# Frank Learning Maths Class 6 Solutions Chapter 8

## Frank Learning Maths Class 6 Solutions Chapter 8 Simple Equations

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 8 Simple Equations. Here students can easily find step by step solutions of all the problems for Simple Equations. Here students will find solutions for Exercise 8.1 and 8.2 and also get solutions for Chapter Check-up Questions. Exercise wise proper solutions. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Simple Equations Exercise 8.1 Solution

Question no – (1)

Solution :

Equation :

Equation is a condition which is to be satisfied by a variable.

Example : 3x – 5 = 16

Question no – (2)

Solution :

(a) 15 + x = 23

= x is the variable

(d) d – 5 = 90

= d is the variable

(e) 2p > 50

= p is the variable

(f) l/9 < 10

= l is the variable

(g) 2f + 3 = 7

= f is the variable

(h) 2m – 3 = 7

= m is the variable

(j) 7 = (11 × 2) + p

= p is the variable

(k) 20 = 5y

= y is the variable

(l) 3q/2 < 5

= q is the variable

(m) z + 12 > 24

= z ix the variable

(o) 7 – x = 5

= x is the variable

Question no – (3)

Solution :

(a) x + 9 = 20

= x = 11

So, the value is = 11

(b) 6 × b = 36

= b = 6

Hence, the value is = 6

(c) x – 5 = 100

= x = 105

So, the value is = 105

(d) 8 + y = 8

= y = 0

So, the value is = 0

(e) 20a = 40

= a = 2

Thus, the value is = 2

(f) p – 8 = 7

= p = 15

So, the value is = 15

(g) t – 19 = 5

= t = 24

Hence, the value is = 24

(h) 8x = 96

= x = 12

So, the value is = 12

(i) d ÷ 7 = 6

= d = 42

Thus, the value is = 42

(j) f ÷ 11 = 6

= f = 66

Therefore, the value is = 66

Question no – (4)

Solution :

(a) Thrice a number added to 9 gives 15.

In equation = 3x + 9 = 15

(b) One-fifth of a number is 8.

In equation = x/5 = 8

(c) 4 subtracted from thrice a number is 20.

In equation = 3x – 4 = 20

Question no – (5)

Solution :

Table : (1)

 S.No. Equation Value of the variable Solution (Yes/No) (a) 4m = 24 m = 2 No (b) 4m = 24 m = 6 Yes (c) 4m = 24 m = 7 No (d) y ÷ 8 = 3 y = 8 No (e) y ÷ 8 = 8 y = 16 No

Table : (2)

 S.No. Equation Value of the variable Solution (Yes/No) (f) y ÷ 8 = 3 y = 24 Yes (g) 2t + 1 = 5 t = 1 No (h) 2t + 1 = 5 l = 2 Yes (i) 3l – 4 = 5 l = 2 No (j) 3l – 4 = 5 l = 3 Yes

Try These (8 (i) :

Question no – (1)

Solution :

(a) The perimeter (p) of an equilateral triangle is three times of its side (a).

In equation, = p = 3a

(b) The diameter (d) of a circle is twice its radius (r).

In equation, = d = 2r

(c) The selling price (s) of an item is equal to the sum of the cost price (c) of an item and the profit (p) earned.

In equation, = S = c + p

(d) Amount (a) is equal to the sum of principal (p) and interest (i).

In equation, = a = p + I

Question no – (2)

Solution :

 Equation Value of the variable Solution (Yes/No) (1)  x + 10 = 30 x = 10 No (2) x + 10 = 30 x = 30 No (3) x + 10 = 30 x = 20 Yes (4) p – 3 = 7 p = 5 No (5) p – 3 = 7 p = 15 No (6) p – 3 = 7 P = 10 Yes (7) 3n = 21 n = 10 No (8) 3n = 21 n = 7 Yes (9) t/5 = 4 t = 25 No (10) t/5 = 4 t = 20 Yes (11) 2l + 3 = 7 l = 5 No (12) 2l + 3 = 7 l = 1 No (13) 2l + 3 = 7 l = 2 Yes

Simple Equations Exercise 8.1 Solution

Question no – (1)

Solution :

