19 May 2017

Practice Quantitative Aptitude Questions For NICL AO Prelims 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.

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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. 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

Explanation With Answer Key:
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.



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