NCERT SOLUTIONS EXERCISE 11.5 CLASS 6 MATHS

Find here step by step NCERT Solutions Exercise 11.5 Class 6 Maths Chapter 11 Algebra. NCERT Solutions Exercise 11.5 Class 6 Maths deals with Equations and Trial and Error method to find their Solutions.

Prerequisite / Revise this:

NCERT Solutions Exercise 11.5 Class 6 Maths – An Equation

• An Equation is a condition on a variable. It is expressed by saying that an expression with a variable is equal to a fixed number, e.g. x + 3 = 7

• An equation has two sides, LHS and RHS, between them is the equal (=) sign.

• The LHS of an equation is equal to its RHS only for a definite value of the variable in the equation. We say that this definite value of the variable satisfies the equation. This value itself is called the Solution of the Equation.

• For getting the solution of an equation, one method is the trial and error method. In this method, we give some value to the variable and check whether it satisfies the equation. We go on giving this way different values to the Variable until we find the right value which satisfies the equation.

EXERCISE 11.5
Solution of An Equation / Trial and Error Method

QUESTION 1
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.

(a) 17 = x + 7 (b) (t – 7) > 5

(c) \frac{4}{2} = 2 (d) (7 × 3) – 19 = 8

(e) 5 × 4 – 8 = 2x (f ) x – 2 = 0

(g) 2m < 30 (h) 2n + 1 = 11

(i) 7 = (11 × 5) – (12 × 4) ( j) 7 = (11 × 2) + p

(k) 20 = 5y ( l) \frac{3q}{2}< 5

(m) z + 12 > 24 (n) 20 – (10 – 5) = 3 × 5

(o) 7 – x = 5

ANSWER

(a) It is an equation with variable as the LHS is equal to the RHS.
Variable in the equation is x.

(b) It is not an equation with variable as the LHS is not equal to the RHS.

(c) It is not an equation with variable as there is no variable in it.

(d) It is not an equation with variable as there is no variable in it.

(e) It is an equation with variable as LHS is equal to the RHS.
Variable in the equation is x.

(f ) It is an equation with variable as LHS is equal to the RHS.
Variable in the equation is x.

(g) It is not an equation with variable as LHS is not equal to the RHS.

(h) It is an equation with variable as LHS is equal to the RHS.
Variable in the equation is n.

(i) It is not an equation with variable as there is no variable in it.

( j) It is an equation with variable as LHS is equal to the RHS.
Variable in the equation is p.

(k) It is an equation with variable as LHS is equal to the RHS.
Variable in the equation is y.

( l) It is not an equation with variable as LHS is not equal to the RHS.

(m) It is not an equation with variable as LHS is not equal to the RHS.

(n) It is not an equation with variable as there is no variable in it.

(o) It is an equation with variable as LHS is equal to the RHS.
Variable in the equation is y.

QUESTION 2
Complete the entries in the third column of the table.

NCERT SOLUTIONS EXERCISE 11.5 CLASS 6 MATHS

ANSWER
(a) No, 10×10 \neq 80.

(b) Yes, 10×8 = 80.

(c) No, 10×5 \neq 80.

(d) No, 4×20 \neq 20.

(e) No, 4×80 \neq 80.

(f) Yes, 4×5 \neq 20.

(g) No, 5+5 \neq 9.

(h) No, 5+5 \neq 9.

(i) Yes, 4+5 = 9.

(j) Yes, 13 – 8 = 5.

(k) No, 8 – 8 \neq 5.

(l) No, 0 – 8 \neq 5.

(m) No, 3 + 3 \neq 1.

(n) No, 1 + 3 \neq 1.

(o) No, 0 + 3 \neq 1.

(p) No, -1 + 3 \neq 1.

(q) Yes, -2 + 3 \neq 1.

QUESTION 3
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.

(a) 5m = 60 (10, 5, 12, 15)

(b) n + 12 = 20 (12, 8, 20, 0)

(c) p – 5 = 5 (0, 10, 5 – 5)

(d) \frac{q}{2} = 7 (7, 2, 10, 14)

(e) r – 4 = 0 (4, – 4, 8, 0)

(f) x + 4 = 2 (– 2, 0, 2, 4)

ANSWER

(a) LHS = 5m
RHS = 60

Putting m = 10 in the LHS,
LHS = 5×10 = 50 \neq RHS,
So, 10 doesn’t satisfy the equation.

Putting m = 5 in the LHS,
LHS = 5×5 = 25 \neq RHS,
So, 5 doesn’t satisfy the equation.

Putting m = 12 in the LHS,
LHS = 5×12 = 60 = RHS,
So, 12 does satisfy the equation
Hence, 12 is the solution of the equation.

Putting m = 15 in the LHS,
LHS = 5×15 = 75 \neq RHS,
So, 15 doesn’t satisfy the equation.

(b) LHS = n + 12
RHS = 20

Putting n = 12 in the LHS,
LHS = 12 + 12 = 24 \neq RHS,
So, 12 doesn’t satisfy the equation.

Putting n = 8 in the LHS,
LHS = 8 + 12 = 20 = RHS,
So, 12 does satisfy the equation,
Hence, 12 is the solution of the equation.

Putting n = 20 in the LHS,
LHS = 20 + 12 = 32 \neq RHS,
So, 20 doesn’t satisfy the equation.

