RD Sharma Solutions for Class 7 Maths Chapter 1 Integers

Mathematics is not just about numbers—it's a powerful tool that sharpens logical thinking, improves problem-solving abilities, and lays the foundation for success in higher studies. For Class 7 students, mastering Chapter 1: Integers from RD Sharma Maths Book is a vital step. This chapter introduces students to the world of positive and negative numbers, helping them understand how to compare, add, subtract, multiply, and divide integers with confidence. By learning integers thoroughly, students build a solid base for advanced topics like algebra, equations, and data handling

Integers are whole numbers that can either be positive, negative, or zero. In this chapter, students will learn how to add, subtract, multiply, and divide integers. The chapter helps build a strong understanding of operations with integers, which are essential for solving real-life problems involving quantities that can be both positive and negative, such as temperature, finance, and elevations.

The RD Sharma Solutions for Class 7 Maths Chapter 1 Integers is designed to make learning these concepts easy and effective. These solutions are explained step-by-step, helping students grasp the logic behind every operation. Whether it's addition or subtraction of integers, or understanding how to multiply and divide positive and negative numbers, this chapter covers everything in a simple and structured manner.

To assist students in strengthening their understanding, the solutions include a variety of exercises with detailed explanations and solved examples. The chapter also offers practice questions that encourage students to apply their knowledge and improve their problem-solving abilities. By solving these exercises, students can easily master the concept of integers and be well-prepared for exams.

With the help of RD Sharma’s solutions, students can easily understand and practice integers, ensuring they have a solid foundation for tackling more complex mathematical topics in the future.

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With RD Sharma Class 7 Chapter Wise Solutions free PDF, students can access clear, step-by-step answers that simplify tough problems and build confidence in handling integers. These free PDF downloads are not only a smart study tool but also a great way to practice anytime, anywhere. RD Sharma Class 7 Chapter 1 PDF includes detailed solutions, examples, and extra questions to help you master real numbers and other topics. Click here to download the RD Sharma Class 7 Chapter 1 PDF.

Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 1 Integers

Q1. Determine the following products:

(i) 12 × 7

(ii) (-15) × 8

(iii) (-25) × (-9)

(iv) 125 × (-8)

Solutions:

(i) 12 × 7 = 84

Since both numbers are positive, their product is positive. Therefore, 12 × 7 = 84.

(ii) (-15) × 8 = -120

When multiplying a positive number by a negative number, the result is negative. Hence, -15 × 8 = -120.

(iii) (-25) × (-9) = +225
Multiplying two negative numbers results in a positive number. Thus, -25 × -9 = 225.

(iv) 125 × (-8) = -1000

Again, multiplying a positive number by a negative one gives a negative result. Hence, 125 × -8 = -1000.

Q2. Find the following products:

(i) 3 × (-8) × 5

(ii) 9 × (-3) × (-6)

(iii) (-2) × 36 × (-5)

(iv) (-2) × (-4) × (-6) × (-8)

Solutions:

(i) 3 × (-8) × 5 = -120

Multiplying first 3 × (-8) gives -24, and then -24 × 5 = -120.

(ii) 9 × (-3) × (-6) = 162

Multiplying two negative numbers gives a positive product: 9 × 18 = 162.

(iii) (-2) × 36 × (-5) = 360

Multiplying the first two numbers, -2 × 36 = -72, and then -72 × -5 = 360.

(iv) (-2) × (-4) × (-6) × (-8) = 384

Multiplying two negative numbers first gives a positive result: (-2) × (-4) = 8, and (-6) × (-8) = 48, so 8 × 48 = 384.

Q3. Find the value of:

(i) 1487 × 327 + (-487) × 327

(ii) 28945 × 99 – (-28945)

Solutions:

(i) 1487 × 327 + (-487) × 327 = 327000

By simplifying, we get 486249 - 159249 = 327000.

(ii) 28945 × 99 – (-28945) = 2894500

Multiplying and simplifying, we get 2865555 + 28945 = 2894500.

Q4. Determine the integer whose product with -1 is:

(i) 58

(ii) 0

(iii) -225

Solutions:

(i) 58 × (-1) = -58

(ii) 0 × (-1) = 0

(iii) -225 × (-1) = 225

Q5. Divide:

(i) 102 ÷ 17

(ii) -85 ÷ 5

(iii) -161 ÷ -23

(iv) 76 ÷ -19

(v) 17654 ÷ -17654

(vi) (-729) ÷ (-27)

(vii) 21590 ÷ -10

(viii) 0 ÷ -135

Solutions:

(i) 102 ÷ 17 = 6

(ii) -85 ÷ 5 = -17

(iii) -161 ÷ -23 = 7

(iv) 76 ÷ -19 = -4

(v) 17654 ÷ -17654 = -1

(vi) (-729) ÷ (-27) = 27

(vii) 21590 ÷ -10 = -2159

(viii) 0 ÷ -135 = 0

Q6. Simplify each of the following expressions:

(i) 36 ÷ 6 + 3

(ii) 24 + 15 ÷ 3

(iii) 120 – 20 ÷ 4

(iv) 32 – (3 × 5) + 4

(v) 3 – (5 – 6 ÷ 3)

(vi) 21 – 12 ÷ 3 × 2

(vii) 16 + 8 ÷ 4 – 2 × 3

(viii) 28 – 5 × 6 + 2

Solutions:

(i) 36 ÷ 6 + 3 = 9

(ii) 24 + 15 ÷ 3 = 29

(iii) 120 – 20 ÷ 4 = 115

(iv) 32 – (3 × 5) + 4 = 21

(v) 3 – (5 – 6 ÷ 3) = 0

(vi) 21 – 12 ÷ 3 × 2 = 13

(vii) 16 + 8 ÷ 4 – 2 × 3 = 12

(viii) 28 – 5 × 6 + 2 = 0

Q7. Simplify:

(i) 3 – (5 – 6 ÷ 3)

(ii) -25 + 14 ÷ (5 – 3)

(iii) 25 – ½ {5 + 4 – (5 – 4)}

(iv) 27 – [38 – {46 – (15 – 11)}]

(v) 36 – [18 – {14 – (11 ÷ 2 × 2)}]

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

Solutions:

(i) 3 – (5 – 6 ÷ 3) = 0

(ii) -25 + 14 ÷ (5 – 3) = -18

(iii) 25 – ½ {5 + 4 – (5 – 4)} = 21

(iv) 27 – [38 – {46 – (15 – 11)}] = 31

(v) 36 – [18 – {14 – (11 ÷ 2 × 2)}] = 21

(vi) 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}] = 26