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.
NCERT Solutions for Class 7 Maths Chapter 1 Integers
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 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) 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) 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) 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 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.
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 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:
Temperature at Srinagar on Monday = – 5°C
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.
Question 4.
A plane is flying at the height of 5000 m above the sea level. At a particular point, it is e×actly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?
Solution:
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 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.
Question 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 positive integer then, how will you represent the distance travelled towards west?
By which integer will you represent her final position from A?
Solution:
The distance towards 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 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 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 com pleting the following;
(a) -3 + 2 – …………. = -8
(b) 4 – 2 + = 8.
In (a) the sum (-8) represent 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 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.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2
Question 1.
Write down a pair of integers whose:
(a) sum is – 7
(b) difference is – 10
(c) 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 score of team A
= (- 40) + 10 + 0 = (- 40) + 10 = – (40 – 10) = – 30
Total score of team B
= 10 + 0 + (- 40) = 10 + (- 40) = – (40 – 10) = – 30
So, both the teams scored equally. Yes ! We can say that we can add integers in any order.
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).
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 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 × (- 1)
3 × (- 1) = – (3 × 1) = – 3.
(b) (- 1) × 225
(- 1) × 225 = – (1 × 225) = – 225.
(c) (- 21) × (- 30)
(- 21) × (- 30) = 21 × 30 = 630.
(d) (- 316) × (- 1)
(- 316) × (- 1) = 316 × 1 = 316.
(e) (- 15) × 0 × (- 18)
(- 15) × 0 × (- 18) = (- 15) × 0) × (- 18) = 0 × (-18) = 0.
(f) (- 12) × (-11) × 10
(- 12) × (- 11) × 10
= [(- 12) × (- 11)] × 10
= (12 × 11) × 10
= 132 × 10 = 1320.
(g) 9 × (- 3) × (- 6)
9 × (- 3) × (- 6) = 9 × [(- 3) × (- 6)]
= 9 × (3 × 6)
= 9 × 18 = 162.
(h) (- 18) × (- 5) × (- 4)
(- 18) × (- 5) × (- 4)
= [(- 18) × (- 5)] × (- 4)
= (18 × 5) × (- 4)
= – (90 × 4) = – 360.
(i) (- 1) × (- 2) × (- 3) × 4
(- 1) × (- 2) × (- 3) × 4
= [(- 1) × (- 2)] × [(- 3) × 4]
= (1 × 2) × [- (3 × 4)]
= 2 × [-(12)] = 2 × ( – 12)
= – ( 2 × 12) = -24.
(j) (- 3) × (- 6) × (- 2) × (- 1)
(- 3) × (- 6) × (- 2) × (- 1)
= [(- 3) × (- 6)] × [(- 2) × (- 1)]
= (3 × 6) × (2 × 1) = 18 × 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) (-1) × a = – a
(ii) (a) 22 (b)-37 (c) 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) 26 × (- 48) + (- 48) × (- 36)
26 × (- 48) + (- 48) × (- 36)
= [- (26 × 48)] + (48 × 36)
a × (- b) = – (a × b) = (- 1248) + 1728
= 480. | (—a) × (—b) = a × b
Aliter. 26 × (- 48) + (- 48) × (- 36)
= 26 ×(- 48) + (-36) × (-48)
= [26 + (-36)] × (-48)
[Distributive property] = (-10) × (-48) = 480.
(b) 8 × 53 × (- 125)
8 × 53 × (- 125) = (8 × 53) × (- 125)
= 424 × (- 125)
= – (424 × 125)
∝ × (- b) = – (a × b) = – [424 × (100 + 25)] = – [424 × 100 + 424 × 25]
Distributivity of multiplication over addition = – [42400 + 10600] = – 53000
Aliter. 8 × 53 × (- 125)
= 8 × (- 125) × 53
Commutativity of multiplication
= [8 × (- 125)] × 53 = [- (8 × 125)] × 53
α × (- b) = – (a × b) = (- 1000) × 53 = – (1000 × 53)
(- α) × b = – (a × b) = – 53000.
