### SSC CHSL- Aptitude Arithmetic Practice Questions (With Solutions) Set-37

SSC CHSL- Aptitude Arithmetic Practice Questions (With Solutions) Set-37:-
List of Practice Aptitude Questions for Upcoming SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1). If books bought at prices ranging from Rs. 200 to Rs. 350 are sold at prices ranging from Rs. 300 to Rs. 425, what is the greatest possible profit that might be made in selling eight books ?
a)    600
b)    1200
c)    1800
d)    None of these

2).The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
a)    Rs. 2000
b)    Rs. 2200
c)    Rs. 2400

3). The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?
a)    10
b)    20
c)    30
d)    40

4). A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
a)    120 m
b)    400 m
c)    240 m
d)    450 m

5). A wire is in  the form of a circle . The radius of the circle is 28 cm. The wire  is then moulded to form a square . find the side of the square formed?
a)    44 cm
b)    66 cm
c)    22 cm
d)    11 cm

6). A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:
a)    Rs. 8400
b)    Rs. 11,900
c)    Rs. 13,600
d)    Rs. 14,700

7). A tree increases annually by 1⁄5 th of its height. If its height today is 50 cm, what will be the height after 2 years?
a)    64 cm
b)    72 cm
c)    66 cm
d)    84 cm

8). A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
a)    126 sq. ft.
b)    64 sq. ft.
c)    100 sq. ft.
d)    102 sq. ft.

9). The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
a)    18 cm
b)    16 cm
c)    40 cm
d)    20 cm

10). The ratio of investment of two partners is 11:12 and the ratio of their profits 2:3 . if A invested the money for 8 months. Find for how much time B invested his money?
a)    11 months
b)    10 months
c)    9 months
d)    8 months

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

1).
Least Cost Price = Rs. (200 * 8) = Rs. 1600.
Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.
Required profit = Rs. (3400 - 1600) = Rs. 1800.
2).
Let C.P. be X
1920-x/x*100 = x-1280/x*100
1920-x = x-1280
2x= 3200, x= 1600
s.p. = 125%of 1600= 125/100*1600 = 2000
3).
Let's take the present age of the elder person = x
and the present age of the younger person = x – 16
(x – 6) = 3 (x-16-6)
=> x – 6 = 3x – 66
=> 2x = 60
=> x = 60/2 = 30
4).
Train crosses the man in 20 sec.
So length of train / speed of train in m/sec = 20 sec
L/(54*5/18)= 20 ; Length of train= 300 m
Now length of train+ platform/ (54*5/18)= 36 sec
On solving Length of Platform = 240 m
5).
Radius of the circle (r) = 28 cm.
Length of the wire (circumference) = 2 pi r = 2 pi 28 = 176 cm
Let side of the square be "a" cm.
Perimeter of square(4a) = Circumference of the circle = 176 cm
or, 4a = 176 cm
or, a = 44 cm
Thus, side of the square is 44 cm.
6).
Let C = x.
Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.
So, x + x + 5000 + x + 9000 = 50000
=>  3x = 36000
=> x = 12000
A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.
A's share = Rs.(35000 x 21/50) = Rs. 14,700.
7).
This problem is similar to the problems we saw in compound interest.
We can use the formulas of compound interest here as well.
Rate of increase = 1/5×100=20%
Height after 2 years = P(1+R/100) T
= 50(1+20/100)2=50(1+1/5)2
=50(6/5)2=(50×6×6)/(5×5)=2×6×6=72 cm
8).
Let l = 9 ft.
Then l + 2b = 37
2b = 37 - l = 37 - 9 = 28
b = 28/2 = 14 ft.
Area = lb = 9 × 14 = 126 sq. ft.
9).
Then length = 2x cm
Area = lb = x × 2x = 2x2
New length = (2x - 5)
New breadth = (x + 5)
New Area = lb = (2x - 5)(x + 5)
But given that new area = initial area + 75 sq.cm.
(2x - 5)(x + 5) = 2x^2 + 75
2x^2 + 10x - 5x - 25 = 2x^2 + 75
5x - 25 = 75
5x = 75 + 25 = 100
x = 100/5 = 20 cm
Length = 2x = 2 × 20 = 40cm