Directions:1-5) In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
1.

I.x2 +11x +18 = 0
II.y2 +16y+ 48 = 0

Ans:5
I)x2 +11x +18 = 0
x2 +2x+9x+18=0
x(x+2)+9(x+2)=0
(x+2) (x+9)=0
x=-2,-9y2 +16 y+ 48 = 0
y2+4y+12y+48=0
y(y+4) + 12(y+4)=0
(y+4)(y+12)=0
y=-4,-12
So relationship between x and y cannot be established

2.

I.3x2 +10x +7 = 0
II.3y2 +4y+ 1 = 0

Ans:4
I.3x2 +10x +7 = 0
3x2 +3x+7x+7=0
3x(x+1) + 7(x+1)=0
(x+1) (3x+7)=0
x=-1,-7/3II.3y2 +4y+ 1 = 0
3y2 + 3y+y+1=0
3y(y+1) + 1(y+1) =0
(y+1) (3y+1)=0
y=-1,-1/3
So x ≤ y

3.

I.5x2 -7x +2 = 0
II.2y2 -7y+ 3 = 0

Ans:5
I.5x2 -7x +2 = 0
5x2-5x-2x+2=0
5x(x-1) -2(x-1)=0
(x-1) (5x-2)=0
x=1,2/5II.2y2 -7 y+ 3 = 0
2y2-6y-y+3=0
2y(y-3) -1(y-3)=0
(y-3) (2y-1)=0
y=3,1/2
Relationship between x and y cannot be established

4.

I.2x2 -3x +1 = 0
II.y2-8 y+ 15 = 0

Ans:2
I.2x2 -3x +1 = 0
2x2 -2x-x +1=0
2x(x-1)-1(x-1)=0
(x-1) (2x-1)=0
x=1,1/2II.y2-8 y+ 15 = 0
y2-3y-5y+15=0
y(y-3)-5(y-3)=0
(y-3)(y-5)=0
y=3,5
So x <y

5.

I)x = (-4)2
II)y2=256

Ans:3
I)x = (-4)2x=16

II)y2=256

y=16 and -16

Hence x ≥ y

Directions:6-10):Two equations (I) and (II) are given in each question. On the basis of these questions, you have to decide the relation between p and q and give answer:

6.

  1. 5p2– 87p + 378 = 0
  2. 3q2– 49q + 200 = 0

Ans:1
I.5p2 – 45p – 42p + 378 = 05p (p – 9) – 42(p- 9) = 0

(5p – 42)(p -9) = 0

p = 9, 42/5 = 9, 8.4

II.3q2 – 24q – 25q + 200=0

3q (q – 8) -25(q – 8) = 0

(q – 8)(3q-25)=0

q = 8, 25/3 = 8, 8.33

Hence, p>q

7.

  1. 10p2– p – 24 = 0
  2. q2– 2q = 0

Ans:5
I.10p2 +15p – 16p – 24 = 05p(2p + 3) – 8(2p+ 3)=0

(2p + 3)(5p – 8) = 0

p = -3/2, 8/5 = -1.5, 1.6

II.q2– 2q = 0

q(q -2) = 0

q = 0, 2

(i.e.) no relationship exists between p and q.

8.

  1. p2– 5p + 6 = 0
  2. 2q2– 15q + 27 = 0

Ans:4
p2 – 2p – 3p + 6 =0p(p -2) – 3(p -2) = 0

(p -2) (p -3) = 0

p =2, 3

2q2  – 6q – 9q + 27 = 0

2q(q – 3)-9(q -3) = 0

(q- 3) (2q -9) = 0

q = 3, 9/2 = 3, 4.5

Hence, p ≤ q

9.

  1. 3p + 2q = 301
  2. 7p – 5q = 74

Ans:2
I.eqn (I) × 5 + eqn (II) × 2[15p + 10q = 1505] + [14p – 10q = 148] = 29p = 1653

p = (1653/29) = 57

And q = 65

Hence, p< q

10.

  1. 14p2– 37p + 24 = 0
  2. 28q2– 53q + 24 = 0

Ans:3
14p2 – 37p + 24 = 014p2 – 21p- 16p + 24 = 0

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

(2p – 3)(7p – 8) = 0

p = (3/2), (8/7)

II.28q2 – 53q + 24 = 0

28q2  -21q – 32q + 24 =0

7q(4q – 3) -8(4q – 3) = 0

(7q – 8) (4q – 3) = 0

q = 8/7, 3/4

p ≥ q

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