# NCERT Solutions for Class 7 Maths Chapter 1 Integers| PDF Download

**Chapter 1 Integers NCERT Solutions for Class 7 Maths**are helpful resources that can help you not only cover the entire syllabus but also provide in depth analysis of the topics. Chapter 1 Integers NCERT Solutions PDF Download available here will give good experience and provide opportunities to learn new things.

**Chapter 1 Class 7 Maths NCERT Solutions**will help you in getting command over the problems present in the exercises.

Through the help of

**NCERT Solutions for Class 7**, you can also complete your homework on time and increase concentrations. It can be used to enrich knowledge and make lessons for learners more exciting also learning efficiently.**Try these**

We have done various patterns with numbers in our previous class. Can you find a pattern for each of the following? If yes, complete them:

(a) 7, 3, – 1, – 5, _____, _____, _____.

(b) – 2, – 4, – 6, – 8, _____, _____, _____.

(c) 15, 10, 5, 0, _____, _____, _____.

(d) – 11, – 8, – 5, – 2, _____, _____, _____.

Make some more such patterns and ask your friends to complete them.

**Answer**

**(a) 7, 3, -1, -5, -9, -13, -17**

(b) -2, -4, -6, -8, -10, -12, -14

(c) 15, 10, 5, 0, -5, -10, -15

(d) -11, -8, -5, -2, -1, -4, -7

Page No: 4

**Exercise 14.1**

1. Following number line shows the temperature in degree celsius (°C) at different places on a particular day.

(a) Observe this number line and write the temperature of the places marked on it.

(b) What is the temperature difference between the hottest and the coldest places
among the above

(c) What is the temperature difference between Lahulspiti and Srinagar?

(d) Can we say temperature of Srinagar and Shimla taken together is less than the
temperature at Shimla? Is it also less than the temperature at Srinagar?

**Answer**

(a) The temperatures of the places marked on number line are:

Lahulspiti : -8⁰C

Srinagar : -2⁰C

Shimla : 5⁰C

Ooty : 14⁰C

Bangalore : 22⁰C

(b) Temperature at the hottest place = 22⁰C

Temperature at the coldest place = -8⁰C

The temperature difference between these places = 22⁰C - (-8⁰C) = 30⁰C

(c) The temperature at Lahulspiti = -8⁰C

The temperature at Srinagar = -2⁰C

The temperature difference between these= -8⁰C - (-2⁰C) = -6⁰C

(d) Temperature of Srinagar= -2⁰C

Temperature of Shimla = 5⁰C

Temperature of Srinagar and Shimla taken together = -2⁰C + 5⁰ = 3⁰C

Therefore: 3⁰C < 5⁰C

Yes, we can say that temperature of Srinagar and Shimla taken together is less than the temperature at Shimla.

However, 3⁰C > -2⁰C

Thus, the temperature at Srinagar is not less than the temperature at Shimla.

2. In a quiz, positive marks are given for correct answers and negative marks are given
for incorrect answers. If Jack’s scores in five successive rounds were 25, –5, –10,
15 and 10, what was his total at the end?

**Answer**

Jack’s scores in five successive rounds were 25, -5, -10, 15 and 10.

His total at the end = 25 – 5 – 10 + 15 + 10 = 35

3. At Srinagar temperature was – 5°C on Monday and then it dropped
by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday?
On Wednesday, it rose by 4°C. What was the temperature on this
day?

**Answer**

Temperature on Monday= -5⁰C

Temperature on Tuesday = -5⁰C - 2⁰C = -7⁰C

Temperature on Wednesday= -7⁰C + 4⁰C = -3⁰C

Hence, the temperature on Tuesday and Wednesday was -7⁰C and -3⁰C.

4. A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

**Answer**

Height of plane is above the sea level = 5000 m

A submarine is floating below the sea level = -1200m

Vertical distance between them = 5000 m - (-1200)m = 5000 m + 1200m = 6200m

5. Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642
from it, the next day. If withdrawal of amount from the account is
represented by a negative integer, then how will you represent the amount
deposited? Find the balance in Mohan’s account after the withdrawal.

**Answer**

If withdrawal of amount from the account is represented by a negative integer, then the amount deposited will be represented by a positive integer.

