# NCERT Solutions for Class 7 Maths Chapter 1 Integers

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 7 Maths Chapter 1 Integers Exercise 1.1

Q.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:

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(A)
(a) Lahulspriti = -8°C
(b) Srinagar = -2°C
(c) Shimla = 5°C
(d) Ooty = 14°C
(e) Bangluru =22°C
(B)
Difference = 22°C-( -8°C) = 22°C+8°C =30°C
(C)
Difference = -2°C-( -8°C) = -2°C+ 8°C = 6°C
(D)
Temperature of the above cities taken together = -2°C + 5°C = 3°C
Temperature of Shimla = 5°C
Hence, the temperature of Srinagar and Shimla taken together is less than that of Shimla by 2°C.
i.e., (5°C – 3°C) = 2°C

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Given scores are 25, -5, -10, 15, 10 Total Mark's = 25 + (-5) +(-10) +15 +10 =
25 +15+10+(-15)= 50-15= 35

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Srinagar temperature on Monday = -5°C
Temperature on Tuesday = -5°C-2°C= -7°
Temperature was increased by 4°C on Wednesday.
$?$Temperature on Wednesday = -7°C + 4°C = -3°C

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Height of the flying plane = 5000 m
Depth of the submarine = -1200 m
$?$Distance between them
= + 5000 m – (-1200 m)
= 5000 m + 1200 m = 6200 m
Hence, the vertical distance = 6200 m

Q.5 Mohan deposits Rs.2,000 in a 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.

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Total amount deposited in bank account by the Mohan = Rs. 2000
Total amount withdrawn from the bank account by the Mohan = – Rs.1642
Balance in Mohan’s account after the withdrawal = amount deposited + amount withdrawn
=Rs. 2000 + (-Rs.1642)
= Rs. 2000 – Rs.1642
= Rs.358

Q.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 her final position from A?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Distances travelled towards east from point A will be represented by positive integer i.e. +20 km.
Distance travelled towards the west from point B will be represented by negative integer, i.e., —30 km.
Final position of Rita from A
= 20 km – 30 km = – 10 km
Hence, the required position of Rita will be presented by a negative number, i.e., -10.

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

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(i) R1 = 5 + (-1) + (—4)
=5 – 1 – 4 = 5 – 5 = 0
R2 = (-5) + (-2) + 7
= -5 – 2 + 7 = -7 + 7 = 0
R3 = 0 + 3 + (-3)
= 0 + 3- 3 = 0
C1 = 5 + (-5) + 0
= 5 – 5 + 0 = 0
C2 = (-1) + (-2) + (3)
=-1 – 2 + 3 = -3 + 3 = 0
C3 = (-4) + 7 + (-3)
= -4 + 7 – 3 = 7 – 7 = 0
D1 = 5 + (-2) + (-3)
= 5 – 2- 3 = 5 – 5 = 0
D2 = (-4) + (-2) + 0
= -4 – 2 + 0 = -6 + 0 = -6
Here, the sum of the integers of diagonal d2 is different from the others.
Hence, it is not a magic square.
(ii) R1 = 1 + (-10) + 0
= 1 – 10 + 0 = -9
R2 = (-4) + (-3) + (-2)
= -4 – 3 – 2 = -9
R3 = (-6) + (4) + (-7)
= -6 + 4 – 7 = -9
C3 = 1 + (-4) + (-6)
= 1 – 4 – 6 = -9
C2 = (-10) + (-3) + 4
= -10 – 3 + 4 = -9
C3 = 0 + (-2) + (-7)
= 0 – 2 -7 = -9
D1 = 1 + (-3) + (-7)
= 1 – 3 – 7 = 1 – 10 = -9
D2 = 0 + (-3) + (-6)
= 0 – 3- 6 = -9
Here, sum of the integers column wise, row wise and diagonally is same i.e. -9.
Hence, (ii) is a magic square.

