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 0∘
is 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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