A/q,

Amount deposited by Mohan = Rs. 2,000

Amount withdrawn = Rs. (-1642)

Balance in Mohan’s account = 2000 + (-1642) = 2000 - 1642 = 358

6. Rita goes 20 km towards east from a point A to the point B. From B,
she moves 30 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 travelled towards west? By which integer will
you represent her final position from A?

**Answer**

If the distance towards east is represented by a positive integer then the distance travelled towards west will be represented by negative integer.

Distance travelled towards east direction = 20 km

Distance travelled towards west direction = 30 km

Distance travelled from A = 20 + (-30) = -10 km

7. In a magic square each row, column and diagonal have the same sum. Check which
of the following is a magic square.

**Answer**

In square (i) each row, column and diagonal have the same sum i.e., 0. However, sum of one of its diagonals is not 0.

Therefore, square (i) is not a magic square.

In square (ii) each row, column and diagonal have the same sum i.e., -9.

Therefore, square (ii) is a magic square.

8. Verify a - (-b) = a + b for the following values of a and b.

(i) a = 21, b = 18

(ii) a = 118, b = 125

(iii) a = 75, b = 84

(iv) a = 28, b = 11

**Answer**

(i) a = 21, b= 18

→ a - (-b) = 21 - (-18) = 21 +18 = 39

→ a + b = 21 + 18 = 39

∴ a - (-b) = a + b = 39

(ii) a= 118, b= 125

→ a - (-b) = 118 - (-125) = 118 + 125 = 243

→ a + b = 118 + 125 = 243

∴ a - (-b) = a + b = 243

(iii) a=75, b= 84

→ a - (-b) = 75 - (-84) = 75 + 84 = 159

→ a + b = 75 + 84 = 159

∴ a - (-b) = a + b = 159

(iv) a=28, b= 11

→ a - (-b) = 28 - (-11) = 28 + 11 = 39

→ a + b = 28 +11 = 39

∴ a - (-b) = a + b = 39

9. Use the sign of >, < or = in the box to make the statements true.

(a) (–8) + (–4) ☐ (–8) – (–4)

(b) (–3) + 7 – (19) ☐ 15 – 8 + (–9)

(c) 23 – 41 + 11 ☐ 23 – 41 – 11

(d) 39 + (–24) – (15) ☐ 36 + (–52) – (–36)

(e) –231 + 79 + 51 ☐ –399 + 159 + 81

**Answer**

(a) LHS = (–8) + (–4) = -8-4 = -12

RHS = (–8) – (–4) = -8+4 = -4

∴ -12 < -4

(b) LHS = (–3) + 7 – (19) = -15

RHS = 15 – 8 + (–9) = 15 - 8 - 9 = -2

∴ -15 < -2

(c) LHS = 23 – 41 + 11 = -7

RHS = 23 – 41 – 11 = 29

∴ -7 > -29

(d) LHS = 39 + (-24) - (15) = 39 - 24 - 15 = 0

RHS = 36 + (-52) - (-36) = 36 - 52 + 36 = 20

∴ 0 < 20

(e) LHS = –231 + 79 + 51 = -101

RHS = –399 + 159 + 81 = -159

∴ -101 > -159

10. A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first
step). The water level is at the ninth step.

(i) He jumps 3 steps down and then jumps back 2 steps up.
In how many jumps will he reach the water level?

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

(iii) If the number of steps moved down is represented by
negative integers and the number of steps moved up by
positive integers, represent his moves in part (i) and (ii)
by completing the following; (a) – 3 + 2 – ... = – 8
(b) 4 – 2 + ... = 8. In (a) the sum (– 8) represents going
down by eight steps. So, what will the sum 8 in (b)
represent?

**Answer**

(a) The monkey was at step = 1

After first jump he will be at the step = 1+3 = 4

After second jump he will at the step = 4 + (-2) = 2

After third jump he will be at the step = 2 + 3 = 5

After fourth jump he will be at the step = 5 + (-2) = 3

After fifth jump he will be at the step = 3 + 3 = 6

After sixth jump he will be at the step = 6 + (-2) = 4

After seventh jump he will be at the step = 4 + 3 = 7

After eighth jump he will be at the step = 7 + (-2) = 5

After ninth jump he will be at the step = 5 + 3 = 8

After tenth jump he will at the step = 8 + (-2) = 6

After eleventh jump he will at the step = 6 + 3 = 9

Therefore, after 11th jump the monkey will reach the water level.

