SSC CHSL- Practice Aptitude Questions (Arithmetic) With Solutions Set-41

Practice Aptitude Questions
SSC CHSL- Practice Aptitude Questions With Solutions Set-41:
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). One litre of pure alcohol is added to 8 litres of 30% alcohol solution. The percentage of water in the solution is
a)    41.74%
b)    37.77%
c)    30%
d)    40%
2). If A : B = 2 : 3, B : C = 4 : 5 and C : D = 7 : 6, find the ratio of A : C.
a)    3 : 7
b)    7 : 4
c)    3 : 5
d)    8 : 15
3). If three-fourths two-thirds of one-sixth of a number is 180, what is 250% of that number?
a)    2160
b)    4800
c)    5400
d)    None of these
4). A shopkeeper sells two articles for Rs.4500 each at no profit no loss on the whole business. If he earned 25% profit on the first item then the loss on the second items will be
a)    15%
b)    20%
c)    25%
d)    16(2/3) %
5). Which equivalent discount of the following is the greatest?
a)    10%, 20%, 30%
b)    20%, 10%, 30%
c)    30%, 20%, 10%
d)    All are the same
6). A sum of money amounts to Rs.1150 in 3 years and Rs.1250 after 5 years at the same rate of simple interest. Find the percentage rate of interest per annum.
a)    6% per annum
b)    5% per annum
c)    10% per annum
d)    8% per annum
7). Find the compound interest on Rs.5000 at 8% per annum for 2 years compounded half-yearly.
a)    Rs.849.29
b)    Rs.1698.58
c)    Rs.424.60
d)    Rs.212.30
8). 40 men working 14 hours per day dig a well in 24 days. How many hours should 48 men work per day to dig double of the well in 28 days?
a)    40 hours/day
b)    10 hours/day
c)    15 hours/day
d)    20 hours/day
9). Three pipes can fill a cistern in 15 minutes, 30 minutes and 45 minutes respectively. These pipes are opened alternately for 1 minute each, beginning with the first pipe. In what time will the cistern be full?
a)    24 (6/11) minutes
b)    9 (1/11)  minute
c)    13 (1/11) minute
d)    14 (4/11) minute
10). The radius of a car’s wheel is70 cm. The car is running at a speed of 32 kmph. Find the number of rotations of the wheel per minute. You are not expected to calculate the approximate value.
a)    60
b)    50
c)    121
d)    100

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

Solutions for the above aptitude Questions:

1). Volume of solution – 8 litre Alcohol = 30%
Amount of alcohol = (8 × 30) / 100 = 2.4 litres
Amount of Water = (8- 2.4) = 5.6  litres
Now, Volume of solution = (8+1) litres = 9 litres
Amount of alcohol = (2.4 + 1)litres = 3.4 litres
Required % = (3.4/9)×100 = 37.77%
Answer: b)
2). A : B = 2:3
B : C = 4:5
C : D = 7:6
A = 2×4×7 = 56
B=3×4×7 = 84
C= 3×5×7 = 105
D=3×5×6 = 90
Thus, A : C = 56 : 105 = 8 : 15
Answer: d)
3). Let the number be x,
Then, x × ¾ × 2/3 × 1/6 = 180
X = (180×4×3×6) / (3×2) = 2160
Now, 250% of 2160 = (2160×250)/100 = 5400
Answer: c)
4).
SP of each item = Rs.4500
Now, % profit on one item = 25%
CP of one item = 4500/125 × 100 = Rs.3600
Sum of both items CP = Sum of both items
SP = (4500 + 4500) = Rs.9000
CP of other item = (9000-3600) = Rs.5400
SP = Rs.4500
Loss = CP – SP = (5400-4500) = Rs.900
Required % loss = loss/ CP × 100 = 900/5400 × 100 = 16 (2/3)%
Answer: d)
5).
Equivalent discount of 10%, 20% and 30% = 100 – 100 × 90/100 × 80/100 × 70/100
= 100- 50.4 = 49.6
Now in case of (B) – 20%, 10% and 30% - equivalent discount is = 100 – 100 × 80/100 × 90/100 × 70/100 =49.6%
Hence, all equivalent discounts are the same.
Answer: d)
6).
Amount after 3 years = Rs.1150
Amount after 5 years = Rs.1250
Interest for 2 years = Rs.(1250-1150) = Rs.100
Interest for 1 year = Rs.50
Simple interest for 3 years = Rs.(50×3) = Rs.=150
Amount = Rs.1150
Principal = (1150-150) = Rs.1000
Rate = (Interest × 100) / (Principal × Time)
= (150×100) / (1000×3) = 5% pa
Answer: b)
7).
Principal = Rs.5000
Rate of interest = 8% pa
= 8%/2 half –yearly = 4% half-yearly
Time = 2 years
Time will be double = 2 × 2 years = 4 years
CI = Principal { [ 1+(r/100)n ] – 1 }
= 5000 { [ 1+(4/100)4 ] -1 }
= 5000 { [ ( (100+4) /100 )4 ] -1 }
=5000 { [ (26/25)4 ] -1 }
=5000 { [ 456976 / 390625 ] -1 }
= 5000 × 66351/390625 = Rs.849.29
Answer: a)
8).
In first case, men (M1) = 40
Hours (H1) = 14 hr/day
Days (D1) = 24 days, work = W1 = 1
In second case, men (M2) = 48, H2 = x hr/day
D2 = 28 days, work (W2) = 2
Now, (M1 × D1 × H1) / W1 = (M2 × D2 × H2) / W2
Or, (40×14×24) / 1 = (48×x×28)/2
Or, 48 × x × 28 = 40 × 14 × 28 = 40 × 14 × 24 × 2
Or, x= (40×14×24×2)/(48×28)
x= 20 hours/day
Answer: d)
9).
In 1 minute first pipe will fill (1/15) part
In 2nd minute second pipe will fill (1/30) part.
In 3rd minute third pipe will fill (1/45) part.
In 3 minute they will fill (1/15 + 1/30 + 1/45) = (6+3+2)/90 part = 11/90 part
11/90 part is filled by three pipes in 3 minutes
11/90 part is filled by three pipes in (3/11) × 90
= 270/11 minutes  = 24(6/11) minutes
Answer: a)
10).
Car’s wheel’s radius = 70 cm
Circumference of the wheel = 2πr = 2 × 22/7 × 70 cm = 440 cm = 4.40 m
Car’s speed = 32 kmph
Distance covered in minute = 32000/60 m
Number of rotations per minute  = (Distance covered in 1 minute) / circumference
 = 32000 / ( 60 × 44 ) = 121.21 rotations/minute
≈ 121

Answer: c)