### 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

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
2). [81 / 169]-1 / 2 = [169 / 81]1 / 2 = (169)1 / 2 / (81)1 / 2 = 13 / 9
3). 3x + 3 + 7 = 250
3x + 3 = 243 = 35
x + 3 = 5
x = 2
4). Since ∠AOB is a straight angle, we have
∠AOC + ∠COD + ∠BOD = 180
(3x – 8) + 50 + (x + 10) = 180
4x = 128
x = 32
5). x  = (90 – x)
2x = 90
x = 45
6). Cos0 = 1
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
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
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
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