbers are from 100 to 999
Find the sum of all natural numbers between 300 and 500 which are divisible by 11.
First number = 300 + (11 – 3) = 308
Last number = 500 – 5 = 495
The AP is
308 + 319 +……+ 495
a = 308, l = 495, d = 111
d
a-ln
111
308495n
117111
187
n = 18
)al(2
nsn
)803(2
18
)495308(2
18s18
= 9 803
= 7227
9
2 7
11 3 0 0
2 2
8 0
7 7
3
4 5
11 5 0 0
4 4
6 0
5 5
5
A sum of Rs. 1000 is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P.? If so, find the total interest at the end of 30 years.
Every year deposit = Rs. 1000
Rate of interest = 8% = 0.08
First year interest = 1000 0.08 = 80
Second year interest = 2000 0.08 = 160
Third year interest = 3000 0.08 = 240
The interest 80, 160, 240,…. Forms an AP.
Total interest = 80 + 160 + 240 +……30terms
= 15 2480
= Rs.37200
})1(2{2n dnan
S
}80)130(802{2
3030 S
}8029160{2
30
}2320160{2
30
24802
30
Find the sum of all numbers between 100 and 200 which are not divisible by 5.
The required sum
= (100 +101 + 102+ ……+ 200) – (100 + 105 + 110 + … + 200)
100 + 101 + 102 + ………… + 200
= (1 + 2 + 3 + ………….+ 200) – (1 + 2 + 3+ ………+ 99)
2
1)n(nn
= 100 201 – 99 50
= 20100 – 4950
= 15150
100 + 105 + …… + 200
Here a = 100, d = 5, l = 200
2
1)99(99
2
1)200(200
2
10099
2
201200
100 50
1n
d
al
15
100200n
15
100
21120
Find the sum of all numbers between 100 and 200 which are not divisible by 5.
)(2
nsn al
)300(2
21
)100200(2
21s21
= 21 150
= 3150
150
The required sum
= 15150 – 3150
= 12000
Calculate the standard deviation of the following data. 10, 20, 15, 8, 3, 4
35.66,6655
25106610
.9
9981
8566118
7
78309
10
20
15
8
3
4
60
x d = x –x d2 n
xx 10
6
60
10 – 10 = 0
20 – 10 = 10
15 – 10 = 5
8 – 10 = -2
3 – 10 = -7
4 – 10 = -6
0
0
100
25
4
49
36
214
n
dDS
2
.
6
214
6666.35
97.5
257
Calculate the standard deviation of the following data. 38, 40, 34 ,31, 28, 26, 34
22.00,0044
166008
.6
6516
840092
9
98361
38
40
34
31
28
26
34
231
x d = x –x d2 n
xx 33
7
231
38 – 33 = 5
40 – 33 = 7
34 – 33 = 1
31 – 33 = -2
28 – 33 = -5
26 – 33 = -7
34 – 33 = 1
0
25
49
1
4
25
49
1
154
n
dDS
2
.
7
154
22
69.4
39
Calculate the standard deviation of the following data.
3
7
8
10
13
15
18
10
23
8
50
x f d = x – 3 d2 fd
fd2 3 – 3 = 0
8 – 3 = 5
13 – 3 = 10
18 – 3 = 15
23 – 3 = 20
0
25
100
225
400
X 3 8 13 18 23
f 7 10 15 10 8
0
50
150
150
160
510
0
250
1500
2250
3200
7200
Standard deviation
22
f
fd
f
fd
2
50
510
50
7200
22.10144
04.104144
96.39
39.9666
36396123369
27
.3
3.6
11. A group of 45 house owners contributed money towards green environment of their street. The amount of money collected is shown in the table below. Calculate the variance and standard deviation.
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
Time f mid x d = (x – 30)/20 d2 fd fd2
Amount 0-20 20-40 40-60 60-80 80-100
No of house owners 2 7 12 19 5
2
7
12
19
5
45
10
30
50
70
90
-2
-1
0
1
2
4
1
0
1
4
-4
-7
0
19
10
28
8
7
0
19
20
58
Standard deviation 20
22
f
fd
f
fd
2045
28
45
582
0.904490.9
8194418925
19
5
2062.02888.1 2
203844.02888.1 209044.0
2095.0 19
5Variance = (S.D)2
= (19)2= 361
Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.
