NCERT Solutions for Class 7 Maths Chapter 1 Integers (New Syllabus)

NCERT Solutions for Class 7 Maths Chapter 1 Integers are available here. We at Study Circle have prepared step by step solutions with detailed explanation.

Contents hide
2 Exercise 1.1
2.15 Exercise 1.2

These solutions will prove to be a boon for all the students, we recommend those students who want to score good marks in maths, they should adopt these solutions and strengthen their knowledge, so that every student can get 100% marks.

NCERT Solutions for Class 7 Maths Chapter 1 – Integers has 3 exercises, and the NCERT solutions provided on this page provide solutions to the questions present in the exercises. All these solutions are based on the new syllabus of NCERT

CLASS – 7 MATHS NCERT CHAPTER -1 NEW SYLLABUS

Exercise 1.1

 Q –1. Write down a pair of integers whose:

(a) sum is -7

Solution:-
= – 4 + (-3)
= – 4 – 3 … [∵ (+ × – = -)]
= – 7

(b) the difference is – 10

Solution:-
= -25 – (-15)
= – 25 + 15 … [∵ (- × – = +)]
= -10

(c) sum is 0

Solution:-
= 4 + (-4)
= 4 – 4
= 0

2. (a) Write a pair of negative integers whose difference gives 8

Solution:-
= (-5) – (- 13)
= -5 + 13 … [∵ (- × – = +)]
= 8

(b) Write a negative integer and a positive integer whose sum is – 5.

Solution:-
= -25 + 20
= -5

(c) Write a negative integer and a positive integer whose difference is – 3.

Solution:-
= – 2 – (1)
= – 2 – 1
= – 3

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:-
From the question, it is given that
The score of team A = -40, 10, 0
Total score obtained by team A = – 40 + 10 + 0
= – 30
The score of team B = 10, 0, -40
Total score obtained by team B = 10 + 0 + (-40)
= 10 + 0 – 40
= – 30
Thus, the score of both the A team and B team is the same.
Yes, we can say that we can add integers in any order.

4. Fill in the blanks to make the following statements true.

(i) (–5) + (– 8) = (– 8) + (…………)

Solution:-

Let us assume the missing integer be x,
Then,
= (–5) + (– 8) = (– 8) + (x)
= – 5 – 8 = – 8 + x
= – 13 = – 8 + x
By sending – 8 from RHS to LHS, it becomes 8
= – 13 + 8 = x
= x = – 5
Now, substitute the x value in the blank place.
(–5) + (– 8) = (– 8) + (- 5) … [This equation is in the form of the Commutative Law of Addition]

(ii) –53 + ………… = –53

Solution:-
Let us assume the missing integer be x,
Then,
= –53 + x = –53
By sending – 53 from LHS to RHS, it becomes 53
= x = -53 + 53
= x = 0
Now, substitute the x value in the blank place.
= –53 + 0 = –53 … [This equation is in the form of the Closure Property of Addition]

(iii) 17 + ………… = 0

Solution:-
Let us assume the missing integer be x,
Then,
= 17 + x = 0
By sending 17 from LHS to RHS, it becomes -17
= x = 0 – 17
= x = – 17
Now, substitute the x value in the blank place.
= 17 + (-17) = 0 … [This equation is in the form of Closure Property of Addition]
= 17 – 17 = 0

(iv) [13 + (– 12)] + (…………) = 13 + [(–12) + (–7)]

Solution:-

Let us assume the missing integer be x,
Then,
= [13 + (– 12)] + (x) = 13 + [(–12) + (–7)]
= [13 – 12] + (x) = 13 + [–12 –7]
= [1] + (x) = 13 + [-19]
= 1 + (x) = 13 – 19
= 1 + (x) = -6
By sending 1 from LHS to RHS, it becomes -1.
= x = -6 – 1
= x = -7
Now, substitute the x value in the blank place.
= [13 + (– 12)] + (-7) = 13 + [(–12) + (–7)] … [This equation is in the form of Associative Property of Addition]

(v) (– 4) + [15 + (–3)] = [– 4 + 15] +…………

Solution:-
Let us assume the missing integer be x.
Then,
= (– 4) + [15 + (–3)] = [– 4 + 15] + x
= (– 4) + [15 – 3)] = [– 4 + 15] + x
= (-4) + [12] = [11] + x
= 8 = 11 + x
By sending 11 from RHS to LHS, it becomes -11,
= 8 – 11 = x
= x = -3
Now, substitute the x value in the blank place.
 
