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NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

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NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3 - FREE PDF Download

In Class 7 Maths Chapter 1 Integers, Exercise 1.3 focuses on understanding and performing various operations on integers. This exercise is crucial as it helps students grasp the fundamental concepts of adding and subtracting integers, which are essential for solving more complex mathematical problems.

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Table of Content
1. NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3 - FREE PDF Download
2. Glance on NCERT Solutions Class 7 Maths Chapter 1 Exercise 1.3 | Vedantu
3. Access NCERT Solutions Class 7 Maths Chapter 1 Integers Exercise 1.3
    3.1Exercise 1.3
4. Conclusion
5. Class 7 Maths Chapter 1: Exercises Breakdown
6. CBSE Class 7 Maths Chapter 1 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 7 Maths
8. Important Related Links for NCERT Class 7 Maths
FAQs


What is important in class 7 maths Exercise 1.3 solutions is to pay close attention to the rules of integer operations, particularly the signs associated with addition and subtraction. Students should focus on practicing these operations thoroughly to build a strong foundation in integers. By mastering these class 7 maths 1.3 concepts, students will be well-prepared for future topics in mathematics.  To boost your exam preparations, you can download the FREE PDF for NCERT Solutions for Class 7 Maths from Vedantu’s website. 


Glance on NCERT Solutions Class 7 Maths Chapter 1 Exercise 1.3 | Vedantu

  • In class 7 maths chapter 1 Exercise 1.3 solutions pd is all about Integers - Division is the Inverse operation of Multiplication. 

  • When we divide a negative integer by a positive integer, we divide them as whole numbers and then put a negative sign(-) before the quotient.

  • Dividing a negative number by a negative number, we divide them as a whole number and then put a positive sign(+) before the quotient.

  • Properties of Division of Integers in class 7 maths chapter 1 Ex 1.3

    • If an integer a is not equal to 0, then a ÷ a = 1.

    • For every integer a, you have a ÷ 1= a.

    • If an integer a is non-zero, then 0 ÷ a = 0.

  • This article integer class 7 contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter  - Integer, which you can download as PDFs.

  • In class 7 maths chapter 1 Exercise 1.3 solutions pdf there are 7 fully solved questions that help you to understand Integers.

Access NCERT Solutions Class 7 Maths Chapter 1 Integers Exercise 1.3

Exercise 1.3

1. Evaluate each of the following:

Ans:

(a) $( - 30) \div 10 = ( - 30) \times \dfrac{1}{{10}} = \dfrac{{ - 30 \times 1}}{{10}} =  - 3$

(b) $(50) \div ( - 5) = 50 \times \left( {\dfrac{{ - 1}}{5}} \right) = \dfrac{{50 \times ( - 1)}}{5} =  - 10$

(c) $( - 36) \div ( - 9) = ( - 36) \times \left( {\dfrac{{ - 1}}{9}} \right) = \dfrac{{( - 36) \times ( - 1)}}{9} = \dfrac{{36}}{9} = 4$

(d) $( - 49) \div (49) = ( - 49) \times \left( {\dfrac{1}{{49}}} \right) = \dfrac{{ - 49}}{{49}} =  - 1$

(e) $13 \div \left[ {( - 2) + 1} \right] = 13 \div \left( { - 1} \right) = 13 \times \left( {\dfrac{{ - 1}}{1}} \right) =  - 13$

(f) $0 \div \left( { - 12} \right) = 0 \times \left( {\dfrac{{ - 1}}{{12}}} \right) = \dfrac{0}{{12}} = 0$

(g) $( - 31) \div \left[ {( - 30) \div ( - 1)} \right] = ( - 31) \div ( - 30 - 1) = ( - 31) \div ( - 31) = ( - 31) \times \left( {\dfrac{{ - 1}}{{31}}} \right) = \dfrac{{31}}{{31}} = 1$

