# NCERT Solutions For Class 7 Maths Chapter 1

### Ncert Solutions For Class 7 Maths Chapter 1 PDF Free Download

NCERT Solutions For Class 7 Maths Chapter 1 (integers) are given in a very detailed manner. The stepwise solutions given here can be of great help to the CBSE students who face any doubts while solving the questions from the class 7 maths book. Also, the class 7 NCERT Solutions for chapter 1 maths PDF is also provided here which the students can download and check at any time they want. The solutions for integers (i.e. chapter 1) given here are extremely easy to understand and any student can easily clear their doubts from here.

It is always important for the students to solve the questions from the NCERT textbooks as they help the students to have a better understanding of the respective concepts. The NCERT solutions are of great help to the students who pile up their doubts for later. Similarly, the integers chapter of class 7 maths book includes several practice questions for the students. The detailed exercise-wise NCERT solutions for class 7 maths chapter 1 PDF can be checked from below.

### NCERT Class 7 Solutions For Maths Chapter 1 Exercises

EXERCISE 1.1

Question 1:

The temperature (in degree Celsius) of different places on a particular day is represented using a number line.

i) Write the temperature of the places marked on the number line by observing it.

ii) Calculate temperature difference between the hottest place and the coldest place on the number line.

iii) Calculate the temperature difference between Lachung and Gangtok.

iv) Is the temperature of Gangtok and Majhitar taken together is less than the temperature at Majhitar? Is it also less than the temperature at Gangtok?

i) Following are the temperature of the places marked on the number line:

 Places Temperature Shillong 22oC Siliguri 14oC Majhitar 5oC Gangtok -2oC Lachung -8oC

ii) Temperature difference between the hottest and the coldest place =

Temperature of the hottest place (Shillong) – Temperature of the coldest place (Lachung)

= 22oC – (-80C) = 22oC + 8oC = 30oC

iii) Difference between Lachung and Gangtok = Temperature at Gangtok – Temperature at Lachung

= -2oC – (-8oC) = -2oC + 8oC = 6oC

iv) Temperature of Gangtok and Majhitar taken together = -2oC + 5oC = 3oC

Temperature at Majhitar = 5oC

Therefore, 3oC < 5oC

Yes, the temperature of Gangtok and Majhitar taken together is less than the temperature at Majhitar.

Now, the temperature of Gangtok = -2oC

Therefore, 3oC > -2oC

No, it is not less than the temperature at Gangtok.

Question 2:

In a quiz, positive marks and negative marks are given for correct answers and incorrect answers respectively. If Sangam’s scores in five successive round were 20, -2, -8, 12, and 15, what was his total at the end?

Given, Sangam’s scores in five successive rounds are 20, -2, -8, 12, and 15.

Total marks obtained by Sangam = 20 + (-2) + (-8) + 12 + 15 = 37

Thus, Sangam obtained 37 marks in a quiz.

Question 3:

At Gangtok, temperature on Monday was -5oC. On Tuesday, the temperature dropped by 3oC and on Wednesday it rose by 3oC. Calculate the temperature on Tuesday and Wednesday.

At Gangtok, temperature on Monday = -5oC

On Tuesday, temperature dropped by 3oC

Therefore, Temperature on Tuesday = -5oC – 3oC = -8oC

On Wednesday, temperature rose up by = 3oC

Therefore, Temperature on Wednesday = -8oC + 3oC = -5oC

Thus, temperature on Tuesday was -8oC and on Wednesday was -5oC.

Question 4:

The plane is flying at the height of 5000 m above sea level. A submarine is floating 1200 m below the sea level. Calculate the vertical distance between them when at a particular point the plane is exactly above the submarine.

Height of the plane above the sea level = 5000 m

Position of submarine below sea level = 1200 m

Therefore, the vertical distance between the plane and the submarine = 5000 + 1200 = 6200 m.

Question 5:

Reuben deposits Rs 3,000 in his bank account. On the next day, he withdraws Rs 1,560. Find the balance in Reuben’s account after the withdrawal.

Deposit amount = Rs 3,000 and Withdrawal amount = Rs 1,560

Therefore, Balance = 3,000 – 1,560 = Rs 1,440

Thus, the balance in Reuben’s account after withdrawal is Rs 1,440.

Question 6:

From a point P, Priyanka goes 40 km towards east, to the point Q. From Q, she moves 60 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance traveled towards west? By which integer will you represent her final position from P?

According to the number line, Priyanka moving towards east is represented by a positive integer. When she moves in the opposite direction, i.e. when she moves west, it is represented by a negative integer.

