NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT
SOLUTIONS
CLASS - 7TH

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Class : 7th
Subject : Maths
Chapter : 1
Chapter Name : Integers

Exercise 1.1

Q1 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) By observing the given data, the temperatures of these cities are as follows.
Lahulspiti : −8°C
Srinagar : −2°C
Shimla : 5°C
Ooty : 14°C
Bangalore : 22°C
(b) Temperature at the hottest place, i.e., Bangalore = 22°C
Temperature at the coldest place, i.e., Lahulspiti = −8°C
Temperature difference = 22°C − (−8°C)
= 30°C
Therefore, the temperature difference between the hottest and the coldest places is 30ºC.
(c) Temperature at Lahulspiti = −8°C
Temperature at Srinagar = −2°C
Temperature difference = −2°C − (−8°C)
= 6°C
Therefore, the temperature difference between Lahulspiti and Srinagar is 6ºC.
(d) Temperature at Srinagar = −2°C
Temperature at Shimla = 5°C
Temperature of Srinagar and Shimla taken together = −2°C + 5°C

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

= 3°C
3°C < 5°C
3°C < Temperature of Shimla Yes, the temperature of Srinagar and Shimla taken together is
less than the temperature of Shimla. However, 3°C > −2°C
Hence, the temperature of Srinagar and Shimla taken together is not less than the
temperature of Srinagar.

Page : 4 , Block Name : Exercise 1.1

Q2 In a quiz, positive marks are given for correct answers and negative marks are given for
incorrect answers. If Jack’s scores in ve successive rounds were 25, – 5, – 10, 15 and 10, what
was his total at the end?

Answer. Jack’s scores in ve successive rounds are 25, −5, −10, 15, and 10. Total score of Jack at
the end will be the sum of these scores.
Therefore, Jack’s total score at the end = 25 − 5 − 10 + 15 + 10 = 35

Page : 4 , Block Name : Exercise 1.1

Q3 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?

Answer. Temperature on Monday = −5°C
Temperature on Tuesday = Temperature on Monday −2°C
= −5°C − 2°C = −7°C
Temperature on Wednesday = Temperature on Tuesday + 4°C
= −7°C + 4°C = −3°C
Therefore, the temperature on Tuesday and Wednesday was −7ºC and −3ºC respectively.

Page : 4 , Block Name : Exercise 1.1

Q4 A plane is ying at the height of 5000 m above the sea level. At a particular point, it is
exactly above a submarine oating 1200 m below the sea level. What is the vertical distance
between them?

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Answer. Height of plane = 5000 m
Depth of submarine = −1200 m
Distance between plane and submarine = 5000 m − (−1200) m
= 5000 m + 1200 m = 6200 m

Page : 4 , Block Name : Exercise 1.1

Q5 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.

Answer. Since the amount withdrawn is represented by a negative integer, the amount
deposited will be represented by a positive integer.
Amount deposited = Rs 2000
Amount withdrawn = −Rs 1642
Balance in Mohan’s account = Money deposited + Money withdrawn
= 2000 + (−1642) = 2000 − 1642 = 358
Therefore, balance in Mohan’s account after withdrawal is Rs 358.

Page : 4 , Block Name : Exercise 1.1

Q6 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 nal position from A?

Answer. Since the distance towards east is represented by a positive integer, the distance
travelled towards west will be represented by a negative integer.
Distance travelled in east direction = 20 km
Distance travelled in west direction = −30 km
Distance travelled from A = 20 + (−30) = −10 km
Therefore, we will represent the distance travelled by Rita from point A by a negative integer,
i.e., −10 km (i.e., Rita is now in west direction).

Page : 5 , Block Name : Exercise 1.1

Q7 In a magic square each row, column and diagonal have the same sum. Check which of the
following is a magic square.

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Answer. It can be observed that in square (i), every row and column add up to give 0. However,
the sum of one of its diagonals is not 0.
As − 4 − 2 = −6 ≠ 0,
Therefore, (i) is not a magic square.
Similarly, in square (ii), each row, column, and diagonal add up to give −9. Therefore,
(ii) is a magic square.

Page : 5 , Block Name : Exercise 1.1

Q8 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

Answer. (i) a = 21, b = 18
a − (−b) = 21 − (−18) = 21 + 18 = 39
a + b = 21 + 18 = 39
∴ a − (−b) = a + b = 39
(ii) a = 118, b = 125
a − (−b) = 118 − (−125) = 118 + 125 = 243
a + b = 118 + 125 = 243
∴ a − (−b) = a + b = 243
(iii) a = 75, b = 84
a − (−b) = 75 − (−84) = 75 + 84 = 159
a + b = 75 + 84 = 159
∴ a − (−b) = a + b = 159
(iv) a = 28, b = 11
a − (−b) = 28 − (−11) = 28 + 11 = 39
a + b = 28 + 11 = 39
∴ a − (−b) = a + b = 39

Page : 5 , Block Name : Exercise 1.1

Q9 Use the sign of > , < or = in the box to make the statements true.