(h) $\left[ {\left( { - 36} \right) \div 12} \right] \div 3 = \left[ {\left( { - 36} \right) \times \dfrac{1}{{12}}} \right] \times \dfrac{1}{3} = \left( {\dfrac{{ - 36}}{{12}}} \right) \times \dfrac{1}{3} = \left( { - 3} \right) \times \dfrac{1}{3} = \dfrac{{ - 3}}{3} =  - 1$

(i) $\left[ {\left( { - 6} \right) + 5} \right] \div \left[ {\left( { - 2} \right) + 1} \right] = \left( { - 6 + 5} \right) \div \left( { - 2 + 1} \right) = \left( { - 1} \right) \div \left( { - 1} \right) = \left( { - 1} \right) \times \dfrac{{\left( { - 1} \right)}}{1} = 1$


2.  Verify that a ÷(b + c) ≠ (a ÷b) + (a ÷c) for each of the following values of a,b and c.

(a) a = ${\mathbf{12}}$ , b =$ - {\mathbf{4}}$ ,  c = ${\mathbf{2}}$                        

(b) $ a = \mathbf{(-10)} , b ={\mathbf{1}}, c ={\mathbf{1}}$

Ans:

(a) Given: $a \div (b + c) \ne (a \div b) + (a \div c)$

a = $1$$2$ , b = $ - 4$,  c = $2$

Putting the given values in L.H.S. = $12 \div ( - 4 + 2)$

$ = 12 \div \left( { - 2} \right) = 12 \times \left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{{ - 12}}{2} =  - 6$

Putting the given values in R.H.S. = $\left[ {12 \div \left( { - 4} \right)} \right] + \left( {12 \div 2} \right)$

$ = \left( {12 \times \dfrac{{ - 1}}{4}} \right) + 6 =  - 3 + 6 = 3$

Since,

L.H.S. ≠ R.H.S.

Hence verified. 

(b) Given: $\;a \div (b + c) \ne (a \div b) + (a \div c)$

a = −$10$, b =$1$ ,  c =$1$

Putting the given values in L.H.S. = $\; - 10 \div (1 + 1)$

= −$10 \div (2) = {\text{ }} - 5$

Putting the given values in R.H.S. =$\;\left[ { - 10 \div 1} \right] + \left[ { - 10 \div 1} \right]$

=$\; - 10 - 10 = {\text{ }} - 20$

Since,

L.H.S. ≠ R.H.S.

Hence verified.


3.  Fill in the blanks:

(a) ${\mathbf{369}}{\text{ }} \div {\text{ }}\_\_\_\_\_\_\_{\text{ }} = {\text{ }}{\mathbf{369}}$                             

(b) $\left( { - {\mathbf{75}}} \right){\text{ }} \div {\text{ }}\_\_\_\_\_\_\_{\text{ }} = {\text{ }}\left( { - {\mathbf{1}}} \right)$

(c) $\left( { - {\mathbf{206}}} \right){\text{ }} \div {\text{ }}\_\_\_\_\_\_\_{\text{ }} = {\mathbf{1}}$                              

(d) $\left( { - {\mathbf{87}}} \right){\text{ }} \div {\text{ }}\_\_\_\_\_\_\_{\text{ }} = {\text{ }}{\mathbf{87}}$

(e) $\_\_\_\_\_\_\_ \div {\mathbf{1}}{\text{ }} = {\text{ }} - {\mathbf{87}}$ 

(f) $\;\_\_\_\_\_\_\_ \div {\text{ }}{\mathbf{48}}{\text{ }} = {\text{ }} - {\mathbf{1}}$

(g) ${\mathbf{20}}{\text{  }} \div \_\_\_\_\_\_\_{\text{ }} = {\text{ }} - {\mathbf{2}}$   

(h) $\;\_\_\_\_\_\_\_ \div \left( {\mathbf{4}} \right){\text{ }} = {\text{ }} - {\mathbf{3}}$

Ans:

