# NCERT Solutions for Class 7 Maths chapter 1 Integers | integers class 7

NCERT Solutions for Class 7 Maths chapter 1 Integers Class 7th maths chapter 1 Class 7 maths. Here We learn what is in Class 7 maths solution chapter 1 Integers and how to solve questions with easiest method. In this chapter we solve the question of NCERT Solutions for integers class 7 Exercise 1.1, Class 7 maths chapter 1 Exercise 1.2, Class 7 maths chapter 1 Exercise 1.3 and Class 7 maths chapter 1 Exercise 1.4. chapter 1 Class 7 maths class 7th maths chapter 1
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# ncert solutions for Class 7 maths chapter 1 Integersintegers class 7th maths chapter 1

## Exercise – 1integers class 7

### ncert solutions for Class 7 maths chapter 1 IntegersClass 7th maths solution ex 1.1

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

 Places Temperature Lahulspiti − 8°C Srinagar − 2°C Shimla + 5°C Ooty + 15°C Bengaluru + 22°C

(b) What is the temperature difference between the hottest and the coldest places among the above?
The hottest places Bengaluru temperature = + 22°C
The coldest places Lahulspiti temperature = − 8°C
The temperature difference between the hottest and the coldest places is = + 22 − (− 8)
= + 22 + 8
= + 30°C

(c) What is the temperature difference between Lahulspiti and Srinagar?
the temperature of Lahulspiti = − 8°C
the temperature of Srinagar = − 2°C
the temperature difference between Lahulspiti and Srinagar is = − 2 − (− 8)
= − 2 + 8
= 6°C

(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?
the temperature at Shimla is = 5°C
the temperature at Srinagar = − 2°C
The sum of the temperature at Shimla and Srinagar = 5 + (− 2)
= 5 − 2
= 3°C
The sum of the temperature at Shimla and Srinagar < the temperature at Shimla
3°C < 5°C
Yes, We can say that temperature of Srinagar and Shimla taken together is less than the temperature at Shimla

The sum of the temperature at Shimla and Srinagar < the temperature at Srinagar
3°C > − 2°C
No, The sum of the temperature at Shimla and Srinagar is not less than Srinagar temperature.

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?
Solution :
Jack’s scores = 25, – 5, – 10, 15 and 10
The sum of Jack’s scores = 25 + (− 5) + (− 10) + 15 + 10
= 25 − 5 − 10 + 15 + 10
= 25 + 15 + 10 − 5 − 10
= 50 − 15
= 35 Marks

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?
Solution :
Srinagar temperature on Monday = − 5°C
Temperature dropped at Tuesday = − 2°C
So, Tuesday temperature is = − 5 + (− 2)
= − 5 − 2
= − 7°C
Srinagar temperature on Monday = − 7°C
Temperature rose on Wednesday = + 4°C (Temperature rise show in positive sign)
Temperature on Wednesday = − 7 + 4
= − 3°C

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?
Solution :
Plane height from the sea level = + 5000 m
submarine below the sea level = − 1200 m
the vertical distance between them = 5000 − (− 1200)
= 5000 + 1200
= + 6200 m

5. Mohan deposits rupees 2,000 in his bank account and withdraws rupees 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.
Solution :
Deposits money in Mohan bank account = rupees + 2000
Withdraws money from Mohan bank account = rupees − 1642
We will represent the deposited amount in positive form.
Balance in Mohan’s account after withdrawal = 2000 − 1642
= rupees 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?
Solution :
Distance covered by Rita from point A towards east to point B = 20 km (positive integer)
Distance covered along the road from point B to the west = − 30 km
Note : If the distance covered towards East is represented by a positive integer, then we will express the distance covered towards West as a negative integer.
its final position from point A = 20 + (− 30)
= 20 − 30
= − 10

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

In square (i) sum of first row = 5 + (− 1) + (− 4) = 5 − 1 − 4 = 5 − 5 = 0
In square (i) sum of second row = − 5 + (− 2) + 7 = − 5 − 2 + 7 = − 7 + 7 = 0
In square (i) sum of third row = 0 + 3 + (− 3) = 0 + 3 − 3 = 3 − 3 = 0
In square (i) sum of first column = 5 + (− 5) + 0 = 5 − 5 = 0
In square (i) sum of second column = − 5 + (− 5) + 0 = − 5 − 5 + 0 = 0
In square (i) sum of third column = − 1 + (− 2) + 3 = − 1 − 2 + 3 = − 3 + 3 = 0
In square (i) sum of first diagonal = 5 + (− 2) + (− 3) = 5 − 2 − 3 = 5 − 5 = 0
In square (i) sum of second diagonal = − 4 + (− 2) + 0 = − 4 − 2 + 0 = − 6
In square (i) each row, column and diagonal have not the same sum. So, this is not a magical square.