Q.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, 6 = 11

NCERT Solutions for Class 7 Maths Chapter 1 Integers

In question
a- (-b) = a+b
(i) If a=21 , b= 18
Then L.H.S = a-(-b) = 21-(-18) = 21+18 = 39
R.H.S = a+b= 21+18 = 39
L.H.S= R.H.S proved
(ii) If a=118 , b= 125
Then L.H.S = a-(-b) = 118-(-125) = 118+125 = 243
R.H.S = a+b= 118+125 = 243
L.H.S= R.H.S proved
(iii) If a=75 , b= 84
Then L.H.S = a-(-b) = 75-(-84) = 75+84 = 159
R.H.S = a+b= 75+85 = 159
L.H.S= R.H.S proved
(iv) If a=28 , b= 11
Then L.H.S = a-(-b) = 28-(-11) = 28+11= 39
R.H.S = a+b= 28+11 = 39
L.H.S= R.H.S proved

Q.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

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) (-8) + (-4) [ ] (-8) – (-4)
LHS = (-8) + (-4) = -8 – 4 = – 12
RHS = (-8) – (-4) = -8 + 4 = -4
Here – 12 < -4
Hence, (-8) + (-4) [<] (-8) – (-4)

(b) (-3) + 7 – (19) [ ] 15 – 8 + (-9)
LHS = (-3) + 7 – (19) =-3 + 7-19
= -3 – 19 + 7 = -22 + 1 = -15
RHS = 15 – 8 + (-9) = 15-8-9
= 15 – 17 = -2
Here -15 < -2
Hence, (-3) + 7 – (19) [<] 15 – 8 + (-9)

(c) 23 – 41 + 11 [ ] 23 – 41 – 11
LHS = 23 – 41 + 11 = 23 + 11 – 41 = 34 – 41 = -7
RHS = 23 – 41 – 11 = 23 – 52 = -29 Here, -7 > -29
Hence, 23 – 41 + 11 [>] 23 – 41 – 11

(d) 39 + (-24) – (15) [ ] 36 + (-52) – (-36)
LHS = 39 + (-24) – (15)
= 39 – 24 – 15 = 39 – 39 = 0
RHS = 36 + (-52) – (-36) = 36 – 52 + 36
= 36 + 36 – 52 = 72 – 52 = 20
Here 0 < 20
Hence, 39 + (-24) – (15) [<] 36 + (-52) – (-36)

(e) -231 + 79 + 51 [ ] -399 + 159 + 81
LHS = -231 + 79 + 51 = -231 + 130 = -101
RHS = -399 + 159 + 81 = -399 + 240 = -159
Here, -101 > -159
Hence, -231 + 79 + 51 [>] -399 + 159 + 81

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(i) Let us consider steps moved down are represented by positive integers and then, steps moved up are represented by negative integers.

Initially monkey is sitting on the top most step i.e., first step
In 1st jump monkey will be at step = 1 + 3 = 4 steps

In 2nd jump monkey will be at step = 4 + (-2) = 4 – 2 = 2 steps

In 3rd jump monkey will be at step = 2 + 3 = 5 steps

In 4th jump monkey will be at step = 5 + (-2) = 5 – 2 = 3 steps

In 5th jump monkey will be at step = 3 + 3 = 6 steps

In 6th jump monkey will be at step = 6 + (-2) = 6 – 2 = 4 steps

In 7th jump monkey will be at step = 4 + 3 = 7 steps

In 8th jump monkey will be at step = 7 + (-2) = 7 – 2 = 5 steps

In 9th jump monkey will be at step = 5 + 3 = 8 steps

In 10th jump monkey will be at step = 8 + (-2) = 8 – 2 = 6 steps

In 11th jump monkey will be at step = 6 + 3 = 9 steps

?$?$ Monkey took 11 jumps (i.e., 9th step) to reach the water level

(ii) Let us consider steps moved down are represented by positive integers and then, steps moved up are represented by negative integers.
Initially monkey is sitting on the ninth step i.e., at the water level
In 1st jump monkey will be at step = 9 + (-4) = 9 – 4 = 5 steps

In 2nd jump monkey will be at step = 5 + 2 = 7 steps

In 3rd jump monkey will be at step = 7 + (-4) = 7 – 4 = 3 steps

In 4th jump monkey will be at step = 3 + 2 = 5 steps

In 5th jump monkey will be at step = 5 + (-4) = 5 – 4 = 1 step

?Monkey took 5 jumps to reach back the top step i.e., first step.

