RBSE Solutions for Class 6 Maths Chapter 12 Algebra Ex 12.1 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 12 Algebra Exercise 12.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 12 |

Chapter Name |
Algebra |

Exercise |
Ex 12.1 |

Number of Questions |
9 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 6 Maths Chapter 12 Algebra Ex 12.1

Question 1.

Make the matchstick pattern of the letters given below. Draw the (RBSESolutions.com) figures of the patterns in your note book. Create rules to find out number of matchsticks required for each pattern. (You can use literals like a, b, x, y e.t.c. to create rules).

(i) T pattern of T, T, TT …………….

(ii) N pattern of N, N, NN, …………….

(iii) W pattern of W, W, WW, …………….

Solution:

(i) Pattern of T

T, TT, TTT, TTTT, …………….

It is clear from figure, that we need two matchsticks to (RBSESolutions.com) make one pattern of T. Let we has to make n, T, then numbers of matchsticks are as follow

Number of matchsticks required for one T = 2

Similarly, number of matchsticks required for nT = 2 × n = 2n.

(ii) Pattern of N

N, NN, NNN, NNNN, …………….

It is clear from figures, that we need three (RBSESolutions.com) matchsticks to make on pattern of N. Let we has to make n, N, then

Number of matchsticks required will be = n × 3 = 3n.

(iii) Pattern of W

W, WW, WWW, WWWW, …………….

It is clear from figure, that we need four matchsticks to make one pattern of W, let we has to make n, W, then

Number of matchsticks required will be = n × 4 = 4n.

Question 2.

Tree plantation program was held in a school. 4 Trees were (RBSESolutions.com) planted in each row. Write the number of trees planted in terms of the number of rows.

Solution:

Number of trees in one row = 4

Let total number of rows = x

∴ Total number of trees in x rows

= 4 × x = 4x.

Question 3.

Ranu is 5 years (RBSESolutions.com) younger than leela.

(i) Let Leela’s age be years. Write the age of Ranu in terms of .

(ii) Let Ranu’s age of be P years. Write the age of Leela in terms of P.

Solution:

(i) Leela’s age in x years.

∵ Ranu is 5 years younger than Leela.

∴ Ranu’s age = (x – 5) years.

(ii) Ranu’s age is P years.

∵ Leela’s is 5 years older than Ranu.

∴ Leela’s age = (P + 5) years.

Question 4.

Cost of a Pen is Rs.5 Madan has some money with him. He (RBSESolutions.com) pays all of that money to buy those pens. Write the number of pens purchased in terms of the money he had.

Solution:

Cost of a pen = Rs. 5

Let madan has Rs. x, and he buy pen of Rs. x.

Let n pens are purchased in Rs x.

∴ Total spent money = cost of one pen × total pen

⇒ x = 5 × n

⇒ n = x / 5.

Question 5.

Complete the table (RBSESolutions.com) given below

Solution:

Given equation = 2x + 3

For x = 3

2x + 3 = 2 (3) + 3 = 6 + 3 = 9

For x = 4

2x + 3 = 2 (4) + 3 = 8 + 3 = 11

For x = 5

2x + 3 = 2 (5) + 3 = 10 + 3 = 13

If 2x + 3 = 15 than 2x = 15 – 3

⇒ 2x = 12

⇒ x = 12/2 = 6

for x = 7

2x + 3 = 2 (7) + 3 = 17

∴ Given table of question will be

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