(c) 15 × (- 25) × (- 4) × (- 10)
15 × (- 25) × (- 4) × (- 10)
= 15 × (- 25) × (- 10) × (-4)
Commutativity of multiplication = 15 × (- 10) × (- 25) × (- 4)
Commutativity of multiplication = [(15) × (- 10)] × [(- 25) × (- 4)]
= [-(15 × 10)] × [25 × 4]
α × (- b) = -(a × b) (- a) × (- b) = a × b
= (- 150) × 100 = -(150 × 100) = – 15000.
(d) (- 41) × 102
(- 41) × 102 = – (41 × 102)
(- α) × b = – (a × b) = – [41 × (100 + 2)] = – [41 × 100 + 41×2]
Distributivity of multipli-cation over addition
= – [4100 + 82] = – 4182.
(e) 625 × (- 35) + (- 625) × 65
625 × (- 35) + (- 625) × 65
= 625 × (- 35) + 625 × (- 65)
(- α) × b = a × (- b) = 625 × [(-35) + (- 65)]
Distributivity of multipli-cation over addition
= 625 × (- 100) = – (625 × 100)
α × (- b) = – (a × b) = – 62500.
(f) 7 × [50 – 2]
7 × [50 – 2] = 7 × 50 – 7 × 2
Distributivity of multiplication over subtraction = 350 – 14 = 336.
(g) (- 17) × (- 29)
(- 17) × (- 29) = 17 × 29
(-α) × (- b) = a × b = 17 × (30-1) = 17 × 30- 17 × 1
Distributivity of multiplication over subtraction = 510 – 17 = 493.
(h) (- 57) × (- 19) + 57
(- 57) × (- 19) + 57 = 57 × 19 + 57
(-α) × (- b) = a × b = 57 × 19 + 57 ×1 I a × 1 = a = 57 × (19 + 1)
Distributivity of multipli-cation over addition = 57 × 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, company is in 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 = \(\frac { 6400\times 5 }{ 8 }\) = 800 × 5
= 4,000 bags.
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) × (- 9) = 27
(b) 5 × (- 7) = – 35
(c) 7 × (- 8) = – 56
(d) (- 11) × (- 12) = 132.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 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:
- [9, (- 3)]
- [12, (- 4)]
- [15, (-5)]
- [-9, 3]
- [-12, 4].
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:
Temperature at 12 noon = + 10°C
Temperature at 1 P.M.= 10°C – 2°C = 8°C
Temperature at 2 P.M. = 8°C – 2°C = 6°C
Temperature at 3 P.M. = 6°C – 2°C = 4°C
Temperature at 4 P.M. = 4°C – 2°C = 2°C
Temperature at 5 P.M. = 2°C – 2°C = 0°C
Temperature at 6 P.M. = 0°C – 2°C = – 2°C
Temperature at 7 P.M. = – 2°C – 2°C = – 4°C
Temperature at 8 P.M. = – 4°C – 2°C = – 6°C
Temperature at 9 P.M. = – 6°C – 2°C = – 8°C
So, the temperature was 8°C below zero at 9 P.M.
Temperature at 10 P.M. = – 8°C – 2°C = – 10°C
Temperature at 11 P.M. = – 10°C – 2°C = – 12°C
Temperature at 12 P.M. = – 12°C – 2°C = – 14°C
So, the temperature at mid-night was 14 degrees below zero.
Aliter:
Difference between the temperatures 10°C above zero and 8°C below zero
= 10°C – (- 8°C)
= 10°C + 8°C = 18°C
Time for this degrees
= 18 ÷ 2 = 9 hours
So, the temperature was 8°C below zero at 9 P.M.
Difference between the times 12 noon and mid-night = 12 hours
Degrees in temperature in 12 hours = 12 × 2 = 24°C
So, temperature at mid night
= 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 question attempted by Radhika.
According to the question, we get
(+ 3) × 12 + x × (-2) = 20
⇒ 36 – 2x = 20
⇒ 2x = 36 – 20
⇒ x = \(\frac { 16 }{ 2 } \) = 8
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 = \(\frac { 26 }{ 2 } \) = 13
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:
Let ‘x’ minutes be the time taken to reach – 350 m.
According to the question, we get
x = \(\frac { (-10)+(-350) }{ 6 } \) = \(\frac { -360 }{ 6 } \) = -60
Negative sign indicates descent.
