NCERT Solutions for Class 7 Maths Chapter 1: Integers - Exercise 1.4

Exercise 1.4

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

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.

NCERT Solutions for Class 7 Maths

• Chapter 1 - Integers

• Chapter 2 - Fractions and Decimals

• Chapter 3 - Data Handling

• Chapter 4 - Simple Equations

• Chapter 5 - Lines and Angles

• Chapter 6 - The Triangle and Its Properties

• Chapter 7 - Congruence of Triangles

• Chapter 8 - Comparing Quantities

• Chapter 9 - Rational Numbers

• Chapter 10 - Practical Geometry

• Chapter 11 - Perimeter and Area

• Chapter 12 - Algebraic Expressions

• Chapter 13 - Exponents and Powers

• Chapter 14 - Symmetry

• Chapter 15 - Visualising Solid Shapes

NCERT Solution Class 7 Maths of Chapter 1 All Exercises

NCERT Solutions for Class 7 Maths Chapter 1 – Exercise 1.4 Questions

Ex 1.4 Class 7 included in CBSE NCERT books aims to provide students profound knowledge on integers and how its preceding sign is affected considering the operations implemented on it. Look at the below-listed question from this exercise. 

(a) Exercise 1.4 Class 7 Question 1 

The very first question in this exercise in NCERT books online requires students to solve the integer expressions for its value. It consists of 9 part-questions that require solving integer expressions. 

Candidates need to understand the traits of an integer and how they are affected because of various mathematical operations like division, multiplication, addition, etc. Students struggling with the concept can refer our 7th Maths Exercise 1.4 Solutions and cross-refer to the framed out solutions.

(b) Exercise 1.4 Class 7 Question 2 

In this question, an expression with three variables, namely, a, b, and c, is given. Students need to verify if this equation is true by substituting different integer values offered in the question. Candidates can practice these questions with the shortcut techniques provided in our NCERT Solutions for Class 7 Maths Chapter 1 and self-assess for their upcoming exam. 

(c) Exercise 1.4 Class 7 Question 3 

NCERT books download for Class 7 Maths Integers have this question where students need to fill in the blanks with appropriate integer values to derive a meaningful integer equation. This question comprises of 8 sub-questions, each requiring to be filled with the right integer value. 

(d) Exercise 1.4 Class 7 Question 4 

Class 7 Maths Chapter 1 Exercise 1.4 question 4 tests a student’s understanding of the topic by inquiring about five possible combinations of integers that will give the same result. Candidates can analyze and answer the questions in the same manner. In this regard, our NCERT solutions are deemed the most preferred, considering the appropriate structuring of the answers and its explanation as framed by the pros in our team.  

(e) Exercise 1.4 Class 7 Question 5 

NCERT books PDF Maths Class 7 Chapter 1 help students learn the integer concepts thoroughly and practice questions based on it. As for this question, students are requested to answer the word problem, which involves the temperature at various times of the day. To find the accurate answer to such questions, candidates can refer our NCERT Solutions for Class 7 Maths Chapter 1 that prepares students, not only for this question but the likes of it too. 

(f) Exercise 1.4 Class 7 Question 6 

For incorporating a deeper understanding of this Mathematical concept, the latest edition of NCERT Solutions for Class 7 Maths Chapter 1 PDF has questions like question 6. Candidates have to analyze and apply their logical and reasoning skills on a test to be able to answer these questions. 

(g) Exercise 1.4 Class 7 Question 7 

Students can solve this question with their basic understanding of Mathematical concepts like integers, as explained in Exercise 1.4 Class 7. Our study guide can be referred to as a source to acquire an in-depth understanding of the topic to solve these questions. 

NCERT Solutions for Class 7 Maths Chapter 1 – Overview of Other Exercise Questions 

NCERT Solution for Class 7 Maths Chapter 1 is as per the syllabus set by the CBSE board and caters to the understanding of the students of Standard 7. Distinct types of questions are included in the other exercises as well to help students understand the importance of learning integers in day to day life. It runs true for Exercises 1.1, 1.2, and 1.3.

Students are taught from basic concepts of how an integer is given a positive or negative sign, to an intricate understanding of Mathematical operations on integers. They are also asked to compare two integer values and state, which has a higher value than the other. 

Acquiring a deeper understanding of these concepts can, however, be tiring without study aid like these solutions. These solutions can be read to practice questions and even prepare for upcoming exams.  

Vedantu – The Much-Needed Exam Preparation Study Material 

Vedantu is a single-stop solution for all your study aid that can be accessed from anywhere and anytime. We provide NCERT Solutions for all classes and all subjects on a chapter-wise basis for the ease of student’s learning. Get access to NCERT Solutions for Class 7 Maths Chapter 1 and practice sums related to integers. 

Vedantu helps students prepare for the upcoming exam or self-assess their knowledge through these solutions. Our subject professionals implement shortcut techniques to help students with both board questions as well as get them prepared for any competitive exams as well. 

Are you struggling with concepts related to your syllabus? Get your free PDF solutions right away!