24.00,0044
168008
.8
8704
960096
9
98721
20
18
32
24
26
120
x d = x –x d2 n
xx 24
5
120
20 – 24 = -4
18 – 24 = -6
32 – 24 = 8
24 – 24 = 0
26 – 24 = 2
0
16
36
64
0
4
120
n
dDS
2
.
5
120
24
%100x
CV
879
%10024
89.4
89.4
%24
489
%375.20
The time (in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below. Calculate the variance and standard deviation of the data.
5 – 10
10 – 15
15 – 20
20 – 25
25 – 30
Time f mid x d = x –17.5 d2 fd fd2
Time (sec) 5-10 10-15 15-20 20-25 25-30
No of people 4 8 15 12 11
4
8
15
12
11
50
7.5
12.5
17.5
22.5
27.5
-10
-5
0
5
10
100
25
0
25
100
-40
-40
0
60
110
90
400
200
0
300
1100
2000
Variance
22
f
fd
f
fd
2
50
90
50
2000
24.340)8.1(40 2
76.36
76.36. DS36.7666
367600120
6 7236374
.06
06.6
Show that the points A(2 , 3), B(4 , 0) and C(6, -3) are collinear.
Area of the ABC =
½ {(x1y2 + x2 y3 + x3y1) – (x1y3 + x3y2+ x2 y1)}sq. units
2 4 6
2
3 0 -3
3
= ½{(2)(0) + (4)(–3) + (6)(3)} – {(2)(–3) + (6)(0) + (4)(3)}
= ½ {(0 – 12 + 18) – (–6 + 0 + 12)}
= ½ {(18 – 12) – (12 – 6)}
= ½ (6 – 6)
= 0
The given points are collinear.
find the value of k for which the given points are collinear. (2, - 5),(3, - 4) and (9, k)
If the points are collinear, Area of the ABC = 0
{(x1y2 + x2 y3 + x3y1) – (x1y3 + x3y2+ x2 y1)} = 0
2 3 9 2
-5 -4 k -5
{(2)(-4) + (3)(k) + (9)(-5)} – {(2)(k) + (9)(-4) + (3)(-5)} = 0
{(–8 + 3k – 45) – (2k – 36 – 15)} = 0
{(3k – 53) – (2k – 51)} = 0
3k – 2k – 53 + 51 = 0
k – 2 = 0
k = 2
Find the area of the quadrilateral whose vertices are(6, 9), (7, 4), (4,2) and (3,7)
Area of the quadrilateral ABCD
= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)
– (x2y1 + x3 y2 + x4y3 + x1y4)}
4 7 6 3 4
2 4 9 7 2
= ½ {(4)(4) + (7)(9) + (6)(7) + (3)(2)} –
{(2)(7) + (4)(6) + (9)(3) + (7)(4)}
= ½ {(16 + 63 + 42 + 6) – (14+ 24 + 27 + 28)}
A(4,2)
D(3,7)
B(7,4)
C(6,9)
= ½ {127 – 93}
= ½ (34)
= 17 sq. units
Find the area of the quadrilateral whose vertices are(-3, 4), (-5,- 6), (4,- 1) and (1, 2)
Area of the quadrilateral ABCD =
= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)
– (x2y1 + x3 y2 + x4y3 + x1y4)}
-5 4 1 -3
-5
-6 -1 2 4
-6
= ½ {(-5)(-1) + (4)(2) + (1)(4) + (-3)(-6)} –
{(-5)(4) + (-3)(2) + (1)(-1) + (4)(-6)}
= ½ {(5 + 8 + 4 + 18) – (-20 – 6 – 1 – 24)}
A(-5,-6)
D(-3,4)
B(4,-1)
C(1,2)
= ½ {35 – (–51)}
= ½ (35 + 51)
= ½ (86)
= 43 sq. units
Find the area of the quadrilateral whose vertices are(-4, 5), (0, 7), (5,- 5) and (-4,- 2)
Area of the quadrilateral ABCD =
= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)
– (x2y1 + x3 y2 + x4y3 + x1y4)}
-4
-2
5
-5
0
7
-4
5
-4
-2
= ½ {(-4)(-5) + (5)(7) + (0)(5) + (-4)(-2)} –
{(-4)(5) + (-4)(7) + (0)(-5) + (5)(-2)}
= ½ {(20 + 35 + 0 + 8) – (-20 – 28 – 0 – 10)}
= ½ {63 – (–58)}
= ½ (63 + 58)
= ½ (121)
= 60.5 sq. units
A(-4,-2)
D(-4,5)
B(5,-5)
C(0,7)