In square (ii) sum of first row = 1 + (− 10) + 0 = 1 − 10 = − 9
In square (ii) sum of second row = − 4 + (− 3) + (− 2) = − 4 − 3 − 2 = − 9
In square (ii) sum of third row = − 6 + 4 + (− 7) = − 6 + 4 − 7 = − 13 + 4 = − 9
In square (ii) sum of first column = 1 + (− 4) + (− 6) = 1 − 4 − 6 = 1 − 10 = − 9
In square (ii) sum of second column = − 4 + (− 3) + (− 2) = − 4 − 3 − 2 = − 9
In square (ii) sum of third column = − 6 + 4 + (− 7) = − 6 + 4 − 7 = − 13 + 4 = − 9
In square (ii) sum of first diagonal = 1 + (− 3) + (− 7) = 1 − 3 − 7 = 1 − 10 = − 9
In square (ii) sum of second diagonal = 0 + (− 3) + (− 6) = − 3 − 6 = − 9
In square (ii) each row, column and diagonal have the same sum. So, this is a magical 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

Solution :
(i) a = 21, b = 18
Put the a and b value in a − (− b) = a + b
21 − (− 18) = 21 + 18
21 + 18 = 21 + 18
39 = 39
L.H.S = R.H.S

(ii) a = 118, b = 125
Put the a and b value in a − (− b) = a + b
118 − (− 125) = 118 + 125
118 + 125 = 118 + 125
243 = 243
L.H.S = R.H.S

(iii) a = 75, b = 84
Put the a and b value in a − (− b) = a + b
75 − (− 84) = 75 + 84
75 + 84 = 75 + 84
159 = 159
L.H.S = R.H.S

(iv) a = 28, b = 11
Put the a and b value in a − (− b) = a + b
28 − (− 11) = 28 + 11
28 + 11 = 28 + 11
39 = 39
L.H.S = R.H.S

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 + (− 19) (c) 23 − 41 + 11 . . . . 23 − 41 − 11 (d) 39 + (− 24) − (15) . . . . 23 − 41 − 11 (e) 39 + (− 24) − (15) . . . . 36 + (− 52) − (− 36)

Solution :
(a) (− 8) + (− 4) . . . . (− 8) − (− 4)
− 8 − 4 . . . . − 8 + 4
− 12 < − 4

(b) (− 3) + 7 − (19) . . . . 15 − 8 + (− 19)
− 3 + 7 − 19 . . . . 15 − 8 − 19
− 22 + 7 . . . . 15 − 27
− 15 < − 12

(c) 23 − 41 + 11 . . . . 23 − 41 − 11
23 + 11 − 41 . . . . 23 − 52

34 − 41 . . . . − 29
− 7 > − 29

(d) 39 + (− 24) − (15) . . . . 36 + (− 52) − (− 36)
39 − 24 − 15 . . . . 36 − 52 + 36
39 − 39 . . . . 36 + 36 − 52
0 . . . . 72 − 52
0 < 20

(e) − 231 + 79 + 51 . . . . − 399 + 159 + 81
− 231 + 130 . . . . − 399 + 240
− 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?
Solution :
(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?
monkey is sitting on the topmost step
The position of the monkey after the jumps will be as follows:
In the first jump he will reach the step = 4 step
In the second jump he will reach the step = 4 − 2 = 2 step
In the third jump he will reach the step = 2 + 3 = 5 step
In the four jump he will reach the step = 5 − 2 = 3 step
In the five jump he will reach the step = 3 + 3 = 6 step
In the six jump he will reach the step = 6 − 2 = 4 step
In the seven jump he will reach the step = 4 + 3 = 7 step
In the eight jump he will reach the step = 7 − 2 = 5 step
In the nine jump he will reach the step = 5 + 3 = 8 step
In the ten jump he will reach the step = 8 – 2 = 6 step
In the eleven jump he will reach the step = 6 + 3 = 9 step
Hence, the ninth step which is at the water level will be able to reach in 11 jumps.

(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?
Here, the monkey is at the water level i.e. the ninth step.
After the jumps, the position of the monkey from the top step will be as follows:
In the first jump he will reach the step = 9 – 4 = 5 steps
In the second jump he will reach the step = 5 + 2 = 7 steps
In the third jump he will reach the step = 7 – 4 = 3 steps
In the fourth jump he will reach the step = 3 + 2 = 5 steps
In the fifth jump he will reach the step = 5 – 4 = 1 step
Number of jumps required = 5

(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

(a) − 3 + 2 + . . . = − 8 (b) 4 − 2 + . . . = 8.