(b) The monkey was at the step 9.

After first jump he will be at the step = 9 + (-4) = 5

After second jump he will be at the step = 5 + 2 = 7

After third jump he will be at the step = 7 + (-4) = 3

After fourth jump he will be at the step = 3 + 2 = 5

After fifth jump he will be at the step = 5 + (-4) = 1

Therefore, after fifth jump he will reach back the top step.

(c) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, then his moves in part (i):

-3 + 2 -3 + 2 -3 + 2 -3 + 2 -3 + 2 -3 = -8

moves in part (ii):

4 -2 +4 -2 +4 = 8

Moves in part (ii) represent going up by 8 steps.

Page No: 8

**Try these**

1. Write a pair of integers whose sum gives

(a) a negative integer

(b) zero

(c) an integer smaller than both the integers.

(d) an integer smaller than only one of the integers.

(e) an integer greater than both the integers.

**Answer**

(a) 2 + (-8) = -6

(b) 4 + (-4) = 0

(c) -2 + -4 = -6

(d) 4 + (-2) = 2

(e) 2 + 5 = 7

(c) -2 + -4 = -6

(d) 4 + (-2) = 2

(e) 2 + 5 = 7

2. Write a pair of integers whose difference gives

(a) a negative integer.

(b) zero.

(c) an integer smaller than both the integers.

(d) an integer greater than only one of the integers.

(e) an integer greater than both the integers.

**Answer**

(a) 2 - 8 = -6

(b) 4 - 4 = 0

(c) 2 - 7 = -5

(d) 9 - 3 = 6

(e) 4 - (-3) = 7

Page No: 9

**Exercise 1.2**

1. Write down a pair of integers whose:

(a) sum is -7

(b) difference is –10

(c) sum is 0

**Answer**

(a) -8 + (+1) = -7

(b) -12 – (-2) = -10

(c) 5 + (-5) = 0

2. (a) Write a pair of negative integers whose difference gives 8.

(b) Write a negative integer and a positive integer whose sum is –5.

(c) Write a negative integer and a positive integer whose difference is –3.

**Answer**

(a) -2 - (-10) = 8

(b) -8 +3 = -5

(c) -2 - (+1) = -3

3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, - 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

**Answer**

Team A scored = -40, 10, 0

Total scored by Team A= -40 +10 +0 = -30

Team B scored = 10, 0, - 40

Total scored by Team B = 10 +0 + (-40) = -30

Hence, both teams scored equal.

4. Fill in the blanks to make the following statements true:

(i) (–5) + (............) = (– 8) + (............)

(ii) –53 + ............ = –53

(iii) 17 + ............ = 0

(iv) [13 + (– 12)] + (............) = ............ + [(–12) + (–7)]

(v) (– 4) + [............ + (–3)] = [............ + 15] + ............

**Answer**

(i) (-5) +

__(-8)__= (-8) +__(-5)__
(ii) -53 +

__0__= -53
(iii) 17 +

__(-17)__= 0
(iv) [13 + (-12)] +

__(-7 )__=__13__+ [(-12) + (-7)]
(v) (-4) + [

__15__+ (-3)] = [__-4__+ 15] +__(-3)__
Page No: 10

**Try these**

Find:

4 × (– 8),

8 × (–2),

3 × (–7),

10 × (–1) using number line.

4 × (– 8) means -8 + (-8) + (-8) + 8) = -32

1. Evaluate each of the following:

(a) (−30)÷10

(b) 50÷(−5)

(c) (−36)÷(−9)

(d) (−49)÷49

(e) 13÷[(−2)+1]

(f) 0÷(−12)

(g) (−31)÷[(−30)+(−1)]

(h) [(−36)÷12]÷3

(i) [(−6)+5]÷[(−2)+1]

(a) (−30)÷10

(b) 50÷(−5)

(c) (−36)÷(−9)

(d) (−49)÷49

(e) 13÷[(−2)+1]

(f) 0÷(−12)

(g) (−31)÷[(−30)+(−1)]

(h) [(−36)÷12]÷3

(i) [(−6)+5]÷[(−2)+1]

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

(a) a=12,b=−4,c=2

(b) a=(−10),b=1,c=1

(a) Given: a÷(b+c) ≠ (a÷b)+(a÷c)

a=12,b=−4,c=2

Putting the given values in L.H.S. = 12÷(−4+2)

= 12÷(−2)

Hence, verified.

