NCERT Solutions for Class 7 Maths - Integers (Exercise 1.1, 1.2 and 1.3)

Exercise 1.1

Q1: Write down a pair of integers whose

(a) Sum is -7.

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

(b) Difference is -10.

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

(c) Sum is 0.

Ans:
= 4 + (-4)
= 4 – 4
= 0

Note: You can also think other combinations, it completely depends on you.

Q2: (a) Write a pair of negative integers whose difference gives 8.

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

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

Ans:
= -25 + 20
= -5

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

Ans: = – 2 – (1)
= – 2 – 1
= –3

Q3: In a quiz, team A scored -40, 10, 0 and team B scores 10, 0, -40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order? 

Ans: 

• Team A scored -40, 10, 0
  ⇒ Total score of Team A = -40 + 10 + 0 = - 30
• Team B scored 10, 0, -40
  ⇒ Total score of Team B = 10 + 0+ (-40) - 10 + 0 - 40 = -30
  Thus, the scores of both teams are same.
• Yes, we can add integers in any order due to the commutative property of addition of integers which says: A + B = B + A

Q4: Fill in the blanks to make the following statements true: 
(i) (-5) + (-8) = (-8) + (....)

Ans: 

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 Commutative law of Addition]

(ii) -53 +... = -53

Ans: 

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 Closure property of Addition]

(iii) 17 +... = 0

Ans: 

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)]

Ans: 

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] + .....

Ans: 

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]

Exercise 1.2

Q1: Find the each of the following products
(a) 3 x (-1)

Ans: 

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

(b) (-1) x 225

Ans: 

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

(c) (-21) x (-30) 

Ans: 

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

(d) (-316) x (-1)

Ans: 

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

(e) (-15) x 0 x (-18) 

Ans: 

By the rule of Multiplication of integers,
= (–15) × 0 × (–18) = 0
∵ Any integer is multiplied with zero and the answer is zero itself.

(f) (-12) x (-11) x (10)

Ans: 

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

(g) 9 x (-3) x (-6)

Ans: 

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

(h) (-18) x (-5) x (-4) 

Ans: 

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

(i) (-1) x (-2) x (-3) x 4

Ans: 

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

(j) (-3) x (-6) x (2) x (-1)

Ans: 

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

Q2: Verify the following 
(a) 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]

Ans: 

18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]
⇒ 18 x 4 = 126 + (-54)
⇒ 72 = 72
⇒ L.H.S. = R.H.S.
Hence, Verified.

(b)(-21) x [(-4) + (-6)] = [(-21) x (-4)] + [(-21) x (-6)] 

Ans: 

(-21) x [(-4) + (-6)] = [(-21)x (-4)] + [(-21) x (-6)]
⇒ (-21) x (-10) = 84 + 126
⇒ 210 = 210
⇒ L.H.S. = R.H.S.
Hence, Verified.

Q3: (i) For any integer a, what is (-1) x a equal to?

Ans: = (-1) × a = -a
Because, when we multiplied any integer a with -1, then we get additive inverse of that integer.

(ii) Determine the integer whose product with (-1) is
(a) -22

Ans: 

Now, multiply -22 with (-1), we get
= -22 × (-1)
= 22
Because, when we multiplied integer -22 with -1, then we get additive inverse of that integer. 

(b) 37

Ans: 

Now, multiply 37 with (-1), we get
= 37 × (-1)
= -37
Because, when we multiplied integer 37 with -1, then we get additive inverse of that integer.

(c) 0

Ans: 

Now, multiply 0 with (-1), we get
= 0 × (-1)
= 0
Because, the product of negative integers and zero give zero only. 

Q4: Starting from (-1) x 5, write various products showing some patterns to show (-1) x (-1) = 1

Ans: 

According to the pattern,
(−1) × (5) = −5
(−1) × (4) = −4
(−1) × (3) = −3
(−1) × (2) = −2
(−1) × (1) = −1
(−1) × (0) = 0
(−1) × (−1) = 1
Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer is a negative integer whereas the product of two negative integers is a positive integer.

Exercise 1.3 

Q1: Evaluate each of the following
(a) (-30) ÷ 10

Ans:

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

(b) 50 ÷ (-5)

Ans:

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

(c) (-36) ÷ (-9)

Ans: 

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.   

= (-36) ÷ (-9)
= 4 

(d) (-49) ÷ 49

Ans: 

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.

= (–49) ÷ 49
= – 1

(e) 13 + [(-2) + 1]

Ans:

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

(f) 0 ÷ (-12)

Ans:

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

(g) (-31) ÷ [(-30) ÷ (-1)]

Ans:

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

(h) [(-36) ÷ 12] ÷ 3

Ans: 

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.

First we have to solve the integers with in the bracket,
= [(–36) ÷ 12]
= (–36) ÷ 12
= – 3
Then, = (-3) ÷ 3 = -1

(i) [(-6) + 5] ÷ [(-2) + 1]

Ans: 

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.

The given question can be written as,
= [-1] ÷ [-1]
= 1

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  

Ans:
From the question,
a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given that, a = 12, b = – 4, c = 2
Now, consider
LHS = a ÷ (b + c)

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.
= 12 ÷ (-4 + 2)
= 12 ÷ (-2) = -6
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

Ans:
From the question,
a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given, a = (-10), b = 1, c = 1
Now, consider

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.
LHS = a ÷ (b + c) 

= (-10) ÷ (1 + 1)
= (-10) ÷ (2) = -5
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.

Q3: Fill in the blanks
(a) 369 ÷ _____   = 369

Ans: 

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)

Ans: 

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

Ans: 

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

Ans: 

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

Ans: 

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

Ans: 

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

Ans: 

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

Ans: 

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

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

Ans: 

(i)  (-6) ÷ 2 = -3    
(ii) 9 ÷ (-3) = -3
(iii) 12 ÷ (-4) = -3    
(iv) (-9) ÷ 3 = -3
(v) (-15) ÷ 5 = -3

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 midnight?

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

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

Ans: 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

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?

Ans: 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) ÷ (-6)
= 60 minutes
= 1 hour