Distance from P to Q = 40 km

Distance from Q to R = 60 km

Distance from P to R = 40 – 60 = -20 km

Thus, Priyanka is at a final position from P to R is -20 km.

Question 7:

Check whether the following is a magic square (each row, column and diagonal have the same sum) or not.

Taking the first square box,

a) Taking rows

1 + (-10) + 0 = 1 – 10 = -9

(-4) + (-3) + (-2) = -4 – 3 – 2 = -9

(-6) + 4 + (-7) = -6 + 4 – 7 = -9

b) Taking columns

1 + (-4) + (-6) = -9

(-10) + (-3) + 4 = -9

0 + (-2) + (-7) = -9

c) Taking diagonals

1 + (-3) + (-7) = -9

0 + (-3) + (-6) = -9

This box is a magic square because all the sums are equal, i.e., -9.

Taking the second square box,

a) Taking rows

8 + 0 + (-8) = 0

(-5) + 5 + 0 = 0

(-3) + (-5) + 8 = 0

b) Taking columns

8 + (-5) + (-3) = 0

0 + 5 + (-5) = 0

(-8) + 0 + 8 = 0

c) Taking diagonals

8 + 5 + 8 = 21

(-8) + 5 + (-3) = -6

This box is not a magic square because not all the sums are equal.

Question 8:

Verify x – (-y) = x + y for the following values of x and y:

i) x = 22, y = 17

ii) x = 119, y = 124

iii) x = 76, y = 83

iv) x = 29, y = 10

i) Given: x = 22, y = 17

We have x – (-y) = x + y

Putting the values in L.H.S. = x – (-y) = 22 – (-17) = 22 + 17 = 39

Putting the values in R.H.S. = x + y = 22 + 17 = 39

Since, L.H.S = R.H.S.

Hence, verified.

ii) Given: x = 119, y = 124

We have x – (-y) = x + y

Putting the values in L.H.S. = x – (-y) = 119 – (-124) = 119 + 124 = 243

Putting the values in R.H.S. = x + y = 119 + 124 = 243

Since, L.H.S = R.H.S.

Hence, verified.

iii) Given: x = 76, y = 83

We have x – (-y) = x + y

Putting the values in L.H.S. = x – (-y) = 76 – (-83) = 76 + 83 = 159

Putting the values in R.H.S. = x + y = 76 + 83 = 159

Since, L.H.S = R.H.S.

Hence, verified.

iv) Given: x = 29, y = 10

We have x – (-y) = x + y

Putting the values in L.H.S. = x – (-y) = 29 – (-10) = 29 + 10 = 39

Putting the values in R.H.S. = x + y = 29 + 10 = 39

Since, L.H.S = R.H.S.

Hence, verified.

Question 9:

Use the sign >, < or = in the blank spaces to make the statement correct:

i) (-7) + (-5) ______ (-7) – (-5)

ii) (-2) + 6 – (-18) ______ 14 – 6 – (-8)

iii) 22 – 43 + 11 ______ 22 – 43 – 11

iv) 40 + (-23) – 15 ______ 20 + (-22) + 54

v) (-230) + 80 + 50 ______ (-299) + 160 + 80

i) (-7) + (-5) ______ (-7) – (-5)

-7 – 5 ______ -7 + 5

-12 ______ -2

-12 < -2

ii) (-2) + 6 – (-18) ______ 14 – 6 – (-8)

-2 +6 +18 ______ 14 – 6 + 8

22 ______ 16

22 > 16

iii) 22 – 43 + 11 ______ 22 – 43 – 11

-10 ______ -32

-10 > -32

iv) 40 + (-23) – 15 ______ 20 + (-22) + 54

40 – 23 – 15 ______ 20 – 22 +54

2 ______ 52

2 < 52

v) (-230) + 80 + 50 ______ (-299) + 160 + 80

-230 + 80 +50 ______ -299 +160 +80

-100 ______ -59

-100 < -59

Question 10:

A monkey is sitting on the topmost step (i.e., the first step) of a water tank. The level of water is at the ninth step:

a) The monkey jumps 3 steps down and then jumps back 2 steps up. In how many jumps he can reach the water level.

b) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps 2 steps back down in every move. In how many jumps will he reach back the top step?

c) If the number of steps moved down and the number of steps moved up is represented by negative integers and positive integers respectively, represent his moves in parts (a) and (b) by completing the following:

i)-3 + 2 – ______ = -8

ii) 4 – 2 + ______ = 8

In (i) the sum (-8) represent going down by eight steps. So, what will the sum 8 in (ii) represent?

a) Monkey jumps 3 steps down and jumps back 2 steps up.

