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NCERT Solutions for Class 7 Maths Chapter 1 Integers are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers.
Class 7 Maths NCERT Solutions Chapter 1 Integers Ex 1.1
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 1 |
| Chapter Name | Integers |
| Exercise | Ex 1.1, Ex 1.2, Ex 1.3, Ex 1.4 |
| Number of Questions Solved | 30 |
| Category | NCERT Solutions |
NCERT Solutions for Class 7 Maths Chapter 1 Integers
Chapter 1 Integers Exercise 1.1
Ex 1.1 Class 7 Maths Question 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) From the given number line, we find that the temperature of the indicated places as under :
(b) Temperature difference between the hottest and the coldest places
= Temperature of Bangalore – Temperature of Lahul-Spiti
= 22°C-(-8°C)
= 22°C + 8°C
= 30°C
(c) The temperature difference between Lahulspiti and Srinagar
= -2 °C – (-8 °C)
= -2 °C +8 °C = 6 °C
(d) Temperature of Srinagar and Shimla together
= Temperature of Srinagar + Temperature of Shimla
= -2° + 5°C
= 3°C
Temperature at Shimla = 5°C
Temperature at Srinagar = – 2°C.
Therefore, we can say that the temperature of Srinagar and Shimla together is less than the temperature at Shimla but the temperature of Srinagar and Shimla together is not less than the temperature at Srinagar.
Ex 1.1 Class 7 Maths Question 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 in five successive rounds were given as 25, -5, -10, 15 and 10.
Jack’s total score
=25 +(-5)+ (-10)+ 15+ 10
= 25-5-10+15 + 10 = 50-15 = 35
Ex 1.1 Class 7 Maths Question 3.
At Srinagar, the 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.
Temperature at Srinagar on Monday = – 5°C
The drop-in temperature at Srinagar on Tuesday = 2°C
∴Temperature at Srinagar on Tuesday = – 5°C – 2°C = – 7°C
Rise in temperature at Srinagar on Wednesday = 4°C
Temperature at Srinagar on Wednesday
= – 7°C + 4°C
= – (7 – 4)°C
= -3°C.
Ex 1.1 Class 7 Maths Question 4.
A plane is flying at the height of 5000 m above sea level. At a particular point, it is exactly above a submarine floating 1200 m below sea level. What is the vertical distance between them?
Solution.
Vertical distance between the plane and the submarine
= 5000 m + 1200 m = 6200 m
Ex 1.1 Class 7 Maths Question 5.
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹ 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.
Amount deposited = + ₹2000
Balance in Mohan’s account after the withdrawal
= ₹ 2000 – ₹ 1642
= ₹ (2000 – 1642)
= ₹ 358.
Ex 1.1 Class 7 Maths Question 6.
Rita goes 20 km towards the east from point A to the point B. From B, she moves 30 km towards the 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 the west? By which integer will you represent her final position from A?
Solution.
The distance towards west will be represented by a negative integer.
Rita’s movement is shown as under:
Since, Rita moves 20 km towards east from a point A, so she reaches B, and then from B she moves 30 km towards the west along the same road and reaches C. Thus, her final position from A will be represented by the integer -10.
Ex 1.1 Class 7 Maths Question 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) :
Row 1 : 5 + (-1) + (-4) = 5-1-4 =0
Row 2 : (-5) + (-2) + 7 =-5-2 + 7 = 0
Row 3 : 0 + 3 + (-3) =0+3-3 =0
Column 1 : 5 + (-5) + 0= 5- 5 + 0= 0
Column 2 : (-1) + (-2) + 3 = -1 -2 + 3 = 0
Column 3 : (-4) + 7 + (-3) = -4 + 7- 3 = 0
Diagonal 1 : 5 + (-2) + (-3) = 5 – 2 – 3 =0
Diagonal 2 : (-4) + (-2) + 0 = -4 – 2 + 0 = -6
∵ The sum of digits along with the diagonal 2 ≠ 0.
Thus, it is not a magic square.
In square (ii) :
Row 1 : 1 + (-10) + 0 = 1-10+0 = -9
Row 2 : (-4) +(-3) +(-2) = -4-3-2 = -9
Row 3 : (-6) + 4 + (-7) = -6 + 4 – 7 = -9
Column 1 : 1 + (-4) + (-6) = 1- 4- 6 = -9
Column 2 : (-10) + (-3) + 4 = -10-3 + 4 = -9
Column 3 : 0 + (-2) + (-7) = 0-2-7 =-9
Diagonal 1 : 1 + (-3) + (-7) = -9
Diagonal 2: 0 + (-3) + (-6) = 0- 3- 6 = -9
∵ Each row, column, and diagonal have the same sum.