∴ x = 60 minutes = 1 hour
NCERT Solutions for Class 7 Maths Chapter 1 Integers MCQ
Multiple Choice Questions
1. On a number line, when we add a positive integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these.
2. On a number line, when we add a negative integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these.
3. On a number line, when we subtract a positive integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these.
4. On a number line, when we subtract a negative integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these.
5. When two positive integers are added, we get
(a) a positive integer
(b) a negative integer
(c) sometimes a positive integer, sometimes a negative integer
(d) none of these.
6. When two negative integers are added, we get
(a) a positive integer
(b) a negative integer
(c) sometimes a positive integer, sometimes a negative integer
(d) none of these.
7. Which of the following statements is wrong?
(а) When a positive integer and a negative integer are added, we always get a negative integer
(b) Additive inverse of 8 is (- 8)
(c) Additive inverse of (- 8) is 8
(d) For subtraction, we add the additive inverse of the integer that is being subtracted, to the other integer.
8. Which of the following is true?
(а) (- 8) + (- 4) > (- 8) – (- 4)
(b) (- 8) + (- 4) < (- 8) – (- 4)
(c) (- 8) + (- 4) = (- 8) – (- 4)
(d) none of these.
9. The product of two negative integers is
(a) a positive integer
(b) a negative integer
(c) either a positive integer or a negative integer
(d) none of these.
10. The product of three negative integers is
(a) a positive integer
(b) a negative integer
(c) either a positive integer or a nega¬tive integer
(d) none of these.
11. (- 1) × (- 1) × (- 1) × ……. 10 times is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these.
12. (- 1) × (- 1) × (- 1) × …… 5 times is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these.
13. (- 1) × (- 1) × (- 1) × …… 2m times, where m is a natural number, is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these.
14. (- 1) x (- 1) x (- 1) × …………….. (2m + 1) times, where m is a natural number, is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these.
15. (- 20) × (- 5) is equal to
(a) 100
(b) – 100
(c) 20
(d) 5.
16. (- 30) × 20 is equal to
(a) 600
(b) – 600
(c) 50
(d) 10.
17. 10 × (- 20) is equal to
(a) 200
(b) -200
(c) 30
(d) 10
18. 3 × 0 is equal to
(a) 0
(b) 3
(c) 1
(d) -3
19. 0 × (- 5) is equal to
(a) 0
(b) 5
(c) – 5
(d) 1.
20. (- 2) × 1 is equal to
(a) 2
(b) – 2
(c) 1
(d) – 1.
21. 1 × 6 is equal to
(a) 6
(b) – 6
(c) 1
(d) – 1.
22. 4 × (- 1) is equal to
(a) 4
(b) – 4
(c) 1
(d) – 1.
23. (- 10) × 0 × (- 15) is equal to
(a) 0
(b) 10
(c) 15
(d) 150.
24. The integer whose product with (- 1) is 0, is
(a) 1
(b) -1
(c) 0
(d) none of these.
25. The integer whose product with (- 1) is
(a) -1
(b) 1
(c) 0
(d) none of these.
26. 10 ÷ (- 5) is equal to
(a) 1
(b) 2
(c) -2
(d) 5.
27. (- 6) ÷ (- 3) is equal to
(a) 1
(b) 2
(c) 3
(d) 6.
28. (- 5) ÷ 1 is equal to
(a) 5
(b) -5
(c) 1
(d) -1.
29. (-9) ÷ (-1)ìsequalto
(a) 1
(b) 9
(c) -1
(d) -9.
30. If 16 ÷ x = 16, then x is equal to
(a) 1
(b) 2
(c) 4
(d) 16.
31. If (- 50) ÷ x = 1, then x is equal to
(a) 1
(b) 50
(c) – 50
(d) 10.
32. If x ÷ 1 = 8, then x is equal to
(a) 8
(b) 1
(c) – 8
(d) – 1.
33. The additive identity for integers is
(a) 1
(b) – 1
(c) 0
(d) none of these.
34. The multiplicative identity for integers is
(a) 1
(b) – 1
(c) 0
(d) none of these.
35. a × (5 + c) = a × 6 + a × c is called
(a) commutative property
(b) associative property
(c) distributive property
(d) closure property.
36. Manish deposits ₹ 2000 in his bank account and withdraws ₹ 1000 from it, the next day. Find the balance in Manish’s account
(a) ₹ 2000
(b) ₹ 3000
(c) ₹ 1000
(d) none of these
Answer
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