= (– 4) + [15 + (–3)] = [– 4 + 15] + -3 … 
 
[This equation is in the form of Associative Property of Addition]

 

                                  CLASS – 7 MATHS NCERT CHAPTER -1 NEW SYLLABUS 

 Exercise 1.2

 

1. Find each of the following products:

(a) 3 × (–1)

Solution:-
By the rule of Multiplication of integers,
= 3 × (-1)
= -3 … [∵ (+ × – = -)]

(b) (–1) × 225

Solution:-
By the rule of Multiplication of integers,
= (-1) × 225
= -225 … [∵ (- × + = -)]

(c) (–21) × (–30)

Solution:-
By the rule of Multiplication of integers,
= (-21) × (-30)
= 630 … [∵ (- × – = +)]

(d) (–316) × (–1)

Solution:-
By the rule of Multiplication of integers,
= (-316) × (-1)
= 316 … [∵ (- × – = +)]

(e) (–15) × 0 × (–18)

Solution:-
By the rule of Multiplication of integers,
= (–15) × 0 × (–18)
= 0
∵ If any integer is multiplied by zero, the answer is zero itself.

(f) (–12) × (–11) × (10)

Solution:-
By the rule of Multiplication of integers,
= (–12) × (-11) × (10)
First, multiply the two numbers having the same sign.
= 132 × 10 … [∵ (- × – = +)]
= 1320

(g) 9 × (–3) × (– 6)

Solution:-
By the rule of Multiplication of integers,
= 9 × (-3) × (-6)
First, multiply the two numbers having the same sign.
= 9 × 18 … [∵ (- × – = +)]
= 162

(h) (–18) × (–5) × (– 4)

Solution:-
By the rule of Multiplication of integers,
= (-18) × (-5) × (-4)
First, multiply the two numbers having the same sign.
= 90 × -4 … [∵ (- × – = +)]
= – 360 … [∵ (+ × – = -)]

(i) (–1) × (–2) × (–3) × 4

Solution:-
By the rule of Multiplication of integers,
 
= [(–1) × (–2)] × [(–3) × 4]
= 2 × (-12) … [∵ (- × – = +), (- × + = -)] = – 24

(j) (–3) × (–6) × (–2) × (–1)

Solution:-
By the rule of Multiplication of integers,
= [(–3) × (–6)] × [(–2) × (–1)]
First, multiply the two numbers having the same sign.
= 18 × 2 … [∵ (- × – = +)
= 36

2. Verify the following:

(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]

Solution:-
From the given equation,
Let us consider the Left Hand Side (LHS) first = 18 × [7 + (–3)]
= 18 × [7 – 3]
= 18 × 4
= 72
Now, consider the Right Hand Side (RHS) = [18 × 7] + [18 × (–3)]
= [126] + [-54]
= 126 – 54
= 72
By comparing LHS and RHS,
72 = 72
LHS = RHS
Hence, the given equation is verified.

(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]

Solution:-
From the given equation,
Let us consider the Left Hand Side (LHS) first = (–21) × [(– 4) + (– 6)]
= (-21) × [-4 – 6]
= (-21) × [-10]
= 210
Now, consider the Right Hand Side (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]
= [84] + [126]
= 210
By comparing LHS and RHS,
210 = 210
LHS = RHS
Hence, the given equation is verified.

3. (i) For any integer a, what is (–1) × a equal to?

