28 Jun 2017

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.




00:00:00

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

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

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

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

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

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

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

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

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

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

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

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




        
Click here for Topic wise Discussion on Quantitative Aptitude Day-1 

English Learning Session :– (Day-1)
English Learning Session :– (Day-2)
English Learning Session :– (Day-3)
English Learning Session :– (Day-4)
English Learning Session :– (Day-5)
English Learning Session :– (Day-6)
English Learning Session :– (Day-7)

More Practice Aptitude Questions for Upcoming Exams  - Click Here