(iii) From the question, it is given that
If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers. Monkey moves in part (i)
Then LHS = – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3
= – 8
RHS = -8
$?$ Moves in part (i) represents monkey is going down 8 steps. Because negative integer.
Now,
Monkey moves in part (ii)
Then LHS = 4 – 2 + 4 – 2 + 4
= 8
RHS = 8
$?$ Moves in part (ii) represents monkey is going up 8 steps. Because positive integer.

## NCERT solutions for class 7 Maths Chapter 1 Integers Exercise 1.2

Q.1 Write down a pair of integers whose: (a) sum is -7 (b) difference is -10 (c) sum is 0.

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) -5 +(-2) = -5-2 = -7 hense -5 and -2

(b) 6 -16 = -10 hense 6 and 16

(c) -5 +5 =0 hense -5 and 5

Q.2 (a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and positive integer whose sum is -5.
(c) Write a negative integer and a positive integer whose difference is -3.

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) (-7)- (-15) = -7+15 = 8 , hense -7 and -15
(b) (-18) + (13) = -5 , hense -18 and 13
(c) (-2) – (1) = – 2 – 1 = -3 ,hense -2 and -1

Q.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 you say that we can add integers in any order?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Total score of team A = (-40)+(10)+(0) = -30
Total score of team B = (10) + (0)+(-40)= -30
The scores of both the teams are same i.e. -30. Yes, we can add the integers in any order.

Q.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] + …

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(i) -5 + (-8) – (-8) + (-5) [Commutative law of additions]
(ii) -53 + 0 = -53 [Additive Identity]
[Adding 0 to any integer, it gives the same value]
(iii) 17 + (-17) = 0 [Additive inverse]
(iv) [13 + (-12)] + (-7) – 13 + [(-12) + (-7)] [Associative law of addition]
(v) (-4) + [15 + (-3)] – [-4 + 15] + (-3) [Associative law of addition]

## NCERT solutions for class 7 Maths Chapter 1 Integers Exercise 1.3

Q.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)

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) 3 × (-1) = -3 × 1 = -3
(b) (-1) × 225 = -1 × 225 = -225
(c) (-21) × (-30) = (-) × (-) × 21 × 30 = 630[(-) ×(-)=+]
(d) (-316) × (-1) = (-) × (-) × 316 × 1 = 316
(e) (-15) × 0 × (-18) = 0 [? a × 0 = a]
(f) (-12) × (-11) × (10)
= (-) × (-) × 12 × 11 × 10 = 1320
(g) 9 × (-3) × (-6) = (-3) × (-6) × 9
= (—) × (-) × 3 × 6 × 9 = 162
(h) (-18) × (-5) × (-4)
= (-) × (-) × (-) × 18 × 5 × 4 = -360
(i) (-1) × (-2) × (-3) × 4
= (-) × (-) × (-) × 1 × 2 × 3 × 4 = -24
(j) (-3) × (-6) × (-2) × (-1)
= (-) × (-) × (-) × (-) × 3 × 6 × 2 × 1 = 36

Q.2 Verify the following:
(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]
(b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]
LHS = 18 × [7 + (-3)] = 18 × 4 = 72
RHS = [18 × 7] + [18 × (-3)] = 126 + (-54)
= 126 – 54 = 72
LHS = RHS Hence, verified.

(b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]
LHS = (-21) × [(-4) + (-6)]= (-21) × (-10)
= (-) × (-) × 21 × 10 = 210
RHS = [(-21) × (-4)] + [(-21) × (-6)]
= (84) + (126) = 84 + 126 = 210
LHS = RHS
Hence, verified.

Q.3 (i) For any integer a, what is (-1) × a equal to?
(ii) Determine the integer whose product with (-1) is 0.
(a) -22
(b) 37
(c) 0

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(i) (-1) × a = -a
(ii) (a) (-1) ×( -22) = (-) × (-) × 1× 22= 22
(b) (-1)× 37= -37
(c) (-1) × 0 = 0 [$?$a × 0 = 0]
Hence (c) 0 is the required integer.