$\left( a \right)$ $369 \div 1 = 369$                                

$\left( b \right)( - 75) \div 75 = ( - 1)$        

$\begin{array}{*{20}{l}}{\;\;\;\;\;\;\left( c \right)( - 206) \div ( - 206) = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( d \right)( - 87) \div ( - 1) = 87} \\ {\;\;\;\;\;\;\left( e \right)( - 87) \div 1 =  - 87\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( f \right)( - 48) \div 48 = {\text{ }} - 1} \\ {\;\;\;\;\;\;\left( g \right)20 \div ( - 10) =  - 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( h \right)( - 12) \div (4) = {\text{ }} - 3} \end{array}$


4. Write five pairs of integers $\;\left( {a,b} \right)$ such that $\;\;a \div b = {\text{ }} - {\mathbf{3}}$. One such pair is $\left( {{\mathbf{6}}, - {\mathbf{2}}} \right)$ because

${\mathbf{6}}{\text{ }} \div \left( { - {\mathbf{2}}} \right){\text{ }} = {\text{ }}\left( { - {\mathbf{3}}} \right)$.

Ans:

$\begin{array}{*{20}{l}}{\left( i \right)( - 6) \div 2 = {\text{ }} - 3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {ii} \right)9 \div ( - 3) = {\text{ }} - 3} \\ {\left( {iii} \right)12 \div ( - 4) =  - 3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {iv} \right)( - 9) \div 3 = {\text{ }} - 3} \\ {\left( v \right)( - 15) \div 5 = {\text{ }} - 3} \end{array}$

5. The temperature at noon was $10^\circ {\mathbf{C}}$ above zero. If it decreases at the rate of  $2^\circ C$ per  hour  until mid-night, at what time would the temperature be ${\mathbf{8}}^\circ {\mathbf{C}}$ below zero? What would be the temperature at mid-night?

Ans:

Following number line is representing the temperature:


Number Line Representing the Temperature


Number Line Representing the Temperature


The temperature decreases $2^\circ C$ = 1 hour

The temperature decreases $1^\circ C$ =$\dfrac{1}{2}$hour

The temperature decreases $18^\circ {\mathbf{C}}$ =$\dfrac{1}{2} \times 18 = 9$ hours

Total time = $12$ noon + $9$ hours = $21$ hours = $9$ pm

Thus, at $9$ pm the temperature would be $8^\circ {\mathbf{C}}$ below  zero.


6. In a class test (+${\mathbf{3}}$) marks are given for every correct answer and (-${\mathbf{2}}$) marks are given for every incorrect answer and no marks for not attempting any question.

I. Radhika scored ${\mathbf{20}}$ marks. If she has got ${\mathbf{12}}$ correct answers, how many questions has she attempted incorrectly?

II. Mohini scores (-${\mathbf{5}}$) marks in this test, though she has got ${\mathbf{7}}$ correct answers. How manyquestions has she attempted incorrectly?

Ans:

I. Marks given for one correct answer = $3$

Marks given for $12$correct answers = $3{\text{ }} \times {\text{ }}12{\text{ }} = {\text{ }}36$

Radhika scored $20$ marks.

Therefore, Marks obtained for incorrect answers =$\;20{\text{ }}--{\text{ }}36{\text{ }} = {\text{ }}--16$

Now, marks given for one incorrect answer = –$2$

Therefore, number of incorrect answers =$\;( - 16) \div ( - 2) = 8$

Thus, Radhika has attempted $8$ incorrect questions.

II. Marks given for seven correct answers = $3{\text{ }} \times {\text{ }}7{\text{ }} = {\text{ }}21$

Mohini scores = –$5$

Marks obtained for incorrect answers = –$5{\text{ }}--{\text{ }}21{\text{ }} = {\text{ }}--26$

Now, marks given for one incorrect answer = $--2$

Therefore, number of incorrect answers =$\;( - 26) \div ( - 2) = 13$

Thus, Mohini has attempted $13$ incorrect questions.

      

7. An elevator descends into a mine shaft at the rate of ${\mathbf{6}}$ m/min. If the descent starts from   ${\mathbf{10}}$ above the ground level, how long will it take to reach -${\mathbf{350}}$ m?

