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NCERT Solutions for class 7 Maths chapter 1 – Integers
Back Exercise
Exercise 1.1
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 Lahul-Spiti 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)
(b) The hottest place is Bangalore (22 °C) and the coldest place is Lahulspiti (-8°C). The temperature difference between the hottest and the coldest places
= 22 °C – (-8 °C)
= 22 °C + 8 °C
= 30 °C
(c) Temperature difference between Lahul-Spiti and Srinagar
= Temperature of Srinagar
– Temperature of Lahul-Spiti
= – 2°C – (- 8°C)
= – 2°C + 8°C
= 6°C
(d) Yes, we can say that the temperature of Srinagar and Shimla took together is less than the temperature at Shimla as -2 + 5 = 3 and 3 < 5.
This temperature is not less than the temperature at Srinagar.
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:
Total at the end
= 25 + (- 5) + (- 10) + 15 + 10
= (25 + 15 + 10) + {(- 5) + (- 10))
= 50 + (- 15) = 35
Question 3.
At Srinagar, the temperature was – 5°C on Monday and then it dropped by 2°C on Tues¬day. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?
Solution:
At Srinagar, the temperature was on Monday = -5 °C
Since the temperature was dropped by 2 °C on Tuesday, therefore, the temperature was on Tuesday = (-5-2) °C = -7°C
Also, on Wednesday the temperature rose by 4 °C.
∴ Temperature on Wednesday= (-7 + 4) °C = -3 °C
Question 4.
A plane is flying at the height of 5000 m above sea level. At a particular point, it is e×actly above a submarine floating 1200 m below sea level. What is the vertical distance between them?
Solution:
The vertical distance between the plane and submarine.
= 5000 m – (- 1200 m)
= 5000 m + 1200 m
= 6200 m.
Question 5.
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹1,642 from it, the next day. If the withdrawal of the 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:
The amount deposited will be represented by a positive integer.
Balance in Mohan’s account after withdrawal
= (+ ₹ 2000) + (- ₹ 1642)
= ₹ (2000-1642)
= ₹ 358
Question 6.
Rita goes 20 km towards the east from point A to point B. From B, she moves 30 km towards the west along the same road. If the distance towards the 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 the west = -30 km
Her final position from A
= + 20 km + (- 30) km
= – (30 – 20) km
= – 10 km.
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:
I. Row, Sum = 5 + (- 1) + (- 4)
= 5 + (- 5) = 0
II. Row, Sum = (- 5) + (- 2) + 7
= (- 7) + 7 = 0
III. Row, Sum = 0 + 3 + (- 3)
= 0 + 0 = 0
I. Column, Sum = 5 + (- 5) + 0
= 0 + 0 = 0
II. Column, Sum = (- 1) + (- 2) + 3
= (- 3) + 3 = 0
III. Column, Sum = (- 4) + 7 + (- 3)
= 7 + (- 4) + (- 3)
= 7 + (- 7) = 0
One Diagonal, Sum = 5 + (- 2) + (- 3) = 5 + (- 5) = 0
Other Diagonal, Sum = 0 + (- 2) + (- 4) = 0 + (- 6) = -6 ≠ 0
Therefore, the given square is not a magic square.
I. Row, Sum = 1 + (- 10) + 0 = – 9
II. Row, Sum = (- 4) + (- 3) + (- 2) = – 9
III. Row, Sum = (- 6) + 4 + (- 7)
= 4 + (- 6) + (- 7)
= 4 + (- 13)
= – (13-4) = -9
I. Column, Sum = 1 + (- 4) + (- 6)
= 1 + (- 10)
= -(10 – 1) = – 9
II. Column, Sum = (- 10) + (- 3) + 4
= (-13)+ 4
= – (13-4) = -9
III. Column, Sum = 0 + (- 2) + (- 7)
= 0 + (- 9) = – 9
One Diagonal, Sum = 1 + (- 3) + (- 7)
= 1 + (- 10)
= -(10-1) = -9
Other Diagonal, Sum = (- 6) + (- 3) + 0
= (- 9) + 0 = – 9
Since each row, column, and diagonal ‘ have the same sum, therefore, the given square is a magic square.
