# NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths in Hindi and English Medium PDF format updated for the academic session 2020-21, Download Exemplar problems book, NCERT solutions of Science and answers and other subject’s solutions also free for session 2020-21.

Complete description of each question is given in the solutions which is based on latest NCERT Books 2020-2021. Practice Maths with Vedic Maths to improve your calculations and make it faster. Download Class 7 Maths Offline App in English Medium and 7 Ganit Offline App in Hindi Medium. These Apps work without internet, once downloaded.

## NCERT Solutions for Class 7 Maths

• ### Class 7 Maths Solutions in Hindi, English and Videos

 Class: 7 Maths (Ganit) Medium: English and Hindi Medium

### Class 7 Maths Solutions in Hindi and English Medium

NCERT Solutions for Class 7 Maths are now available in both English Medium and Hindi Medium free to download. These solutions are updated for current session 2020-21.

### Class 7 Maths Solutions & Main Points of All Chapters

#### Chapter 1: Integers

We have done on a number line when we add a positive integer, and we move to the right. To add a negative integer, we move to the left. Similarly, subtracting a positive integer, we move to the left and to subtract a negative integer, we move to the right. In class 7 Maths Chapter 1, we have to use these facts in applications of daily life.

Class 7 Maths Chapter 1 Exercise 1.1 Solution
Class 7 Maths Chapter 1 Exercise 1.2 Solution

#### Chapter 2: Fractions and Decimals

In Class 7 Maths Chapter 2, we will learn about fractions and decimals along with the operations of addition and subtraction on them. We also learnt how to multiply two decimal numbers. While multiplying two decimal numbers, first multiply them as whole numbers. Count the number of digits to the right of the decimal point in both the decimal numbers. Add the number of digits counted. Put the decimal point in the product by counting the digits from its rightmost place. The count should be the sum obtained earlier. We will also study the operations of multiplication and division on fractions as well as on decimals. There are methods for how two fractions are multiplied by multiplying their numerators and denominators separately and writing the product. How to obtain a reciprocal of a fraction is obtained by inverting it upside down. This concept is also used in Chapter 4 Simple Equations also.

Class 7 Maths Chapter 2 Exercise 2.1 Solution
Class 7 Maths Chapter 2 Exercise 2.2 Solution

#### Chapter 3: Data Handling

In Chapter 3 of Class 7 Maths, we will know about the collection, recording and presentation of data. Before collecting data, we need to know that the data that is collected needs to be organised in a proper table so that it becomes easy to understand and interpret. We will discuss the average that is a number that represents or shows the central tendency of a group of observations or data along with arithmetic mean, Mode, which is another form of central tendency or symbolic value and Medium. Median is also a form of symbolic value. It refers to the value which lies in the middle of the data with half of the observations above it and the other half below it. Bar graph and using bars of uniform widths and Double bar graphs to compare two collections of data at a glance is also given in 7th Maths Chapter 3 Data Handling. In the end, there are questions based on the probability that describes the situations in our daily life, that are certain to happen, some that are impossible and some that may or may not happen.

Class 7 Maths Chapter 3 Exercise 3.1 Solution
Class 7 Maths Chapter 3 Exercise 3.2 Solution

#### Chapter 4: Simple Equations

Chapter 4 of Class 7 Maths describes an equation which is a condition on a variable such that two expressions in the variable should have equal value. It also tells about the value of the variable for which the equation is satisfied is called the solution of the equation. We know that an equation remains the same if the LHS and the RHS are interchanged. Addition or subtraction on both the sides equally in an equation remains it unchanged. This is the property which is used to solve the equation. Transposition of a number has the same effect as adding or subtracting the same number to both sides of the equation. Class 7 Chapter 4 Simple Equations describes here, how to construct simple algebraic expressions corresponding to practical situations and using the technique of doing the same mathematical operation on both sides, we can build an equation starting from its solution.

Class 7 Maths Chapter 4 Exercise 4.1 Solution
Class 7 Maths Chapter 4 Exercise 4.2 Solution

#### Chapter 5: Lines and Angles

We already know that a line segment has two endpoints and a ray has only one endpoint. In class 6 Maths, we have studied that a line has no endpoints on either side. Now in Class 7 Maths Chapter 5, we have to discuss pairs of angles condition like two complementary angles (Measures add up to 90°), two supplementary angles (Measures add up to 180°), two adjacent angles (Have a common vertex and a common), Linear pair, Adjacent and supplementary angles. Do you know? When two lines intersect, the meeting point is called the point of intersection and when lines are drawn on a sheet of paper do not meet, however far produced, we call them to be parallel lines. In Chapter 5 of class 7 Maths, we have to study about two pairs of opposite angles, called vertically opposite angles. We also go through about a transversal is a line that intersects two or more lines at distinct points. Chapter 5 will also helpful in Chapter 6 Triangles and its properties.