Ans:

Starting position of mine shaft is $10$ m above the ground but it moves in opposite        direction so it travels the distance (–$350$) m below the ground.

So total distance covered by mine shaft = $10$ m – (–$350$) m = $10$ + $350$ = $360$ m

Now, time taken to cover a distance of $6$ m by it = $1$  minute

So, time taken to cover a distance of $1$ m by it =$\dfrac{1}{6}$minute

Therefore, time taken to cover a distance of $360$ m =$\dfrac{1}{6} \times 360 = 60$ minutes = $1$ hour

(Since $60$ minutes = $1$ hour)

Thus, in one hour the mine shaft reaches –$350$ below the ground.


Conclusion

NCERT Solutions for Class 7th Maths Exercise 1.3 Chapter 1 Integers by Vedantu provides a thorough understanding of division of integers. This class 7 Exercise 1.3 helps students grasp the rules for handling positive and negative numbers, which are crucial for more advanced math topics. It's important to focus on the sign rules: the product or quotient of two numbers with the same sign is positive, while that of two numbers with different signs is negative.


Typically, around 1-2 questions related to these topics are asked in previous year exams, highlighting their importance. By practicing the problems in this exercise and understanding the detailed solutions provided by Vedantu, students can build confidence and proficiency in working with integers, ensuring better performance in their exams.


Class 7 Maths Chapter 1: Exercises Breakdown

Exercise

Number of Questions

Exercise 1.1

4 Questions and Solutions

Exercise 1.2

4 Questions and Solutions



CBSE Class 7 Maths Chapter 1 Other Study Materials



Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.




Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.


FAQs on NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

1. What does Class 7 Maths Chapter 1 deal with?

A. Class 7 Maths Chapter deals with Integers and all of its relevant concepts. Take a look at the overview of all the topics that are being discussed in this chapter.

  • Introduction of Integers.

  • Properties of Addition and Subtraction of Integers.

  • Multiplication of Integers.

  • Multiplication of a Positive and Negative Integer.

  • Multiplication of two Negative Integer.

  • Properties of Multiplication of Integers.

  • Division of Integers.

  • Properties of Division of Integers.

Along with the in-depth concepts, there are various types of questions being asked in the exercises that are given in between and at the end of the chapter. There are a total of four exercises included in the chapter of Class 7 Maths.

2. How many questions are there in Class 7 Maths Chapter 1 Exercise 1.3 of NCERT textbook?

A. There are a total of seven questions in Class 7 Maths Chapter 1 Exercise 1.3 of NCERT textbook. Answers to these questions have been provided in the solutions PDF offered by Vedantu. It is one of the leading ed-tech portals in India. These solutions have been created by Maths experts who have years of experience in the respective industry.

3. What are the advantages of referring to NCERT Solutions Class 7 Maths Chapter 1 Exercise 1.3 offered by Vedantu?

A. NCERT Solutions Class 7 Maths Chapter 1 Exercise 1.3 offered by Vedantu is available online in order to help all the students for their learning endeavours. These solutions can be accessed both online and offline at absolutely no cost. Any student can access these high-quality study materials at any time as per their convenience.


All these solutions are provided in a step-by-step manner so that it is easier to understand by the students of any IQ level. Not only that but also any student can self assess himself/ herself with the help of these NCERT Solutions Class 7 Maths Chapter 1 Exercise 1.3. Solve the questions from the Maths exercises by yourself and then evaluate every answer with the help of solutions PDF. Our solutions are prepared with an easy approach to help you effortlessly learn complicated topics.


Vedantu, a leading ed-tech platform in India, provides high-quality study materials for all the classes and all the subjects in a chapter-wise manner which are very easy to access without any hassle.

4. How can I download NCERT Solutions from Vedantu app?

A. You have to download the Vedantu application in order to download the NCERT Solutions for Class 7 Maths Chapter 1 Exercise 1.3. Visit Google play store to download the Vedantu app. Therefore you will be able to avail the NCERT solutions for all the classes for absolutely free of cost.