NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

# NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

## NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 are the part of NCERT Solutions for Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5.

### Ex 11.5 Class 6 Maths 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) 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) 3q/2 < 5

(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 × 5
(o) 7 – x = 5

Solution:
(a) 17 = x + 7. This is an equation with a variable x.
(b) (t – 7) > 5. This is not an equation because it does not have ‘=’ sign.
(c) 4/2 = 2. This is not an equation because it has no variable.
(d) (7 × 3) – 19 = 8. This is not an equation because it has no variable.
(e) 5 × 4 – 8 = 2x. This is an equation with a variable x.
(f) x – 2 = 0. This is an equation with a variable x.
(g) 2m < 30. This is not an equation because it does not have ‘=’ sign.
(h) 2n + 1 = 11. This is an equation with a variable n.
(i) 7 = (11 × 5) – (12 × 4). This is not an equation because it does not have a variable.
(j) 7 = (11 × 2) + p. This is an equation with a variable p.
(k) 20 = 5y. This is an equation with a variable y.
(l) 3q/2 < 5. This is not an equation because it does not have ‘=’ sign.

(m) z + 12 > 24. This is not an equation because it does not have ‘=’ sign.
(n) 20 – (10 – 5) = 3 × 5. This is not an equation because it has no variable.
(o) 7 – x = 5. This is an equation with a variable x.

### Ex 11.5 Class 6 Maths Question 2.

Complete the entries in the third column of the table.

 S. No. Equation Value of variable Equations satisfied Yes /No (a) 10y = 80 y = 10 (b) 10y = 80 y = 8 (c) 10y = 80 y = 5 (d) 4l = 20 l = 20 (e) 4l = 20 l = 80 (f) 4l = 20 l = 5 (g) b + 5 = 9 b = 5 (h) b + 5 = 9 b = 9 (i) b + 5 = 9 b = 4 (J) h – 8 = 5 h = 13 (k) h – 8 = 5 h = 8 (l) h – 8 = 5 h = 0 (m) P + 3 = 1 p = 3 (n) p + 3 = 1 p = 1 (o) p + 3 = 1 p = 0 (p) p + 3 = 1 p = -1 (q) p + 3 = 1 p = -2

Solution:

 S. No. Equation Value of variable Equations satisfied Yes /No (a) 10y = 80 y = 10 No (b) 10y = 80 y = 8 Yes (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 (k) 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

### Ex 11.5 Class 6 Maths Question 3.

Pick out the solution from the values given in the brackets next to each equation. Show that the other values do not satisfy the equation.
(Ð°) 5m = 60 (10, 5, 12, 15)
(b) n + 12 = 20 (12, 8, 20, 0)
(c) p – 5 = 5 (0, 10, 5, -5)
(d) q/2 = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4, -4, 8, 0)
(f) x + 4 = 2 (-2, 0, 2, 4)

Solution:
(a) For m = 10, LHS = 5 × 10 = 50, RHS = 60
Here, LHS ≠ RHS
m = 10 is not the solution of the equation.
For m = 5, LHS = 5 × 5 = 25, RHS = 60
Here, LHS ≠ RHS
m = 5 is not the solution of the equation.
For m = 12, LHS = 5 × 12 = 60, RHS = 60
Here, LHS = RHS
m = 12 is the solution of the equation.
For m = 15, LHS = 5 × 15 = 75, RHS = 60
Here, LHS ≠ RHS
m = 15 is not the solution of the equation.

(b) n + 12 = 20 (12, 8, 20, 0)
For n = 12, LHS = 12 + 12 = 24, RHS = 20
Here, LHS ≠ RHS
n = 12 is not the solution of the equation.
For n = 8, LHS = 8 + 12 = 20, RHS = 20
Here, LHS = RHS
n = 8 is the solution of the equation.
For n = 20, LHS = 20 + 12 = 32, RHS = 20
Here, LHS ≠ RHS
n = 20 is not the solution of the equation.
For n = 0, LHS = 0 + 12 = 12, RHS = 20
Here, LHS ≠ RHS
n = 0 is not the solution of the equation.