Question 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
L.H.S. = a – (- b) = 21 – (- 18) = 21 + 18 = 39 …..(1)
R.H.S. = a + b = 21 + 18 = 39 …..(2)
From (1) and (2), we get a -(-b) = a + b
(ii) a = 118, b = 125
L.H.S. = a – (- b) = 118 – (- 125) = 118 + 125 = 243 …(1)
R.H.S. = a + b = 118 + 125 = 243 …(2)
From (1) and (2), we get a -(- b) = a + b
(iii) a = 75, b = 84
L.H.S. = a – (- b) = 75 – (- 84) = 75 + 84 = 159 …(1)
R.H.S. = a + b = 75 + 84 = 159 …(2)
From (1) and (2), we get a – (- b) = a + b
(iv) a = 28, b = 11
L.H.S. = a – (- b) = 28 -(-11) = 28 + 11 = 39 …(1)
R.H.S. = a + b = 28 + 11 = 39 …(2)
From (1) and (2), we get a – (- b) = a + b.
Question 9.
Use the sign of >, < or = in the bo× 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.
Solution:
(a) L.H.S. = (- 8) + (- 4)
= – (8 + 4) = – 12
R.H.S. = (- 8) – (- 4)
= – 8 + 4 = – (8 – 4) = – 4
∴ (-8) + (-4) < (-8)-(-4)
(b) L.H.S. = (-3) + 7 – (19)
= + 4 – (19)
= + 4 – 19
= -15
R.H.S. = 15 – 8 + (- 9)
= 7 + (-9) = 7- 9 = -2
∴ (- 3) + 7 – (19) < 15 – 8 + (- 9)
(c) L.H.S. = 23 – 41 + 11
= 23 + 11- 41
= 34 – 41 = – (41 – 34)
= – 7
R.H.S. = 23 – 41 – 11
= 23 – (41 + 11)
= 23 – 52
= – (52 – 23)
= – 29
∴ 23 – 41 + > 23 – 41 – 11
(d) L.H.S. = 39 + (- 24) – (15)
= 39 – 24 – (15)
= 15 – (15) = 0
R.H.S. = 36 + (- 52) – (- 36)
= – (52 – 36) – (- 36)
= – 16 – (- 36)
= – 16 + 36 = 20
∴ 39 + (- 24) – (15) < 36 + (- 52) – (- 36)
(e) L.H.S. = – 231 + 79 + 51
= – 231 + 130
= – (231 – 130) = – 101
R.H.S. = – 399 + 159 + 81
= – 399 + 240
= – (399 – 240)
= – 159
∴ – 231 + 79 + 51 > – 399 + 159 + 81.
Question 10.
A water tank has stepped 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 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) While going down the monkey jumps 3 steps down and then jumps back 2 steps up. To reach the water level he is to jump as under:
-3+ 2 -3 + 2 – 3 +2 – 3 + 2 – 3 + 2 – 3 = -8 Hence, he takes 11 jumps to reach the water level.
(ii) After drinking water, he jumps back as under to reach the top step as under : 4 – 2+ 4 – 2+ 4 = 8
Hence, he takes 5 jumps to reach back the top.
(iii) (a) – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = -8
(b) 4 – 2 + 4 – 2 + 4 = 8
In (b) the sum 8 represents going up 8 steps.
Exercise 1.2
Question 1.
Write down a pair of integers whose:
(a) the sum is -7
(b) the difference is -10
(c) the sum is 0.
Solution:
(a) (-15) and 8
(b) 15 and 25.
(c) (-49) and 49
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) (-10) and (-18)
(b) (-10) and 5
(c) (-1) and 2
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 has an equal score.
Yes, integers can be added in any order and the result remains unaltered.
For example, 10 + 0 + (-40) = -30 = -40 + 0 + 10
Question 4.
Fill in the blanks to make the following statements true:
- (-5) + (-8) = (+8) + (……)
- -53 + …… = -53.
- 17 + …… = 0
- [13 + (-12)] + (…… ) = 13 + [(-12) + (- 7)]
- (- 4) + [15 + (- 3)] = [(-4) + 15] + …….
Solution:
- (-5) + (-8) = (-8) + (- 5)
- -53 + 0 = -53
- 17 + (- 17) = 0
- [13 + (- 12)] + (- 7) = 13 + [(- 12) + (-7)]
- (- 4) + [15 + (- 3)] = [(- 4) + 15] + (- 3).
Exercise 1.3
Question 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 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
Question 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)]
L.H.S. = 18 × [7 + (- 3)]
= 18 × L(7 – 3)] = 18 × (4) = 18 × 4 = 72
R.H.S. = [18 × 7] + [18 × (- 3)]
= 126 + [- (18 × 3)] = 126 + (- 54) = 126 – 54 = 72
So, 18 × [7 + (- 3)]
= [18 × 7] + [18 × (- 3)]
(b) (- 21) × [(- 4) + (- 6)] = [(- 21) × (- 4)] + [(- 21) × (- 6)]
L.H.S. = (- 21) × [(- 4) + (- 6)]
= (- 21) × (- 10)
= 21 × 10 = 210
R. H.S. = [(- 21) × (- 4)] + [(- 21) × (- 6)]
= (21 × 4) + (21 × 6)
= 84 + 126 = 210
So, (- 21) × [(- 4) + (- 6)]
= [(- 21) × (- 4)] + [(- 21) × (- 6)].
