Practice Aptitude Questions for SSC CGL,CHSL,CPO Exam

Practice Aptitude Questions With Solutions SSC CGL, CHSL Exams Set-26:
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).A sum of money placed at compound interest triples itself in 5 yr. In how many years will it amount to nine times itself?
a)    12
b)    10
c)    15
d)    18

2).A number divided by 56 gives 29 as remainder. If the same number is divided by 8, the remainder will be
a)    4
b)    5
c)    6
d)    7

3).What is the value of √7.84 + √0.0784 + √0.000784 + √0.00000784 ?
a)    3.08
b)    3.108
c)    3.1008
d)    3.1108

4).If 1(2 / 3)÷(2 / 7) × (x / 7) = 1(1 / 4) × (2 / 3) ÷ (1 / 6) then find the value of x.
a)    0.006
b)    1/6
c)    0.6
d)    6

5).A man has 100 kg of sugar, part of which he sold at 7% profit and rest at 17% profit. He gained 10% on the whole. How much did he sell at 7% profit?
a)    65 kg
b)    35 kg
c)    30 kg
d)    70 kg

6).The price of rice is reduced by 2%. How many kilograms of rice can now be bought for the money which was sufficient to buy 49 kg of rice earlier?
a)    48
b)    49
c)    50
d)    51

7).The ages of Vaibhav and Jagat are in the ratio of 12 : 7. After 6 yr, the ratio of their ages will be 3 : 2. What is the difference in their ages?
a)    8 yr
b)    12 yr
c)    9 yr
d)    10 yr

8).If both the radius and height of a right circular cone are increased by 20%, its volume will be increased by
a)    20%
b)    40%
c)    60%
d)    72.8%

9).In an examination, 52% students failed in Hindi and 42% in English. If 17% failed in both the subjects, what percentage of students passed in both the subjects?
a)    38
b)    33
c)    23
d)    18

10).The least number, which must be added to 6709 to make it exactly divisible by 9, is
a)    5
b)    4
c)    7
d)    2

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

Solution:
1). Let Rs. P be the Principal.
3P = P (1 + r / 100)5
3 =  (1 + r / 100)5
On squaring both sides, we get
32 = (1 + r / 100)10                       ….(i)
Let the sum will be nine times in n yr.
9P = P (1 + r / 100)n
32 = (1 + r / 100)n                     ….(ii)
On comparing Eqs (i) and (ii) we get n = 10 yr
Answer:  b)
2). Let the number be x.
Then, according to the question,
x = 56k + 29
Then, x = (8 × 7k) + (8 × 3) + 5
= 8 × (7k + 3) + 5
Therefore, when x is divided be 8, the required remainder = 5
Answer: b)
3). √7.84 + √0.0784 + √0.000784 + √0.00000784
 = √784 / 100 + √ 784 / 10000 + √784 / 1000000 + √784 / 100000000
= (28 / 10) + (28 / 100) + (28 / 1000) + (28 / 10000)
= 2.8 + 0.28 + 0.028 + 0.0028
= 3.1108
Answer: d)
4). Given expression,
(5 / 3) × (7 / 2) × (x / 7) = (5 / 4) × (2 / 3) × 6
x = (2 × 6) / 2
x = 6
Answer: d)
5). By the rule of alligation, we have



Now the quantity of sugar sold at
7% is given by = (7 / 10) × 100 = 70kg
Answer: d)
6). Let the original price per kg be Rs. 100
Reduced price = Rs. 98
Then, the amount sufficient for 49 kg is = 49 × 100 = RS. 4900
Amount to be bought = 4900 / 98 = 50kg
Answer: c)
7). Let the present age of Vaibhav = 12x yr
And present age of Jagat = 7x yr
According to the question,
(12x + 60 / (7x + 6) = 3 / 2
24x + 12 = 21x + 18
24x – 21x = 18 – 12
3x = 6
x = 6 / 3 = 2
Required difference = 12x – 7x = 5x = 5 × 2 = 10 yr
Answer: d)
8). If height and radius both of a cylinder change by x%, then volume changes by
= [ 3x + (3x2 / 100) + (x3 / 1002)]%
= [3 × 20 + (3 × 20 × 20) / 100 + (20 × 20 × 20) / 10000 ]
= 60 + 12 + 0.8 = 72.8%
Answer: d)
9). Let the total number of students = 100
Number of students who failed in Hindi or English or both = 52 + 42 – 17 = 77
Number of students who passed in both subjects = 100 – 77 = 23
Required percentage = 23%
Answer: c)
10). A number is divisible by 9, if the sum of its digits is divisibke by 9.
Here 6 + 7 + 0 + 9 = 22
Now, 22 + 5 = 27, which is divisible by 9. Hence, 5 must be added to 6709.

Answer: a)

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