(b) Given: a÷(b+c) ≠ (a÷b)+(a÷c)

a=−10, b=1, c=1

Putting the given values in L.H.S. = −10÷(1+1)

= −10÷(2)

L.H.S. ≠ R.H.S.

Hence, verified.

3. Fill in the blanks:

(a) 369÷____=369

(b) (−75)÷____=(−1)

(c) (−206)÷____=1

(d) (−87)÷____=87

(e) ____÷1 = −87

(f) ____÷48 = −1

(g) 20÷ ___ = −2

(h) ___ ÷(4) = −3

(a) 369÷1 = 369

(b) (−75)÷75 = (−1)

(c) (−206)÷(−206) =1

(d) (−87)÷(−1) =87

(e) (−87) ÷1=−87

(f) (−48) ÷48=−1

(g) 20÷(−10) =−2

(h) (−12) ÷(4)=−3

4. Write five pairs of integers (a,b) such that a÷b = −3. One such pair is (6,−2) because 6÷(−2)=(−3).

(i) (−6)÷2=−3

(ii) 9÷(−3)=−3

(iii) 12÷(−4)=−3

(iv) (−9)÷3=−3

(v) (−15)÷5=−3

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

The temperature decreases 1°C = 1/2 hour

The temperature decreases 18°C = 1/2×18= 9 hours

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

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

6. In a class test (+3) marks are given for every correct answer and (−2) 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 (−5) marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?

(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 = (−16)÷(−2) = 8(−16)÷(−2)=8

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 = (−26) ÷ (−2) = 13

Thus, Mohini has attempted 13 incorrect questions.

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 −350 m?

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 = 1/6 minute

Therefore, time taken to cover a distance of 360 m = 1/6×360 = 60 minutes = 1 hour

(Since 60 minutes = 1 hour)

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

4 × (– 8),

8 × (–2),

3 × (–7),

10 × (–1) using number line.

**Answer**4 × (– 8) means -8 + (-8) + (-8) + 8) = -32

8 × (–2) means -2 + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) = -16

3 × (–7) means -7 + (-7) + (-7) = -21

10 × (–1) means -1 + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) = -10

Find:

(i) 6 × (-19)

(ii) 12 × (-32)

(iii) 7 × (-22)

**Answer**

(i) 6 × (-19) = -19 + (-19) + (-19) + (-19) + (-19) + (-19)= -114

(ii) 12 × (-32) = -32 + (-32) + (-32) + (-32) + (-32) + (-32) + (-32) + (-32) + (-32) + (-32) + (-32) + (-32) = -144096

(iii) 7 × (-22) = -22 + (-22) + (-22) + (-22) + (-22) + (-22) + (-22)= -154

Page No. 11

**Try these**

1. Find:

(a) 15 × (–16)

(b) 21 × (–32)

(c) (– 42) × 12

(d) –55 × 15

**Answer**

(a) 15 × (–16) = 16 × (–15) = -240

(b) 21 × (–32) = 32 × (–21) = -672

(c) (–42) × 12 = -504

(d) –55 × 15 = -825

2. Check if

(a) 25 × (–21) = (–25) × 21

(b) (–23) × 20 = 23 × (–20)

Write five more such examples.

**Answer**

(a) Yes.

25 × (–21) = (–25) × 21 = -525

(b) Yes.

(–23) × 20 = 23 × (–20) = -460

Five examples: (i) 15 × (–16) = 16 × (–15)

(ii) 14 × (–18) = 18 × (–14)

(iii) 25 × (–16) = 16 × (–25)

(iv) 28 × (–15) = 15 × (–28)

(v) 8 × (–16) = 16 × (–8)

Page No. 21

**Exercise 1.3**

1. Find each of the following products:

(a) 3 × (–1)

(b) (–1) × 225

(c) (–21) × (–30)

(d) (–316) × (–1)

(e) (–15) × 0 × (–18)

(f) (–12) × (–11) × (10)

(g) 9 × (–3) × (– 6)

(h) (–18) × (–5) × (– 4)

(i) (–1) × (–2) × (–3) × 4

(j) (–3) × (–6) × (–2) × (–1)