Putting n = 0 in the LHS,
LHS = 0 + 12 = 12 \neq RHS,
So, 0 doesn’t satisfy the equation.

(c) LHS = p – 5
RHS = 5

Putting p = 0 in the LHS,
LHS = 0 – 5 = -5 \neq RHS,
So, 0 doesn’t satisfy the equation.

Putting p = 10 in the LHS,
LHS = 10 – 5 = 5 = RHS,
So, 10 does satisfy the equation,
Hence, 10 is the solution of the equation.

Putting p = 5 in the LHS,
LHS = 5 – 5 = 0 \neq RHS,
So, 20 doesn’t satisfy the equation.

Putting p = -5 in the LHS,
LHS = 0 – 5 = 12 \neq RHS,
So, 0 doesn’t satisfy the equation.

(d) LHS = \frac{q}{2}
RHS = 7

Putting q = 7 in the LHS,
LHS = \frac{7}{2} = 3.5 \neq RHS,
So, 7 doesn’t satisfy the equation.

Putting q = 2 in the LHS,
LHS = \frac{2}{2} = 1 \neq RHS,
So, 2 doesn’t satisfy the equation,

Putting q = 10 in the LHS,
LHS = \frac{10}{2} = 5 \neq RHS,
So, 10 doesn’t satisfy the equation.

Putting q = 14 in the LHS,
LHS = \frac{14}{2} = 7 = RHS,
So, 7 doesn satisfy the equation,
Hence, 7 is the solution of the equation.

(e) LHS = r – 4
RHS = 0

Putting r = 4 in the LHS,
LHS = 4 – 4 = 0 = RHS,
So, 4 does satisfy the equation,
Hence, 4 is the solution of the equation.

Putting r = -4 in the LHS,
LHS = -4 – 4 = -8 \neq RHS,
So, -4 doesn’t satisfy the equation,

Putting r = 8 in the LHS,
LHS = 8 – 4 = 4 \neq RHS,
So, 8 doesn’t satisfy the equation.

Putting r = 0 in the LHS,
LHS = 0 – 4 = -4 \neq RHS,
So, 0 doesn’t satisfy the equation.

(f) LHS = x + 4
RHS = 2

Putting x = -2 in the LHS,
LHS = -2 + 4 = 2 = RHS,
So, -2 does satisfy the equation,
Hence, -2 is the solution of the equation.

Putting x = 0 in the LHS,
LHS = 0 + 4 = 4 \neq RHS,
So, 0 doesn’t satisfy the equation,

Putting x = 2 in the LHS,
LHS = 2 + 4 = 6 \neq RHS,
So, 2 doesn’t satisfy the equation.

Putting x = 4 in the LHS,
LHS = 4 + 4 = 8 \neq RHS,
So, 4 doesn’t satisfy the equation.

QUESTION 4
(a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.

NCERT Solutions Exercise 11.5 class 6 MATHS

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

(c) Complete the table and find the solution of the equation z/3 = 4 using the table.

(d) Complete the table and find the solution to the equation m – 7 = 3.

ANSWER

(a)

m234567_
m+10121314151617_

m + 10 = 16 at m = 6,

\therefore Solution of the equation is 6.

(b)

t456789_
5t2025 30354045_

5t = 35 at t = 7,

\therefore Solution of the equation is 7.

(c)

z69121518__
z/323456__

z/3 = 4 at z = 12,

\therefore Solution of the equation is 12.

(d)

m7891011__
m-701234__

m – 7 = 3 at m = 10,

\therefore Solution of the equation is 10.

QUESTION 5
Solve the following riddles, you may yourself construct such riddles.

Who am I?

(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!

ANSWER

(i) Number of corners in a square = 4,

Counting every corner thrice, we get
4×3 = 12,

Let, the variable (me) be x,

Then, x + 12 = 34

\implies x = 34 - 12

\implies x = 22

Hence, answer of the riddle is 22.

(ii) Number of days in a week = 7,

Let the variable (me) be y,

Up counting from y will result 23,

\implies y + 7 = 23

\implies y = 23-7

\implies y = 16

Hence, answer of the riddle is 16.

(iii) Let the special number be x,

If we take away 6 from it, we get
x – 6,

Number of players in a cricket team = 11,

A/Q,

x - 6 = 11

x = 11 + 6

x = 17

Hence, answer of the riddle is 17.

(iv) Let, the number be x,

If we take x out of twenty two, we get x,

So, 22 - x = x

\implies 22 = x+x

\implies 22 = 2x

\implies 2x = 22

\implies x = \frac{22}{2}

\implies x = 11.

NCERT SOLUTIONS FOR CLASS 8 MATHS – CHAPTER 11 ALGEBRA

Check NCERT Solutions of other exercises from Class 6 Maths Chapter 11 Algebraic Expressions by clicking the links given below.

Exercise 11.1
Exercise 11.2
Exercise 11.3
Exercise 11.4
Exercise 11.5

We hope Ncert Solutions Exercise 11.5 Class 6 Maths has helped you understand What is an Equation and How to find Solutions of an Equation by Trial and Error Method.

Write in the comment section for any error or any solution related queries from the exercise. [ NCERT Solutions Exercise 11.5 Class 6 Maths ]

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