RD Sharma Solutions for Class 6 Maths Chapter 8 Introduction to Algebra

RD Sharma Solutions for Class 6 Maths Chapter 8: RD Sharma Solutions for Class 6 Maths Chapter 8 – Introduction to Algebra are specially designed to help students build a strong foundation in Algebra. These solutions, created by subject experts as per the latest CBSE syllabus, cover all important topics like integers, basic algebraic language, and solving simple equations using addition, subtraction, multiplication, and division. 

The step-by-step solutions are based on a concept-based approach, making it easier for students to understand and practice algebraic problems. Students can easily download the RD Sharma Class 6 Maths Chapter 8 PDF to strengthen their skills, revise quickly, and stay exam-ready with confidence.

RD Sharma Solutions for Class 6 Maths Chapter 8 - Overview

Download RD Sharma Solutions for Class 6 Maths Chapter 8

RD Sharma Solutions for Class 6 Maths Chapter 8: Introduction to Algebra help students understand the basics of Algebra in an easy and simple way. Designed by subject experts as per the latest CBSE syllabus, these solutions explain key topics like variables, expressions, and simple equations through clear, step-by-step methods.

 To practice more, students can solve Class 6 Maths worksheets along with the RD Sharma Solutions for better problem-solving skills. For quick and easy revision, the Class 6 Maths PDF download is also available. By using RD Sharma Solutions for Class 6 Maths Chapter 8, students can strengthen their Algebra basics and prepare well for their exams with full confidence.  

RD Sharma Textbook Solutions for Class 6 Chapter 8 - Introduction to Algebra

1. Write the following using numbers, literals and signs of basic operations. State what each letter represents:

(i) The diameter of a circle is twice its radius.

(ii) The area of a rectangle is the product of its length and breadth.

(iii) The selling price equals the sum of the cost price and the profit.

(iv) The total amount equals the sum of the principal and the interest.

(v) The perimeter of a rectangle is two times the sum of its length and breadth.

(vi) The perimeter of a square is four times its side.

Solution:

(i) Consider d as the diameter and r as the radius of the circle

Hence, we get d = 2r.

(ii) Consider A as the area, l as the length and b as the breadth of a rectangle

Hence, we get A = l × b.

(iii) Consider S.P as the selling price, C.P as the cost price and P as the profit

Hence, we get S.P = C.P + P

(iv) Consider A as the amount, P as the principal and I as the interest

Hence, we get A = P + I

(v) Consider P as the perimeter, l as the length and b as the breadth of a rectangle

Hence, P = 2 (l + b)

(vi) Consider P as the perimeter and a as the side of a square

Hence, P = 4a

2. Write the following using numbers, literals and signs of basic operations:

(i) The sum of 6 and x.

(ii) 3 more than a number y.

(iii) One-third of a number x.

(iv) One-half of the sum of number x and y.

(v) Number y less than a number 7.

(vi) 7 taken away from x.

(vii) 2 less than the quotient of x and y.

(viii) 4 times x taken away from one-third of y.

(ix) Quotient of x by 3 is multiplied by y.

Solution:

(i) The sum of 6 and x can be written as 6 + x.

(ii) 3 more than a number y can be written as y + 3.

(iii) One-third of a number x can be written as x/3.

(iv) One-half of the sum of number x and y can be written as (x + y)/ 2.

(v) Number y less than a number 7 can be written as 7 – y.

(vi) 7 taken away from x can be written as x – 7.

(vii) 2 less than the quotient of x and y can be written as x/y – 2.

(viii) 4 times x taken away from one-third of y can be written as y/3 – 4x.

(ix) Quotient of x by 3 is multiplied by y can be written as xy/3.

3. Think of a number. Multiply by 5. Add 6 to the result. Subtract y from this result. What is the result?

Solution:

Consider x as the number.

Multiplying the number by 5 = 5x

Again add 6 to the number = 5x + 6

By subtracting y from the above equation = 5x + 6 – y.

Hence, the result is 5x + 6 – y.

4. The number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor. If the first floor has x rooms, how many rooms does the ground floor has?

Solution:

Consider y as the number of rooms on the ground floor

We know that

The number of rooms on the first floor = x

It is given that number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor

So we get

y = 2x – 12

Hence, the rooms on the ground floor is y = 2x – 12.

5. Binny spend Rs a daily and saves Rs b per week. What is her income for two weeks?

Solution:

Amount spent by Binny = Rs a

Amount saved by Binny = Rs b

Amount spent by Binny in one week = 7a

So the total income for one week = Amount spent by Binny in one week + Amount saved by Binny

Substituting the values

Total income for one week = 7a + b

We get Binny’s income for 2 weeks = 2 (7a + b) = Rs 14a + 2b

Hence, the income of Binny for two weeks is Rs 14a + 2b.

6. Rahul scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?

Solution:

Marks scored by Rahul in English = 80

Marks scored by Rahul in Hindi = x

So the total scores in the two subjects = x + 80

Hence, the total score of Rahul in two subjects is x + 80.