In (a) the sum (− 8) represents going down eight steps, then
(b) What does the sum 8 represent?
solution :
(a) − 3 + 2 + . . . = − 8
− 3 + 2 − 3 + 2 − 3 + 2 − 3 + 2 − 3 + 2 − 3 = − 8

(b) 4 − 2 + 4 − 2 + 4 = 8
The number 8 in (b) represents climbing 8 steps uphill.

### ncert solutions for Class 7 maths chapter 1 IntegersClass 7th maths solution ex 1.2

1. Write down a pair of integers whose:

 (a) sum is − 7 (b) difference is − 10 (c) sum is 0

Solution :
(a) sum is − 7
− 4 + (− 3) = − 7
− 10 + 3 = − 7

(b) difference is − 10
− 6 − (+4) = − 10
+ 10 − 20 = − 10

(c) sum is 0
2 + 2 = 0
− 4 + 4 = 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.
Solution :
(a) Write a pair of negative integers whose difference gives 8.
− 4 − (− 12) = − 4 + 12 = 8

(b) Write a negative integer and a positive integer whose sum is – 5.
6 + (− 11) = 6 − 11 = − 5

(c) Write a negative integer and a positive integer whose difference is – 3.
9 − (+12) = 9 − 12 = − 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?
Solution :
team A scored = − 40, 10, 0
sum of team A scored = − 40 + 10 + 0
= − 30
team A scored = 10, 0, − 40
sum of team B scored = 10 + 0 + (− 40)
= 10 + 0 − 40
= − 30
So both the teams got equal marks − 30.
Yes, integers can be added in any order.

4. Fill in the blanks to make the following statements true:
(i) (–5) + (– 8) = (– 8) + (…………)
(ii) –53 + ………… = –53
(iii) 17 + ………… = 0
(iv) [13 + (– 12)] + (…………) = 13 + [(–12) + (–7)]
(v) (– 4) + [15 + (–3)] = [– 4 + 15] + …………
Solution :
(i) (− 5) + (− 8) = (− 8) + (. . . . . . . .)
(− 5) + (− 8) = (− 8) + (. . . . . . . .)
− 5 − 8 = (− 8) + (. . . . . . . .)
− 13 = (− 8) + (− 5)
So adding − 5 to − 8 will make both sides equal.

(ii) – 53 + . . . . . . . . = − 53
− 53 + 0 = − 53
So adding 0 to − 53 will make both sides equal.

(iii) 17 + . . . . . . . . . = 0
17 + (− 17) = 0
0 = 0
So adding − 17 to 17 will make both the sides equal.

(iv) [13 + (– 12)] + (. . . . . . . . .) = 13 + [(– 12) + (– 7)]
13 − 12 + . . . . . . . = 13 − 12 − 7
1 + . . . . . . . = 13 − 19
1 + . . . . . . . = − 6
1 + (− 7) = − 6
So adding − 7 to 1 will make both sides equal.

(v) (– 4) + [15 + (– 3)] = [– 4 + 15] + . . . . . . . . . . .
− 4 + 15 − 3 = − 4 + 15 + . . . . . . . . .
− 7 + 15 = 11 + . . . . . . .
8 = 11 + (− 3)
8 = 8
So adding − 3 to 11 will make both sides equal.

### ncert solutions for Class 7 maths chapter 1 IntegersClass 7th maths solution ex 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)

Solution :
(a) 3 × (– 1)
Since the product of + and − is − .
= − 3

(b) (– 1) × 225
Since the product of + and − is − .
= − 225

(c) (– 21) × (– 30)
Since the product of − and − is + .
= 630

(d) (– 316) × (– 1)
Since the product of − and − is + .
= 316

(e) (– 15) × 0 × (– 18)
Since multiplying zero by any number gives the product zero.
= 0

(f) (– 12) × (– 11) × (10)
Since the product of − and − is + .
= 132 × 10

= 1320

(g) 9 × (– 3) × (– 6)
Since the product of + and − is − .
= − 27 × (− 6)

Since the product of − and − is + .
= 162

(h) (– 18) × (– 5) × (– 4)
Since the product of − and − is + .
= 90 × (− 4)

Since the product of + and − is − .
= − 360

(i) (– 1) × (– 2) × (– 3) × 4
Since the product of − and − is + .
= 2 × (− 3) × 4

Since the product of − and + is − .
= − 6 × 4

= − 24

(j) (– 3) × (– 6) × (– 2) × (– 1)
Since the product of − and − is + .
= 18 × 2

= 36

2. Verify the following:
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]
Solution :
(a) 18 × [7 + (– 3)] = [18 × 7] + [18 × (– 3)]
18 × [7 − 3] = 126 + (− 54)
18 × 4 = 126 − 54
72 = 72
L.H.S. = R.H.S.

