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

Give Answer

1.
I. 15a2 – 10a – 5 = 0;
II. 20b2 + 18b + 4 = 0

Ans:5
15 a2 + 15a – 5a –  5 = 0

15a (a + 1) – 5 (a + 1) = 0

(15a – 5) (a + 1) = 0

a = 1/3, -1

20b2 + 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.

2.
I. 26a2 – 88 = 16;
II. 3√b/8 – √b/4 = 1/√b

Ans:3
26a2 = 16 + 88 = 104

a2 = 104/26 = 4

a = ±2

3√b/8 – 2√b/8 = 1/√b

√b/8 = 1/√b

b = 8

a < b.

3.
I. 2a + 3b = 20;
II. 4a + 2b = 24

Ans:5
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.

4.
I. 10a2 + 9a -7 = 0;
II. 12b + 84 = 122

Ans:3
10a2 + 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.

5.
I. a2 = 196;
II. b2 + 11b + 24 = 0

Ans:5
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.

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.

6.
i) 2x2+11x+12=0
ii) x2+2x-35=0

Ans:5
a) 2x2+11x+12=0

2x2+8x+3x+12=0

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

x= -4 and -3/2

b) x2+2x-35=0

x2+7x-5x-35=0

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

x=-7 or 5

Hence the relationship cannot be determined.

7.
i) x2-20x+99=0
ii) 2y2-15y+28=0

Ans:1
a) x2-20x+99=0

x2-9x-11x+99=0

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

x=9 and 11

b) 2y2-15y+28=0

2y2-8y-7y+28=0

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

y= 7/2 and 4

Thus x>y

8.
i) 2x+5y-√2601=0
ii) 7x+3y-77=0

Ans:1
After solving above equations we have x= 8 and y=7

Hence x>y

9.
i) x2+19y+90=0
ii) 14y2-37y+24=0

Ans:2
a) x2+19y+90=0

x2+9y +10y+90=0

(x+9)(x+10)=0

x= -9 and -10

b) 14y2-37y+24=0

14y2-21y-16y+24=0

7y(2y-3)-8(2y-3)=0

y= 3/2 or 8/7

Hence x<y

10.
i) 5x2+36√3x+192=0
ii) 2y2+9√3y+27=0

Ans:2
a)5x2+36√3x+192=0

5x2+20√3x+16√3x+192=0

(5x+16√3)(x+4√3)=0

x=-4√3 or -16√3/5

b) 2y2+9√3y+27=0

2y2+6√3y+3√3y +27=0

(2y+3√3)(y+3√3)=0

y= -3√3 or -3√3/2

Thus x<y

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