Practice Aptitude Questions With Solutions for SSC CGL,CHSL,CPO Exam

Practice Aptitude Questions With Solutions SSC CGL, CHSL Exams Set-24:
Dear Readers, Important Aptitude Questions for SSC CGL,CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).  [13 (4 / 7) ×13 (4 / 7) - 8 (3 / 7) × 8 (3 / 7) ] / [13 (4 / 7) - 8 (3 / 7)]  to get
a)   18
b)   22
c)   20
d)   24

2). If  x = 1 + √2 + √3, then the value of (2x4 – 8x3 – 5x+ 26x – 28) is
a)   6 √6
b)   0
c)   3 √6
d)   2√6

3). If  x [(3) – 2 / x] = 3 / x, x ≠ 0, then the value of x2 + (1 / x) is
a)   2 (1 / 3)
b)   2 (2 / 3)
c)   2 (4 / 9)
d)   2 (5 / 9)

4). If  x – (1 / x) = 4, then [x + (1 / x)] is equal to
a)   5√2
b)   2√5
c)   4√2
d)   4√5

5).[(50)3 + (-30)3 + (-20)3 ] is equal to
a)   170000
b)   -15000
c)   90000
d)   - 90000

6).The value of (0.98)3 + (0.02)3 + 3 x 0.98 x 0.02 – 1 is
a)   1.98
b)   1.09
c)   1
d)   0

7).If x = 0.5 and y = 0.2, then the value of √0.6 × (3y)x is
a)   1.0
b)   0.5
c)   0.6
d)   1.1

8).(137 × 137 + 137 × 133 + 133 × 133) / (137 × 137 × 137 -  133 × 133 × 133) is equal to
a)   4
b)   270
c)   1 / 4
d)   1 / 270

9).The value of  (1 / 2) + [ (1 / 2) × (1 / 2) ] / { (1 / 2) × (1 / 2) / (1 / 2) + ( ( 1 / 2) / (1 / 2) ) } ] is
a)   2 / 3
b)   1 / 2
c)   1 / 4
d)   1 / 5

10).(8 / 125)-4/3 simplifies to
a)   625 / 16
b)   625 / 8
c)   625 / 32
d)   16 / 625

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

Solution:

1). A = 13(4 / 7) and b = 8(3 / 7)
Expression = (a × a – b × b) / (a – b)
=(a2 – b2) / (a – b) = 13(4 / 7) + 8(3 / 7) = 22
Answer:  b)

2). x = 1 + √2 + √3
x - 1 = √2 + √3
On squaring both sides, we get
x2 – 2x + 1 = 2 + 3 + 2√6
x2 – 2x  = 4 + 2√6                     …(i)
x2 – 2x  - 4 = 2√6
on squaring both sides, we get
x4 + 4x2 + 16 – 4x3 + 16x – 8x2 = 24
x4 – 4x3 - 4x2 + 16x – 8 = 0
2x4 – 8x3 - 8x2 + 32x – 16 = 0
2x4 – 8x3 - 5x2 + 26x – 28 = 0
= 3x– 6x – 12 = 3(x2 – 2x) – 12
= 3(4 + 2√6) – 12
= 12 + 6√6 – 12 =    6√6
Answer: a)

3). Given, x[3 – (2 / x)] = (3 / x) (x ≠ 0)
3x – (3 / x) = 2
3(x – (1 / x) )  = 2
x – (1 / x) = 2 / 3
On squaring both sides, we get
x2 + (1 / x2) – 2 = 4 / 9
x2 + (1 / x2) = (4 / 9) + 2 = 22 / 9
x2 + (1 / x2) = 2(4 / 9)
Answer: c)

4). x – (1 / x) = 4
We know that,
(x + (1 / x))2 = (x - (1 / x) ) 2 + 4 ×( x (1 / x))
(x + (1 / x))2 = (4)2 + 4
(x + (1 / x))2 = 20
x + (1 / x) = 2√5
Answer: b)

5). [(50)3 + (-30)3 + (-20)3 ]
= 103 [ (5)3 + (-3)3 + (-2)3]
=  103 [ 125 – 27 – 8] = 1000 × 90 = 90000
Answer: c)

6). Let a = 0.98, b = 0.02, c = -1
But a + b + c = 0.98 + 0.02 – 1 = 0
If a + b + c = 0 then
a3 + b+ c3 – 3abc = 0
(0.98)3 + (0.02) – 1 +  3 × 0.98 × 0.02  = 0
Answer: d)

7). Here, x = 0.5 and y = 0.2
√0.6 × (3y)x = √0.6 × ( 3 × 0.2)0.5
= √0.6 ×√0.6  = 0.6
Answer: c)

8). (137 × 137 + 137 × 133 + 133 × 133) / (137 × 137 × 137 -  133 × 133 × 133)
We know that,
a - b3 = (a – b) (a2 + ab + b2)
= (137 × 137 + 137 × 133 + 133 × 133) / [ (137 -  133 )(137 × 137 + 137 × 133 + 133 × 133)
= 1 / (137 – 133) = 1 / 4
Answer: c)

9). 1 / 2 + [ (1 / 2) × (1 / 2) ] / { (1 / 2) × (1 / 2) / (1 / 2) + ( ( 1 / 2) / (1 / 2) ) } ]
= 1 / 2 + [ (1 / 2) × (1 / 2) ] / { (1 / 2) × (1 / 2) / (1 / 2) + ( 1 ) } ]
= 1 / 2 + [ (1 / 2) × (1 / 2) ] / {  (1 / 2) + ( 1 ) }
= 1 / 2 + [ (1 / 2) × (1 / 2) ] /  (3 / 2)
= 1 / 2 + [ (1 / 2) × (1 / 2)   × (2 / 3)  ]
= (1 / 2) + (1 / 6)
= (3 + 1) / 6 = 4 / 6  = 2 / 3
Answer: a)

10). Here (8 / 125)-4 / 3 = (23 / 534 / 3
= (2 / 5) -4  = (16 / 625)-1 = 625 / 16
Answer: a)

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