First jump = 1 + 3 = 4 steps

Second jump = 4 – 2 = 2 steps

Third jump = 2 + 3 = 5 steps

Fourth jump = 5 – 2 = 3 steps

Fifth jump = 3 + 3 = 6 steps

Sixth jump = 6 – 2 = 4 steps

Seventh jump = 4 + 3 = 7 steps

Eighth jump = 7 – 2 = 5 steps

Ninth jump = 5 + 3 = 8 steps

Tenth jump = 8 – 2 = 6 steps

Eleventh jump = 6 + 3 = 9 steps

He will reach ninth steps in 11 jumps.

b) The monkey jumps four steps and then jumps down 2 steps. Thus monkey reach back on the first step in fifth jump.

c)

i) -3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 -3 + 2 – 3 + 2 = -8

ii) 4 – 2 + 4 – 2 + 4 – 2 + 4 – 2 = 8

Thus, sum 8 in (ii) represents going up by eight steps.

EXERCISE 1.2

Question 1:

Write down a pair of integers in which:

i) sum is -6

ii) difference is -8

iii) sum is 1

Answer: 5 + (-4) = 1

Question 2:

i) Write a pair of negative integers whose difference gives 7.

Answer: (-2) – (-9) = -2 + 9 = 7

ii) Write a negative integer and a positive integer whose sum is -6.

Answer: (-7) + 1 = -6

iii) Write a negative integer and a positive integer whose difference is -7.

Answer: (-4) – 3 = -4 – 3 = =7

Question 3:

A quiz was held in which team A scored -35, 12, 0 and team B scored 12, 8, -30 in three successive rounds. Who won the quiz by scoring more marks? Can the integers be added in any manner?

Score of Team A = -35, 12, 0

Total score = -35 + 12 + 0 = -23

Score of Team B = 12, 8, -30

Total score = 12 + 8 + (-30) = -10

As -10 > -30, Team B won.

Yes, the integers can be added in any manner.

Question 4:

Fill in the blanks.

i) (-4) + (-6) = (-6) + (__)

ii) -34 + ______ = -34

iii) 18 + _____ = 0

iv) [14 + (-11)] + (___) = 14 + [(-11) + (-9)]

i) (-4) + (-6) = (-6) +(-5)[Commutative property]

ii) -34 +0= -34 [Zero additive property]

iii) 18 + (-18) = 0                                                                  [Additive identity]

iv) [14 + (-11)] + (-9) = 14 + [(-11) +(-9)][Associative property]

EXERCISE 1.3

Question 1:

Find the product of the following:

(a) 4 x (–1)

= -4

(b) (–1) x 345

= -345

(c) (–12) x (–20)

= 240

(d) (–136) x (–1)

= 136

(e) (–13) x 0 x (–17)

= 0

(f) (–11) x (–13) x (10)

= 1430

(g) 10 x (–4) x (–8)

= 320

(h) (–16) x (–5) x (–2)

= – 160

(i) (–2) x (–1) x (–4) x 4

= – 32

(j) (–4) x (–5) x (2) x (–1)

= – 40

Question 2:

Verify:

i) 16 x [4 + (–3)] = [16 x 4] + [16 x (–3)]

ii) (–22) x [(–6) + (–4)] = [(–22) x (-6)] + [(–22) x (–4)]

i) 16 x [4 + (–3)] = [16 x 4] + [16 x (–3)]

16 x 1 = 64 + (-48)

16 = 16

L.H.S. = R.H.S.

Hence, verified.

ii) (–22) x [(–6) + (–4)] = [(–22) x (-6)] + [(–22) x (–4)]

(-22) x (-10) = 132 + 88

220 = 220

L.H.S. = R.H.S.

Hence, verified.

Question 3:

a) For any integer b, what is(-1) x b equal to?

b) Determine the integer whose product with(-1)is:

i) –12

ii) 47

iii) 0

a) (-1) x b = -b, where b is an integer.

b)

i) (-1) x (-12) = 12

ii) (-1) x 47 = -47

iii) (-1) x 0 = 0

Question 4:

Starting from (-1) x 7, write various products showing some patterns to show (-1) x (-1) = 1.

