### Practice Aptitude Questions With Solutions for SSC CGL,CHSL / Railway Exam

Practice Aptitude Questions With Solutions Railway / SSC CGL, CHSL Exams Set-17:
Dear Readers, Important Aptitude Questions for Railway / SSC CGL,CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1). If 9√x = √12 + √147, then x = ?
a)   2
b)   3
c)   4
d)   5

2).  3√1 – (127 / 343) is equal to
a)   5 / 9
b)   1 – (1 / 7)
c)   4 / 7
d)   1 – (2 / 7)

3). √(6 + √(6 + √6 + … )) = ?
a)   2.3
b)   3
c)   6
d)   6.3

4). If √2n = 64, the value of n is
a)   12
b)   6
c)   4
d)   2

5). If x = √3 + √2, then the value of (x + (1 / x)) is
a)   2
b)   3
c)   2√2
d)   2√3

6). A rationalizing factor of (3√9 - 3√3 + 1)  is
a)   3√3 - 1
b)   3√3 + 1
c)   3√9 + 1
d)   3√9 - 1

7). If √5 = 2-24 and √6 = 2.45, then the value of √(2 / 3) + √(5 / 6) is
a)   1.37
b)   1.57
c)   1.73
d)   1.75

8). If a = 0.1039, then the value of √(4a2 – 4a + 1 + 3a ) is
a)   0.1039
b)   0.2078
c)   1.1039
d)   2.1039

9). The largest number of five digits when it is a perfect square is
a)   99967
b)   99764
c)   99856
d)   99999

10). The least number to be added to 269 to make it a perfect square is
a)   31
b)   16
c)   7
d)   20

1). b) 2). b) 3). b) 4). a) 5). d) 6). b) 7). c) 8). c) 9). c) 10). d)

Solution:

1).   9√x = √12 + √147
=> 9√x = √4 × 3 + √49 × 3
=> 9√x = 2√3 + 7√3
=> 9√x = 9√3
=> √x = √3
On squaring both sides, we get x = 3

2). 3√1 – (127 / 343) = 3√(343 – 127) / 343
3√(216 / 343)
= (6 / 7) = 1 – (1 / 7)

3). Let x = √(6 + √(6 + √6 + … ))
=> x = √6 + x
=> x2 = 6 + x
=> x2 - 6 – x = 0
=> x2 – 3x + 2x – 6 = 0
=> x(x – 3) + 2(x – 3) = 0
=> (x + 2) = 0
=> x = -2 (does not exist) and (x – 3) = 0
=> x = 3
Hence ? = 3

4). √2n = 64
On squaring both sides, we get
2n = 64 × 64 = (2)6 × (2)6
2n = 212
:. n = 12

5). x = √3 + √2
(1 / x) = [1 / (√3 + √2)] Rationalizing (1 / x)
= [1 / (√3 + √2)] × [(√3 - √2) / (√3 - √2)]
= [(√3 - √2) / (3 – 2)] = √3 - √2
:. x + (1 / x) = √3 + √2 + √3 - √2 = 2√3

6). 3√9 - 3√3 + 1 = (3)2 / 3 – (3)1 / 3  + (1)2 / 3
:. (3√3 + 1) (3√9 - 3√3 + 1) = (31 / 3 )3 + 1
= 3 + 1 = 4
[∴ (a3 + b3) = (a + b) (a2 – ab + b2)]
Rationalizing factor = 3√3 + 1

7). √(2 / 3) + √(5 / 6) = √(2 / 3) + √(5 / 6)
= [(√2 ×√2) / (√3 ×√2)] + (√5 / √6)
= [(2 + √5) / √6] = [(2 + 2.24) / 2.45]
= (4.24 / 2.25) = 1.73

8). √(4a2 – 4a + 1 + 3a )
=√((1)2 – 2 × 2a + (2a)2 + 3a)
= √(1 – 2a)2 + 3a
= 1 – 2a + 3a = 1 + a
= 1 + 0.1039 = 1.1039