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Answer.

Page : 5 , Block Name : Exercise 1.1

Q10 A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the rst
step). The water level is at the ninth step.

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

(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?

Answer. Let the steps moved down be represented by positive integers and the steps moved up
be represented by negative integers.
(i) Initially, the monkey was at step = 1
After 1st jump, the monkey will be at step = 1 + 3 = 4
After 2nd jump, the monkey will be at step = 4 + (−2) = 2
After 3rd jump, the monkey will be at step = 2 + 3 = 5
After 4th jump, the monkey will be at step = 5 + (−2) = 3
After 5th jump, the monkey will be at step = 3 + 3 = 6
After 6th jump, the monkey will be at step = 6 + (−2) = 4
After 7th jump, the monkey will be at step = 4 + 3 = 7
After 8th jump, the monkey will be at step = 7 + (−2) = 5
After 9th jump, the monkey will be at step = 5 + 3 = 8
After 10th jump, the monkey will be at step = 8 + (−2) = 6
After 11th jump, the monkey will be at step = 6 + 3 = 9
Clearly, the monkey will be at water level (i.e., 9th step) after 11 jumps.
(ii) Initially, the monkey was at step = 9
After 1st jump, the monkey will be at step = 9 + (−4) = 5
After 2nd jump, the monkey will be at step = 5 + 2 = 7
After 3rd jump, the monkey will be at step = 7 + (− 4) = 3
After 4th jump, the monkey will be at step = 3 + 2 = 5
After 5th jump, the monkey will be at step = 5 + (− 4) = 1
Clearly, the monkey will reach back at the top step after 5 jumps.
(iii) If steps moved down are represented by negative integers and steps moved up are
represented by positive integers, then his moves will be as follows.
Moves in part (i)

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

− 3 + 2 − 3 + 2 − 3 + 2 − 3 + 2 − 3 + 2 − 3 = −8
Moves in part (ii)
4−2+4−2+4=8
Moves in part (ii) represent going up 8 steps.

Page : 5 , Block Name : Exercise 1.1

Exercise 1.2

Q1 Write down a pair of integers whose:
(a) sum is –7
(b) difference is –10
(c) sum is 0

Answer. (a) − 8 + (+1) = −7
(b) − 12 − (−2) = −10
(c) 5 + (−5) = 0

Page : 8 , Block Name : Exercise 1.2

Q2 (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.

Answer. (a) −2 − (−10) = 8
(b) −8 + 3 = −5
(c) −2 − (+1) = −3

Page : 8 , Block Name : Exercise 1.2

Q3 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?

Answer. Team A scored − 40, 10, 0.
Total score = − 40 + 10 + 0 = −30
Team B scored 10, 0, −40.
Total score = 10 + 0 + (−40) = −30
∴ The scores of both teams are equal.
Yes, we can add integers in any order. We had observed that the scores obtained by both teams
in successive rounds were numerically equal but different in order. Yet, the total score of both
teams were equal.

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Page : 9 , Block Name : Exercise 1.2

Q4 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] + ............

Answer.

Page : 9 , Block Name : Exercise 1.2

Exercise 1.3

Q1 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)

Answer. (a) 3 × (−1) = −3
(b) (−1) × 225 = −225
(c) (−21) × (−30) = 630
(d) (−316) × (−1) = 316

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

(e) (−15) × 0 × (−18) = 0
(f) (−12) × (−11) × 10 = 1320
(g) 9 × (−3) × (−6) = 162
(h) (−18) × (−5) × (−4) = −360
(i) (−1) × (−2) × (−3) × 4 = −24
(j) (−3) × (−6) × (−2) × (−1) = 36

Page : 21 , Block Name : Exercise 1.3

Q2 Verify the following:
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]

Answer. (a) L.H.S. = 18 × [7 + (− 3)] = 18 × [7 − 3] = 18 × 4 = 72
R.H.S. = [18 × 7] + [18 × (− 3)] = 126 + (− 54) = 72
18 x [7 + (-3)] = [18x7] + [18x(-3)
Hence,
(b) L.H.S. = (−21) × [(−4) + (−6)] = (−21) × [− 4 − 6] = (−21) × [−10] = 210
R.H.S. = [(−21) × (−4)] + [(−21) × (−6)] = 84 + 126 = 210
(-21) x [ (-4) + (-6) ]=[ (-21) x (-4) ]+[ (-21) x (-6) ]

Page : 21 , Block Name : Exercise 1.3

Q3 (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

Answer.