(-1) x 7 = -7                         (-1) x 3 = -3              (-1) x (-1) = 0

(-1) x 6 = -6                          (-1) x 2 = -2

(-1) x 5 = -5                          (-1) x 1 = -1

(-1) x 4 = -4                          (-1) x 0 = 0

Thus, we can conclude that it shows that if we multiply a negative integer with a positive integer the outcome is a negative integer whereas if we multiply a negative integer with another negative integer number then the outcome is a positive integer.

Question 5:

Using suitable properties, find the product of the following:

i) (-48) x 26 + 26 x (-15)

ii) 7 x 53 x (-145)

iii) (-55) x (-18) +57

i) (-48) x 26 + 26 x (-15)

26 x [(-48) + (-15)]                                                       [Distributive property]

26 x (-63) = -1638

ii) 7 x 53 x (-145)

7 x [53 x (-145)]                                                [Commutative property]

7 x (-7685) = -53795

iii) (-55) x (-18) + 55

(-55) x (-18) + 55 x 1

55 x 18 + 55 x 1

55 x (18+1)                                                        [Distributive property]

55 x 19 = 1045

Question 6:

A certain freezing process requires the room temperature to be lowered from 40oC at the rate of 5oC every hour. What will be the room temperature after 10 hours?

Given:

Room temperature at present = 40oC

Rate = 5o per hour

Decrease in temperature in 10 hours = 50oC

Room temperature after 10 hours = 40oC – 50oC = -10oC

Thus, the room temperature after 10 hours, after the process begins, is -10oC.

Question 7:

A class test is taken comprising of 10 questions. For every correct answer 5 marks is awarded, for every incorrect answer (-2) marks is awarded and 0 marks is awarded for questions not attempted. Calculate the total score when:

a) Vipul gets seven correct and three incorrect answers.

a) Marks for correct answers = 7 x 5 = 35

Marks for incorrect answers = 3 x (-2) = -6

Total score = 35 + (-6) = 29

Thus, Vipul gets 29 marks in a class test.

b) Marks for correct answers = 5 x 5 = 25

Marks for incorrect answers = 5 x (-2) = -10

Toatal score = 25 + (-10) = 15

Thus, Ankesh gets 15 marks in a class test.

Question 8:

A cement company earns a profit of Rs 10 per bag of white cement sold and loss of Rs 6 per bag of grey cement sold.

i) What is its profit or loss when the company sells 2,000 bags of white cements and 4,000 bags of grey cement in a month?

ii) The number of grey cement bags sold is 6,500 bags. How many white cement bags the company must sell to have neither profit nor loss?

Given:

Profit of 1 white cement bag = Rs 10

Loss of 1 grey cement bag = Rs 6

i) Profit on selling 2,000 bags of white cement bags = 2000 x 10 = Rs 20000

Loss on selling 4,000 bags of grey cement = 4000 x 6 = Rs 24000

Since, Profit < Loss

Therefore, company’s total loss = Loss – Profit = 24000 – 20000 = Rs 4,000

Thus, company’s lost is Rs 4,000.

ii) Let the number of bags of white cement be x.

Loss on selling 6,500 bags of grey cement = 6500 x 6 = Rs 39000

According to the question, Profit = Loss

Therefore, 10x = 39000

x = 39000/10 = 3,900

Thus, company must sell 3,900 bags of white cement to have neither profit nor loss.

Question 9:

Fill up the blanks with suitable integer.

i) (-4) x ________ = 24

ii) (-5) x _______ = 45

iii) _______ x (-9) = -81

iv) _______ x (-12) = 12

i) (-4) x(-6)= 24

ii) (-5) x(-9)= 45

iii) x (-9) = -81

iv)(-1)x (-12) = 12

EXERCISE 1.4

Question 1:

Evaluate:

i) (-20)¸10

(-20) ¸ 10 = $(-20) \times \frac{1}{10}=-\frac{20\times 1}{10}=-2$

ii) 60¸(-6)

60 ¸ (-6) = $60\times (-\frac{1}{6})=-\frac{60\times 1}{6}=-10$

iii) (-45) ¸ (-9)

(-45) ¸ (-9) = $(-45)\times (-\frac{1}{9})=\frac{45\times 1}{9}=5$

iv) 14¸[(-2) + 1]

14 ¸ [(-2) + 1] = 14 ¸ (-1) = $13\times (\frac{-1}{1})=-13$

v) [(-48)¸12] ¸ 2

[(-48) ¸ 12] ¸ 2 =

$[(-48)\times \frac{1}{12}]\times \frac{1}{2}=(\frac{-36}{12})\times \frac{1}{2}=(-4)\times \frac{1}{2}=-2$

Question 2:

