# NCERT Solutions For Class 6 Maths Chapter 11

## NCERT Solutions Class 6 Maths Algebra

NCERT Solutions for class 6 Maths Chapter 11 Algebra is a comprehensive exercise wise solution resource for NCERT class 6 maths textbooks questions. Algebra is a branch of Maths that started around 3500 years ago (i.e. 1550 BC). Initially, people used symbols for denoting unknown quantities. Later on the use of letters for denoting unknowns and forming expressions from them become quite common. Several great Indian mathematicians like Aryabhata, Brahmagupta, Mahavira and Bhaskara II contributed a lot to the study of algebra. NCERT Solutions for class 6 maths Algebra chapter 11 covers a detailed explanation of concepts like Introduction to Algebra, Variables and constant, operations, linear equation in 1 variable and more.

Brief Summary: Every equation has two sides, namely, LHS and RHS separated by equal to sign. Both LHS and RHS of an equation are equal only for definite values of a variable known as the solution of the equation. Using the trial and error method we can easily find the values of the solutions of an equation. To use this method, provide some random value to the variables and examine whether those values satisfy the given equation. Keep on assigning different values to the variable unless we get the right value which satisfies both LHS and the RHS of the equation.

NCERT Solutions for class 6 Maths Chapter 11 pdf is also made available for better assistance. The students are advised to download and practice them on a regular basis for better results.

### NCERT Solutions For Class 6 Maths Chapter 11 Exercises

Exercise 11.1

Question 1

Let the side of an equilateral triangle be m. Show the perimeter of the triangle with the help of m.

Perimeter = sum of three sides

= m+m+m

= 3m

Question 2

Let the side of a regular hexagon be denoted by m. Show the perimeter of the hexagon with the help of m.

(All sides are equal in regular hexagon)

Side of regular hexagon = m

Perimeter = sum of all sides

= 6m

Question 3

In the given figure a cube, 3 – D cube is shown. It has six identical squares as all the sides are equal so same six faces. Let the length of a side of cube be given by m. Find formula for total length of the cube.

Length of an edge = m

Number of edges = 12

Total length of the edges = total number of edges × Length of each edge

= 12m

Question 4

Diameter of a circle is a line which passes through the centre and joins two points of circle. (In the given figure XY is a diameter of circle; c is centre.) Express the diameter of the circle (d) in form of its radius(r).

= 2r

Exercise 11.2

Question 1

Make as many expressions with no. as you can from three numbers 2, 6 and 9. Each no. should be used only once. Use only addition, multiplication and subtraction.

We can make many expressions using 2, 6 and 9 as given below.

9+ (6 – 2)

9 + (6 × 2)

(6 – 2)  + 9

(6 + 2) × 9

(2 + 9) – 6

(2 × 9) × 6

Question 2

Which of these expressions are formed with numbers only?

1. y + 6
2. (8 × 10) – 6p
3. 9 (15 – 7) + 6 × 4
4. 9n
5. 11
6. 7 – 7m
7. (2 × 40) – (4 × 20) – 25 + q

From above expressions given, it can be observed that option (c) and (e) are formed by using numbers only.

Question 3

Identify the operations and tell how the expressions have been formed.

1. a + 2, a – 2, b + 13, b – 13
2. 7t, t/7, 15t
3. 4x + 13, 4x – 13
4. 8c, -8c + 2, -8m – 2

Subtraction: 2 is subtracted from a

Subtraction: 13 is subtracted from b

2. Multiplication: t is multiplied with 7

Division: t is divided by 7

Multiplication: t is multiplied with 15

3.Multiplication and addition: 4 is multiplied with x and result is added to 13.

Multiplication and subtraction: 4 is multiplied with x and 13 is subtracted from the result.

4. Multiplication: 8 is multiplied with c

Multiplication and addition: -8 is multiplied by c and add 2 to the result.

Multiplication and subtraction: c is multiplied with -8 and 2 is subtracted from the result.

Question 4

Give expressions for the below:

2. 3 subtracted from m
3. m multiplied by 8
4. n divided by 8
5. 9 subtracted from –n
6. –m multiplied by 6
7. –n divided by 6
8. n multiplied by -9

1. m + 3
2. m – 3
3. 8m
4. n/8
5. –n – 9
6. -6m
7. –n/6
8. -9n

Question 5

Give expressions for the following cases

2. 15 subtracted from 3p
3. 7 times x to which 6 is added
4. 7 times x from which 6 is subtracted
5. z is multiplied by -9
6. z is multiplied by -5 and then 3 is added to the result
7. q is multiplied by 8 and the result is subtracted from 12
8. q is multiplied by -8 and the result is added to 12

1. 3p + 15
2. 3p – 15
3. 7x + 6
4. 7x – 6
5. -9z
6. -5z + 3
7. 12 – 8q
8. -8q + 12

Question 6

(a) Form expressions with s and 5. Use only one operation in the expressions. Every expression should have s in it.

(b) Form expressions with t, 3 and 8. Every expression should have t. Use only two operations. These two operations should be different.

(a) s + 5

s – 5

5s

$$\frac{s}{5}$$

$$\frac{5}{s}$$

5 – s

(b) 3t + 8

3t – 8

8t + 3

8t – 3 etc..

The values of variables are not fixed i.e. the length of a rectangle can attain any value. Hence, it is a variable. However, the number of angles of a triangle have fixed value and therefore it is not a variable. Variables are denoted by letters a, b, c, d, m, n, o, p, x, y, z, etc. Using variables we can easily express the relationship between any practical situation. Although variables are not fixed but they are numbers. Arithmetic operations like subtraction, addition, division and multiplication can be performed on them like m + 3, p + 1, 2n, 9m, 11p, 31y + 1, 8l + 95, etc. With the help of variables, common geometric and arithmetic rules can be easily expressed.