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.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 1 |

Chapter Name | Integers |

Exercise | Ex 1.4 |

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

Table of Contents

**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**

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.