SSC CHSL-
Practice Aptitude Questions With Solutions Set-39:
Important
Aptitude Questions for SSC CHSL Exam was given here with solutions, candidates
those who are preparing for those exams can use this material.
1).
The greatest three-digit number which when divided by 10,15 and 20 leaves
remainder 5 in each case, is
a)
965
b)
955
c)
995
d)
945
2).
The ratio of two numbers is 6 : &. Their HCF is 6. Then find their LCM.
a)
292
b)
252
c)
168
d)
84
3).
(9/10) of (3/4) of a number is 405. Find the number.
a)
690
b)
600
c)
666
d)
660
4).
The length of a road is 2 kilometers. The number of plans required for
plantation at a gap of 40 meters on both sides of the road is
a)
50
b)
100
c)
51
d)
102
5).
The digit at until’s place of the number (1699)2 + (1700)3
+ (1681)4 + 1455)3 is
a)
7
b)
4
c)
9
d)
5
6).
The greatest number among 350, 440, 260 and 530
is
a)
350
b)
260
c)
440
d)
530
7).
Find the sum of the series 450 + 460 + 470 …. + 590 + 600
a)
8400
b)
4800
c)
9600
d)
10800
8).Which
of the following will be completely divisible by (99)61 + (99)62
+ (99)63 ?
a)
980199
b)
970299
c)
980099
d)
970799
9).
If a/4 = b/5 = c/18, then (a + b+ c) / (a + b) is equal to
a)
3
b)
2
c)
1
d)
9
10).
Find the value of
a)
(x
+ 5)/4
b)
4/(x+5)
c)
X
d)
1+
[ 1/ (x+5) ]
11).
The product of two numbers is 4056 and their HCF is 26 . The number of such
pairs is
a)
Two
b)
One
c)
Four
d)
Three
12).
The difference between the greatest and the smallest fraction among the
following is : 1/3, 3/4, 4/5, 6/7
a)
13/14
b)
18/19
c)
17/21
d)
11/21
13).
If x:y = 6:7, find the value of (9x-2y)
+ (4x-3y).
a)
49
b)
41
c)
43
d)
52
Answers:
1). a) 2).b)
3).b) 4).d) 5).a) 6).c) 7).a) 8).a) 9).a) 10).a) 11).a) 12).d) 13).c)
Solutions for the above aptitude
Questions:
1).
Answer: a)
2). Ratio of
the numbers = 6 : 7
HCF
of the number = 6
First
number = 6 × 6 = 36
And
the second number = 7 × 6 = 42
LCM
of 36 and 42 = 2 × 3 × 6 × 7 = 252
Answer: b)
3).
Let the number be x.
Then
x × 9/10 × ¾ = 450
X
= (405×10×4) / (9×3) = 600
Answer: b)
4).
Length of the road = 2 kilometres = 2000m
Gap
between plantation = 40m
The
number of plants on one side = (2000/40) + 1
We
added 1 because the last plant will be on the last end of the road = 50 + 1 =
51
Number
of plants on both sides = 51 × 2 = 102
Answer: d)
5).
Unit digit of (1699)2 = (9)2 = 81 = 1
Unit
digit of (1700)3 = (0)3 = 0
Unit
digit of (1681)4 = (1)4 = 1
Unit
digit of (1455)3 = (5)3
= 125 = 5
The
required unit’s digit of the no. = 1 + 1 + 5 = 7
Answer: a)
6).
Given numbers are 350, 440, 260, 530,
Now,
(3)50 = (35) ×10 = (243)10
(4)40
= (44)10 =(256)10
(2)60
= (26)10= (64)10
(5)30
= (53)10 = (125)10
Hence
the greatest number = (256)10= (4)40
Answer: c)
7).
First term of the series = 450
Last
term of series = 600
Common
difference = 460 – 450 = 10
Number
of terms in the series
=
[ (Last term – First term) / Common Difference ] + 1
=
[ (600-450) / 10 ] + 1
=
(150/10) + 1 = 15+1 = 16
Sum
of the series
=
[ (Last term + first Term) / 2 ] × n
=
[ (600 + 450) / 2 ] × 16
=
1050/2 × 16 = 525 × 16
=8400
Answer: a)
8).
(99)61 + (99)62 + (99)63
=
9960(99 + 992 + 993)
=
9960 (99 + 9801 + 970299)
=
9960(980199)
Hence
980199 is completely divisible by (9961
+ 9962 + 9963).
Answer: a)
9).
Given a/4 = b/5 = c/18
Let
a/4 = b/5 = c/18 = R
A=
4R, b=5R and c= 18R
Now,
a+b+c = 4R + 5R + 18R = 27R
Again,
a+b = 4R + 5R = 9R
So,
(a+b+c) / (a+b) = 27R/9R = 3
Answer: a)
10).
Answer: a)
11).
HCF of numbers = 26
Let
the first number be 26a
Second
number = 26b
Now,
26a × 26b = 4056
ab= 4056 / (26×26) = 6
Pair
= (1, 6), (2, 3)
Required
number of pairs = 2
Answer: a)
12).
Given fraction = 1/3, ¾, 4/5, 9/7
LCM
of denominator = 3 × 4 × 5 × 7
So,
1/3 = 1/3 × 3 × 4 × 5 × 7 = 140
3/4=
3/4× 3 × 4 × 5 × 7 = 315
4/5
= 4/5×3 × 4× 5× 7 = 336
6/7
= 6/7× 3×4×5×7 = 360
Hence
greatest no. = 6/7 and smallest no. = 1/3
Required
difference = 6/7 – 1/3 = (18-7) / 21 = 11/21
Answer: d)
13).
x= 6 , y=7
Now,
9x – 2y =9 × 6 – 2 × 7 = 54 – 14 =40
And,
4x – 3y = 4 × 6 - 3 × 7 = 24 – 21 = 3
So,
(9x-2y) + (4x-3y) = 40+3 = 43
Answer: c)
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