25 May 2017

Practice Quantitative Aptitude Questions For NICL AO Prelims 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.

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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

Explanation With Answer Key:
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



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