Q.1. (A) Solve the following :(Any 4)

1. C = { –1, –2, –3, ... }

2. In 2916

, 16 is the denominator and

16 = 2 ´ 2 ´ 2 ´ 2

Thus, denominator has only 2’s as prime factors.

\ 2916

has terminating decimal representation.

3. p(x) = x2 – 5x + 5 x = 0 \ p(0) = (0)2 – 5 × (0) + 5 = 0 – 0 + 5 \ p(0) = 5 \ The value of given polynomial when x = 0 is 5.

4. x + y = 7 if x = 1 then y = 6, if x = 7 then y = 0 if x = –2 then y = 9, if x = –1 then y = 8 if x = 0 then y = 7 (1,6), (7,0), (–2 ,9),(–1,8),(0,7),... are the solutions of the given equation.

5. |15 – 2| = |13| = 13 \|15 – 2| = 13 6. Now, 5.2 = 5.20 and 5.3 = 5.30 \5.20 < 5.21 < 5.22 < 5.23 < ... < 5.30 \Three rational numbers between 5.2 and 5.3 are 5.21, 5.22, 5.23.

Time : 2 Hours (Answer paper) Max. Marks : 40

MT - MATHEMATICS (71) Algebra - SEMI PRELIM - II - PAPER - III

MTSeat No.2018 ____ ____ 1100

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Q.1. (B) Solve the following :(Any2) 1. 2x – 7y = 7 ... (i) 3x + y = 22 ... (ii) Multiplying (ii) by 7, 21x + 7y = 154 ... (iii) Adding (i) and (iii) , 2x – 7y = 7 21x + 7y = 154 23x = 161

∴ x = 16123

∴ Substituting the value of x in (ii), 3(7) + y = 22 ∴ 21 + y = 22 ∴ y = 22 – 21 ∴ ∴

2. 98 2÷

= 982

= 982

= 49 = 7 \ 98 2 7

3. n(A ∪ B) = n(A) + n(B) – n(A ∩ B) \29 = 15 + n(B) – 7 \ 29 = 8 + n(B) \ n(B) = 29 – 8 \ n(B) = 21

2/MT PAPer - III

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x = 7 and y = 1 is the solution of given equations.

x = 7

y = 1

Q.2. (A) Solve the following : 1. (D) 3x2 + 15x + 3 = 0 Working: Sum of roots is –5. For the equation 3x2 + 15x + 5 = 0 Comparing with ax2 + bx + c = 0 a =3, b = 15, c = 3

Product of roots = ca

= 153

= –5

2. (B) 1.5

3. (B) 3 4x + 5y = 19 x = 1 ∴ 4(1) + 5y = 19 ∴ 4 + 5y = 19 ∴ 5y = 19 – 4 ∴ 5y = 15

∴ y = 155

∴ y = 3

4. (B) 17

Working: 2 x2 – 5x + 2 = 0 Comparing with ax2 + bx + c = 0 we get, a = 2 , b = –5, c = 2 D = b2 – 4ac = (–5)2 – 4( 2 )( 2 ) = 25 – 8 =17 Q.2. (B) Solve the following : (Any 2) 1. A die consist of 6 faces \ n(S) = 6

(i) LetAbetheeventthattheletterappearsontheuppermostfaceisA. \ n(A) = 2 (AstherearetwofaceswithletterA)

P(A) = n(A)n(S)

=26=

13

3/MT PAPer - III

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∴ P(A) = 13

(ii)LetBbetheeventthattheletterappearingontheuppermostfaceisD. \ n(B) = 1

P(B) = BS

n( )n( )

=16

∴ P(B) = 16

2. 6x – 3y = –10 ... (i) Expressing the given equation 3x + 5y – 8 = 0 in the form ax + by = c we get, 3x + 5y = 8 ... (ii)

D=6 - 33 5

= (6 ´ 5) – (–3 ´ 3) = 30 – (–9) = 30 + 9 \ D = 39

Dy = 6 103 8

= (6 ´ 8) – (– 10 ´ 3) = 48 – (– 30) = 48 + 30

\ Dy = 78

3. 3 x2 + 2 x – 2 3 = 0

Comparing with ax2 + bx + c = 0, we get, a = 3 , b = 2 , c = – 2 3

D =b2 – 4ac

= 22( ) – 4 × 3 × (– 2 3 )

= 2 + 24 \ D = 26

4/MT PAPer - III

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Q.3. (A) Solve the following activity : (Any2) 1.

∴

2. 5m2 – 4m – 2 = 0

Comparing with am2 + bm + c = 0, we get,

a = 5, b = –4, c = –2. \ b2 – 4ac = (–4)2 – 4 × 5 × (–2) = 16 + 40 = 56

m = − ± −b b aca

2 42

...(Formula)

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(x, y) = (4, 5) is the solution of the given simultaneous equations.

