RD Sharma Class 7 ex 1.4 Solutions Chapter 1 Integers

In this chapter, we provide RD Sharma Class 7 ex 1.4 Solutions Chapter 1 Integers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 1.4 Solutions Chapter 1 Integers Maths pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 1
Chapter NameIntegers
ExerciseEx 1.4

RD Sharma Solutions for Class 7 Chapter 1 Integers Ex 1.4 Download PDF

Question: 1

Simplify

3 – (5 – 6 ÷ 3)

Solution:

3 – (5 – 6 ÷ 3)

= 3 – (5 – 2)

= 3 – 3

= 0

∴ 3 – (5 – 6 ÷ 3) = 0

Question: 2

Simplify

– 25 + 14 ÷ (5 – 3)

Solution:

-25 + 14 ÷ (5 – 3) = – 25 + 14 ÷ (2)

= – 25 + 14/2

= –25 + 7

= –18

∴ – 25 + 14 ÷ (5 – 3) = – 18

Question: 3

Simplify

Solution:

Question: 4

Simplify

Solution:

Question: 5

Simplify

36 – [18 – (14 – (15 – 4 ÷ 2 × 2))]

Solution:

36 – [18 – (14 – (15 – 4 ÷ 2 × 2))]

= 36 – [18 – (14 – (11 ÷ 2 × 2))]

= 36 – [18 – (14 – 11/2 × 2))]

= 36 – [18 – (14 – 11)]

= 36 – [18 – 3]

= 36 – 15

= 21

∴ 36 – [18 – (14 – (15 – 4 ÷ 2 × 2))] = 21

Question: 6

Simplify

45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)]

Solution:

45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)]

= 45 – [38 – (20 – (6 – 3) ÷ 3)]

= 45 – [38 – (20 – 3 ÷ 3)]

= 45 – [38 – (20 – 1)]

= 45 – [38 – 19]

= 45 – [19]

= 26

∴ 45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)] = 26

Question: 7

Simplify

Solution:

= 23 – [23 – (23 – (23 – 0))]

= 23 – [23 – (23 – 23)]

= 23 – [23 – 0]

= 23 – 23

= 0

Question: 8

Simplify

Solution:

= 2550 – [510 – (270 – (90 – 150))]

= 2550 – [510 – (270 – (–60))]

= 2550 – [510 – 330]

= 2550 – [180]

= 2550 – 180

= 2370

Question: 9

Simplify

Solution:

Question: 10

Simplify

Solution:

Question: 11

Simplify

Solution:

Question: 12

Simplify

[29 – ( – 2)(6 – (7 – 3))] ÷ [3 × (5 + ( – 3) × (-2))]

Solution:

[29 – (-2)(6 – (7 – 3))] ÷ [3 × (5 + (- 3) × (- 2))]

= [29 – (- 2)(6 – 4)] ÷ [3 × (5 + (3 × 2))]

= [29 – (-2)(2)] ÷ [3 × (5 + 6)]

= [29 + 4] ÷ [3 × 11]

= [33] ÷ [33]

= 1

∴ [29 – (-2) (6 -(7- 3))] ÷ [3 × (5 + (- 3) × (-2))] = 1

Question: 13

Using brackets, write a mathematical expression for each of the following:

(i) Nine multiplied by the sum of two and five.

(ii) Twelve divided by the sum of one and three.

(iii) Twenty divided by the difference of seven and two.

(iv) Eight subtracted from the product of two and three.

(v) Forty divided by one more than the sum of nine and ten.

(vi) Two multiplied by one less than the difference of nineteen and six.

Solution:

(i) 9 (2 + 5)

(ii) 12 ÷ (1 + 3)

(iii) 20 ÷ (7 – 2)

(iv) 2 × 3 – 8

(v) 40 ÷ [1 + (9 + 10)]

(vi) 2 × [(19 – 6) – 1]

All Chapter RD Sharma Solutions For Class 7 Maths

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