## 13 May 2017

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

[Dated: 13th 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 each question two equations numbered I and II are given. You have to solve both the equations and mark the answer.
a)   If x > y
b)   If x ≥ y
c)   If x < y
d)   If x ≤ y
e)   If x = y or no relation can be established between x and y.
1).
I. x2 + 18x + 72 = 0
II. y2 + 6y + 8 =0

2).
I. 8x2 – 22x + 15 = 0
II. 3y2 – 13y + 14 = 0

3).
I. 9x2 – 26x + 16 = 0
II. 3y2 – 16y + 20 = 0

4).
I. 10x2 – 17x + 7 = 0
II. 15y2 – 19y + 6 = 0

5).
I. 12x2 + 19x + 5 = 0
II. 5y2 + 16y + 3 = 0

6).
I. x2 – 8x + 15 = 0
II. 2y2 – 11y + 14 = 0
a)   x > y
b)   x = y or relationship can’t be established
c)   x ≤ y
d)   x ≥ y
e)   x < y

7).
I. x = √2916
II. y2 = 2916
a)   x < y
b)   x > y
c)   x ≤ y
d)   x ≥ y
e)   x = y or relationship can’t be established

8).
I. 3x2 + 35x + 88 = 0
II. y2 + 787 = 1316
a)   x > y
b)   x ≤ y
c)   x = y or relationship can’t be established
d)   x ≥ y
e)   x < y

9).
I. x2 – 14x + 45 = 0
II. y2 – 9y + 20 = 0
a)   x < y
b)   x > y
c)   x ≤ y
d)   x ≥ y
e)   x = y or relationship can’t be established

10).
I. x2 – 54 = 3x
II. y2 = 36
a)   x > y
b)   x ≤ y
c)   x ≥ y
d)   x > y
e)   x = y or relationship can’t be established

1). C)
I. x2 + 18x + 72 = 0
or, x2 + 12x + 6x + 72 =0
or, x(x + 12) + 6(x + 12) = 0
x = -9, -12
II. y2 + 6y + 8 =0
or, y2 + 4y + 2y + 8 =0
or, y(y + 4) + 2(y + 4) = 0
y = -2, -4
Hence x < y

2). C)
I. 8x2 – 22x + 15 = 0
or, 8x2 - 12x - 10x + 15 = 0
or, 4x(2x - 3) - 5(2x - 3) = 0
or, (4x - 5) (2x - 3) = 0
x = 5/4, 3/2
II. 3y2 – 13y + 14 = 0
or, 3y2 - 6y - 7y + 14 =0
or, 3y(y - 2) - 7(y - 2) = 0
or, (3y - 7) (y - 2) = 0
y = 7/3, 2
Hence x < y

3). D)
I. 9x2 – 26x + 16 = 0
or, 9x2 - 18x - 8x + 16 = 0
or, 9x(x - 2) - 8(x - 2) = 0
or, (9x - 8) (x - 2) = 0
x = 8/9, 2
II. 3y2 – 16y + 20 = 0
or, 3y2 - 6y - 10y + 20 =0
or, 3y(y - 2) - 10(y - 2) = 0
or, (3y - 10) (y - 2) = 0
y = 2, 10/3
Hence x ≤ y

4). A)
I. 10x2 – 17x + 7 = 0
or, 10x2 - 10x - 7x + 7 = 0
or, 10x(x - 1) - 7(x - 1) = 0
or, (10x - 7) (x - 1) = 0
x = 7/10, 1
II. 15y2 – 19y + 6 = 0
or, 15y2 - 10y - 9y + 6 =0
or, 5y(3y - 2) – (3y - 2) = 0
or, (3y - 2) (5y - 3) = 0
y = 3/5, 2/3
Hence x > y

5). E)
I. 12x2 + 19x + 5 = 0
or, 12x2 + 4x + 15x + 5 =0
or, 4x(3x + 1) + 5(3x + 1) = 0
or, (4x + 5) (3x + 1) = 0
x = -5/4, -1/3
II. 5y2 + 16y + 3 = 0
or,5y2 + 15y + y + 3 =0
or, 5y(y + 3) + 1(y + 3) = 0
or, (5y + 1) (y + 3) = 0
y = -1/5, -3
Hence no relationship can be established.

6). B)
I. x2 – 8x + 15 = 0
or, x2 - 5x - 3x + 15 =0
or, x(x - 5) - 3(x - 5) = 0
or, (x - 5) (x - 3) = 0
x = 3, 5
II. 2y2 – 11y + 14 = 0
or, 2y2 – 7y – 4y + 14 =0
or, 2y(y - 2) - 7(y - 2) = 0
or, (2y - 7) (y - 2) = 0
y = 7/2, 2
Hence no relationship can be established.

7). D)
I. x = √2916 = 54
II. y2 = 2916
y = ± 54
hence x ≥ y

8). C)
I. 3x2 + 35x + 88 = 0
or, 3x2 + 24x + 11x + 88 =0
or, 3x(x + 8) +11(x + 8) = 0
or, (3x + 11) (x + 8) = 0
x = -11/3, -8
II. y2 + 787 = 1316
y2 = 1316 – 787 = 529
y = √529 = ±23
Hence no relationship can be established.

9). D)
I. x2 – 14x + 45 = 0
or, x2 - 9x - 5x + 45 =0
or, x(x - 9) -5(x - 9) = 0
or, (x - 5) (x - 9) = 0
x = 5, 9
II. y2 – 9y + 20 = 0
or, y(y - 5) – 4 (y - 5) = 0
y = 5, 4
Hence x ≥ y

10). E)
I. x2 – 54 = 3x
Or, x2 – 9x + 6x – 54 = 0
Or, x (x - 9) +6 (x - 9) = 0
x = 9, -6
II. y2 = 36
y = ± 6

Hence no relationship can be established.