7. Rohit covers x centimetres in one step. How much distance does he cover in y steps?

Solution:

Distance covered by Rohit in one step = x cm

So the distance covered by Rohit in y steps = xy cm

Hence, Rohit covers xy cm in y steps.

8. One apple weighs 75 grams and one orange weighs 40 grams. Determine the weight of x apples and y oranges.

Solution:

Weight of one apple = 75 g

Weight of one orange = 40 g

So the weight of x apples = 75x g

So the weight of y oranges = 40y g

We get the weight of x apples and y oranges = (75x + 40y) g

Hence, the weight of x apples and y oranges is (75x + 40y) g.

9. One pencil costs Rs 2 and one fountain pen costs Rs 15. What is the cost of x pencils and y fountain pens?

Solution:

Cost of one pencil = Rs 2

Cost of one fountain pen = Rs 15

Cost of x pencils = 2x

Cost of y fountain pens = 15y

So the cost of x pencils and y fountain pens = Rs (2x + 15y)

Hence, the cost of x pencils and y fountain pens is Rs (2x + 15y).

Exercise 8.2 page: 8.11

1. Write each of the following products in exponential form:

(i) a × a × a × a × …….. 15 times

(ii) 8 × b × b × b × a × a × a × a

(iii) 5 × a × a × a × b × b × c × c × c

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times

Solution:

(i) a × a × a × a × …….. 15 times is written in exponential form as a15.

(ii) 8 × b × b × b × a × a × a × a is written in exponential form as 8a4b3.

(iii) 5 × a × a × a × b × b × c × c × c is written in exponential form as 5a3b2c3.

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times is written in exponential form as 7a8b5.

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times is written in exponential form as 4a5b12c15.

2. Write each of the following in the product form:

(i) a2 b5

(ii) 8x3

(iii) 7a3b4

(iv) 15 a9b8c6

(v) 30x4y4z5

(vi) 43p10q5r15

(vii) 17p12q20

Solution:

(i) a2 b5 is written in the product form as a × a × b × b × b × b × b.

(ii) 8x3 is written in the product form as 8 × x × x × x.

(iii) 7a3b4 is written in the product form as 7 × a × a × a × b × b × b × b.

(iv) 15 a9b8c6 is written in the product form as 15 × a × a …… 9 times × b × b × … 8 times × c × c × ….. 6 times.

(v) 30x4y4z5 is written in the product form as 30 × x × x × x × x × y × y × y × y × z × z × z × z × z.

(vi) 43p10q5r15 is written in the product form as 43 × p × p …. 10 times × q × q …. 5 times × r × r × …. 15 times.

(vii) 17p12q20 is written in the product form as 17 × p × p …. 12 times × q × q × ….. 20 times.

3. Write down each of the following in exponential form:

(i) 4a3 × 6ab2 × c2

(ii) 5xy × 3x2y × 7y2

(iii) a3 × 3ab2 × 2a2b2

Solution:

(i) 4a3 × 6ab2 × c2 is written in exponential form as 24a4b2c2.

(ii) 5xy × 3x2y × 7y2 is written in exponential form as 105x3y4.

(iii) a3 × 3ab2 × 2a2b2 is written in exponential form as 6a6b4.

4. The number of bacteria in a culture is x now. It becomes square of itself after one week. What will be its number after two weeks?

Solution:

Number of bacteria in a culture = x

It is given that

Number of bacteria becomes square of itself in one week = x2

So the number of bacteria after two weeks = (x2)2 = x4

Hence, the number of bacteria after two weeks is x4.

5. The area of a rectangle is given by the product of its length and breadth. The length of a rectangle is two-third of its breadth. Find its area if its breadth is x cm.

Solution:

It is given that

Area of rectangle = l × b

Breadth = x cm

Length = (2/3) x cm

So the area of the rectangle = (2/3) x × x = 2/3 x2 cm2

Hence, the area of rectangle is (2/3) x2 cm2.

6. If there are x rows of chairs and each row contains x2 chairs. Determine the total number of chairs.

Solution:

Number of rows of chairs = x

Each row contains = x2 chairs

So the total number of chairs = number of rows of chairs × chairs in each row

We get

Total number of chairs = x × x2 = x3

Hence, the total number of chairs is x3.

Advantages of RD Sharma Solutions for Class 6 Maths Chapter 8 (Introduction to Algebra)

• Foundational Concepts - These solutions establish strong algebraic fundamentals by explaining variables, constants, and expressions in easy-to-understand language, helping students grasp these abstract concepts during their first introduction to algebra.
• Step-by-Step Approach - The solutions break down complex algebraic problems into manageable steps, making it easier for young learners to follow the logical sequence needed to solve equations and form expressions.
• Varied Practice Problems - RD Sharma provides diverse question types with increasing difficulty levels, allowing students to build confidence gradually while developing their algebraic thinking skills.
• Real-Life Applications - The solutions connect algebraic concepts to everyday situations, helping students understand why algebra matters and how mathematical language can represent real-world scenarios.
• Self-Assessment Support - These comprehensive solutions enable students to check their work independently, identify mistakes in their problem-solving process, and learn from errors without always needing teacher assistance.