Solution:-
= (-1) × a = -a
When we multiply any integer a with -1, then we get the additive inverse of that integer.

(ii). Determine the integer whose product with (–1) is

(a) –22

Solution:-
Now, multiply -22 with (-1), and we get
= -22 × (-1)
= 22
When we multiply integer -22 with -1, then we get the additive inverse of that integer.

(b) 37

Solution:-
Now, multiply 37 with (-1), and we get
= 37 × (-1)
= -37
When we multiply integer 37 with -1, then we get the additive inverse of that integer.

(c) 0

Solution:-
Now, multiply 0 by (-1), and we get
= 0 × (-1)
= 0
Because the product of negative integers and zero gives zero only.

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

(–1) × (–1) = 1.

Solution:-
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
We concluded that the product of one negative integer and one positive integer is a negative integer. Then, the product of two negative integers is a positive integer.

CLASS – 7 MATHS NCERT CHAPTER -1 NEW SYLLABUS   

  Exercise 1.3

 

Q–1. Evaluate each of the following.
(a) (–30) ÷ 10
Solution:-
= (–30) ÷ 10
= – 3
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(b) 50 ÷ (–5)

Solution:-
= (50) ÷ (-5)
= – 10
When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(c) (–36) ÷ (–9)

Solution:-
= (-36) ÷ (-9)
= 4
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put the positive sign (+) before the quotient.

(d) (– 49) ÷ (49)

Solution:-
= (–49) ÷ 49
= – 1
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(e) 13 ÷ [(–2) + 1]

Solution:-
= 13 ÷ [(–2) + 1]
= 13 ÷ (-1)
= – 13
When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(f) 0 ÷ (–12)

Solution:-
= 0 ÷ (-12)
= 0
When we divide zero by a negative integer, it gives zero.

(g) (–31) ÷ [(–30) + (–1)]

Solution:-
= (–31) ÷ [(–30) + (–1)]
= (-31) ÷ [-30 – 1]
= (-31) ÷ (-31)
= 1
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put the positive sign (+) before the quotient.

(h) [(–36) ÷ 12] ÷ 3

Solution:-
First, we have to solve the integers within the bracket.
= [(–36) ÷ 12]
= (–36) ÷ 12
= – 3
Then,
= (-3) ÷ 3
= -1
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(i) [(– 6) + 5)] ÷ [(–2) + 1]

Solution:-
The given question can be written as,
= [-1] ÷ [-1]
= 1
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put the positive sign (+) before the quotient.

Q–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

Solution:-
From the question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given, a = 12, b = – 4, c = 2
Now, consider LHS = a ÷ (b + c)
= 12 ÷ (-4 + 2)
= 12 ÷ (-2)
= -6
When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.
Then, consider RHS = (a ÷ b) + (a ÷ c)
= (12 ÷ (-4)) + (12 ÷ 2)
= (-3) + (6)
= 3
By comparing LHS and RHS,
= -6 ≠ 3
= LHS ≠ RHS
Hence, the given values are verified.

(b) a = (–10), b = 1, c = 1

Solution:-
From the question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given, a = (-10), b = 1, c = 1
Now, consider LHS = a ÷ (b + c)
= (-10) ÷ (1 + 1)
= (-10) ÷ (2)
= -5
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.
Then, consider RHS = (a ÷ b) + (a ÷ c)
= ((-10) ÷ (1)) + ((-10) ÷ 1)
= (-10) + (-10)
= -10 – 10
= -20
By comparing LHS and RHS,
= -5 ≠ -20
= LHS ≠ RHS
Hence, the given values are verified.