Thus, it is a magic square.
Ex 1.1 Class 7 Maths Question 8.
Verify a – (-b) = a + b for the following values of a and b:
- a = 21, b = 18
- a = 118, b = 125
- a = 75, b = 84
- a = 28, b = 11
Solution.
- L.H.S. = a – (-b) = 21 – (-18) = 21 +18 = 39
R.H.S. = a + b = 21 +18 =39
∴ L.H.S. = R.H.S. - L.H.S. = a – (-b) = 118 – (-125) = 118 +125 = 243
R.H.S. = a + b = 118+125 = 243
∴ L.H.S. = R.H.S. - L.H.S. = a – (-b) = 75 – (-84) = 75+ 84 = 159
R.H.S. = a + b =75 + 84 = 159 - L.H.S. = a – (-b) = 28 – (-11) = 28+ 11 = 39
R.H.S. = a + b = 28 + 11 = 39
Ex 1.1 Class 7 Maths Question 9.
Use the sign of >, < or = in the box to make the statements true.
(a) (-8) + (-4) 
(b) (-3) + 7 – (19)
(c) 23 – 41 + 11
(d) 39+ (-24) – (15)
(e) -231 + 79 + 51
Solution.
Ex 1.1 Class 7 Maths Question 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) To reach the water level his jump will be as follows:
(- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) = -8.
Hence in 11 jumps, he will reach the water level.
(ii) To reach back to the top step his jumps will be as follows:
4 + (-2) + 4 + (-2) + 4 = 8
Therefore, he will be out of the tank in 5 jumps.
(iii) (a) – 3 + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (T 3) + 2 + (- 3) = -8
(6) 4 – 2 + 4 – 2 + 4 = 8
The sum 8 in (b) will represent going up.
Chapter 1 Integers Exercise 1.2
Ex 1.2 Class 7 Maths Question 1.
Write down a pair of integers whose:
(a) sum is -7
(b) difference is -10
(c) sum is 0
Solution.
(a) A pair of integers whose sum is -7 can be (-1) and (-6).
∵ (-1) + (-6) = -7
(b) A pair of integers whose difference is -10 can be (-11) and (-1)
∵ -11 – (-1) = -11+1 = -10
(c) A pair of integers whose sum is 0 can be 1 and (-1).
∵ (-1) + (1) = 0.
Ex 1.2 Class 7 Maths Question 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) A pair of negative integers whose difference gives 8 can be -12 and -20.
∵ (-12) – (-20) = -12+20 =8 .
(b) A negative integer and a positive integer whose sum is -5 can be -13 and 8.
∵ (-13) + 8 = -13 +8 = -5
(c) A negative integer and a positive integer whose difference is -3 can be -1 and 2.
∵ (-1) – 2 = – 1 -2 = -3
Ex 1.2 Class 7 Maths Question 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.
Total scores of team A = (-40) + 10 +0
= -40 + 10 + 0 = -30
and, total scores of team B = 10 + 0 + (-40)
= 10 + 0 – 40 = -30
Since, the total scores of each team are equal.
∴ No team scored more than the other but each have equal score.
Yes, integers can be added in any order and the result remains unaltered. For example, 10 +0 +(-40) = -30 = -40 +0 +10
Ex 1.2 Class 7 Maths Question 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)
(ii) -53 + 0 = -53
(iii) 17 + (-17) = 0
(iv) [13 + (-12)] + (-7) = (13) + [(-12) + (-7)]
(v) (-4) + [15 + (-3)] = [(-4) + 15] + (-3)
Chapter 1 Integers Exercise 1.3
Ex 1.3 Class 7 Maths Question 1.
Find each of the following products :
(a) 3 x (-1)
(b) (-1) x 225
(c) (-21) x (-30)
(d) (-316) x (-1)
(e) (-15) x O x (-18)
(f) (-12) x (-11) x (10)
(g) 9 x (-3) x (-6)
(h) (-18) x (-5) x (-4)
(i) (-1) x (-2) x (-3) x 4 Sol. (a) 3 x (-1) = – (3 x 1) = -3
(j) (-3) x (-6) x (-2) x (-1)
Solution.
