## 19 May 2017

### Practice Quantitative Aptitude Questions For NICL AO Prelims 2017

[Dated: 19th May] Practice Quantitative Aptitude Questions For NICL AO Prelims 2017 (Quadratic Equation):
Dear Readers, Important Practice Aptitude Questions for  Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.

Directions (Q. 1-10): In the following questions two equations numbered I and II are given. Solve both the equations and give answer.
a)   If x < y
b)   If x ≥ y
c)   If x ≤ y
d)   If x = y or no relationship can be established
e)   If x > y
1).
I. 2x2 - 6√2 x + 8 = 0
II. 2y2 - 7√2 y + 12 = 0

2).
I. x2 – 2x – 35 = 0
II. y2 – 15y + 26 = 0

3).
I. 2x2 + 17x + 36 = 0
II. y2 + 21y +108 =0

4).
I. x2 – 7x -60 = 0
II. 4y2 -13y -12 = 0

5).
I. x2 – 54 = 3x
II. y2 = 36

6).
I. x = √7921
II. y = 3√704969

7).
I. x2 + 9x + 20 = 0
II. y2 + 5y + 6 = 0

8).
I. 2x + 4y = 13
II. 4x + 2y = 14

9).
I. x2 + 11x + 30 = 0
II. y2 + 13y +42 = 0

10).
I. x2 + 29x = -210
II. y2 + 28y = -195

1). D)
I. 2x2 - 6√2 x + 8 = 0
or, x2 - 3√2 x + 4 = 0
or, x2 – 2 √2 x - √2 x + 4 = 0
or, x (x - 2√2) -√2 (x -2√2) = 0
or, (x - √2) (x - 2√2)
x = √2, 2√2
II. 2y2 - 7√2 y + 12 = 0
or, 2y2 - 4√2y - 3√2y + 12 = 0
or, 2y (y - 2√2) - 3√2 (y - 2√2) = 0
or, (2y - 3√2) (y – 2√2) = 0
or, y = 3√2 / 2, 2√2
Hence no relation can be established.

2). D)
I. x2 – 2x – 35 = 0
or, x2 – 7x + 5x - 35 = 0
or, x (x - 7) + 5 (x - 7) = 0
or, (x - 7) (x + 5) = 0
x = 7, x = - 5
II. y2 – 15y + 26 = 0
or, y2 – 13y – 2y +26 = 0
or, y (y - 13) - 2 (y - 13) = 0
or, (y - 2) (y - 13) = 0
y = 2, 13
Hence relationship can't be established.

3). E)
I. 2x2 + 17x + 36 = 0
or, 2x2 + 9x + 8x + 36 = 0
or, 2x (x + 4) + 9 (x + 4) = 0
or, (x + 4) (2x + 9) = 0
x = -9/2,  -4
II. y2 + 21y +108 =0
or, y2 + 12y + 9y +108 = 0
or, y (y + 12) + 9 (y + 12) = 0
or, (y + 9) (y + 12) = 0
y = -9,  -12.
Hence x > y

4). D)
I. x2 – 7x -60 = 0
or, x2 – 12x + 5x – 60 = 0
or, x(x - 12) + 5 (x - 12) = 0
or, (x+5) (x-12) = 0
x = -5,  12
II. 4y2 -13y -12 = 0
or, 4y2 – 16y + 3y – 12 = 0
or, 4y (y - 4) + 3(y - 4) = 0
or, (4y+3)(y-4) = 0
y = -3/4,  4
Hence relationship can't be established.

5). D)
I. x2 – 54 = 3x
or, x2 – 3x – 54 = 0
or, x2 – 9x + 6x – 54 = 0
or, x (x - 9) + 6(x - 9) = 0
or, (x + 6) (x - 9) = 0
x = 9,   -6
II. y2 = 36
y = ±6
Hence relationship can't be established.

6). D)
I. x = 89
II. y = 89
Therefore x = y.
Note: x = √7921 gives only +ve value.
Do not confuse it with x2 = 7921. In this case, when we find value of x, then x = ±81.
But x = √7921 means only +ve value is given.

7). A)
I. x2 + 9x + 20 = 0
à x2 + 9x + 20 = (x+4) (x+5) = 0
x = -4, -5
II. y2 + 5y + 6 = 0
à y2 +5y +6 = (y+2) (y+3) = 0
y = -2, -3
Therefore x < y.

8). E)
I. 2x + 4y = 13
II. 4x + 2y = 14
To solve both the equations
x = 2.5 and y = 2
x > y.

9). B)
I. x2 + 11x + 30 = 0
à (x+5) (x+6) = 0
X = -5, -6
II. y2 + 13y +42 = 0
y = -6, -7
Therefore x ≥ y.

10). C)
I. x2 + 29x = -210
= x2 + 29x + 210 = 0
= (x +14) (x+15) = 0
x = -14, -15
II. y2 + 28y = -195
= y2 +28y + 195 = 0
= (y+13) (y+15) = 0
y = -13, -15
Therefore, x ≤ y.