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

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

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

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.