(a) 3 x (-1) = – (3 x 1) = -3
(b) (-1) x 225 = – (1 x 225) = -225
(c) (-21) x (-30) = 21 x 30 =630
(d) (-316) x (-1) = 316 x 1 = 316
(e) (-15) x 0 x (-18) = [(-15) x 0] x (-18) = 0 x (-18) = 0
(f) (-12) x (-11) x (10) = [(-12) x (-11)] x (10)
= (132) x (10) =1320
(g) 9 x (-3) x (-6) = [9 x (-3)] x (-6) = (-27) x (-6) = 162
(h) (-18) x (-5) x (-4) = [(-18) x (-5)] x (-4)
= 90 x (-4) – -360
(i) (-1) x (-2) x (-3) x 4 = [(-1) x (-2)] x [(-3) x 4]
= (2)x (-12) = -24
(j) (-3) x (-6) x (-2) x (-1) = [(-3) x (-6)] x [(-2) x (-1)] = (18) x (2) = 36
Ex 1.3 Class 7 Maths Question 2.
Verify the following:
(a) 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]
(b) (-21) x [(-4) + (-6)] = [(-21) x (-4)] + [(-21) x (-6)]
Solution.
(a) We have,
18 x [7 + (-3)] = 18 x 4 = 72
and, [18 x 7] + [18 x (-3)] = 126 – 54 =72
18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]
(b) We have,
(-21) x [(-4) + (-6)] = (-21) x (-4 -6)
= (-21)(-10) = 210 and, [(-21) x (-4) + [(-21) x (-6)]
= 84+126 =210
∴ (-21) x [(-4) + (-6)] = [(-21) x (-4)] + [(-21) x (-6)]
Ex 1.3 Class 7 Maths Question 3.
(i) For any integer a, what is (-1) x a equal to?
(ii) Determine the integer whose product with (-1) is
(a) -22
(b) 37
(c) 0
Solution.
(i) For any integer a, (-1) x a = -a.
(ii) We know that the product of any integer and (-1) is the additive inverse of integer.
The integer whose product with (-1) is
(a) additive inverse of -22, t. e., 22.
(b) additive inverse of 37, i.e., -37.
(c) additive inverse of 0, i.e., 0.
Ex 1.3 Class 7 Maths Question 4.
Starting from (-1) x 5, write various products showing some pattern to show (-1) x (-1) = 1.
Solution.
(-1) x 5 = -5
(-1) x 4 = -4 = [-5 – (-1)] = -5 +1
(-1) x 3 = -3 = [-4 – (-1)] = -4 +1
(-1) x 2 = -2 = [-3 – (-1)] = -3 +1
(-1) x 1 = -1 = [-2 – (-1)] = -2 +1
(-1) x 0 = 0 = [-1 – (-1)] = -1 +1
(-1) x (-1) =[0 – (-1)] = 0 + 1 = 1
Ex 1.3 Class 7 Maths Question 5.
Find the product, using suitable properties :
(a) 26 x (-48) + (-48) x (-36)
(b) 8 x 53 x (-125)
(c) 15 x (-25) x (-4) x (-10)
(d) (-41) x 102
(e) 625 x (-35) +(-625) x 65
(f) 7 x (50-2)
(g) (-17) x (-29)
(h) (-57) x (-19) + 57
Solution.
(a) We have, 26 x (-48) + (-48) x (-36)
= (-48) x 26 + (-48) x (-36)
= (-48) x [26 + (-36)]
= (-48) x (26 – 36)
=(-48) x (-10)= 480
(b) We have,
8 x 53 x (-125) = [8 x (-125)] x 53
= (-1000) x 53 = -53000
(c) We have,
15 x (-25) x (-4) x (-10)
=15 x [(-25) x (-4)] x (-10)
= 15 x (100) x (-10)
= (15 x 100) x (-10)
= 1500 x (-10) = -15000
(d) We have,
(-41) x 102 = (-41) x (100 +2)
= (-41) x 100 + (-41) x 2 = -4100 – 82 = -4182
(e) We have, 625 x (-35) + (-625) x 65
= 625 x (-35) + (625) x (-65)
= 625 x [(-35)+ (-65)]
= 625 x (-100) = -62500
(f) 7 x (50-2) = 7 x 50 – 7 x 2
= 350 -14 =336
(g) (-17) x (-29) = (-17) x [(-30) + 1]
= (-17) x (-30) + (-17) x 1 = 510 – 17 = 493
(h) (-57) x (-19)+ 57 =57 x 19 + 57 x 1
= 57 x (19 +1)
= 57 x 20 = 1140
Ex 1.3 Class 7 Maths Question 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.