Q–3. Fill in the blanks:

(a) 369 ÷ _____ = 369

Solution:-
Let us assume the missing integer be x,
Then,
= 369 ÷ x = 369
= x = (369/369)
= x = 1
Now, put the valve of x in the blank.
= 369 ÷ 1 = 369

(b) (–75) ÷ _____ = –1

Solution:-
Let us assume the missing integer be x.
Then,
= (-75) ÷ x = -1
= x = (-75/-1)
= x = 75
Now, put the valve of x in the blank.
= (-75) ÷ 75 = -1

(c) (–206) ÷ _____ = 1

Solution:-
Let us assume the missing integer be x.
Then,
= (-206) ÷ x = 1
= x = (-206/1)
= x = -206
Now, put the valve of x in the blank.
= (-206) ÷ (-206) = 1

(d) – 87 ÷ _____ = 87

Solution:-
Let us assume the missing integer be x.
Then,
= (-87) ÷ x = 87
= x = (-87)/87
= x = -1
Now, put the valve of x in the blank.
= (-87) ÷ (-1) = 87

(e) _____ ÷ 1 = – 87

Solution:-
Let us assume the missing integer be x.
Then,
= (x) ÷ 1 = -87
= x = (-87) × 1
= x = -87
Now, put the valve of x in the blank.
= (-87) ÷ 1 = -87

(f) _____ ÷ 48 = –1

Solution:-
Let us assume the missing integer be x.
Then,
= (x) ÷ 48 = -1
= x = (-1) × 48
= x = -48
Now, put the valve of x in the blank.
= (-48) ÷ 48 = -1

(g) 20 ÷ _____ = –2

Solution:-
Let us assume the missing integer be x.
Then,
= 20 ÷ x = -2
= x = (20)/ (-2)
= x = -10
Now, put the valve of x in the blank.
= (20) ÷ (-10) = -2

(h) _____ ÷ (4) = –3

Solution:-
Let us assume the missing integer be x.
Then,
= (x) ÷ 4 = -3
= x = (-3) × 4
= x = -12
Now, put the valve of x in the blank.
= (-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).

Solution:-
(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 10oC above zero. If it decreases at the rate of 2oC per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight?

Solution:-
From the question, it is given,
The temperature at the beginning, i.e., at 12 noon = 10oC
Rate of change of temperature = – 2oC per hour
Then,
Temperature at 1 PM = 10 + (-2) = 10 – 2 = 8oC
Temperature at 2 PM = 8 + (-2) = 8 – 2 = 6oC
Temperature at 3 PM = 6 + (-2) = 6 – 2 = 4oC
Temperature at 4 PM = 4 + (-2) = 4 – 2 = 2oC
Temperature at 5 PM = 2 + (-2) = 2 – 2 = 0oC
Temperature at 6 PM = 0 + (-2) = 0 – 2 = -2oC
Temperature at 7 PM = -2 + (-2) = -2 -2 = -4oC
Temperature at 8 PM = -4 + (-2) = -4 – 2 = -6oC
Temperature at 9 PM = -6 + (-2) = -6 – 2 = -8oC
∴ At 9 PM, the temperature will be 8oC below zero.
Then,
The temperature at midnight, i.e., at 12 AM
Change in temperature in 12 hours = -2oC × 12 = – 24oC
So, at midnight temperature will be = 10 + (-24)
= – 14oC
So, at midnight, the temperature will be 14oC below 0.

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 scored –5 marks on this test, though she got 7 correct answers. How many questions has she attempted incorrectly?

Solution:-
From the question,
Marks awarded for 1 correct answer = + 3
Marks awarded for 1 wrong answer = -2
(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 So, the number of incorrect answers made by Radhika = (-16) ÷ (-2)

= 8

(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 12 correct

Answers

= – 5 – 21

= – 26

So, the number of incorrect answers made by Mohini = (-26) ÷ (-2)

= 13

Q–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?

Solution:-

From the question,
The initial height of the elevator = 10 m
The final depth of the elevator = – 350 m … [∵distance descended is denoted by a negative
integer]
The total distance to descended by the elevator = (-350) – (10)
= – 360 m
Then,
Time taken by the elevator to descend -6 m = 1 min
So, the time taken by the elevator to descend – 360 m = (-360) ÷ (-60)
= 60 minutes
= 1 hour
Sharing Is Caring:

Leave a comment