(b) (– 21) × [(– 4) + (– 6)] = [(– 21) × (– 4)] + [(– 21) × (– 6)]
− 21 × [− 4 − 6] = 84 + 126
− 21 × (− 10) = 210
210 = 210
L.H.S. = R.H.S.

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

Solution :
(i) For any integer a, what is (– 1) × a equal to?
(− 1) × a
= − a

(ii) Determine the integer whose product with (– 1) is
(a) − 22
− 22 × (− 1)
= 22

(b) 37
37 × (− 1)
= − 37

(c) 0
0 × (− 1)
= 0

4. Starting from (– 1) × 5, write various products showing some pattern to show (– 1) × (– 1) = 1.
Solution :
(– 1) × 5 = − 5
(– 1) × 4 = − 4
(– 1) × 3 = − 3
(– 1) × 2 = − 2
(– 1) × 1 = − 1
(– 1) × 0 = 0
(–1) × (− 1) = 1

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

Solution :
(a) 26 × (– 48) + (– 48) × (– 36)
= (− 48) (26 − 36)
= (− 48) × (− 10)
= 480

(b) 8 × 53 × (–125)
= 8 × (− 125) × 53
= − 1000 × 53
= − 53000

(c) 15 × (– 25) × (– 4) × (– 10)
= 15 × (– 10) × (– 25) × (– 4)
= − 150 × 100
= − 15000

(d) (– 41) × 102
= (– 41) × (100 + 2)
= (− 41) × 100 + (− 41) × 2
= − 4100 + (− 82)
= − 4100 − 82
= − 4182

(e) 625 × (– 35) + (– 625) × 65
= 625 × (− 35 − 65)
= 625 × (− 100)
= − 62500

(f) 7 × (50 – 2)
= 7 × 50 − 7 × 2
= 350 − 14
= 336

(g) (− 17) × (− 29)
= (− 17) × (− 30 + 1)
= (− 17) × (− 30) + (− 17) × 1
= 510 − 17
= 493

(h) (− 57) × (− 19) + 57
= − 57 × (− 19 − 1)
= − 57 × (− 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?
Solution :
Room temperature = 40°C
Decrease per hour = 5C
Decrease in temperature after 10 hours = 10 × (− 5)
= − 50°C
Hence temperature after 10 hours = 40 + (− 50)
= 40 − 50
= − 10°C

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?
Solution :
(i) Mohan gets four correct and six incorrect answers. What is his score?
Sum of 4 correct questions attempted by Mohan = 4 × 5 = 20
Sum of 6 wrong questions attempted by Mohan = 6 × (− 2) = − 12
Total marks obtained by Mohan = 20 − 12 = 8

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
Sum of 5 correct questions attempted by Reshma = 5 × 5 = 25
Sum of 5 wrong questions scored by Reshma = 5 × (− 2) = − 10
Total marks obtained by Reshma = 25 − 10 = 15

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Sum of 2 correct questions attempted by Heena = 2 × 5 = 10
Sum of 5 wrong questions attempted by Heena = 5 × (− 2) = − 10
Total marks obtained by Heena = 10 − 10 = 0

8. A cement company earns a profit of rupees 8 per bag of white cement sold and a loss of rupees 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.
Solution :
(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?
Profit on selling one white bag of cement = Rs.8
Profit on selling 3000 white bags of cement = 8 × 3000 = Rs 24000
Loss on selling one gray bag of cement = Rs 5
Loss on selling 5000 gray bags of cement = 5 × 5000 = Rs 25000
Loss = 24000 − 25000
= − 1000
Hence he will have a loss of Rs 1000.

(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.
Loss on selling one gray bag of cement = Rs 5
Loss on selling 6400 gray bags of cement = 5 × 6400 = Rs 32000
Since profit of Rs.8 = 1 bag
Hence profit of Rs 1 = 1/8 on the bag
Number of white bag sold for a profit of Rs 32000 = 32000/8
= 4000 bags

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

Solution :
(a) (– 3) × _____ = 27

(− 3) × (− 9) = 27

(b) 5 × _____ = – 35
5 × (− 7) = − 35

(c) _____ × (– 8) = – 56
7 × (– 8) = – 56

(d) _____ × (– 12) = 132
(− 11) × (– 12) = 132

## Ncert class 7 maths chapter 1class 7th mathEx 1.4

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]

Solution :