Initial room temperature = 40 X
Temperature lowered every hour = (-5) °C
Temperature lowered in 10 hours = (-5) x 10 °C = -50 °C
∴ Room temperature after 10 hours = 40 X – 50 X = -10 °C
Ex 1.3 Class 7 Maths Question 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) Marks awarded for one correct answer = 5
Marks scored for 4 correct answer = 5 x 4 = 20
Marks awarded for one incorrect answer = (-2)
Marks scored for 6 incorrect answer = (-2) x 6 = -12
Hence, Mohan’s score = 20 – 12 = 8 marks.
(ii) Reshma’s score for 5 correct answers = 5 x 5 = 25 marks
Reshma’s score for 5 incorrect answers = (-2) x 5 = -10 marks
Hence, Reshma’s score = 25-10 =15 marks
(iii) Heena’s score for 2 correct and 5 incorrect answers
= (5 x 2) + {(-2) x 5}
= 10+ (-10) = 10 – 10 =0.
Ex 1.3 Class 7 Maths Question 8.
A cement company earns a profit of ? 8 per bag of white cement sold and a loss of ? 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.
Profit on sale of 1 bag of white cement = ₹ 8
Loss on sale of 1 bag of grey cement = – ₹ 5
(a) Profit on sale of 3000 bags of white cement
= ₹ (3000 x 8)
= ₹ 24,000
Loss on sale of 5000 bags of grey cement = ₹ (5000 x -5)
= – ₹ 25,000
Difference between the two = ₹ 24,000 – ₹ 25,000 = – ₹ 1,000
Hence, there is a loss of ₹ 1000.
(b) Loss on the sale of 6400 bags of grey cement = ₹ (6400 x 5) = ₹ 32,000
In order to have neither profit nor loss, the profit on the sale of white cement should be ? 32,000.
Number of white cement bags sold
Hence, 4000 bags of white cement should be sold to have neither profit nor loss.
Replace the blank with an integer to make it a true statement.
(a) (-3) x = 27
(b) 5 x = -35
(c) 7 x (-8) = -56
(d) (-11) x (-12) = 132
Solution.
(a) (-3) x (-9) = 27
(b) 5 x (-7) = (-35)
(c) 7 x (-8) = (-56)
(d) (-11) x (-12) =132
Chapter 1 Integers Exercise 1.4
Ex 1.4 Class 7 Maths Question 1.
Evaluate each of the following:
Solution.
Ex 1.4 Class 7 Maths Question 2.
values of a, b and c.
(a) a = 12, b = -4, c = 2
(b) a = (-10), b = 1, c = 1
Solution.
Ex 1.4 Class 7 Maths Question 3.
Fill in the blanks :
Solution.
Ex 1.4 Class 7 Maths Question 4.
Write five pairs of integers (a, b) such that 
Solution.
Five pairs of integers (a, b) such that a + b = -3 are : (-6,2), (-9, 3), (12,-4), (21,-7), (-24, 8)
Note : We may write many such pairs of integers.
Ex 1.4 Class 7 Maths Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2 °C per hour until mid-night, at what time would the temperature be 8 °C below zero? What would be the temperature at mid-night?
Solution.
Difference in temperatures +10 °C and -8
= [10 – (-8)] °C = (10 + 8)° C = 18 °C
Decrease in temperature in one hour = 2°C
Number of hours taken to have temperature 8 °C below zero
So, at 9 P.M., the temperature will be 8 °C below zero
Temperature at mid-night = 10 °C – (2 x 12) °C
= 10°C – 24 °C = -14 °C
Ex 1.4 Class 7 Maths Question 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) Marks given for 12 correct answers at the rate of + 3 marks for each answer = 3 x 12 = 36 Radhika’s score = 20 marks
∴ Marks deducted her for incorrect answers = 20 – 36 = -16
Marks given for one incorrect answer = -2
Number of incorrect answers
(ii) Marks given for 7 correct answers at the rate of + 3 marks for each answer = 3 x 7 = 21 Mohini’s score = -5
∴ Marks deducted for incorrect answers
= – 5 – 21 = -26
Marks given for one incorrect answer = -2
∴ Number of incorrect answers
Ex 1.4 Class 7 Maths Question 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.
Answer.
Difference in heights at two positions = 10 m – (- 350 m) = 360 m
Rate of descent = 6 m/minute
∴ Time taken 
= 60 minutes = 1 hour
Hence, the elevator will take 1 hour to reach – 350 m.
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