## 15 Jun 2017

### Practice Aptitude Questions For NICL AO & RBI Grade B 2017 (Quadratic Equation)

Practice Aptitude Questions For NICL AO & RBI Grade B 2017 (Quadratic Equation):
Dear Readers, Important Practice Aptitude Questions for NICL AO & RBI Grade B 2017 Exams was given here with Solutions. Aspirants those who are preparing for Bank exams & all other Competitive examination can use this material.

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Directions (Q. 1-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give 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. 3x^2 – 29x + 56 = 0
II. 3y^2 – 5y – 8 = 0

(a)
(b)
(c)
(d)
(e)

1). I. 3x^2 - 29x + 56 = 0
or 3x^2 - 21x - 8x + 56 = 0
or 3x(x - 7) - 8(x - 7) = 0
or (3x - 8) (x - 7) = 0
x = 8/3, 7
II. 3y^2 - 5y - 8 = 0
or 3y^2 + 3y - 8y - 8 = 0
or 3y(y + 1) - 8(y + 1) = 0
or (3y - 8) (y + 1) = 0
or (3y - 8) (y + 1) = 0
y = -1, 8/3
x ≥ y
2.I. 15x^2 – 41x + 14 = 0
II. 2y^2 – 13y + 20 = 0

(a)
(b)
(c)
(d)
(e)
2).I. 15x^2 - 41x + 14 = 0
or 15x^2 - 6x - 35x + 14 = 0
or 3x(5x - 2) - 7(5x - 2) = 0
or (3x - 7)(5x - 2) = 0
x = 7/3, 2/5
II. 2y^2 - 13y + 20 = 0
or 2y^2 - 8y - 5y + 20 = 0
or 2y(y - 4) - 5(y - 4) = 0
or (2y - 5) (y - 4) = 0
y = 4, 5/ 2
x < y

3.I. 7x – 4y = 40
II. 8x + 8y = 8

(a)
(b)
(c)
(d)
(e)
3).7x - 4y = 40 ...(i)
and 8x + 8y = 8
or x + y = 1 ...(ii)
Solving (i) and (ii), we have
x = 4, y = - 3
x > y

4.I. x^2 – 7x = 0
II. 2y^2 + 5y + 3 = 0

(a)
(b)
(c)
(d)
(e)
4).I. x^2 - 7x = 0
or x (x - 7) = 0
x = 0, 7
II. 2y^2 + 5y + 3 = 0
or 2y^2 + 2y + 3y + 3 = 0
or 2y(y + 1) + 3(y + 1) = 0
or (2y + 3) (y + 1) = 0
y = - 1, - 3/2
x > y
5.I. 5x^2 + 26x – 24 = 0
II. 5y^2 – 34y + 24 = 0

(a)
(b)
(c)
(d)
(e)
5).I. 5x^2 + 26x - 24 = 0
or 5x^2 + 30x - 4x - 24 = 0
or 5x(x + 6) - 4(x + 6) = 0
or (5x - 4) (x + 6) = 0
x = 4/5, - 6
II. 5y^2 - 30y - 4y + 24 = 0
or 5y(y - 6) - 4(y - 6) = 0
or (5y - 4) (y - 6) = 0 4
y = 4/5, 6
x ≤ y

6.I. 6x - 5y = -47
II. 5x + 3y = 11

(a)
(b)
(c)
(d)
(e)
6). Equation (I) × 3 = 18x - 15y = -141 Equation (II) × 5 = 25x + 15y = 55
43x = -86
X = - 86/43 = - 2
5x + 3y = 11
3y = 11 - 5x
3y = 11 + 10
3y = 21 y = 7
x < y
7. I. x^2 - 26x + 168 = 0
II. y^2 - 25y + 156 = 0

(a)
(b)
(c)
(d)
(e)
7).I. x^2 - 26x + 168 = 0
x^2 - 12x - 14x + 168 = 0
x (x - 12) - 14(x - 12) = 0
(x - 12) (x - 14) = 0
x = 12, 14
II. y^2 - 25y + 156 = 0
y^2 - 13y - 12y + 156 = 0
y (y - 13) - 12(y - 13) = 0
(y - 12) (y - 13) = 0
y = 12, 13
Hence, no relation can be established between x and y
8. I. x = √1764
II. y^2 = 1764

(a)
(b)
(c)
(d)
(e)
8).I. x = √1764
x = 42
II. y^2 = 1764
y = ± 42
x ≥ y
9.I. 2.3x - 20.01 = 0
II. 2.9y - x = 0

(a)
(b)
(c)
(d)
(e)
9).I. 2.3x - 20.01 = 0
X = 20.01 / 2.3 = 8.7
II. 2.9y - x = 0
or, x = 2.9y
y = 8.7 /2.9 = 3
x > y

10.I. x^2 - 13x + 42 = 0
II. y^2 + y - 42 = 0

(a)
(b)
(c)
(d)
(e)
10).I. x^2 - 13y + 42 = 0
x^2 - 6x - 7x + 42 = 0
x(x - 6) - 7(x - 6) = 0
(x - 6) (x - 7) = 0
x = 6, 7
II. y^2 + y - 42 = 0
y^2 + 7y - 6x - 42 = 0
y(y + 7) - 6(y + 7) = 0
(y - 6)(y + 7) = 0
y = 6, - 7
x ≥ y

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