## 25 May 2017

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

[Dated: 25th 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. 3x2 + 11x + 6 = 0
II. 3y2 + 10y + 8 = 0

2).
I. 3x2 - 7x + 2 = 0
II. 2y2 - 9y + 10 = 0

3).
I. x2 = 9
II. 2y2 - 19y + 44 = 0

4).
I. 2x2 - 15x + 28 = 0
II. 4y2 - 23y + 30 = 0

5).
I. 2x2 - 15x + 27 = 0
II. 5y2 - 26y + 33 = 0

6).
I. 6x2 + 25x + 24 = 0
II.12y2 + 13y + 3 = 0

7).
I. 12x2 – x - 1 = 0
II. 20y2 - 41y + 20 = 0

8).
I. 10x2 + 33x + 27 = 0
II. 5y2 + 19y + 18 = 0

9).
I. 15x2 - 29x + 14 = 0
II. 6y2 - 5y - 25 = 0

10).
I. 3x2 - 22x + 7 = 0
II. y2 - 20y + 91 = 0

1). D)
I. 3x2 + 11x + 6 = 0
or, 3x2 + 9x + 2x + 6 = 0
or, 3x (x +3 ) + 2 (x + 3) = 0
or, (3x + 2) (x + 3) = 0
x = -2/3, x = - 3
II. 3y2 + 10y + 8 = 0
or, 3y2 + 6y + 4y + 8 = 0
or, 3y (y +2) + 4 (y + 2) = 0
or, (3y + 4) (y + 2) = 0
y = -4/3, y=- 2
x = y or no relationship can be established.

2). C)
I. 3x2 - 7x + 2 = 0
or, 3x2 - 6x - x + 2 = 0
or, 3x (x -2) - 1 (x - 2) = 0
or, (3x - 1) (x - 2) = 0
x = 1/3, x = 2
II. 2y2 - 9y + 10 = 0
or, 2y2 – 5y - 4y + 10 = 0
or, y (2y - 5)  -2 (2y - 5) = 0
or, (2y - 5) (y - 2) = 0
y = 5/2, y= 2
Hence x ≤ y

3). A)
I. x2 = 9
x = ±3
II. 2y2 - 19y + 44 = 0
or, 2y2 – 11y - 8y + 44 = 0
or, y (2y - 11)  -4 (2y - 11) = 0
or, (2y - 11) (y - 4) = 0
y = 11/2, y= 4
Hence x < y

4). D)
I. 2x2 - 15x + 28 = 0
or, 2x2 - 7x - 8x + 28 = 0
or, x (2x -7) - 4 (2x - 7) = 0
or, (2x - 7) (x - 4) = 0
x = 7/2 à 3.5, x = 4
II. 4y2 - 23y + 30 = 0
or, 4y2 – 15y - 8y + 30 = 0
or, y (4y - 15)  -2 (4y - 15) = 0
or, (4y - 15) (y - 2) = 0
y = 15/4 à3.8, y= 2
Hence x = y or no relationship can be established.

5). B)
I. 2x2 - 15x + 27 = 0
or, 2x2 - 9x - 6x + 27 = 0
or, x (2x -9) - 3 (2x - 9) = 0
or, (2x - 9) (x - 3) = 0
x = 9/2 à 4.5, x = 3
II. 5y2 - 26y + 33 = 0
or, 5y2 – 15y - 11y + 33 = 0
or, 5y (y - 3)  -11 (y - 3) = 0
or, (5y - 11) (y - 3) = 0
y = 11/5 à2.2, y= 3
Hence x ≥ y

6). A)
I. 6x2 + 25x + 24 = 0
or, 6x2 + 9x + 16x + 24 = 0
or, 3x (2x +3) + 8 (2x + 3) = 0
or, (2x + 3) (3x + 8) = 0
x = - 3/2 à - 1.5, x = - 8/3 à - 2.6
II.12y2 + 13y + 3 = 0
or, 12y2 + 9y +4y + 3 = 0
or, 3y (4y +3) +1 (4y + 3) = 0
or, (3y +1) (4y + 3) = 0
y = - 1/3  à -0.9,  y= - 3/4 à -0.75
Hence x < y

7). A)
I. 12x2 – x - 1 = 0
or, 12x2 -4x + 3x -1 = 0
or, 4x (3x -1) + 1 (3x-1) = 0
or, (4x + 1) (3x - 1) = 0
x = - 1/4 à - 0.25, x = 1/3 à 0.3
II. 20y2 - 41y + 20 = 0
or, 20y2 - 16y -25y + 20 = 0
or, 4y (5y -4) -5 (5y-3) = 0
or, (5y - 4) (4y - 5) = 0
y = 4/5  à 0.8,  y= 5/4 à 1.25
Hence x < y

8). B)
I. 10x2 + 33x + 27 = 0
or, 10x2 +15x + 18x +27 = 0
or, 5x (2x +3) + 9 (2x + 3) = 0
or, (5x + 9) (2x +3) = 0
x = - 9/5 à - 1.8, x = 3/2 à -1.5
II. 5y2 + 19y + 18 = 0
or, 5y2 + 10y +9y + 18 = 0
or, 5y (y + 2) +9 (y + 2) = 0
or, (5y + 9) (y +2) = 0
y = - 9/5  à 0.8,  y= - 2
Hence x ≥ y

9). D)
I. 15x2 - 29x + 14 = 0
or, 15x2 - 15x - 14x + 14 = 0
or, 15x (x  - 1) - 14 (x - 1) = 0
or, (15x - 14) (x - 1) = 0
x = 14/15 à 0.9,  x = 1
II. 6y2 - 5y - 25 = 0
or, 6y2 - 15y + 10y - 25 = 0
or, 3y (2y - 5) +5 (2y - 5) = 0
or, (3y + 5) (2y - 5) = 0
y = -5/3  à -1.6,  y= 5/2 à 2.5
Hence x = y or no relationship can be established.

10). C)
I. 3x2 - 22x + 7 = 0
or, 3x2 - 21x - x + 7 = 0
or, 3x (x  - 7) - 1 (x - 7) = 0
or, (3x - 1) (x - 7) = 0
x = 1/3 à 0.3,  x = 7
II. y2 - 20y + 91 = 0
or, y2 - 13y -7y + 91 = 0
or, y (y - 13) -7 (y - 13) = 0
or, (y - 13) (y - 7) = 0
y = 13 , y= 7
Hence x ≤ y