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NCERT Solutions for Class 7 Maths 1.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 7 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

**NCERT solutions for Class 7 Integers Maths**** Download as PDF**

## NCERT Solutions for Class 7 Maths Integers

**Class –VII Mathematics (Ex. 1.4)**

**Question** 1.Evaluate each of the following:

(a) {tex}\left( { – 30} \right) \div 10{/tex}

(b) {tex}50 \div \left( { – 5} \right){/tex}

(c) {tex}\left( { – 36} \right) \div \left( { – 9} \right){/tex}

(d) {tex}\left( { – 49} \right) \div 49{/tex}

(e) {tex}13 \div \left[ {\left( { – 2} \right) + 1} \right]{/tex}

(f) {tex}0 \div \left( { – 12} \right){/tex}

(g) {tex}\left( { – 31} \right) \div \left[ {\left( { – 30} \right) \div \left( { – 1} \right)} \right]{/tex}

(h) {tex}\left[ {\left( { – 36} \right) \div 12} \right] \div 3{/tex}{tex}\left( { – 206} \right) \div \_\_\_\_\_\_\_ = 1{/tex}

(i) {tex}\left[ {\left( { – 6} \right) + 5} \right] \div \left[ {\left( { – 2} \right) + 1} \right]{/tex}

**Answer:**

(a) {tex}\left( { – 30} \right) \div 10{/tex} = {tex}\left( { – 30} \right) \times \frac{1}{{10}} = \frac{{ – 30 \times 1}}{{10}} = – 3{/tex}

(b) {tex}50 \div \left( { – 5} \right){/tex} = {tex}50 \times \left( {\frac{{ – 1}}{5}} \right) = \frac{{50 \times \left( { – 1} \right)}}{5} = – 10{/tex}

(c) {tex}\left( { – 36} \right) \div \left( { – 9} \right){/tex} = {tex}\left( { – 36} \right) \times \left( {\frac{{ – 1}}{9}} \right) = \frac{{\left( { – 36} \right) \times \left( { – 1} \right)}}{9} = \frac{{36}}{9} = 4{/tex}

(d) {tex}\left( { – 49} \right) \div 49{/tex} = {tex}\left( { – 49} \right) \times \frac{1}{{49}} = \frac{{ – 49}}{{49}} = – 1{/tex}

(e) {tex}13 \div \left[ {\left( { – 2} \right) + 1} \right]{/tex} = {tex}13 \div \left( { – 1} \right) = 13 \times \left( {\frac{{ – 1}}{1}} \right) = – 13{/tex}

(f) {tex}0 \div \left( { – 12} \right){/tex} = {tex}0 \times \left( {\frac{{ – 1}}{{12}}} \right) = \frac{0}{{12}} = 0{/tex}

(g) {tex}\left( { – 31} \right) \div \left[ {\left( { – 30} \right) \div \left( { – 1} \right)} \right]{/tex} = {tex}\left( { – 31} \right) \div \left( { – 30 – 1} \right) = \left( { – 31} \right) \div \left( { – 31} \right) = \left( { – 31} \right) \times \left( {\frac{{ – 1}}{{31}}} \right) = \frac{{31}}{{31}} = 1{/tex}

(h) {tex}\left[ {\left( { – 36} \right) \div 12} \right] \div 3{/tex} = {tex}\left[ {\left( { – 36} \right) \times \frac{1}{{12}}} \right] \times \frac{1}{3} = \left( {\frac{{ – 36}}{{12}}} \right) \times \frac{1}{3} = \left( { – 3} \right) \times \frac{1}{3} = \frac{{ – 3}}{3} = – 1{/tex}

(i) {tex}\left[ {\left( { – 6} \right) + 5} \right] \div \left[ {\left( { – 2} \right) + 1} \right]{/tex} = {tex}\left( { – 6 + 5} \right) \div \left( { – 2 + 1} \right) = \left( { – 1} \right) \div \left( { – 1} \right) = \left( { – 1} \right) \times \frac{{\left( { – 1} \right)}}{1} = 1{/tex}

NCERT Solutions for Class 7 Maths 1.4

**Question** 2.Verify that {tex}a \div \left( {b + c} \right) \ne \left( {a \div b} \right) + \left( {a \div c} \right){/tex} for each of the following values of {tex}a,b{/tex} and {tex}c.{/tex}

(a) {tex}a = 12,b = – 4,c = 2{/tex}

(b) {tex}a = \left( { – 10} \right),b = 1,c = 1{/tex}

**Answer:**

(a) Given: {tex}a \div \left( {b + c} \right) \ne \left( {a \div b} \right) + \left( {a \div c} \right){/tex}

{tex}a = 12,b = – 4,c = 2{/tex}

Putting the given values in L.H.S. = {tex}12 \div \left( { – 4 + 2} \right){/tex}

= {tex}12 \div \left( { – 2} \right) = 12 \div \left( {\frac{{ – 1}}{2}} \right) = \frac{{ – 12}}{2} = – 6{/tex}

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

= {tex}\left( {12 \times \frac{{ – 1}}{4}} \right) + 6 = – 3 + 6 = 3{/tex}

Since, L.H.S. {tex} \ne {/tex} R.H.S.

Hence verified.

