## 22 Jul 2016

### IBPS PO/Upcoming Exams 2016-Aptitude Questions (Quadratic Equations)

IBPS PO/Upcoming Exams 2016-Aptitude Questions (Quadratic Equations) Set-3:

Dear Readers, Important Practice Aptitude Questions for IBPS PO and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

Directions (1-5): For the two given Equations I and II. You have to solve both the equations.

a)   If p is greater than q
b)   If p is smaller than q
c)   If p is equal to q
d)   If p is either equal to or greater than q
e)   If p is either equal to or smaller than q

1).
I. p2 +5p + 6 = 0
II.q2 + 3q + 2 = 0

2).
I. p2 = 4
II.q2 + 4q = -4

3).
I. p2 + p = 56
II.q2 – 17q + 72 =0

4).
I. 3p + 2q – 58 = 0
II. 4q + 4p = 92

5).
I. 3p2 + 17p + 10 =0
II. 10q2 + 9q + 2 = 0

Directions (6-10): For the two given Equations I and II. You have to solve both the equations.
a)   If x>y
b)   If x≥y
c)   If x<y
d)   If x≤y
e)   If x=y or the relationship cannot be established

6).
I. √x - √6/√x=0
II. y3 – 6(3/2) = 0

7).
I. 3x – 2y = 10
II. 5x – 6y = 6

8).
I. x2 + x – 12 = 0
II.y2 – 5y + 6 = 0

9).
I. x2 + 9x + 18 = 0
II.y2 – 13y + 40 = 0

10).
I. √(x+6) = √121 - √36
II. y2 + 112 = 473

1).e)   2).d)   3).b)   4).a)   5).b)   6).e)  7).a)   8).e)  9).c)   10).b)

SOLUTION:
1).
I.p2 + 3p + 2p + 6 = 0
à p (p+3) + 2 (p+3) = 0
à (p+3)(p+2) = 0
à p = -2 or -3
II. q2 + q + 2q + 2 = 0
à q(q+1) + 2 (q+1) = 0
à (q+1)(q+2)=0
à q=-1 or -2
Clearly, p≤q

2).
I. p = ±2
II. q2 + 2q + 2q + 4 = 0
à q(q+2) + 2(q+2)=0
à (q+2)(q+2) = 0
à q = -2
Clearly, p≥q

3). I. p2 + p – 56 = 0
à p2 + 8p – 7p – 56 = 0
à p(p+8)-7(p+8)=0
à (p+8)(p-7)=0
à p = 7 or -8
II. q2-8q-9q+72 = 0
à q(q-8) – 9(q-8) = 0
à (q-8)(q-9) = 0
q = 8 or 9
Clearly, p<q

4). We have,
3p + 2q = 58 ……….. (i)
4p + 4q = 92
à 2p+2q = 46  ………(ii)
By Eq. (i) – Eq. (ii), we get p = 12
From eq. (i), 3 × 12 + 2q = 58
à 2q = 58 – 36 = 22
q = 11
Hence, p>q

5). I. 3p2 + 15p + 2p + 10 = 0
à 3p (p+5) + 2 (p+5)=0
à (p+5) (3p+2) = 0
à p = -5 or -2/3
II. à10q2 + 5q + 4q + 2 = 0
5p (2q+1) + 2 (2q+1) = 0
(2q+1) (5q+2) = 0
q= -1/2 or -2/5
clearly, p<q

6). I. √x - √6/√x = 0
x-√6 = 0
x=√6
II. y3 = 63/2 = (√6)3 à y =√6
Clearly x=y

7). I. By Eq. I × 3 – Eq. II,
9x - 6y – 5x + 6y = 30 – 6
4x = 24 à x = 6
From Eq. I, 3 × 6 – 2y = 10
à 2y = 18 – 10 = 8 à y =4
Clearly, x>y

8). I. x2 + x – 12 = 0
à x2 + 4x – 3x – 12 = 0
x(x+4) – 3 (x+4) = 0
(x-3) (x+4) = 0
x=3 or -4
II. y2 – 5y + 6 =0
y2 – 3y – 2y + 6 =0
y(y-3) – 2(y-3) = 0
y = 2 or 3

9). I. x2 + 9x + 18 = 0
x2 + 6x + 3x +18 = 0
x(x+6) + 3(x+6) = 0
(x+3) (x+6) =0
x=-3 or -6
II. y2 – 13y + 40 = 0
y2 – 8y – 5y + 40 = 0
y(y-8) – 5(y-8) = 0
(y-5)(y-8) = 0
y=5 or 8
Clearly x<y