(c) p – 5 = 5 (0, 10, 5,  -5)
For p = 0, LHS = 0 – 5 =  -5, RHS = 5
Here, LHS ≠ RHS
p = 0 is not the solution of the equation.
For p = 10, LHS = 10 – 5 = 5, RHS = 5
Here, LHS = RHS
p = 10 is the solution of the equation.
For p = 5, LHS = 5 – 5 = 0, RHS = 5
Here LHS ≠ RHS
p = 5 is not the solution of the equation.
For p = -5, LHS = -5 – 5 = -10, RHS = 5
Here, LHS ≠ RHS
p = -5 is not the solution of the equation.

(d) q/2 = 7 (7, 2, 10, 14)
For q = 7, LHS = 7/2 , RHS = 7
Here LHS ≠ RHS
q = 7 is not the solution of the equation.
For q = 2, LHS = 2/2 = 1, RHS = 7
Here, LHS ≠ RHS
q = 2 is not the solution of the equation.
For q = 10, LHS = 10/2 = 5, RHS = 7
Here, LHS ≠ RHS

q = 10 is not the solution of the equation.
For q = 14, LHS = 14/2 = 7, RHS = 7
Here, LHS = RHS
q = 14 is the solution of the equation.

(e) r – 4 = 0 (4, -4, 8, 0)
For r = 4, LHS = 4 – 4 = 0, RHS = 0
Here, LHS = RHS
r = 4 is the solution of the equation.
For r = -4, LHS = -4 – 4 = -8, RHS = 0
Here, LHS ≠ RHS
r = -4 is not the solution of the equation.
For r = 8, LHS = 8 – 4 = 4, RHS = 0
Here, LHS ≠ RHS
r = 8 is not the solution of the equation.
For r = 0, LHS = 0 – 4 = –4, RHS = 0
Here, LHS ≠ RHS
r = 0 is not the solution of the equation.

(f) x + 4 = 2 (-2, 0, 2, 4)
For x = -2, LHS = -2 + 4 = 2, RHS = 2
Here, LHS = RHS
x = -2 is the solution of the equation.
For x = 0, LHS = 0 + 4 = 4, RHS = 2
Here, LHS ≠ RHS
x = 0 is not the solution of the equation.
For x = 2, LHS = 2 + 4 = 6, RHS = 2
Here, LHS ≠ RHS
x = 2 is not the solution of the equation.
For r = 4, LHS = 4 + 4 = 8, RHS = 2
Here, LHS ≠ RHS
x = 4 is not the solution of the equation.

### Ex 11.5 Class 6 Maths Question 4.

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

Solution:
(a) By inspections method, we have the following table.

So, m = 6 is the solution of the equation.

(b) By inspections method, we have the following table. Given that 5t = 35.

So, t = 7 is the solution of the equation.

(c) By inspections method, we have the following table. Given that z/3 = 4.

So, z = 12 is the solution of the equation.

(d) By inspections method, we have the following table. Given that m – 7 = 3.

So, m = 10 is the solution of the equation.

### Ex 11.5 Class 6 Maths 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!
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 you a pretty clue!
you will get me back
If you take me out of twenty-two!

Solution:
(i) According to the condition,
We have, x + 12 = 34
By inspection method, we have
22 + 12 = 34
So, I am 22.

(ii) Let I am ‘x’.
We know that there are 7 days in a week.
upcounting from x for 7, the sum = 23, i.e., x + 7 = 23
By inspections method, we have
16 + 7 = 23
x = 16
Thus, I am 16.

(iii) Let the special number be x and there are 11 players in a cricket team.
Special Number – 6 = 11
x – 6 = 11
By inspection method, we have
17 – 6 = 11
x = 17
Thus, I am 17.

(iv) Suppose I am ‘x’.
22 – x = x
By inspection method, we have
22 – 11 = 11
x = 11
Thus, I am 11.

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