RBSE Solutions for Class 6 Maths Chapter 12 Algebra In Text Exercise

RBSE Solutions for Class 6 Maths Chapter 12 Algebra In Text Exercise is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 12 Algebra In Text Exercise.

 Board RBSE Textbook SIERT, Rajasthan Class Class 6 Subject Maths Chapter Chapter 12 Chapter Name Algebra Exercise In Text Exercise Number of Questions 9 Category RBSE Solutions

Rajasthan Board RBSE Class 6 Maths Chapter 12 Algebra In Text Exercise

(Page No. 165)
Question 1.
Let y denote age in year. Write down (RBSESolutions.com) the value of y for your 10 friends.
Solution:
Let age of one friend = y years. then age of 10 friends are as follows:

1. (y – 2) years
2. (y + 3) years
3. (y + 1) years
4. (y – 1) years
5. y years
6. (2 y – 10) years
7. (y + 2) years
8. (y – 3) years
9. (y + 3) years
10. (y – 2.5) years

(Page No. 168)
Question 1.
Write algebraic expression through (RBSESolutions.com) given instructions about how to form it.
(i) Sum of 5 and a variable …………. .
(ii) Difference between 7 and a number …………. .
(iii) 3 times of a variable …………. .
(iv) 12 less than 6 times of a variable …………. .
(v) Half of a variable …………. .
(vi) 200 less than one third of a variable …………. .
Solution:
Statement in the form of algebraic expression
(i) 5 + 2 (let variable is 2)
(ii) 9 – 7 (let variable is 9)
(iii) 3 × 3 (let variable is 3)
(iv) 6 × 4 – 12 (let variable is 4)
(v) $$\frac { 8 }{ 2 }$$  (let variable is 8)
(vi) $$\frac { 900 }{ 3 }$$– 200 (let variable is 900).

Question 2.
Shweta secured 75 marks in mathematics. Her (RBSESolutions.com) score in Science is not known. Let her science score be x. What is her total score?
Solution:
Marks of shweta in mathematics = 75
Marks of shweta in science = x (let)
∴ Total marks = x + 75.

Question 3.
Sakshi has some candies with her. Ashu has 4 times as many candies as Sakshi. How many candies are there in total ?
Solution:
Sakshi has = x candies (let)
According to question,
Ashu has = 4 × x candies = 4x candies
∴ Total candies = x + 4x = 5x.

(Page No. 169)
Question 1.
Match the algebraic expressions with (RBSESolutions.com) appropriate situations in the following
(i) x + 4 = (A) Prashant has 4 times as many wealth as Kamli.
(ii) x – 4 = (B) Malti has Rs. 4 more than Seema.
(iii) 4 – x = (C) My weight is 4 kgs less than Nancy.
(iv) 4y = (D) I had Rs. 4 from which I spend some money. How much I am left with ?
(v) $$\frac { y }{ 4 }$$ = (E) Banshi had some marbles. He distributed them between his 4 friends equally. How marbles each friend get ?
Solution:
Correct matching are as follow

Text Book Questions
(Page No. 166)
Question 1.
Fill in the table with the (RBSESolutions.com) help of given figures

Solution:
It is clear from figure, that each child has one – one ribbon (RBSESolutions.com) in their both hands. Thus, each child has two ribbons. If numbers of children are 6, than number of ribbons will be double of it, i.e. 6x × 2 = 12 Similarly, number of ribbons for 6, 7, 8, 9, 10 and 11 students are shown in following table

∴ Table given in question is as below

(Page No. 167)
Question 1.
Some containers are given below, look (RBSESolutions.com) them carefully and answer the following questions.

how many match sticks are required to form 8 containers ? Make a table.

Solution:
It is clear from figure that to make container, the (RBSESolutions.com) numbers of match sticks required can be find by following pattern.

Similarity, (RBSESolutions.com) for n containers
Numbers of matchsticks
= (2 + 2 + 2 + ….. n) + 1
= 2 × n + 1
= 2 n + 1
∴ Following is the table given in questions

(Page No. 173)
Question 1.
Adding 15 in twice of a number results 51 Form (RBSESolutions.com) equations form this statement.
Solution:
Let number x.
double of number + 15 = 51
⇒ 2 × x + 15 = 51
⇒ 2x + 15 = 51.

Question 2.
Derive statement from 3x – 7 = 11.
Solution:
Subtracting 7 from thrice of a number gives 11.

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