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

NCERT Solutions for Class 7 Maths Chapter 1 Integers

The various products are,
= -1 × 5 = -5
= -1 × 4 = -4
= -1 × 3 = -3
= -1 × 2 = -2
= -1 × 1 = -1
= -1 × 0 = 0
= -1 × -1 = 1

Q.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

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) 26 × (-48) + (-48) × (-36)
= -48 × [26 + (-36)] = -48 × [26 – 36] = -48 × -10 = 480
[Distributive property of multiplication over addition]

(b) 8 × 53 × (-125) = 53 × [8 × (-125)]
[Associative property of multiplication] = 53 × (-1000) = -53000

(c) 15 × (-25) × (-4) × (-10)
= [(-25) × (-4)] × [15 × (-10)]
[Regrouping the terms] = 100 × (-150) = -15000

(d) (-41) × 102 = (-41) × [100 + 2]
= (-41) × 100 + (-41) × 2
[Distributive property of multiplication over addition] = -4100 – 82 = -4182

(e) 625 × (-35) + (-625) × 65
= 625 × [(-35) + (-65)]
[Distributive property of multiplication over addition] = 625 × (-100) = -62500

(f)7 × (50 – 2) = 7 × 50 -7 × 2 = 350 – 14 = 336
[Distributive property of multiplication over addition]

(g) (-17) × (-29) = (-17) × [30 + (-1)]
= (-17) × 30 + (-17) × (-1)= -510 + 17 = -493
[Distributive property of multiplication over addition]

(h) (-57) × (-19) + 57 = 57 × 19 + 57
= 57 × 19 + 57 × 1[Distributive property of multiplication over addition]
= 57 × (19 + 1) = 57 × 20 = 1140

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Temperature of the room in the beginning = 40°C
Temperature after 1 hour
= 40°C – 1 × 5°C = 40°C – 5°C – 35°C Similarly, temperature of the room after 10 hours = 40°C – 10 × 5°C = 40°C – 50°C = -10°C

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(i) Awarded for correct answer =(4×5) marks
Awarded for incorrect answer =6× (-2) marks
Total marks obtained by Mohan
= (4×5)+{6× (-2)}= 20-12=8
(ii) Awarded for correct answer = 5×5 = 25 Marks
Awarded for incorrect answer = 5 ×(-2) = -10
Total Marks obtained by Reshma =25-10 = 15 marks
(iii) Awarded for correct answer = 2×5 = 10Marks
Awarded for incorrect answer = 5×(-2) = -10
Total Marks obtained by Reshma =10-10= 0marks

Q.8 A cement company earns a profit of ? 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.

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) Profit on one white cement bag = Rs. 8
loss on one grey cement bag = Rs– 5
Profit on 3,000 bags of white cement
= Rs(8 × 3,000) = Rs 24,000
Loss on 5,000 bags of grey cement
= Rs (-5 × 5000) =Rs – 25,000
Total loss = – Rs25,000 + Rs24,000
= – Rs1000 i.e. Rs 1000
(b) Selling price of grey bags at a loss of ? 5
= Rs(5 × 6,400) – Rs 32,000
For no profit and no loss, the selling price of white bags = Rs32,000
Rate of selling price of white bags at a profit of Rs 8 per bag.
Therefore Number of white cement bags sold= 32,000÷8 = 4000 bags
Hence, the required number of bags = 4,000

Q.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

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(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 solutions for class 7 Maths Chapter 1 Integers Exercise 1.4

Q.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]

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(a) (-30) ÷ 10 = -3 [ (-) ÷(+) = (-) ]
(b) 50 ÷(-5) = -10 [(+)÷(-)= (-)]
(c) (-36) ÷ (-9) = 4 [(-) ÷(-) = (+)]
(d) (-49) ÷ (49) = -1
(e) 13 ÷ [(-2) + 1] = 13 ÷ (-1) = -13
(f) 0 ÷ (-12) = 0
(g) (-31) ÷ [(-30) + (-1)]= (-31) ÷ (-31) = 1
(h) [(-36) ÷ 12] ÷ 3= (-3) ÷ 3 = -1
(i) [(-6) + 5] ÷ [(-2) + 1]= (-1) ÷(-1)= 1