Verify that x ¸ (y + z)  (x ¸ y) + (x ¸ z) for the following values of x, y and z.

i) x = 18, y = 6, z = 3

ii) x = (-6), y = 1, z = 1

i) Given: x ¸ (y + z) ≠ (x ¸ y) + (x ¸ z) and x = 18, y = 6, z = 3

Putting the given values in L.H.S. = 18 ¸ (6 + 3) = $18\times \frac{1}{9}=\frac{18\times 1}{9}=\frac{18}{9}=2$

Putting the given values in R.H.S. = (18 ¸ 6) + (18 ¸ 3) = $(18\times \frac{1}{6})+(18\times \frac{1}{3})=3+6=9$

Since, L.H.S. ≠ R.H.S.

Hence, verified.

ii) Given: x ¸ (y + z) ≠ (x ¸ y) + (x ¸ z) and x = (-6), y = 1, z = 1

Putting the given values in L.H.S. = (-6) ¸ (1 + 1) = -3

Putting the given values in R.H.S. = [(-6) ¸ 1] + [(-6) ¸ 3] = -8

Since, L.H.S. ≠ R.H.S.

Hence, verified.

Question 3:

Fill up the blanks:

i) 460¸________ = 460

ii) (-69)¸_______ = (-1)

iii) (-123) ¸ ________ = 1

iv) (-34)¸________ = 34

v) _______¸1 = -45

vi) _______¸34 = -1

vii) 40 ¸ ________ = -4

viii) ________ ¸ (3) = -3

i) 460 ¸1= 460

ii) (-69) ¸69= (-1)

iii) (-123) ¸ (-123) = 1

iv) (-34) ¸(-1)= 34

v)(-45)¸ 1 = -45

vi)(-34)¸ 34 = -1

vii) 40 ¸ (-10) = -4

viii) (-9) ¸ (3) = -3

Question 4:

Write four pairs of integers (x,y) such that x ¸ y = -2.

i) (2,-1)

ii) (-2,1)

iii) (4,-2)

iv) (-4,2)

Question 5:

The temperature at noon was 10oC. Until midnight, it decreases at the rate of 2oC per hour. At what time the temperature will be -8oC?

The temperature is represented in the following number line:

The temperature decreases 2oC = 1 hour

The temperature decreases 1oC = ½ hour

The temperature decreases 18oC = ½ x 18 = 9 hours

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

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

Question 6:

A class test is taken in which (+3) marks are given for every correct answer, (-2) marks are given for every incorrect answer and no marks for not attempting any question.

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

ii) Radhika scored (-5) marks. She got 7 correct answers. Calculate the number of questions she attempted incorrectly.

Given:

Marks obtained for one correct answer = 3

Marks obtained for one incorrect answer = -2

i) Marks obtained for 12 correct answers = 3 x 12 = 36

Rohini scored 20 marks.

Therefore, marks obtained for incorrect answers = 20 – 36 = -16

Number of incorrect answers = (-16) ¸ (-2) = 8

Thus, Rohini has attempted 8 incorrect questions.

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

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

Number of incorrect answers = (-26) ¸ (-2) = 13

Thus, Radhika attempted 13 incorrect questions.

Question 7:

The rate of descending of an elevator into a mine shaft is 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350 m?

Total distance covered by mine shaft = 10 m – (-350 m) = 10 + 350 = 360 m

According to the question, time taken to cover a distance of 6m = 1 min

So, time taken to cover a distance of 1 m = 1/6 min

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

= 60 minutes

= 1 hour

Thus, the elevator will reach -350 m, from 10 m, in one hour.

In chapter 1 of CBSE class 7 maths book, the concepts of integers are first recalled along with the basics of the number line. Then, students are introduced to different important topics like the properties of addition and subtraction of integers, multiplication of integers and properties of multiplication of integers, and division of integers along with their properties.

All the sub-topics included in the integers chapter are extremely crucial for the students as these basic concepts are also included in several higher level topics in the later grades. So, students need to be thorough with all the concepts in this chapter to be able to comprehend topics in the higher grades.

In the NCERT maths book of class 7, chapter 1 is integers and several activities are included along with numerous practice questions. Students need to solve these questions to be able to have a proper understanding of the concepts and develop a deeper knowledge about integers. Students can always refer to these solutions for chapter 1 here.

Apart from these solutions, students can also get the detailed NCERT Solutions For Class 7 Maths for all the chapters at BYJU’S. Students are also provided with several notes, sample papers and question papers here to help them to prepare in a more effective way.