Class 7 Maths Exercise 1.4

Integers Class 7 Exercise 1.1 Integers Class 7 Exercise 1.2 Integers Class 7 Exercise 1.3 Integers Class 7 Exercise 1.4

NCERT Answers for Class 7 Maths Chapter 1 Integers Exercise 1.4

Page 26

Ex 1.4 Class 7 Maths 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]

(a) (–30) ÷ 10
= (–30) ÷ 10
= – 3
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.

(b) 50 ÷ (–5)
= (50) ÷ (-5)
= – 10
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.

(c) (–36) ÷ (–9)
= (-36) ÷ (-9)
= 4
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.

(d) (– 49) ÷ (49)
= (–49) ÷ 49
= – 1
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.

(e) 13 ÷ [(–2) + 1] = 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 minus sign (-) before the quotient.

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

(g) (–31) ÷ [(–30) + (–1)] = (–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 positive sign (+) before the quotient.

(h) [(–36) ÷ 12] ÷ 3
First we have to solve the integers with in 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 minus sign (-) before the quotient.

(i) [(– 6) + 5)] ÷ [(–2) + 1] 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 positive sign (+) before the quotient.

Ex 1.4 Class 7 Maths 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 = 1

Answer

(a) a = 12, b = – 4, c = 2
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 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
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 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.

Ex 1.4 Class 7 Maths 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

Answer

(a) 369 ÷ _____ = 369
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
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
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
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
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
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
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
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

Ex 1.4 Class 7 Maths Question 4. Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).

Answer

(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)

Ex 1.4 Class 7 Maths Question 5. 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

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

Ex 1.4 Class 7 Maths 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?

Answer

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 Radhika = (-26) ÷ (-2)
= 13

Ex 1.4 Class 7 Maths Question 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

Answer

From the question,
The initial height of the elevator = 10 m
Final depth of 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, time taken by the elevator to descend – 360 m = (-360) ÷ (-60)
= 60 minutes
= 1 hour