Practice Aptitude Questions With Solutions for Railway / SSC Exam

Practice Aptitude Questions With Solutions Railway / SSC Exams Set-4:
Dear Readers, Important Aptitude Questions for RRB/SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).83  × 82  × 8-5   is equal to
a)    0
b)    1
c)    2
d)    None of the above

2).[ 81 / 169 ]-1 / 2 is equal to
a)    3 / 169
b)    9 / 169
c)    9 / 13
d)    13 / 9

3).If 3x + 3 + 7 = 250, then x is equal to
a)    1
b)    2
c)    3
d)    5

4).√(a-1 b) . √(b-1 c) √(c-1 a)
a)    abc
b)    √abc
c)    1 / abc
d)    1

5).In the given figure, AOB is a straight line, ∠AOC = (3x – 8), ∠COD = 50, ∠BOD = (x + 10), The value of x is










a)    32
b)    42
c)    36
d)    52

6).If an angle is its own complementary angle, then its measure is
a)    80
b)    45
c)    60
d)    90

7).cos 0is equal to
a)    0
b)    1
c)    Not defined
d)    None of these

8).The sum and product of two numbers are 11 and 18 respectively. The sum of their reciprocal is
a)    2 / 11
b)    11 / 2
c)    18 / 11
d)    11 / 18

9).If 19 is subtracted from a two digit number then resulting number is two – third of the original number. What is the sum of two digits of the number?
a)    12
b)    15
c)    17
d)    14

10).If 2 / 3 part of a number is greater by 4 than 3 / 5th of it, what is the number?
e)    30
f)     40
g)    60
h)    20


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

Solution :
1).   83 × 82 × 8-5 = 8(3 + 2 – 5) = 80 = 1
Answer:  b)
2). [81 / 169]-1 / 2 = [169 / 81]1 / 2 = (169)1 / 2 / (81)1 / 2 = 13 / 9
Answer: d)
3). 3x + 3 + 7 = 250
3x + 3 = 243 = 35
x + 3 = 5
x = 2
Answer: b)
4). Since ∠AOB is a straight angle, we have
∠AOC + ∠COD + ∠BOD = 180
(3x – 8) + 50 + (x + 10) = 180
            4x = 128
            x = 32
Answer: d)
5). x  = (90 – x)
2x = 90
x = 45
Answer: a)
6). Cos0 = 1
Answer: b)
7). Let the numbers be x and y.
According to the question, x + y = 11        ..(i)
            xy = 18                                               ..(ii)
Dividing Eq. (i) by (ii)
            (x + y) / xy = (1 / y) + (1 / x) = 11 / 18
Answer: b)
8). Let the two digits number = 10x + y
According to problem,
(10x + y) – 19 = (10x + y) × (2 / 3)
10x + y – 19 = (20x + 2y) / 3
(10x + y) + 57 = 0
(10x + y) = 57
∴ Number = 57 and sum of the digits = 5 + 7 = 12
Answer: d)
9). Let the number be 1.
2 / 3 of 1 = 2 / 3 and 3 / 5 of 1 = 3 / 5
(2 / 3) – (3 / 5) = (10 – 9) / 15 = 1 / 15
∴ Required number = 4 + (1 / 15) = 60
Answer: a)
10). The LCM of 6, 9, 12, 15 and 18 = 180 and the largest number of 4 digits = 9999
                        180) 9999 (55
                                 900
                               
                                999
                                900       
                                  99
                                
            9999 – 99 = 9900
Required number = 9900 + 1 = 9901
Answer: c) 

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