Q.2 Verify that a ÷ (b + c) $?$(a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(?) a = 12, b = – 4, c = 2
(b) a = (-10), b = 1, c = 1

NCERT Solutions for Class 7 Maths Chapter 1 Integers

a ÷ (b + c) $?$(a ÷ b) + (a ÷ c)
(a) If a = 12, b = – 4, c = 2
Then L.H.S.= 12÷{(-4)+2} = 12÷ (-2)= -6
R.H.S.= {12÷(-4)}+(12÷2) = (-3)+ 6 = 3
Hense L.H.S.$?$R.HS verified

(b) If a = (-10), b = 1, c = 1
Then L.H.S.= (-10)÷(1+1)= -10÷ 2 = -5
R.H.S.= {(-10)÷1}+{(-10)÷1}= (-10)+(-10) =-20
Hense L.H.S.$?$R.HS verified

Q.3 Fill in the blanks:
(a) 369 ÷ ___ = 369
(b) (-75) ÷ ___ = -1
(c) (-206) ÷ ___ =1
(d) -87 ÷ ___ = -87
(e) ___ ÷ 1 = -87
(f) _____ ÷ 48 = –1

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(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

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

NCERT Solutions for Class 7 Maths Chapter 1 Integers

(i) (15, -5) Because, 15 ÷ (–5) = (–3)
(ii) (-15, 5) Because, (-15) ÷ (5) = (–3)
(iii) (18, -6) Because, 18 ÷ (–6) = (–3)
(iv) (-18, 6) Because, (-18) ÷ 6 = (–3)
(v) (21, -7) Because, 21 ÷ (–7) = (–3)

Q.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 midnight?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Temperature at 12 noon was 10°C above zero i.e. +10°C
Rate of decrease in temperature per hour = 2°C
Number of hours from 12 noon to midnight = 12
Therefore Change in temperature in 12 hours = 12 × (-2°C) = -24°C Therefore Temperature at midnight = +10°C + (-24°C) = -14°C Hence, the required temperature at midnight =-14°C Difference in temperature between + 10°C and -8°C = +10°C – (-8°C) = +10°C + 8°C = 18°C Number of hours required = 18?C ÷ 2?C= 9 hours Therefore Time after 9 hours from 12 noon = 9 pm.

Q.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?

NCERT Solutions for Class 7 Maths Chapter 1 Integers

Given that:
+3 marks are given for each correct answer. (-2) marks are given for each incorrect answer. Zero marks for not attempted questions.
(i) Radhika scored 20 marks Then, Total marks awarded for 12 correct answers = 12 × 3 = 36
Marks awarded for incorrect answers = Total score – Total marks awarded for 12 correct Answers = 20 – 36 = – 16

(ii) Mohini scored -5 marks Then, Total marks awarded for 7 correct answers = 7 × 3 = 21
Marks awarded for incorrect answers = Total score – Total marks awarded for 7 correct Answers = -5 - 21 = -26

Q.7 An elevator descends into a nine 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.

NCERT Solutions for Class 7 Maths Chapter 1 Integers

The present position of the elevator is at 10 in above the ground level.
Distance moved by the elevator below the ground level = 350 m
Therefore Total distance moved by the elevator = 350 m + 10 m = 360 m
Rate of descent = 6 m/min.
Total time taken by the elevator
=360m ÷ 6m/min = 60 minutes = 1 hour
Hence, the required time = 1 hour.

##### FAQs Related to NCERT Solutions for Class 7 Maths Chapter 1 Integers
There are total 30 questions present in ncert solutions for class 7 maths chapter 1 integers
There are total 4 long question/answers in ncert solutions for class 7 maths chapter 1 integers
There are total 4 exercise present in ncert solutions for class 7 maths chapter 1 integers