Practice Aptitude Questions For NICL AO Mains 2017(Quadratic Equation)

Practice Aptitude Questions For NICL AO Mains 2017 (Data Interpretation)

Practice Aptitude Questions For NICL AO Mains 2017 (Quadratic Equation):

Dear Readers, Important Practice Aptitude Questions for NICL AO Mains 2017 Exams was given here with Solutions. Aspirants those who are preparing for Bank exams & all other Competitive examination can use this material.

Directions (Q.1 - 10): In the following questions two equations numbered I and II are given. You have to solve both the equations and

Give answer a) if x > y
Give answer b) if x ≥ y
Give answer c) if x < y
Give answer d) if x ≤ y
Give answer e) if x = y or the relationship cannot be established
1.I. 3/(x+y) + 2/(x-y) = 2
II. 9/(x+y) – 4/(x-y) = 1

1). I. 3/(x+y) + 2/(x-y) = 2
II. 9/(x+y) – 4/(x-y) = 1
Let, x + y = P and x- y = Q
3/P + 2/Q = 2 ---- (i)
9/P – 4/Q = 1 ---- (ii)
On solving (i) × 3 and (ii) we get,
-10/Q = - 5
Q = 2.
Substituting the value of Q in (i), we get
3/P + 2/2 = 2
3/P = 2 – 1
P = 3
x + y = P = 3;
x - y = Q = 2
∴ x = 5/2 =2.5
y= 1/2 = 0.5
Hence, x > y.
Answer: A

2.I. x^7 – (28×7)^7.5 / (x)^(1/2) = 0
II. ³√y – 27 = 0

2).I. x^7 – [(28 × 7)^7.5 / x^(1/2)] = 0
x^7 × x^(1/2) - (28 × 7)^7.5 = 0
x^7.5 – (196)^7.5 = 0
x = 196
II. ³√y – 27 = 0
³√y = 27
∴ y = (27)^3
y = 19683
Hence, x < y.
Answer: C

3.I. (x^3 –6x^2 +11x -6)/(x -1) = 0
II. (2y^3 – 3y^2 -3y +2)/(y +1) = 0

3).I. (x^3 –6x^2 +11x -6)/(x -1) = 0
(x-1) (x^2 – 5x +6) = 0
x^2 – 5x +6 = 0
(x -3) (x -2) = 0
x = 3, 2
II. (2y^3 – 3y^2 -3y +2)/(y +1) = 0
( y+1) (2y^2- 5y +2) = 0
(2y-1) (2y-4) = 0
y = ½, 2
Hence, x ≥ y.
Answer: B

4.I. (4x^3 -7x +3)/(x -1) = 0
II. (y^3 -10y^2 –y +10)/(y +1) =0

4).I. (4x^3 -7x +3)/(x -1) = 0
(x-1) (4x^2 +4x-3) = 0
(4x-2) (4x+6) = 0
x=1/2, -3/2
II. (y^3 -10y^2 –y +10)/(y +1) =0
(y + 1) (y2 -11y +10) = 0
(y-1)(y-10) = 0
y=1,10
Hence, x < y.
Answer: c

5. I. (2x +3)^2 – 81 = 0
II. 3y^2 – 5y – 12 = 0

5). I. (2x +3)^2 – 81 = 0
(4x^2 +12x +9) -81 =0
4x^2 +12x -72 = 0
Divided by 4 we get,
x^2 +3x -18 = 0
(x+6) (x-3)= 0
x= -6, 3
II. 3y^2 – 5y – 12 = 0
(3y-9) (3y+4) = 0
y = 3, - 4/3
Hence, the relationship cannot be established.
Answer: E

6.I. x^2 – 4x – 8 = 0
II. y^2 – (√3 +1)y +√3 = 0

6). I. x^2 – 4x – 8 = 0

II. y^2 – (√3 +1)y +√3 = 0
y = √3, 1
Hence, the relationship cannot be established.
Answer: B

7. I. √(24-10x) = 3 - 4x
II. 6y^2 – 5y - 25 = 0

7). I. √(24-10x) = 3 - 4x
(24 -10x) = (3 - 4x)^2
24 -10x = 9 +16x^2 - 24x
16x^2 – 14x – 15 = 0
(2x-3) (8x+5) = 0
x= 3/2, -5/8
II. 6y^2 – 5y - 25 = 0
(3y+5) (2y-5) = 0
y= -5/3, 5/2
Hence, the relationship cannot be established.
Answer: E

8.I. (1296)^(1/4) x + √1024 = 188
II. y^2 – 595 = 81

8). I. (1296)^(1/4) x + √1024 = 188
(1296)^(1/4) x + 32 = 188
x=156/ (6^4)^(1/4)
x = 156/6 = 26
II. y^2 – 595 = 81
y^2 = 81+595
y^2 = 676
∴ y = √676 =± 26
Hence, x ≥ y.
Answer: B

9.I. 18x^2 + 6x – 4 = 0 II. 12y^2 – 10y- 8 = 0

9). 18x^2 + 6x – 4 = 0
9x^2 + 3x – 2 = 0
9x^2 + 6x – 3x – 2 = 0
(3x+2) (3x-1) = 0
x= - 2/3, 1/3
II. 12y^2 – 10y- 8 = 0
6y^2 – 5y – 4 = 0
6y^2 – 8y + 3y – 4 = 0
2y(3y - 4) +1 (3y - 4) = 0
(3y – 4) (2y + 1)
y = 4/3, -1/2
Hence, the relationship cannot be established.
Answer: E

10. I. √(625x^2) – 375 = 0
II. √(2116) y – 736 = 0

10). I. √(625x^2) – 375 = 0
25x – 375 = 0
25x=375
∴ x=15
II. √(2116) y – 736 = 0
46 y = 736
∴ y=16
Hence, x < y.
Answer: C