**Answer**

(a) 3 x (–1) = –3

(b) (–1) x 225 = –225

(c) (–21) x (–30) = 630

(d) (–316) x (–1) = 316

(e) (–15) x 0 x (–18) = 0

(f) (–12) x (–11) x (10) = 132 x 10 = 1320

(g) 9 x (–3) x (–6) = 9 x 18 = 162

(h) (–18) x (–5) x (–4) = 90 x (–4) = –360

(i) (–1) x (–2) x (–3) x 4 = (–6 x 4) = –24

(j) (–3) x (–6) x (2) x (–1) = (–18) x (–2) = 36

2. Verify the following:

(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]

(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]

**Answer**

(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]

⇒ 18 x 4 = 126 + (–54)

⇒ 72 = 72

⇒ L.H.S. = R.H.S. Hence verified.

(b) (–21) x [(–4) + (–6)] = [(–21) x (–4)] + [(–21) x (–6)]

⇒ (–21) x (–10) = 84 + 126

⇒ 210 = 210

⇒ L.H.S. = R.H.S. Hence verified.

3. (i) For any integer a, what is (–1) × a equal to?

(ii) Determine the integer whose product with (–1) is

(a) –22

(b) 37

(c) 0

**Answer**

(i) (–1)×a=–a, where aa is an integer.

(ii) (a) (–1)×(–22)=22

(b) (–1)×37=–37

(c) (–1)×0=0

4. Starting from (–1) × 5, write various products showing some pattern to show
(–1) × (–1) = 1.

**Answer**

(–1)×5=–5(–1)×4=–4

(–1)×3=–3(–1)×2=–2

(–1)×1=–1(–1)×0=0

(–1)×(–1)=1

Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer is negative integer whereas the product of two negative integers is a positive integer.

5. Find the product, using suitable properties:

(a) 26 × (– 48) + (– 48) × (–36)

(b) 8 × 53 × (–125)

(c) 15 × (–25) × (– 4) × (–10)

(d) (– 41) × 102

(e) 625 × (–35) + (– 625) × 65

(f) 7 × (50 – 2)

(g) (–17) × (–29)

(h) (–57) × (–19) + 57

**Answer**

(a) 26×(–48)+(–48)×(–36)

⇒ (–48)×[26+(–36)] [Distributive property]

⇒ (–48)×(–10)

⇒ 480

(b) 8×53×(–125)

⇒ 53×[8×(–125)] [Commutative property]

⇒ 53×(–1000)

⇒ –53000

(c) 15×(–25)×(–4)×(–10)

⇒ 15×[(–25)×(–4)×(–10)] [Commutative property]

⇒ 15×(–1000)

⇒ –15000

(d) (–41)×(102)

⇒ –41×[100+2] [Distributive property]

⇒ [(–41)×100]+[(–41)×2]

⇒ –4100+(–82)

⇒ –4182

(e) 625×(–35)+(–625)×65

⇒ 625×[(–35)+(–65)] [Distributive property]

⇒ 625×(–100)

⇒ –62500

(f) 7×(50–2)

⇒ 7×50–7×2 [Distributive property]

⇒ 350–14=336

(g) (–17)×(–29)

⇒ (–17)×[(–30)+1] [Distributive property]

⇒ (–17)×(30)+(–17)×1

⇒ 510+(–17)

⇒ 493

(h) (–57)×(–19)+57

⇒ (–57)×(–19)+57×1

⇒ 57 x 19 + 57 x 1

⇒ 57 x (19 + 1) [Distributive property]

⇒ 57 x 20 = 1140

6. A certain freezing process requires that room temperature be lowered from 40°C at
the rate of 5°C every hour. What will be the room temperature 10 hours after the
process begins?

**Answer**

Given: Present room temperature = 40°C

Decreasing the temperature every hour = 5°C

Room temperature after 10 hours = 40°C + 10 x (–5°C)

= 40°C – 50°C

= –10°C

Thus, the room temperature after 10 hours is –10°C after the process begins.

7. In a class test containing 10 questions, 5 marks are awarded for every correct answer
and (–2) marks are awarded for every incorrect answer and 0 for questions not
attempted.

(i) Mohan gets four correct and six incorrect answers. What is his score?

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?

(iii) Heena gets two correct and five incorrect answers out of seven questions she
attempts. What is her score?