(a) (–30) ÷ 10
= $$\displaystyle \frac{-30}{10}$$= − 3

(b) 50 ÷ (– 5)
= $$\displaystyle \frac{{50}}{{-5}}$$= − 10

(c) (–36) ÷ (–9)
= $$\displaystyle \frac{{-36}}{{-9}}$$= 4

(d) (– 49) ÷ (49)
= $$\displaystyle \frac{{-49}}{{49}}$$= − 1

(e) 13 ÷ [(–2) + 1]
= $$\displaystyle \frac{{13}}{{-1}}$$= − 13

(g) (– 31) ÷ [(– 30) + (– 1)]
= $$\displaystyle \frac{{-31}}{{(-30-1)}}$$= $$\displaystyle \frac{{-31}}{{-31}}$$= 1

(h) [(– 36) ÷ 12] ÷ 3
= $$\displaystyle \left( {\frac{{-36}}{{12}}} \right)\div 3$$= $$\displaystyle \frac{{-3}}{{3}}$$
= − 1

(i) [(– 6) + 5)] ÷ [(–2) + 1]
= $$\displaystyle \frac{{-6+5}}{{-2+1}}$$= $$\displaystyle \frac{{-1}}{{-1}}$$
= 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

Solution :
(a) a = 12, b = – 4, c = 2
put the value of a, b and c in a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
12 ÷ (− 4 + 2) ≠ 12 ÷ (− 4) + (12 ÷ 2)
12 ÷ (− 2) ≠ − 3 + 6
− 6 ≠ 3
L.H.S ≠ R.H.S.

(b) a = (– 10), b = 1, c = 1
put the value of a, b and c in a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
(– 10) ÷ (1 + 1) ≠ (− 10 ÷ 1) + (− 10 ÷ 1)
(– 10) ÷ 2 ≠ − 10 + (− 10)
− 5 ≠ 1

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

Solution :
(a) 369 ÷ _____ = 369
369 ÷ 1 = 369

(b) (– 75) ÷ _____ = – 1
(– 75) ÷ 75 = – 1

(c) (– 206) ÷ _____ = 1
(– 206) ÷ (– 206) = 1

(d) – 87 ÷ _____ = 87
– 87 ÷ − 1 = 87

(e) _____ ÷ 1 = – 87
− 87 ÷ 1 = – 87

(f) _____ ÷ 48 = – 1
− 48 ÷ 48 = – 1

(g) 20 ÷ _____ = – 2
20 ÷ (− 10) = – 2

(h) _____ ÷ (4) = – 3
− 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).
Solution :
(– 6, 2)
because 6 ÷ (–2) = (– 3)

(– 12, 4)
because − 12 ÷ 4 = (– 3)

(12, – 4)
because ÷ (– 4) = (– 3)

(9, – 3)
because 9 ÷ (– 3) = (– 3)

(– 9, 3)
because − 9 ÷ 3 = (– 3)

(− 45, 15)
because − 45 ÷ 15 = (– 3)

5. The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?
Solution :
Temperature at 12 noon = + 10°C

Decrease in temperature till 12 noon = − 2°C
Decrease in temperature to be − 8 °C = − 8 − (+ 10)
= − 18°C
Time taken to lower the temperature by − 2 °C = 1 hour
Time taken for the temperature to drop to −18°C = 18/2 = 9 hours

Decrease in temperature in 12 hours = 12 × (− 2)
= − 24°C Decrease in temperature from 12 noon to 12 noon = − 24 + 10
= − 14°C
So at 12 o’clock the temperature will be −14°C.

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?
Solution :

(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
Marks obtained for correct answer = + 3
Total marks scored by Radhika for correct answer = 12 × 3
= 36 points
Marks obtained by Radhika = 20 marks
Marks of wrong answer = 36 − 20 = 16 marks
Marks for one wrong answer = − 2
Hence 16 marks will be deducted = 16/2 = 8 Answer

(ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Marks obtained by Mohini = (− 5) marks
Correct answer given by Mohini = 7
Marks of correct answer given by Mohini = 7 × 3 = 21 marks
Marks of wrong answer obtained by Mohini = − 5 − (+ 21)
= − 5 − 21 = − 26 points
Hence, wrong answer given for − 26 marks = $$\displaystyle \frac{{26}}{2}$$ = 13 answer

7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution :
Total distance = − 350 − (+ 10)
= − 350 − 10
= − 360 m
Since time taken to walk 6 m = 1 minute
Time taken to walk 1 m = $$\displaystyle \frac{{1}}{6}$$ minute
Time taken to walk 360 m =  $$\displaystyle \frac{{1}}{6}$$ × 360
= 60 minutes or 1 hour

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