Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra InText Questions to get the best methods of solving problems.

## TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Try These

Question 1.

Can you now write the rule to form the following pattern with match-sticks ?

Answer:

The rule for the above figures is 3n.

[∵ 3 × 1, 3 × 2, 3 × 3, ………………]

Question 2.

Find the rule for required number of matchsticks to form a pattern repeating ‘H’. How would the rule be for repeating the shape ‘L’ ?

Answer:

The required pattern of repeating ‘H’ is

‘3n’. (3 × 1, 3 × 2, ………………..)

The rule for ‘L’ is ‘2n’.

Question 3.

A line of shapes is constructed using matchsticks.

(i) Find the rule that shows how many sticks are needed to make a group of such shapes ?

(ii) How many matchsticks are needed to form a group of 12 shapes ?

Answer:

(i) Number of matchsticks that are used in each figure are

3 ; 5 ; 7 ; 9 ………………

Shape-1; Shape-2; Shape-3; Shape-4

2 × 1 + 1; 2 × 2 + 1 ; 2 × 3 + 1 ; 2 × 4 + 1 ………

In general (2n + 1) is the rule for the above shapes.

(ii) 2n + 1 = 2 × 12 + 1 = 25 [ ∵ n = 12]

(Try These)

Question 1.

Find the general rule for the perimeter of a rectangle. Use variables ‘l’ and ‘b’ for length and breadth of the rectangle respectively.

Answer:

Perimeter of a rectangle

= l + b + l + b

= 2l + 2b = 2 (l + b)

Question 2.

Find the general rule for the area of a square by using the variable ’s’ for the side of a square ?

Answer:

The rule for the area of a square is A = side × side

= s × s

A = s^{2}

Question 3.

What would be the rule for perimeter of an Isosceles triangle ?

Answer:

The rule for the perimeter of an Isosceles triangle

= 3 × side

= 3 × a

P = 3a

Do This

Question 1.

Find the n^{th} term in the following sequences.

(i) 3, 6, 9, 12, …………………

(ii) 2, 5, 8, 11, …………………

(iii) 1, 8, 27, 64, 125, …………………

Answer:

(i) The n^{th}> term of the above series is

3 × 1, 3 × 2, 3 × 3, ……………….. = 3n.

(ii) The n^{th} term is 3 × 1 – 1, 3 × 2 – 1, 3 × 3 – 1, 3 × 4 – 1,

i.e., 3n – 1

(iii) The nth term is 1 × 1 × 1, 2 × 2 × 2, 3 × 3 × 3, ……. is

1^{3}, 2^{3}, 3^{3} ………………….

i.e., n^{3}

Question 2.

Write LHS and RHS of the following simple equations:

(i) 2x + 1 = 10

Answer:

2x + 1 = 10

LHS = 2x + 1, RHS = 10

(ii) 9 = y – 2

Answer:

9 = y – 2 ;

LHS = 9, RHS = y – 2

(iii) 3p + 5 = 2p + 10

Answer:

3p + 5 = 2p + 10

LHS = 3p + 5, RHS = 2p + 10

Question 3.

Write any two simple equations and give their LHS and RHS.

Answer:

Simple equation | LHS | RHS |

1. 4x – 10 = 7x + 6 | 4x – 10 | 7x + 6 |

2. t – 2= 5 | t – 2 | 5 |

Question 4.

Find the solution of the equation ‘x – 4 = 2’ by Trail and Error method.

Answer:

Given equation is x – 4 = 2

If x = 1 ⇒ x – 4 = 2

⇒ 1 – 4 = 2

– 3 ≠ 2

If x = 3 ⇒ 3 – 4 = 2

– 1 ≠ 2

If x = 5 ⇒ 5 – 4 = 2 .

1 ≠ 2

If x = 6 ⇒ 6 – 4 = 2

2 = 2

x = 6 satisfies the given equation.

∴ x = 6 is the solution of the equation.