Page : 21 , Block Name : Exercise 1.3

Q4 Starting from (–1) × 5, write various products showing some pattern to show (–1) × (–1) =
1.

Answer. −1 × 5 = −5

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

−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
Therefore, −1 × (−1) = 0 + 1 = 1

Page : 21 , Block Name : Exercise 1.3

Q5 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

Answer. (a) 26 × (−48) + (−48) × (−36)
= (−48) × 26 + (−48) × (−36) (b × a = a × b)
= (−48) [26 − 36] (a × b + a × c) = a (b + c)
= (−48) × (−10) = 480
(b) 8 ×53 × (−125) = 8 × [53 × (−125)]
= 8 × [(−125) × 53] (b × a = a × b)
= [8 × (−125)] ×53 a × (b × c) = (a × b) × c
= [−1000] × 53 = −53000
(c) 15 × (−25) × (−4) × (−10)
= 15 × [(−25) × (−4)] × (−10)
= 15 × [100] × (−10)
= 15 × (−1000) = −15000
(d) (−41) × 102
= (−41) × (100 + 2)
= (−41) × 100 + (−41) × 2 a × (b + c) = (a × b) + (a × c)
= − 4100 − 82 = −4182
(e) 625 × (−35) + (−625) × 65
= 625 × [(−35) + (−65)] (a × b) + (a × c) = a × (b + c)
= 625 × [−100] = −62500
(f) 7 × (50 − 2)
= (7 × 50) − (7 × 2) a × (b − c) = (a × b) − (a × c)
= 350 − 14
= 336
(g) (−17) × (−29)
= (−17) × [−30 + 1]
= [(−17) × (−30)] + [(−17) × 1] a × (b + c) = (a × b) + (a × c)
= [510] + [−17] = 493

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

(h) (−57) × (−19) + 57
= 57 × 19 + 57 × 1
= 57 [19 + 1] (a × b) + (a × c) = a × (b + c)
= 57 × 20 = 1140

Page : 21 , Block Name : Exercise 1.3

Q6 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?

Answer. Initial temperature = 40°C
Change in temperature per hour = −5°C
Change in temperature after 10 hours = (−5) × 10 = −50°C
Final temperature = 40ºC + (−50ºC) = −10°C

Page : 21 , Block Name : Exercise 1.3

Q7 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 ve correct answers and ve incorrect answers, what is her score?
(iii) Heena gets two correct and ve incorrect answers out of seven questions she attempts.
What is her score?

Answer. (i) Marks given for 1 correct answer = 5
Marks given for 4 correct answers = 5 × 4 = 20
Marks given for 1 wrong answer = −2
Marks given for 6 wrong answers = −2 × 6 = −12
Score obtained by Mohan = 20 − 12 = 8
(ii) Marks given for 1 correct answer = 5
Marks given for 5 correct answers = 5 × 5 = 25
Marks given for 1 wrong answer = −2
Marks given for 5 wrong answers = −2 × 5 = −10
Score obtained by Reshma = 25 − 10 = 15
(iii) Similarly,
Marks given for 2 correct answers = 5 × 2 = 10
Marks given for 5 wrong answers = −2 × 5 = −10
Score obtained by Heena = 10 − 10 = 0

Page : 21 , Block Name : Exercise 1.3

Q8 A cement company earns a pro t 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 pro t or loss?

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

(b) What is the number of white cement bags it must sell to have neither pro t nor loss, if the
number of grey bags sold is 6,400 bags.

Answer. Pro t is denoted by a positive integer and loss is denoted by a negative integer.
(a) Pro t earned while selling 1 bag of white cement = Rs 8
Pro t earned while selling 3000 bags of white cement = 8 × 3000
= 24000
Loss incurred while selling 1 bag of grey cement = −Rs 5
Loss incurred while selling 5000 bags of grey cement = −5 × 5000
= −25000
Total pro t/loss earned = Pro t + Loss
= 24000 + (−25000) = −1000
Therefore, a loss of Rs 1000 will be incurred by the company.
(b) Loss incurred while selling 1 bag of grey cement = −Rs 5
Loss incurred while selling 6400 bags of grey cement = (−5) × 6400
= −32000
Let the number of bags of white cement to be sold be x.
Pro t earned while selling 1 bag of white cement = Rs 8
Pro t earned while selling x bags of white cement = x × 8
= 8x
In condition of no pro t no loss,
Pro t earned + Loss incurred = 0
8x + (−32000) = 0
8x = 32000
x = 4000
Therefore, 4000 bags of white cement must be sold.

Page : 21 , Block Name : Exercise 1.3

Q9 Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 27
(b) 5 × _____ = –35
(c) _____ × (– 8) = –56
(d) _____ × (–12) = 132

Answer.