**Answer**

(i) Mohan gets marks for four correct questions = 4 x 5 = 20

He gets marks for six incorrect questions = 6 x (–2) = –12

Therefore, total scores of Mohan = (4 x 5) + [6 x (–2)] = 20 – 12 = 8

Thus, Mohan gets 8 marks in a class test.

(ii) Reshma gets marks for five correct questions = 5 x 5 = 25

She gets marks for five incorrect questions = 5 x (–2) = –10

Therefore, total score of Resham = 25 + (–10) = 15

Thus, Reshma gets 15 marks in a class test.

(iii) Heena gets marks for two correct questions = 2 x 5 = 10

She gets marks for five incorrect questions = 5 x (–2) = –10

Therefore, total score of Resham = 10 + (–10) = 0

Thus, Reshma gets 0 marks in a class test.

8. A cement company earns a profit of Rs 8 per bag of white cement sold and a loss of
Rs 5 per bag of grey cement sold.

(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement
in a month. What is its profit or loss?

(b) What is the number of white cement bags it must sell to have neither profit
nor loss, if the number of grey bags sold is 6,400 bags.

**Answer**

Given: Profit of 1 bag of white cement = Rs. 8

And Loss of 1 bag of grey cement = Rs. 5

(a) Profit on selling 3000 bags of white cement = 3000 × 8 = Rs. 24,000

Loss of selling 5000 bags of grey cement = 5000 × Rs. 5 = Rs. 25,000

Since Profit < Loss

Therefore, his total loss on selling the grey cement bags = Loss – Profit

= 25,000 – 24,000 = Rs. 1,000

Thus, he has lost of Rs. 1,000 on selling the grey cement bags.

(b) Let the number of bags of white cement be x.

According to question, Loss = Profit

∴ 5 x 6,400 = x × 8

⇒ x = (5×6400)/8 = 5000 bags

Thus, he must sell 4000 white cement bags to have neither profit nor loss.

9. Replace the blank with an integer to make it a true statement.

(a) (–3) × _____ = 27

(b) 5 × _____ = –35

(c) _____ × (– 8) = –56

(d) _____ × (–12) = 132

**Answer**

(a) (−3)×(−9)−−−−=27

(b) 5×(−7)−−−−=−35

(c) 7–×(−8)=−56

(d) (−11)−−−−−×(−12)=132

Page No. 26

**Exercise 1.4**

(a) (−30)÷10

(b) 50÷(−5)

(c) (−36)÷(−9)

(d) (−49)÷49

(e) 13÷[(−2)+1]

(f) 0÷(−12)

(g) (−31)÷[(−30)+(−1)]

(h) [(−36)÷12]÷3

(i) [(−6)+5]÷[(−2)+1]

**Answer**(a) (−30)÷10

= (−30)×1/10

= {(−30)×1}/10

= −3

(b) 50÷(−5)

= 50×(−1/5)

= {50×(−1)}/5

=−10

(c) (−36)÷(−9)

= (−36)×(−1/9)

= {(−36)×(−1)}/9

=36/9

=4

(d) (−49)÷49

= (−49)× (1/49)

= −49/49

=−1

(e) 13÷[(−2)+1]

= 13÷(−1)

=13×(−1/1)

=−13

(f) 0÷(−12)

= 0×(−1/12)

=0/12

=0

(g) (−31)÷[(−30)+(−1)]

= (−31)÷(−30−1)

=(−31)÷(−31)

=(−31)×(−1/31)

=31/31

=1

(h) [(−36)÷12]÷3

= [(−36)×1/12]×1/3

=(−36/12)×1/3

=(−3)×1/3

=−3/3

=−1

(i) [(−6)+5]÷[(−2)+1]

= (−6+5)÷(−2+1)

=(−1)÷(−1)

=(−1)×(−1)/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=12,b=−4,c=2

(b) a=(−10),b=1,c=1

**Answer**(a) Given: a÷(b+c) ≠ (a÷b)+(a÷c)

a=12,b=−4,c=2

Putting the given values in L.H.S. = 12÷(−4+2)

= 12÷(−2)

=12÷(−1/2)

= −12/2

=−6

Putting the given values in R.H.S. = [12÷(−4)]+(12÷2)

= (12×−14)+6
=−3+6

=3

Since, L.H.S. ≠ R.H.S.Hence, verified.

