### SSC CHSL- Practice Aptitude Questions With Solutions Set-39

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
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

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).

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
3). Let the number be x.
Then x × 9/10 × ¾ = 450
X = (405×10×4) / (9×3) = 600
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
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
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
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
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).
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
10).

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
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