Class 7 Maths Chapter 5 Exercise 5.1 Solution
Class 7 Maths Chapter 5 Exercise 5.2 Solution

#### Chapter 6: The Triangles and Its Properties

Chapter 6 of Class 7 Maths describes the six elements of a triangle are its three angles and the three sides. It also briefs about three medians and three altitudes of the triangle. We know that an exterior angle of a triangle is formed when a side of a triangle is produced. So, we can have two ways of forming an exterior angle. We have to learn by heart that the measure of any exterior angle of a triangle is equal to the sum of the measures of its interior opposite angles and the angle sum property of a triangle, which tells that the total measure of the three angles of a triangle is 180°. In chapter 6 of Class 7 Maths, we have to read about an equilateral triangle in which each angle has measure 60° and an isosceles if at least any two of its sides are of the same length. We will use the property of the lengths of sides of a triangle to solve various sums quickly. There is the introduction of Pythagoras property also which may be useful in further classes also.

Class 7 Maths Chapter 6 Exercise 6.1 Solution
Class 7 Maths Chapter 6 Exercise 6.2 Solution

#### Chapter 7: Congruence of Triangles

Do you know that congruent objects are exact copies of one another? In class 7 Maths Chapter 7, we will study about the congruency of triangles and their theorems also. In this chapter, it is explained how the method of superposition examines the congruence of plane figures. There are four main rules of congruence of two triangles. SSS, SAS, ASA and RHS congruency rules. Here, S denotes the side, and A denotes the angle of the triangle. There is no such thing as AAA Congruence of two triangles. This chapter is much useful to solve geometrical questions, even in higher classes.

Class 7 Maths Chapter 7 Exercise 7.1 Solution
Class 7 Maths Chapter 7 Exercise 7.2 Solution

#### Chapter 8: Comparing Quantities

Comparing quantities means to find the relative ratio between two or more quantities. We are often required to compare two quantities in our daily life. So, NCERT Books of Class 7 Maths Chapter 8 Comparing Quantities will help to find these comparisons. Quantities may be heights, weights, salaries, marks etc. Two ratios can be compared by converting them to like fractions. If the two fractions are equal, we say the two given ratios are equivalent. If two ratios are equivalent, then the four quantities are said to be in proportion. We may also use percentage for comparing quantities. Percentages are numerators of fractions with denominator 100. Percentages are widely used in our daily life.

Class 7 Maths Chapter 8 Exercise 8.1 Solution
Class 7 Maths Chapter 8 Exercise 8.2 Solution

#### Chapter 9: Rational Numbers

Rational numbers are very common to us as it is already used in previous classes. We know that a number that can be expressed in the form p/q, where p and q are integers and q is not equal to 0, is called a rational number. The numbers ½, 2/5, 7/2, etc. are the example of rational numbers. All integers and fractions are rational numbers. Class 7 Maths Book Chapter 9 Rational Numbers provides a perfect practice of sum based on Rational numbers. When the numerator and denominator, both, are positive integers, it is a positive rational number. When either the numerator or the denominator is a negative integer, it is a negative rational number. The number 0 is neither a positive nor a negative rational number. Here, we will also learn about the standard form of a rational number and how to find the sum, product, division and difference of two rational numbers using numerator and denominator adjustment.

Class 7 Maths Chapter 9 Exercise 9.1 Solution
Class 7 Maths Chapter 9 Exercise 9.2 Solution

#### Chapter 10: Practical Geometry

In Chapter 10 of Class 7 Maths, we looked into the methods constructing figures using ruler and compasses. Class 7 Maths Chapter 7 Congruence of Tringle is used in this chapter to draw triangular figures. During the construction, if a line is given and a point not on it, we used the idea of ‘equal alternate angles’ in a transversal diagram to draw a line parallel to the given line. We could also have used the idea of ‘equal corresponding angles’ to do the construction. Properties of congruence triangles like SSS, ASA, SAS and RHS may also be useful during the study of Chapter 10 of 7th Mathematics.

Class 7 Maths Chapter 10 Exercise 10.1 Solution
Class 7 Maths Chapter 10 Exercise 10.2 Solution

#### Chapter 11: Perimeter and Area

Class 7 Maths Chapter 11 includes the mensuration of 2 – D figures. We know that perimeter is the distance around a closed figure whereas area is the part of the plane occupied by the closed figure. Here, we have to find the perimeter of various objects used in daily life. To find perimeter or area, we may use various formulae like the perimeter of a square = 4 × side, perimeter of a rectangle = 2 × (length + breadth), area of a square = side × side, area of a rectangle = length × breadth, etc. to reach the final results. Based on the conversion of units for lengths, studied earlier classes, the units of areas can also be used like 1 square cm = 100 square mm, 1 square m = 10000 square cm, and 1 hectare = 10000 square m.