(b) Given: {tex}a \div \left( {b + c} \right) \ne \left( {a \div b} \right) + \left( {a \div c} \right){/tex}

{tex}a = – 10,b = 1,c = 1{/tex}

Putting the given values in L.H.S. = {tex} – 10 \div \left( {1 + 1} \right){/tex}

= {tex} – 10 \div \left( 2 \right) = – 5{/tex}

Putting the given values in R.H.S. = {tex}\left[ { – 10 \div 1} \right] + \left( { – 10 \div 1} \right){/tex}

= {tex} – 10 – 10 = – 20{/tex}

Since, L.H.S. {tex} \ne {/tex} R.H.S.

Hence verified.

NCERT Solutions for Class 7 Maths 1.4

**Question** 3.Fill in the blanks:

(a) {tex}369 \div \_\_\_\_\_\_\_ = 369{/tex}

(b) {tex}\left( { – 75} \right) \div \_\_\_\_\_\_\_ = \left( { – 1} \right){/tex}

(c) {tex}\left( -206 \right)\div \_\_\_\_\_\_\_=1{/tex}

(d) {tex}\left( { – 87} \right) \div \_\_\_\_\_\_\_ = 87{/tex}

(e) {tex}\_\_\_\_\_\_\_ \div 1 = – 87{/tex}

(f) {tex}\_\_\_\_\_\_\_ \div 48 = – 1{/tex}

(g) {tex}20 \div \_\_\_\_\_\_\_ = – 2{/tex}

(h) {tex}\_\_\_\_\_\_\_ \div \left( 4 \right) = – 3{/tex}

**Answer:**

(a) {tex}369 \div \underline 1 = 369{/tex}

(b) {tex}\left( { – 75} \right) \div \underline {75} = \left( { – 1} \right){/tex}

(c) {tex}\left( { – 206} \right) \div \underline {\left( { – 206} \right)} = 1{/tex}

(d) {tex}\left( { – 87} \right) \div \underline {\left( { – 1} \right)} = 87{/tex}

(e) {tex}\underline {\left( { – 87} \right)} \div 1 = – 87{/tex}

(f) {tex}\underline {\left( { – 48} \right)} \div 48 = – 1{/tex}

(g) {tex}20 \div \underline {\left( { – 10} \right)} = – 2{/tex}

(h) {tex}\underline {\left( { – 12} \right)} \div \left( 4 \right) = – 3{/tex}

NCERT Solutions for Class 7 Maths 1.4

**Question** 4.Write five pairs of integers {tex}\left( {a,b} \right){/tex} such that {tex}a \div b = – 3.{/tex} One such pair is {tex}\left( {6, – 2} \right){/tex} because {tex}6 \div \left( { – 2} \right) = \left( { – 3} \right).{/tex}

**Answer:**

(i) {tex}\left( { – 6} \right) \div 2 = – 3{/tex}

(ii) {tex}9 \div \left( { – 3} \right) = – 3{/tex}

(iii) {tex}12 \div \left( { – 4} \right) = – 3{/tex}

(iv) {tex}\left( { – 9} \right) \div 3 = – 3{/tex}

(v) {tex}\left( { – 15} \right) \div 5 = – 3{/tex}

NCERT Solutions for Class 7 Maths 1.4

**Question** 5.The temperature at noon was 10oC above zero. If it decreases at the rate of 2oC per hour until mid-night, at what time would the temperature be 8oC below zero? What would be the temperature at mid-night?

**Answer:**

Following number line is representing the temperature:

The temperature decreases 2oC = 1 hour

The temperature decreases 1oC = {tex}\frac{1}{2}{/tex} hour

The temperature decreases 18oC = {tex}\frac{1}{2} \times 18{/tex} = 9 hours

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

Thus, at 9 pm the temperature would be 8oC below 0oC.

NCERT Solutions for Class 7 Maths 1.4

**Question** 6.In a class test (+3) marks are given for every correct answer and {tex}\left( { – 2} \right){/tex} marks are given for every incorrect answer and no marks for not attempting any question.

(i)Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?

(ii)Mohini scores {tex}\left( { – 5} \right){/tex} marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectlt?

**Answer:**

(i) Marks given for one correct answer = 3

Marks given for 12 correct answers = 3 x 12 = 36

Radhika scored 20 marks.

Therefore, Marks obtained for incorrect answers = 20 – 36 = –16

Now, marks given for one incorrect answer = –2

Therefore, number of incorrect answers = {tex}\left( { – 16} \right) \div \left( { – 2} \right) = 8{/tex}

Thus, Radhika has attempted 8 incorrect questions.

(ii) Marks given for seven correct answers = 3 x 7 = 21

Mohini scores = –5

Marks obtained for incorrect answers = = –5 – 21 = –26

Now, marks given for one incorrect answer = –2

Therefore, number of incorrect answers = {tex}\left( { – 26} \right) \div \left( { – 2} \right) = 13{/tex}

Thus, Mohini has attempted 13 incorrect questions.

NCERT Solutions for Class 7 Maths 1.4

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

**Answer:**

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 = {tex}\frac{1}{6}{/tex} minute

Therefore, time taken to cover a distance of 360 m = {tex}\frac{1}{6} \times 360{/tex} = 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 1.4

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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