Question 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, (-1) x a = -a.
(ii) We know that the product of any integer and (-1) is the additive inverse of an 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.
Question 4.
Starting from (- 1) × 5, write various products showing some pattern to show (- 1) × (-1) – 1.
Solution:
(- 1) × 5 = – 5
(- 1) × 4 = – 4 [= (- 5) + 1]
(- 1) × 3 = – 3 [= (- 4) + 1]
(- 1) × 2 = – 2 [= (- 3) + 1]
(- 1) × 1 = – 1 [= (- 2) + 1]
(- 1) × 0 = 0 [= (- 1) + 1]
(- 1) × (- 1) = 1 [= 0 + 1]
Question 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) 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
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:
Room temperature 10 hours after the process begins
= 40°C – 10 × 5°C
= 40°C – 50°C
= – (50 – 40)°C = – 10°C
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 si× 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 for four correct answers 4 × 5 = 20 marks
He also gets for si× incorrect answers. 6 × (- 2) = – 12 marks.
Therefore, Mohan’s score = 20 + (- 12) = 20-12 = 8 marks.
(ii) Reshma gets for five correct answers 5 × 5 = 25 marks
She also gets for five incorrect answers 5 × (- 2) = – 10 marks Therefore, Reshma’s score = 25 + (- 10) = 25-10 = 15 marks.
(iii) Heena gets for two correct answers
2 × 5 = 10 marks.
She also gets for five incorrect answers 5 × (- 2) = – 10 marks
She didn’t attempt three questions. For these, she gets 3×0 = 0 marks
Therefore, Heena’s score = 10 + (- 10) + 0 = 10 – 10 + 0 = 0 marks.
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:
(a) The company sells 3,000 bags of white cement. So her profit = 3,000 × 8 = ₹ 24,000
Also, the company sells 5,000 bags of grey cement. So her loss = 5,000 × 5 = ₹ 25,000
Since 25,000 > 24,000
Therefore, the company is at a loss and the loss is = 25000 – 24000 = ₹ 1000
(b) Let ‘×’ be the number of white cement bags sold.
According to the question, we get
x × 8 = 6400 × 5
⇒ x =
Therefore, 4,000 bags of white cement must be sold to have neither profit nor loss.
Question 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) x (- 9) = 27
- (b) 5 x (-7) = (-35)
- (c) 7 x (-8) = (-56)
- (d) (-11) x (-12) = 132
Exercise 1.4
Question 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 = – 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.
Question 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 = l.
Solution:
(a) a + (b + c) = 12 ÷ [(- 4) + 2] = 12 + (- 2) = – 6
(a ÷ b) + (a ÷ c) = 12 ÷ (- 4) + 12 ÷ 2 = -3 + 6 = 3
So, a + (b + c) ≠ (a + b) + (a + c)
(b) a ÷ (b + c) = (- 10) + (1 + 1) = (- 10) + 2 = – 5
a ÷ b + a ÷ c = (- 10) ÷ 1 + (- 10) ÷ 1 = (- 10) + (- 10) = – 20
So, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).
Question 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 ÷ 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.
Question 4.
Write five pairs of integers (a, b) such that a + b = -3. One such pair is (6, -2) because 6 +(-2) = (-3).
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.
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 degrees 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
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) Let ‘x’ be the number of incorrect questions attempted by Radhika.
According to the question, we get
(+ 3) × 12 + x × (-2) = 20
⇒ 36 – 2x = 20
⇒ 2x = 36 – 20
⇒ x =
Therefore, Radhika attempted 8 incorrect questions.
(ii) Let ‘x’ be the number of incorrect question attempted by Mohini.
According to the question, we get
(+ 3) × 7 + x × (- 2) = – 5
⇒ 21 – 2x = -5
⇒ 2x = 21 + 5
⇒ x =
Therefore, Mohini attempted 13 incorrect questions.
Question 7.
An elevator descends into a mine shaft at the rate of 6m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution:
Difference in heights at two positions = 10 m – (-350 m) = 360 m
Rate of descent = 6 m/minute
∴ Time taken
Hence, the elevator will take 1 hour to reach = 350 m.