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# SSC CGL 2020: Quantitative Aptitude

Published on 15 February 2020   The quantitative Aptitude section is one of the most scoring parts of the SSC CGL Tier-I examination. But this section will need the detailed preparation of the concepts related to each one of the topics and regular practice of them to score really well in the upcoming SSC CGL 2020 examination. In this article, I am going to explain some of the topics from the quantitative aptitude SSC CGL 2020 syllabus with a suitable example and their corresponding solution for your clarity.

## Speed, Distance and Time

If the distance between two points is constant, then the speed of an object will be inversely proportional to the time required to cover that distance.

Basic Formula:

Speed=Distance/Time

If Distance = k (constant)

Then, Speed α 1/Time

Conversion of unit of Speed:

km/hr to m/ sec = speed in km/hr* 5/1

km/sec to km/hr= speed in m/sec* 18/5

Example: A train is running at a rate of 90 Km/hour and crosses a bridge in 5 minutes. Find the length of the bridge?

Solution:   Speed of the train= 90 Km/hour

Time required to cross the bridge= 5 minute

Length of the bridge=?

Conversion of unit of speed from Km/ hour to m/sec:

Speed in m/ sec= speed in km/ hr * 5/18

= 90*5/18

= 5*5

= 25 m/sec

Conversion of unit of time from minute to second:

Time in sec= time in minute * 60

= 5* 60

= 300 sec

Length of the bridge= Speed * time

= 25* 300

= 7500 meter

## Number System

The number system deals with the representation of different types of numbers in writing. The numbers may be of different types like Whole Number, Natural Number, Integer, Even Number, Odd Number, Prime Numbers, Relative Prime, Composite Number, etc.

Example: Find an alternative which is an even prime number?

1. 3
2. 5
3. 2
4. 9

Solution: ‘Option c’ i.e. 2 is the right answer. 2 is an even prime because it is either divisible by 1 or itself which satisfies the property of a prime number and is divisible by 2 which satisfies the property of an even number.

## Mensuration

This technique in mathematics is used to calculate area, perimeter, volume and other parameters of different geometrical structures like Triangle, Polygons, Circle, cube, cuboids, cylinder, Spheres, Hemispheres, etc. Here I am discussing the different formulas related to Triangle:

If a, b and c are three different sides of a triangle, then

Area of the triangle = √s(s-a) (s-b) (s-c) Where s= (a + b + c)/2

Perimeter of the triangle = a + b + c

1. If the height and base of a triangle is given, then

Area of the triangle =12*b*h

1. For equilateral triangle

Area of the equilateral triangle= √34a2

Perimeter of the equilateral triangle =3a

Example: The three sides of a triangle is 40cm, 24cm, and 32cm. Find the area and perimeter of this triangle?

Solution: Three sides of the triangle are

a= 40 cm

b= 24 cm

c= 32 cm

Area of the triangle=?

Perimeter of the triangle=?

s= (a+b+c)/2

= (40+24+32)/2

= 96/2

= 48 cm

Area= √s(s-a) (s-b) (s-c)

= √48(48-40)(48-24)(48-32)

= √48*8*24*16

= √147456

= 384 cm2

Perimeter= a + b + c

= 40+24+32

= 96 cm

All The To All The SSC CGL 2020 Aspirants !

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