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Page : 22 , Block Name : Exercise 1.3

Exercise 1.4

Q1 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]

Answer. (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

Page : 26 , Block Name : Exercise 1.4

Q2 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

Answer. (a) a = 12, b = −4, c = 2
a ÷ (b + c) = 12 ÷ (− 4 + 2) = 12 ÷ (−2) = −6
(a ÷ b) + (a ÷ c) = [12 ÷ (−4)] + [12 ÷ 2] = −3 + 6 = 3
Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
(b) a = −10, b = 1, c = 1
a ÷ (b + c) = (−10) ÷ (1 + 1) = (−10) ÷ 2 = −5
(a ÷ b) + (a ÷ c) = [(−10) ÷ 1] + [(−10) ÷ 1] = − 10 − 10 = −20
Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

Page : 26 , Block Name : Exercise 1.4

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Q3 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

Answer.

Page : 26 , Block Name : Exercise 1.4

Q4 Write ve pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–
2) = (–3).

Answer. (i) (3, −1)
Because 3 ÷ (−1) = −3
(ii) (−3, 1)
Because (−3) ÷ 1 = −3
(iii) (9, −3)
Because 9 ÷ (−3) = −3
(iv) (−9, 3)
Because (−9) ÷ 3 = −3
(v) (12, −4)
Because 12 ÷ (−4) = −3

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Book : Mathematics Ncert Solutions | Chapter-1 Maths

Page : 26 , Block Name : Exercise 1.4

Q5 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?

Answer. Initial temperature i.e., at 12 noon = 10°C
Change in temperature per hour = −2°C
Temperature at 1:00 PM = 10ºC + (−2ºC) = 8ºC
Temperature at 2:00 PM = 8ºC + (−2ºC) = 6ºC
Temperature at 3:00 PM = 6ºC + (−2ºC) = 4ºC
Temperature at 4:00 PM = 4ºC + (−2ºC) = 2ºC
Temperature at 5:00 PM = 2ºC + (−2ºC) = 0ºC
Temperature at 6:00 PM = 0ºC + (−2ºC) = −2ºC
Temperature at 7:00 PM = −2ºC + (−2ºC) = −4ºC
Temperature at 8:00 PM = −4ºC + (−2ºC) = −6ºC
Temperature at 9:00 PM = −6ºC + (−2ºC) = −8ºC
Therefore, the temperature will be 8°C below zero at 9:00 PM.
It will take 12 hours to be midnight (i.e., 12:00 AM) after 12:00 noon.
Change in temperature in 12 hours = −2°C × 12 = −24ºC
At midnight, the temperature will be = 10 + (−24)
= −14°C
Therefore, the temperature at midnight will be 14ºC below 0.

Page : 26 , Block Name : Exercise 1.4

Q6 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?

Answer. Marks obtained for 1 right answer = +3
Marks obtained for 1 wrong answer = −2
(i) Marks scored by Radhika = 20
Marks obtained for 12 correct answers = 12 × 3 = 36
Marks obtained for incorrect answers = Total score − Marks obtained for 12 correct answers
= 20 − 36 = −16
Marks obtained for 1 wrong answer = −2
Thus, number of incorrect answers = (−16) ÷(−2) = 8
Therefore, she attempted 8 questions wrongly.
(ii) Marks scored by Mohini = −5
Marks obtained for 7 correct answers = 7 × 3 = 21
Marks obtained for incorrect answers = Total score − Marks obtained for 12 correct answers
= − 5 − 21 = −26

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Marks obtained for 1 wrong answer = −2
Thus, number of incorrect answers = (−26) ÷(−2) = 13
Therefore, she attempted 13 questions wrongly.
(iii) Total marks scored by Rakesh = 18
Number of questions attempted = 16
(Number of correct answers)(3) + (Number of incorrect answers)(−2) = 18
⇒(Number of correct answers)(3) + (16 − Number of correct answers)(−2) = 18
⇒(Number of correct answers)(3) + −32 + 2(Number of correct answers) = 18
⇒(Number of correct answers)(5) + −32 = 18
⇒(Number of correct answers)(5) = 18 + 32 = 50
⇒ Number of correct answers = 10
∴Number of incorrect answers = 16 − 10 = 6
∴Total number of correct and incorrect answers scored by Rakesh is 10 and 6 respectively.

Page : 26 , Block Name : Exercise 1.4

Q7 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. Distance descended is denoted by a negative integer.
Initial height = +10 m
Final depth = −350 m
Total distance to be descended by the elevator = (−350) − (+10) = −360 m
Time taken by the elevator to descend −6 m = 1 min
Thus, time taken by the elevator to descend −360 m = (−360) ÷ (−6)
= 60 minutes = 1 hour

Page : 26 , Block Name : Exercise 1.4

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