1 12 3

= − y

Replacing 11

− x

by m and 1

2 − y

by n

New equation on solving

m = 13

and n = 13

Replacing m, n by their original valves.

on

solving

x = 4 and y = 5

5m + n = 2 ; 6m –3n = 1

1131

= − x

5 1 6 32 ; 11 2 1 2+ = − =

− − − −x y x y

\ m = − − ±

×( )4 56

2 5

\ m = 4 4 14

10± ×

\ m = 4 2 14

10±

\ m = 2 2 1410

( )±

\ m = 2 145

±

\ m = 2 145

+ or m = 2 145

−

\

3. S : Area of the garden.

Length of the garden = 77 m, Breadth of the garden = 50 m

∴ Area of the garden = 77 × 50 = 3850 sq m. Let A be the event that the towel fell in a lake. Diameterofthelake =14m ∴Radius of the lake = 7 m ∴Area of the lake = pr2

= 227

× 7 × 7

= 154 sq m.

P(A) = n(A)n(S)

=1543850

=125=0.04

∴ P(Towel fell in the lake) = 0.04 Q.3. (B) Solve the following : (Any 2) 1. 5m – 3n = 19 ...(i) m – 6n = –7 ...(ii) Multiplying (i) by 2, 10m – 6n = 38 ...(iii) Subtracting (iii) from (ii)

6/MT PAPer - III

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The roots of given quadratic equation are 2 145

−− and 2 145

++ .

m – 6n = –7 10 m – 6n = 38 (–) (+) (–) – 9m = – 45

\ m =

459

\ m = 5 Substituting m = 5 in (i) \ 5(5) – 3n = 19 \ 25 – 3n = 19 \ –3n = 19 – 25 \ –3n = –6

\ n =

63

\ n = 2 \ (m , n) = (5, 2) is the solution of the given simultaneous equations.

2. Let a and bare the roots of the quadratic equation Let a= 0 and b= 4 \ a + b=0 + 4 = 4 and a×b= 0 × 4 = 0 Then, the quadratic equation is x2 – (a + b)x + ab= 0 \ x2 – 4x + 0= 0 \ x2 – 4x = 0

3. Number of Red pens = 5, Number of Blue pens = 8, Number of green pens = 3 \Total no. of pens = 16 S = {R1, R2, R3, R4, R5, B1, B2, B3, B4, B5, B6, B7, B8,G1, G2,G3} \ n(S) = 16

Let A be the event that the pen picked up is blue. A = {B1, B2, B3, B4, B5, B6, B7, B8} \ n(A) = 8

P(A) = n(A)n(S)

=816=

12

∴ P(A) = 12

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Q.4 Solve the following : (Any 3) 1. Let the he present age of Manish be x years and age of Savita y years. Accordingtothefirstcondition x + y = 31 ...(i) Three years ago, Manish’s Age = (x – 3) years and Savita’s age = (y – 3) years According to second condition, x – 3 = 4 (y – 3) \ x – 3 = 4y – 12 \ x – 4y = – 12 + 3 \ x – 4y = – 9 ...(ii) Subtracting equation (ii) from equation (i) we get, x + y = 31 x – 4y = –9 (–) (+) (+) 5y = 40

\ y = 405

\ y = 8 Substituting y = 8 in equation (i) we get, \ x + y = 31 \ x + 8 = 31 \ x = 31 – 8 \x = 23 \

2. m2 – 5m = –3 m2 – 5m + 3 = 0 m2 – 5m + k = (m– a)2

m2 – 5m + k = m2– 2am + a2

Comparingthecoefficients,weget 5 = 2a k = a2

52

= a k = 25

2

= 254

8/MT PAPer - III

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The present age of Manisha is 23 years and the present age of Savita is 8 years.

\ m2 – 5m + 254

– 254

+ 3 = 0

\ m2 – 5m + 254

= 254

+ 3

\ m −

52

2 = 25

4 – 3

\ m −

52

2 =

134

Taking square root on both sides

\ m – 52

= ± 132

\ 5 132 2

= ±m

\5 13

2±

=m

\ m = 5 132

+ or m = 5 132

−

\ The roots of given quadratic equation are 5 132

++ and 5 132

−− .

