RD Sharma Solutions for Class 6 Chapter 8: Introduction to Algebra Exercise 8.2

RD Sharma Class 6 Solutions Chapter 8 Ex 8.2 PDF Free Download

The students who aspire to perform well in the exam can solve problems using RD Sharma Solutions Class 6. The solutions consist of explanations in simple language which matches the understanding ability of students. The problems of RD Sharma textbook can be solved by using the solutions as a reference guide. With the help of RD Sharma Solutions for Class 6 Maths Chapter 8 Introduction to Algebra Exercise 8.2 PDF students can expertise the concepts which are important for the exam preparation.

RD Sharma Solutions for Class 6 Chapter 8: Introduction to Algebra Exercise 8.2 Download PDF

 

rd sharma solution class 6 maths chapter 8 ex 2
rd sharma solution class 6 maths chapter 8 ex 2

 

Access RD Sharma Solutions for Class 6 Chapter 8: Introduction to Algebra Exercise 8.2

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 numbers 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.

Access other exercise solutions of Class 6 Maths Chapter 8: Introduction to Algebra

Exercise 8.1 Solutions

Objective Type Questions

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