(b) Given: a÷(b+c) ≠ (a÷b)+(a÷c)

a=−10, b=1, c=1

Putting the given values in L.H.S. = −10÷(1+1)

= −10÷(2)

=−5

Putting the given values in R.H.S. = [−10÷1]+(−10÷1)

= −10−10
=−20

Since,L.H.S. ≠ R.H.S.

Hence, verified.

3. Fill in the blanks:

(a) 369÷____=369

(b) (−75)÷____=(−1)

(c) (−206)÷____=1

(d) (−87)÷____=87

(e) ____÷1 = −87

(f) ____÷48 = −1

(g) 20÷ ___ = −2

(h) ___ ÷(4) = −3

**Answer**(a) 369÷1 = 369

(b) (−75)÷75 = (−1)

(c) (−206)÷(−206) =1

(d) (−87)÷(−1) =87

(e) (−87) ÷1=−87

(f) (−48) ÷48=−1

(g) 20÷(−10) =−2

(h) (−12) ÷(4)=−3

4. Write five pairs of integers (a,b) such that a÷b = −3. One such pair is (6,−2) because 6÷(−2)=(−3).

**Answer**(ii) 9÷(−3)=−3

(iii) 12÷(−4)=−3

(iv) (−9)÷3=−3

(v) (−15)÷5=−3

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

**Answer**
Following number line is representing the temperature:

The temperature decreases 2°C = 1 hourThe temperature decreases 1°C = 1/2 hour

The temperature decreases 18°C = 1/2×18= 9 hours

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

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

6. In a class test (+3) marks are given for every correct answer and (−2) 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 (−5) marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?

**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 = (−16)÷(−2) = 8(−16)÷(−2)=8

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 = (−26) ÷ (−2) = 13

Thus, Mohini has attempted 13 incorrect questions.

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 −350 m?

**Answer**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 = 1/6 minute

Therefore, time taken to cover a distance of 360 m = 1/6×360 = 60 minutes = 1 hour

(Since 60 minutes = 1 hour)

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

**Go Back To NCERT Solutions for Class 7 Maths**

## NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT Solutions for Chapter 1 Integers is available on this page which is useful in grasping the important points inside the chapter properly. An integer may comprise a set of whole numbers that include zero, positive number and negative number however it do not include decimals, percent, and fractions.

• Integers are closed for addition and subtraction both. That is, a + b and a – b are again integers, where a and b are any integers.

• Addition is commutative for integers, i.e., a + b = b + a for all integers a and b.

• Addition is associative for integers, i.e., (a + b) + c = a + (b + c) for all integers a, b and c.

• Integer 0 is the identity under addition. That is, a + 0 = 0 + a = a for every integer a.

• Integers are closed under multiplication. That is, a × b is an integer for any two integers a and b.

• Multiplication is commutative for integers. That is, a × b = b × a for any integers a and b.

• The integer 1 is the identity under multiplication, i.e., 1 × a = a × 1 = a for any integer a.

• Multiplication is associative for integers, i.e., (a × b) × c = a × (b × c) for any three integers a, b and c.

We have provided exercisewise Class 7 Maths NCERT Solutions that will inculcate correct learning habits among students and useful for competitive exams and higher grades. It will help in clearing the basic concepts and finding the right questions.

NCERT Solutions for Class 7 will serve as beneficial tool that can be used to recall various questions any time. It will help you in identify, analyze, and then rectify the mistakes.

### NCERT Solutions for Class 7 Maths Chapters:

**FAQ on Chapter**

**1 Integers**

#### How many exercises are there in Chapter 1 Integers Class 7 Maths NCERT Solutions?

There are total 4 exercises in the whole chapter that will give good experience and provide opportunities to learn new things. These NCERT Solutions are accurate and detailed through which you can prepare yourself well before examinations.

#### Evaluate: (a) (–1) × (–2) × (–3) × (–4) × (–5)

–120 is the answer.

#### Find 2 consecutive numbers whose sum is 129.

Assume the numbers are ‘x’ and ‘x+1’

x + x + 1 = 129

so, 2x = 129 – 1

2x = 128

x = 64

if x = 64 then, x+1 = 65.

Hence, the numbers are 64 and 65.

#### What is Associative Property of Integers?

Integer numbers also abide by the associative property of multiplication. This implies that like whole numbers the grouping of integers does not affect the product of integers. Thus, (a×b)×c = (a)×(b×c).