A coin is tossed three times. Find the probability of getting (i) head and tail alternatively (ii) at least two heads (iii) exactly two heads (iv) no heads
Tossing three coins once is as same as a coin is tossed three times
Sample space S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
n(S) = 8
(i) Let the event of getting head and tail alternatively be A
A = {HTH, THT} n(A) = 2
The probability of getting head and tail alternatively is
4
1
8
2
n(S)
n(A) P(A)
(ii) Let the event of getting at least two heads be B
B = {HHH, HHT, HTH, THH} n(B) = 4
The probability of getting at least two heads is 2
1
8
4
n(S)
n(B) P(B)
(iii) Let the event of getting exactly two heads be C
C = {HHT, HTH, THH} n(C) = 3
The probability of getting exactly two heads is
8
3
n(S)
n(C) P(C)
(iv) Let the event of getting no head be D
D = {TTT} n(D) = 3
The probability of getting no head is
8
1
n(S)
n(D) P(D)
A card is drawn from a pack of 52 cards. Find the probability that it is either red card or king card.
Total number of cards = 52
n(S) = 52
Let A be the event drawing a red card
n(A) = 26
The probability of drawing a red card is
2
1
52
26
n(S)
n(A) P(A)
Let B be the event drawing a king card
n(B) = 4
The probability of drawing a king card is
13
1
52
4
n(S)
n(B) P(B)
n(A B) = 2
26
1
52
2
n(S)
B)n(A B)P(A
P(AB) = P(A) + P(B) – P(A B)
52
2
52
4
52
26
52
2-426
13
7
52
28
Two dice are rolled once. Find the probability of getting an even number on the second die or the total of face numbers 10.
When two dice are rolled the sample space
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n(S) = 36
Let the event of getting even in the second die be A
A = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6) (3, 2), (3, 4), (3, 6), (4, 2), (4, 4), (4, 6) (5, 2), (5, 4), (5, 6), (6, 2), (6, 4), (6, 6)}
n(A) = 18
2
1
36
18
n(S)
n(A) P(A)
Let the event of getting the total of the face numbers is 10 be B
B = {(4, 6), (5, 5), (6, 4)} n(B) = 3
12
1
36
3
n(S)
n(B) P(B)
A B = {(4, 6), (6, 4)} n(A B) = 2
18
1
36
2
n(S)
B)n(A B)P(A
P(AB) = P(A) + P(B) – P(A B)
36
2
36
3
36
18
36
2-318
36
19
16. Two persons X and Y appeared in an interview for two vacancies in an office. The chance for X’s selection is 1/5 and the chance for Y’s selection is 1/7. Find the chance that (i) both of them are selected (ii) only one of them is selected, (iii) none of them is selected.
The probability of X’S selection = P(X) = 1/5
The probability of Y’S selection = P(Y) = 1/7
5
4
5
11 )XP(
7
6
7
11 )YP(
(i) The probability of both of them to be selected
35
1
7
1
5
1P(Y)P(X) Y)P(X
(ii) The probability of only one of them to be selected
P(Y))XP( )YP(P(X) Y)XP()YP(X
7
2
35
10
35
4
35
6
7
1
5
4
7
6
5
1
35
24
7
6
5
4)YP()XP( )YXP(
(iii) The probability of none of them to be selected