31 May 2017

Practice Quantitative Aptitude Questions For NICL AO & Upcoming Exams 2017 (Quadratic Equation & Simplification)

Practice Quantitative Aptitude Questions For NICL AO Prelims 2017
Practice Quantitative Aptitude Questions For NICL AO & Upcoming Exams 2017 (Quadratic Equation & Simplification):
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.

Directions (Q. 1-10): in each question two equations numbered I and II are given. You have to solve both the equations and mark the 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. x2 – 16 = 0
II. y4 –  {[(8x2)9/4] / √y } = 0

2).
I. x2 - 2√3x – 72 = 0
II. y2 + 9√3x + 60 = 0

3).
I. x6 – [(√(54×6))6.5 / √x] = 0
II. y – (36 × 81)1/2 = 0

4).
I. x2 – 17x – 234 = 0
II. 4y2 + 32y – 192 = 0

5).
I. x2 = 784
II. y2 – 11y – 102 = 0

Directions (Q. 6-10): What will come in place of question mark (?) in the following questions?
6). 3/8 of 4/9 of 1575 + (√?) – 32% of 786 = 66.98
a)   √56        
b)   56          
c)   (56)2      
d)   36          
e)   (36)2 

 7). 3√97336 + 2√6889 ÷ 2? – 3/4 of 65 + 52 = 43
a)   4            
b)   16          
c)   8            
d)   2            
e)   2.5         

8).  (23)3 × (22)2 × (82)3/2 / (22)6 = (4)?
a)   8            
b)   10          
c)   2.5                  
d)   6            
e)   5

9). (4 × 18)3.35 × (8)5.2 × (64)7.3 × (27 × 3) 9.9 = (72)?
a)   23.15     
b)   24.15     
c)   3.35       
d)   7.3                  
e)   20.75

10). 4 51/13 × 8 3/37 × 1 7/12 × [(16)2 + 3 ] = 23 × ?
a)   1141.584
b)   1151.584
c)   11415.84
d)   11515.84
e)   None of these

Explanation With Answer Key:
1). D)
I. x2 – 16 = 0
x2 = 16
x = ±4
II. y4 –  {[(8x2)9/4] / √y } = 0
y4 √y  –  [(42)9/4]  = 0
y (4+1/2) – (4)9/2 = 0
y9/2 – 49/2 = 0
y = 4
Hence x ≤ y

2). B)
I. x2 - 2√3x – 72 = 0
x2 - 6√3x + 4√3x – 72 = 0
x (x - 6√3) + 4√3 (x - 6√3) = 0
x = -4√3,  6√3
II. y2 + 9√3y + 60 = 0
y2 + 4√3y + 5√3y + 60 = 0
y(y + 4√3) + 5√3 (y + 4√3) = 0
y = -4√3,  -5√3
Hence x ≥ y

3). C)
I. x6 – [(√(54×6))6.5 / √x] = 0
x6+(1/2) – (18)6.5 = 0
x6.5 = 186.5
x = 18
II. y – (36 × 81)1/2 = 0
y – (6 × 9) = 0
y – 54 = 0
 y = 54
Hence x < y

4). E)
I. x2 – 17x – 234 = 0
x2 – 26x + 9x – 234 = 0
x(x - 26) + 9 (x – 26) = 0
(x + 9) (x - 26) = 0
x = -9, 26
II. 4y2 + 32y – 192 = 0
y2 + 32y – 768 = 0
y2 + 48y – 16y – 768 = 0
y (y + 48) -16 (y + 48)
y = -48, 16
Hence relation cannot be determined

5). E)
I. x2 = 784
x = ±28
II. y2 – 11y – 102 = 0
y2 – 17y + 6y – 102 = 0
y (y - 17) +6 (y - 17) = 0
y = 17, -6
Hence Relation cannot be determined

6). C)
3/8 × 4/9 × 1575 + (√?) – 32% of 786 = 66.98
1/2 × 525 + (√?) – (32/100) × 786 = 66.98
262.5 (100) + (√?) (100) – 25152 = 66.98(100)
26250 – 25152 + (√?) (100) = 6698
(√?) (100) = 5600
? = (56)2

7). D)
3√97336 + 2√6889 ÷ 2? – 3/4 of 65 + 52 = 43
46 + 83/2? – 48.75 + 25 = 43
83/20.75 = 2?
4 = 2?
? = 2

8). E)
(23)3 × (22)2 × (82)3/2 / (22)6 = (4)?
(4)? = 29 × 24 × 83 / 212
(4)? = 213 – 12 ×83
(4)? = (4)5
? = 5

9). A)
 (72)? = (4 × 18)3.35 × (8)5.2 × (64)7.3 × (27 × 3) 9.9
(72)? = (4  × 2  × 9)3.35  × (8)5.2 × (8 ×8)7.3  × (9 ×9)9.9
(72)? = (8)3.35+5.2+7.3+7.3 × (9)3.35+9.9+9.9
(72)? = (72)23.15

10). A)
23 x = 103/ 13 x 299/ 37 x 19/12 x 259

x=1141.584




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