NCERT Solutions for Class 7 Maths Exercise 1.4 Chapter 1 Integers

NCERT Solutions For Class 7 Maths Ex 1.4 PDF Free Download

NCERT Solutions for Class 7 Maths Exercise 1.4 Chapter 1 Integers are available here in simple PDF. This exercise of NCERT Solutions for Class 7 Maths Chapter 1 has topics related to the division of Integers. The division is the inverse operation of multiplication. This exercise also deals with the properties of the division of integers. Students are suggested to try solving the questions from NCERT Solutions for Class 7 Maths Chapter 1 Integers.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 1 Integers – Exercise 1.4

 

ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4
ncert solution class 7 maths chapter 1 exercise 1.4

 

Access answers to Maths NCERT Solutions for Class 7 Chapter 1 – Integers Exercise 1.4

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 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 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 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 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 minus sign (-) before the quotient.

(f) 0 ÷ (–12)

Solution:-

= 0 ÷ (-12)

= 0

When we divide zero by a negative integer 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 positive sign (+) before the quotient.

(h) [(–36) ÷ 12] ÷ 3

Solution:-

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]

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 positive sign (+) before the quotient.

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 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 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.

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

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)

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 mid-night?

Solution:-

From the question is given that,

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 mid-night 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 temperature will be 14oC below 0.

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:-

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

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

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


Access other exercises of NCERT Solutions For Class 7 Chapter 1 – Integers

Exercise 1.1 Solutions

Exercise 1.2 Solutions

Exercise 1.3 Solutions

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