3. Two digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5

without repeating digits are as follows : S = {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30,31,32,

34, 35, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54} ∴ n (S) = 25 Condition for event A : The number formed is even A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54} ∴ n (A) = 13 Condition for event B : The number formed is divisible by 3. B = {12, 15, 21, 24, 30, 42, 45, 51, 54} ∴ n (B) = 9 Condition for event C : The number formed is greater than 50. C = {51, 52, 53, 54} ∴ n (C) = 4

9/MT PAPer - III

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4. 148 231 527x y xy

+ = ...(i)

231 148 610x y xy

+ = ...(ii)

Multiplying both the sides of equation (i) and equation (ii) by xy we get, 148y + 231x = 527 i.e. 231x + 148y = 527 ...(iii) 231y + 148x = 610 i.e. 148x + 231y = 610 ...(iv) Adding (iii) and (iv) we get

231x + 148y = 527

148x + 231y = 610

379x + 379y = 1137

Dividingthroughoutby379,

\ x + y = 1137379

\ x + y = 3 ...(v) Subtracting (iv) from (iii) we get, 231x + 148y = 527 148x + 231y = 610 (–) (–) (–) 83x – 83y = – 83 Dividingthroughoutby83 \ x – y = –1 ...(vi)

Adding (v) and (vi) we get, x + y = 3 x – y = –1 2x = 2 \ x = 1 Substituting x = 1 in equation (v) we get x + y = 3 \ 1 + y = 3 \ y = 3 – 1 \ \

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1x = 1 and y = 2 is the solution of the given simultaneous equations.

y = 2

Q.5 Solve the following : (Any 1) 1. 3x – y = 2 i.e. y = 3x – 2

x 0 1 2y –2 1 4

(x, y) (0, –2) (1, 1) (2, 4)

when x = 0 when x = 1 \ y = 3(0) – 2 \ y = 3(1) – 2 \ y = 0 – 2 \ y = 3 – 2 \ y = –2 \ y = 1

when x = 2 \ y = 3(2) – 2

\ y = 6 – 2 \ y = 4

2x – y = 3 i.e. y = 2x – 3

x 0 1 2y –3 –1 1

(x, y) (0, –3) (1, –1) (2, 1)

when x = 0 when x = 1

\ y = 2(0) – 3 \ y = 2(1) – 3

\ y = 0 – 3 \ y = 2 – 3

\ y = – 3 \ y = – 1

when x = 2

\ y = 2(2) – 3

\ y = 4 – 3

\ y = 1

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The lines of the two given simultaneous equations intersect each other at (–1 , –5) \

2. Letthebreadthofrectangularfieldisx m. \ Its length = (2x + 10) Area of the farm = (2x + 10) × x sq.m

Now, side of a square pond = 13

x m \ Area of square pond = side2

= 13

2

x

= 19

2x sq.m

12/MT PAPer - III

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Scale1cm = 1 Uniton both axes

Y

Y’

X’ X0–1 1–2–3–4–5 2 3 4 5 6

–2

–1

–3

–4

–5

–6

2

1

3

4

5

6

(0, –3)

(0, –2)

(2, 1)(1, 1)

(2x –

y = 3

)

(–1, –5)

(3x

– y

= 2)

(2, 4)

(1, –1)

The solution of given simultaneous equations is (–1, –5) i.e. x = –1, y = –5

According to given condition,

(2x + 10) × x = 20 × 19

2x

\ 2x2 + 10x = 209

2x

\ 18x2 + 90x = 20x2

\ 90x = 20x2 – 18x2

\2x2 – 90x = 0 \2x(x – 45) = 0 \2x = 0 or x – 45 = 0 \ x = 0 or x = 45 x ¹ 0asbreadthoffieldcannotbezero. \ x = 45 and (2x + 10) = 2 × 45 + 10 = 90 + 10 = 100

Also, 13

x = 13

× 45 = 15

\

Q.6 Solve the following : (Any 1) 1. y2 – 2y – 7 = 0

Comparing with ay2 + by + c = 0 we get, a = 1, b = –2, c = –7

\a + b= −ba= − −( )2

1= 2 and

a × b= ca= −7

1= –7

(i) a2 + b2 = (a+b)2 – 2ab = (2)2 – 2 × (–7) =4 + 14

\a2 + b2 = 18 (ii) a3 + b3 = (a+b)3 – 3ab(a+b) = (2)3 – 3 × (–7) × (2) =8 + 42

\a3 + b3 = 50

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The length of rectangular field is 100 m and breadth is 45 m. Also, length of each side of square pond is 15 m.

14/MT PAPer - III 2. When two dice are thrown 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 (i) Let A be the event that the sum of the digits on the upper faces is atleast 10 A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} \ n(A) = 6

P(A) = n(A)n(S)

=636=

16

∴ P(A) = 16

(ii) Let B is the event that the sum of the digits on the upper faces is 33. B = { } \ n(B) = 0

P(B) = BS

n( )n( )

=036=0

∴ P(B) = 0

(iii)LetCistheeventthatthedigitsonthefirstdieisgreaterthanthedigits

on second die.

C = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

\ n(C) = 15

P(C) = CS

n( )n( )

=1536=

512

∴ P(C) = 512

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