__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

d)
Data
inadequate

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

__Answers:__**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.

**Answer: c)**

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

**Answer: a)**

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

**Answer: c)**

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

**Answer: c)**

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.

**Answer: a)**

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.

**Answer: d)**

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**Answer: b)**

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.

**Answer: a)**

9).

Let
breadth = x cm

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

**Answer: c)**

10).

Suppose
A invested Rs.11 for 8 months and B invested Rs.12 for y months.

Then
ratio of investment of A and B = (11x8) : (12xy)

=
88 : 12y

88/12y
= 2/3

24y
= 88x3

y
= 264/24 =11 months.

**Answer: a)**

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