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Question 1 of 10

1. Question

Directions:1-5) In the following questions, two equations I and II are given. You have to solve both the equations.

Give Answer

I. 15a^{2} – 10a – 5 = 0;

II. 20b^{2} + 18b + 4 = 0

Correct

15 a^{2} + 15a – 5a – 5 = 0

15a (a + 1) – 5 (a + 1) = 0

(15a – 5) (a + 1) = 0

a = 1/3, -1

20b^{2} + 10b + 8b + 4 = 0

10b (2b +1) + 4 (2b + 1) = 0

(10b + 4) (2b + 1) = 0

b = -2/5, -1/2

Hence there is no relation.

Incorrect

15 a^{2} + 15a – 5a – 5 = 0

15a (a + 1) – 5 (a + 1) = 0

(15a – 5) (a + 1) = 0

a = 1/3, -1

20b^{2} + 10b + 8b + 4 = 0

10b (2b +1) + 4 (2b + 1) = 0

(10b + 4) (2b + 1) = 0

b = -2/5, -1/2

Hence there is no relation.

Question 2 of 10

2. Question

I. 26a^{2 }– 88 = 16;

II. 3√b/8 – √b/4 = 1/√b

Correct

26a^{2} = 16 + 88 = 104

a^{2} = 104/26 = 4

a = ±2

3√b/8 – 2√b/8 = 1/√b

√b/8 = 1/√b

b = 8

a < b.

Incorrect

26a^{2} = 16 + 88 = 104

a^{2} = 104/26 = 4

a = ±2

3√b/8 – 2√b/8 = 1/√b

√b/8 = 1/√b

b = 8

a < b.

Question 3 of 10

3. Question

I. 2a + 3b = 20;

II. 4a + 2b = 24

Correct

4a + 6b = 40 —-I

4a + 2b = 24 —-II

Solve I and II we get

4b = 16

b = 4

Apply the b = 4 in eqn I

a = 4

a= b.

Incorrect

4a + 6b = 40 —-I

4a + 2b = 24 —-II

Solve I and II we get

4b = 16

b = 4

Apply the b = 4 in eqn I

a = 4

a= b.

Question 4 of 10

4. Question

I. 10a^{2} + 9a -7 = 0;

II. 12b + 84 = 12^{2}

Correct

10a^{2} + 14a – 5a – 7 = 0

2a (5a + 7) – 1 (5a + 7) = 0

(2a – 1) (5a + 7) = 0

a = 1/2, -7/5

12b = 144 – 84 = 60

b = 60/12 = 5

a < b.

Incorrect

10a^{2} + 14a – 5a – 7 = 0

2a (5a + 7) – 1 (5a + 7) = 0

(2a – 1) (5a + 7) = 0

a = 1/2, -7/5

12b = 144 – 84 = 60

b = 60/12 = 5

a < b.

Question 5 of 10

5. Question

I. a^{2} = 196;

II. b^{2} + 11b + 24 = 0

Correct

a = ±14

b2 + 8b + 3b + 24 = 0

b (b + 8) + 3 (b + 8) = 0

(b + 3) (b + 8) = 0

b = -3, -8

Hence there is no relation.

Incorrect

a = ±14

b2 + 8b + 3b + 24 = 0

b (b + 8) + 3 (b + 8) = 0

(b + 3) (b + 8) = 0

b = -3, -8

Hence there is no relation.

Question 6 of 10

6. Question

Directions:6-10) In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.

i) 2x^{2}+11x+12=0

ii) x^{2}+2x-35=0

Correct

a) 2x^{2}+11x+12=0

2x^{2}+8x+3x+12=0

(x+4)(2x+3)=0

x= -4 and -3/2

b) x^{2}+2x-35=0

x^{2}+7x-5x-35=0

(x+7)(x-5)=0

x=-7 or 5

Hence the relationship cannot be determined

Incorrect

a) 2x^{2}+11x+12=0

2x^{2}+8x+3x+12=0

(x+4)(2x+3)=0

x= -4 and -3/2

b) x^{2}+2x-35=0

x^{2}+7x-5x-35=0

(x+7)(x-5)=0

x=-7 or 5

Hence the relationship cannot be determined

Question 7 of 10

7. Question

i) x^{2}-20x+99=0

ii) 2y^{2}-15y+28=0

Correct

a) x^{2}-20x+99=0

x^{2}-9x-11x+99=0

(x-9)(x-11)=0

x=9 and 11

b) 2y^{2}-15y+28=0

2y^{2}-8y-7y+28=0

(y-4)(2y-7)=0

y= 7/2 and 4

Thus x>y

Incorrect

a) x^{2}-20x+99=0

x^{2}-9x-11x+99=0

(x-9)(x-11)=0

x=9 and 11

b) 2y^{2}-15y+28=0

2y^{2}-8y-7y+28=0

(y-4)(2y-7)=0

y= 7/2 and 4

Thus x>y

Question 8 of 10

8. Question

i) 2x+5y-√2601=0

ii) 7x+3y-77=0

Correct

After solving above equations we have x= 8 and y=7

Hence x>y

Incorrect

After solving above equations we have x= 8 and y=7