(a) x + 4 = 10

L.H.S = 1 + 4 = 5

R.H.S = 10

1 + 4 = 5 ≠ 10

2 + 4 = 6 ≠ 10

3 + 4 = 7 ≠ 10

4 + 4 = 8 ≠ 10

5 + 4 = 9 ≠ 10

6 + 4 = 10 = 10

6 is the solution

(b) n – 7 = 5

L.H.S = 8 – 7 = 1

R.H.S = 5

8 – 7 = 1 ≠ 5

9 – 7 = 2 ≠ 5

10 – 7 = 3 ≠ 5

11 – 7 = 4 ≠ 5

12 – 7 = 4 ≠ 5

12 – 7 = 5 = 5

12 is the solution

(c) 4m = 20

L.H.S = 4 × 1 = 4

R.H.S = 20

4 × 1 = 4 ≠ 20

4 × 2 = 8 ≠ 20

4 × 3 = 12 ≠ 20

4 × 4 = 16 ≠ 20

4 × 5 = 20 = 20

5 is the solution

(d) y/5 = 6

L.H.S = 5/5 = 1

R.H.S = 6

5/5 = 1 ≠ 6

10/5 = 2 ≠ 6

15/5 = 3 ≠ 6

20/5 = 4 ≠ 6

25/5 =5 ≠ 6

30/5 = 6 = 6

30 is the Solution

(e) 10 – x = 4

L.H.S = 10 – 1 = 9

R.H.S = 4

10 – 2 = 8 ≠ 4

10 – 3 = 7 ≠ 4

10 – 4 = 6 ≠ 4

10 – 5 = 5 ≠ 4

10 – 6 = 4 = 4

6 is the Solution

(f) r/2 = 8

L.H.S = 2/2 = 1

R.H.S = 8

2/2 = 1 ≠ 8

4/2 = 2 ≠ 8

6/2 = 3 ≠ 8

16/2 = 8 = 8

16 is the Solution

(g) 6x + 1 = 19

L.H.S = 6 × 1 + 1 = 7

R.H.S = 19

6 × 1 + 1 = 7 ≠ 19

6 × 2 + 1 = 13 ≠ 19

6 × 3 + 1 = 19 ≠19

3 is the Solution

(h) 5x – 3 = 4x

L.H.S = 5 × 1 – 3 = 2

R.H.S = 4 × 1 = 8

5 × 1 – 3 = 2 ≠ 4 × 1= 8

5 × 2 – 3 = 7 ≠ 4 × 2 = 8

5 × 3 – 3 = 12 ≠ 4 × 3 = 12

∴ 3 is the Solution

Question no – (2)

Solution :

Given, t + 12 = 20

 t 1 2 3 4 5 6 7 8 t + 12 13 14 15 16 17 18 19 20

Question no – (3)

Solution :

Solution to the equation n/5 + 3 = 8

 n 5 10 15 20 15 30 35 40 n/5 + 3 4 5 6 7 8

Question no – (4)

Solution :

(a) 5m = 60

= 5m/5

= m = 12

(b) n + 12 = 20

= n + 12 – 12 = 20

= n = 8

(c) p – 5 = 5

= p – 5 + 5

= 5 + 5

= p = 10

(d) q/2 = 7

= q/2 × 2

= 7 × 2

= q = 14

(e) r – 4 = 0

= r – 4 + 4

= 0 + 4

= r = 4

(f) x + 4 = 2

= x + 4 – 4

= 2 – 4

= x – 2

Question no – (5)

Solution :

(a) 11 added to a number gives 35

x + 11 = 35

= x + 11- 1 = 35 – 11

x = 24

(b) One fifth of a number is 7

x/5 = 7

= x/5 × 5

= 7 × 5

= x = 35

(c) 7 subtracted from a number gives 31

x – 7 = 31

= x – 7 + 7

= 31 + 7

= x = 38

(d) A number multiplied by 7 gives 56

x × 7 = 56 (x × 7)/7

= 56/7

= x = 8

(e) The difference of a number and 5 is 45

x – 5 = 45

= x – 5 + 5

= 45 + 5

= x = 50

(f) Three times a number added to 8 gives 11

3x + 8 = 11

= 3x + 8 – 8

= 11 – 8

= 3x = 3

= 3x/3 = 3/1

= x = 1

Question no – (6)

Solution :

(a) Think of any number. Double it. Multiply the product by 5. Divide the product by 10. What do you get?

So now,

x → 2x → 10 x → x

I got the same number that I thought.

(b) Think of any number.

Add 40 and divide by 2

Take away the number you thought of your answer is five less than half a century.

So according to the riddle,

= x → 2x (x + 25) → (2x + 50 + 40)

= x + 45 → x + 45 – x

= 45

(c) According to the riddle,

You are Two!

Simple Equations Chapter Check-up Solution :

Question no – (1)

Solution :

(a) 17 = x + 7

= x is the variable.

(b) (t – 7) > 5

= t is the variable.

(e) 5 × 4 – 8 = 2x

= x is the variable.

(f) x – 2 = 0

= x is the variable.

(g) 2m < 30

= m is the variable.

(h) 2n + 1 = 11

= n is the variable.

Question no – (2)

Solution :

 S. No. Equation Value of the variable Equation satisfied Yes/No (a) 10y = 80 y = 10 Not satisfied (b) 10y = 80 y = 8 Satisfied (c) 10y = 80 y = 5 No (d) 4l = 20 l = 20 No (e) 4l = 20 l = 80 No (f) 4l = 20 l = 5 Yes (g) b + 5 = 9 b = 5 No (h) b + 5 = 9 b = 9 No (i) b + 5 = 9 b = 4 Yes (j) h – 8 = 5 h = 13 Yes (j) h – 8 = 5 h = 8 No (l) h – 8 = 5 h = 0 No (m) p + 3 = 1 p = 3 No (n) p + 3 = 1 p = 1 No (o) p + 3 = 1 p = 0 No (p) p + 3 = 1 p =- 1 No (q) p + 3 = 1 p = – 2 Yes

Question no – (3)

Solution :

(a) 13 subtracted from twice a number gives 3.

= 2x – 13 = 3

(b) One fifth of a number is 5 less than that number.

= x – x/5 = 5

(c) Two-third of number is 12.

= 2x/3 = 12

(d) 9 added to twice a number gives 13.

= 2x + 9 = 13

(e) 1 subtracted from one-third of a number gives 1.

= x/3 – 1 = 1

Question no – (5)

Solution :

As per the question,

longest side is double the shortest side and the third side is 3 cm less than the longest side perimeter of the triangle is 72 cm.

Let the length of the shortest side = x cm

Length of the longest side = 2x cm

Question no – (6)

Solution :

 Column – I Column – II (i) The number of corners of a quadrilateral (b) constant (ii) The variable in the equation 2p + 3 = 5 (e) p (iii) The solution of the equation x + 2 = 1 (d) – 1 (iv) solution of the equation 2p + 3 = 5 (c) + 1 (v) A sign used in an equation (a) +

Or,

(i) → (b) constant

(ii) → (e) p

(iii) → (d) – 1

(iv) → (c) + 1

(v) → (a) +

Previous Chapter Solution :

Updated: June 6, 2023 — 5:49 am