Class 7 Maths Chapter 11 Exercise 11.1 Solution
Class 7 Maths Chapter 11 Exercise 11.2 Solution

#### Chapter 12: Algebraic Expressions

In Class 7 Maths Chapter 12, we will learn how algebraic expressions are formed from variables and constants. We will also use the operations of addition, subtraction, multiplication and division on the variables and constants to form expressions. We should know that expressions are made up of terms, and terms are added to make an expression. Normally, a term is a product of factors and factors containing variables are said to be algebraic factors. The coefficient is the numerical factor in the term, and any expression with one or more terms is called a polynomial. Polynomials may be in the form of binomial or trinomials. Terms which have the same algebraic factors are like terms, whereas terms which have different algebraic factors are unlike terms.

Class 7 Maths Chapter 12 Exercise 12.1 Solution
Class 7 Maths Chapter 12 Exercise 12.2 Solution

#### Chapter 13: Exponents and Powers

We know that vast numbers are difficult to read, understand, compare and operate upon. Chapter 13 of Class 7 Maths states the way to make all these easier. We use exponents, converting many of the large numbers in a shorter form. Exponents and Powers are useful for later on classes also. Numbers in exponential form obey certain laws, which are unique. The only practice provides a good understanding of Chapter 13 of Grade 7 Maths.

Class 7 Maths Chapter 13 Exercise 13.1 Solution
Class 7 Maths Chapter 13 Exercise 13.2 Solution

#### Chapter 14: Symmetry

Chapter 14 of Grade 7 Maths is based on Symmetry. A figure has line symmetry if there is a line about which the figure may be folded so that the two parts of the figure will coincide. We know that a regular polygons have equal sides and equal angles, so, they have multiple lines of Symmetry. Each regular polygon has as many lines of Symmetry as it has sides. Mirror reflection leads to Symmetry, under which the left-right orientation has to be taken care of. It can be obtained by turning an object about a fixed point, where the fixed point is the centre of rotation. Similarly, we can find the angle of rotation, left turn or right turn. If, after a rotation, an object looks exactly the same, we say that it has a rotational symmetry. The study of Symmetry is important because of its frequent use in day-to-day life and more because of the beautiful designs it can provide us.

Class 7 Maths Chapter 14 Exercise 14.1 Solution
Class 7 Maths Chapter 14 Exercise 14.2 Solution

#### Chapter 15: Visualising Solid Shapes

We know that the circle, the square, the rectangle, the quadrilateral and the triangle are examples of plane figures. Similarly, the cube, the cuboid, the sphere, the cylinder, the cone and the pyramid are examples of solid shapes. Chapter 15 of Standard 7 Maths, tell about the shapes and sketches of these figures. If a plane figure is given, it will be considered as two-dimensions or 2-D and similarly 3-D for solid figures. The corners of a solid shape are called its vertices, the line segments of its skeleton are its edges and its flat surfaces are its faces. The Class 7 Maths Chapter 15 Visualising solid shapes is a very useful skill. Different sections of a solid can be viewed in many ways.

Class 7 Maths Chapter 15 Exercise 15.1 Solution
Class 7 Maths Chapter 15 Exercise 15.2 Solution
• ### Chapter wise Class 7 Maths Solution in English and Hindi Medium

#### Feedback & Suggestions

If you are still facing problem to understand the solution any question, please notify us through “Discussion Forum” section. NCERT Solutions 2020-21 of all exercises are given as separate PDF files. Important questions from U-like, R S Aggarwal, P K Garg, R D Sharma, Assignments, Notes, Sample Papers, Chapter test and other study material will be uploaded very soon.
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##### Hindi Medium & English Medium Solutions

Hindi Medium NCERT Solutions for class 7 Maths is now available for the current academic year 2020-21 onward. यदि विद्यार्थियों या अभिभावकों की तरफ से कोई सलाह वेबसाइट को सुधारने के लिए हो तो अवश्य दे। आपका योगदान अन्य विद्यार्थिओं के लिए मददगार होगा। इस वर्ष 2020-2021 में लगभग सभी विषयों के हिंदी और अंग्रेजी माध्यम के हल उपलब्ध होंगे।
Your suggestion is always valuable for us in improving website as well as contents. We are working for UP Board and Bihar Board also. The contents will be online April onward. NCERT Books 2020-2021 for all Subjects are available in PDF as well as ZIP format. Download 7 Maths Offline App in English Medium and 7 गणित Offline App in हिंदी मीडियम.

