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.

Direction (Q. 1-10): In each of these questions, two equations (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
e) if x = y or no relation can be established between ‘x’ and ‘y’.

1. I. 63x -194 √x +143 = 0
II. 99y - 255 √y +150 = 0

1). ReI. 63x -194 √x +143 = 0
or 63x -117 √x - 77 √x +143 = 0
or (7 √x -13)(9 √x -11) = 0
x = 169/49, 121/81
II. 99y - 255 √y +150 = 0
or 99y - 90 √y -165 √y +150 = 0
or (11 √y -10)(9 √y -15) = 0
y = 100/121, 225/81
Therefore relation cannot be established between x and y.
Answer: E

2.I. x - 7 √(3x) + 36 = 0
II. y -12 √(2y) + 70 = 0

2).I. x - 7 √(3x) + 36 = 0
or x - 7 √3. √x + 36 = 0
or x - 3 √3 √x - 4 √3 √x + 36 = 0
or ( √x - 3 √3)(√ x - 4 √3) = 0
x = 27, 48
II. y - 5 √(2y) - 7 √(2y) + 70 = 0
or y - 5 √2 √y - 7 √2 √y + 70 = 0
or ( y - 5 √2)( y - 7 √2) = 0
y = 50, 98
x < y
Answer: B

3.I. x^2 - 7 √7x + 84 = 0
II. y^2 - 5 √5y + 30 = 0

3).I. x^2 - 7 √7x + 84 = 0
or (x - 4 √7)(x - 3 √7) = 0
x = 4 √7, 3 √7
II. y^2 - 5 √5y + 30 = 0
or (y - 2 √5)(y - 3 √5) = 0
y = 2 √5, 3 √5
x > y
Answer: A

4.I. 10x + 6y = 13
II. 45x + 24y = 56

4).I. 10x + 6y = 13
II. 45x + 24y = 56
On solving both equations,
x = 4/5, y = 5/6
x < y
Answer: B

5. I. 16x^2 – 40x – 39 = 0
II. 12y^2 – 113y + 255 = 0

5).I. 16x^2 - 40x - 39 = 0
or 16x^2 - 52x + 12x - 39 = 0
or (4x- 13) (4x + 3)
x = 13/4, -3/4
II. 12y^2 - 113y + 255 = 0
or 12y^2 - 45y - 68y + 255 = 0
or (4y - 15) (3y - 17) = 0
y = 15/4, 17/3
Therefore y > x or, x < y
Answer: B

6.I. 6x^2 + 13x = 12 – x
II. 1 + 2y^2 = 2y + 5y/6

6). I. 6x^2 + 14x = 12
3x^2 + 7x – 6 = 0
(x + 3) (3x – 2) = 0
x = – 3, 2/3
II. 1 + 2y^2 = 17y / 6
12y^2 – 17y + 6 = 0
12y^2 – 8y – 9y + 6 = 0
4y (3y – 2) – 3 (3y – 2) = 0
(3y – 2) (4y – 3) = 0
y = 2/3, 3/4
Hence, y ≥ x
Answer: D

7.I. 2x^2 + 5x + 1 = x^2 + 2x – 1
II. 2y^2 – 8y + 1 = – 1

7). I. 2x^2 + 5x + 1 = x^2 + 2x – 1
X^2 + 3x + 2 = 0
X^2 + 2x + x + 2 = 0
x (x + 2) + 1 (x – 2) = 0
(x + 2) (x + 1) = 0
x = –2, –1
II. 2y^2 – 8y + 1 = –1
2y^2 – 8y + 2 = 0
Y^2 – 4y + 1 = 0
Wkt, x = - b ± √(b2 – 4ac) / 2a
= +4 ± √ (16 – 4 × 1 × 1)/2 × 1
= 2 ± √12
= 2 ± 2 √3
Therefore relation cannot be established between x and y.
Answer: E

8.I. Y^2 + y – 1 = 4 – 2y – y^2
II. x^2/2 – 3x/2 = x - 3

8). I. 2y^2 + 3y – 5 = 0
2y^2 + 5y – 2y – 5 = 0
y (2y + 5) –1 (2y + 5) = 0
(2y + 5) (y – 1) = 0
y = -5/2, 1
II. x^2 – 3x = 2x – 6
X^2 – 5x + 6 = 0
X^2 – 3x – 2x + 6 = 0
x (x – 3) –2 (x – 3) = 0
(x – 3) (x – 2) = 0
x = 3, 2
Hence, x > y.
Answer: A

9.I. x^2/2 + x – 1/2 = 1
II. 3y^2 – 10y + 8 = y^2 + 2y – 10

9).I. x^2 + 2x – 1 = 2
X^2 + 2x – 3 = 0
x + 3x – x – 3 = 0
x (x + 3) – 1 (x + 3) = 0
(x + 3) (x – 1) = 0
x = –3 , 1
II. 2y^2 – 12y + 18 = 0
Y^2 – 6y + 9 = 0
(y – 3)^2 = 0
y = 3, 3
Hence, y > x
Answer: B

10.I. 4x^2 – 20x + 19 = 4x – 1
II. 2y^2 = 26y + 84

10).I. 4x^2 – 24x + 20 = 0
X^2 – 6x + 5 = 0
X^2 – 5x – x + 5 = 0
x (x – 5) –1 (x – 5) = 0
(x – 5) (x – 1) = 0
x = 5, 1
II. y^2 – 13y + 42 = 0
Y^2 – 7y – 6y + 42 = 0
y (y – 7) – 6 (y – 7) = 0
(y – 7) (y – 6) = 0
y = 7, 6
Hence, y > x
Answer: B