### Important questions in Class 7 Maths (FAQ)

A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?
Height of a place above the sea level = 5000 m
Floating a submarine below the sea level = 1200 m
The vertical distance between the plane and the submarine = 5000 + 1200 = 6200 m
Thus, the vertical distance between the plane and the submarine is 6200 m.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 and 3/4 litres of petrol?
In 1 litre of pertrol, car covers the distance = 16 km
In litres of petrol, car covers the distance = 2 and 3/4 of 16 km = 11/4 x 16 = 44 km
Thus, the car will cover 44 km distance.
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14.
Arranging the given data in ascending order,
12, 12, 13, 13, 14, 14, 14, 16, 19
Mode is the observation occurred the highest number of times = 14
Median is the middle observation = 14
लक्ष्मी के पिता की आयु 49 वर्ष है। उनकी आयु लक्ष्मी की आयु के तीन गुने से 4 वर्ष अधिक है। लक्ष्मी की आयु क्या है?
माना, लक्ष्मी की आयु = y वर्ष
इसलिए, पिता की आयु = 3 y + 4 वर्ष
प्रश्न के अनुसार,
3 y + 4 = 49
So, 3y = 49 – 4 = 45
and y = 45/3 = 15
अतः, लक्ष्मी की आयु 15 वर्ष है।
क्या दो ऐसे कोण संपूरक हो सकते हैं यदि उनमे से दोनों अधिक कोण हैं?
नहीं, क्योंकि दो अधिक कोणों का योग सदैव 180 से अधिक होता है।
त्रिभुज PQR के अभ्यंतर में कोई बिंदु O लीजिए। क्या यह सही है कि OP + OQ > PQ?
हाँ, क्योंकि POQ एक त्रिभुज है और त्रिभुज की कोई दो भुजाओं के मापों का योग, तीसरी भुजा की माप से अधिक होती है।
Give any two real time examples for congruent shapes.
Two footballs with same size and different colour.
I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?
The cost price of T.V. = ₹ 10,000
Profit percent = 20%
Now, Profit = Profit% of C.P.
= 20/100 x 1000 = ₹ 2,000
Selling price = C.P. + Profit = ₹10,000 + ₹2,000 = ₹ 12,000
Hence, he gets ₹12,000 on selling his T.V.
How can two different rational numbers be added?
Two rational numbers with different denominators are added by first taking the LCM of the two denominators and then converting both the rational numbers to their equivalent forms having the LCM as the denominator and adding them as above.
Construct DEF such that DE = 5 cm, DF = 3 cm and EDF = 90.
To construct: DEF where DE = 5 cm, DF = 3 cm and EDF = 90
Steps of construction:
(a) Draw a line segment DF = 3 cm.
(b) At point D, draw an angle of 90 with the help of compass i.e., XDF = 90.
(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.
(d) Join EF.
(e) It is the required right angled triangle DEF.
एक वृत्ताकार फूलों के बगीचे का क्षेत्रफल 314 m2 है। बगीचे के केंद्र में एक घूमने वाला फव्वारा (sprinkler) लगाया जाता है, जो अपने चारों ओर 12 m त्रिज्या के क्षेत्रफल में पानी का छिड़काव करता है। क्या फव्वारा पूरे बगीचे में पानी का छिड़काव कर सकेगा? (π=3.14 लीजिए)
फव्वारे द्वारा छिड़काव किए गए भाग का क्षेत्रफल = πr^2
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m^2
वृत्ताकार फूलों के बगीचे का क्षेत्रफल = 314 m2
यहाँ, वृत्ताकार फूलों के बगीचे का क्षेत्रफल, फव्वारे द्वारा छिड़काव किए गए क्षेत्रफल से कम है।
अतः, फव्वारा पूरे बगीचे में पानी का छिड़काव कर सकेगा।
How are the mathematical operations done in algebraic expressions?
When we add (or subtract) two algebraic expressions, the like terms are added (or subtracted) and the unlike terms are written as they are.
What is the standard form or scientific notation?
Any number can be expressed as a decimal number between 1.0 and 10.0 (including 1.0) multiplied by a power of 10. Such form of a number is called its standard form or scientific notation.
What other name can you give to the line of symmetry of an isosceles triangle?
The line of symmetry of an isosceles triangle is median or altitude.
जाँच कीजिए कि क्या ये कथन सत्य हैं। एक घन एक षट्भुज के आकार की छाया दे सकता है।
असत्य