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.
 NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1
 NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2
 NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3
 NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4
 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.
(b) By inspections method, we have the following
table. Given that 5t = 35.
(c)
By inspections method, we have the following table. Given that z/3 = 4.
(d) By inspections method, we have the following
table. Given that m – 7 = 3.
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!
Add the count to me
To get exactly thirtyfour!
(ii) For each day of the week
Make an upcount from me
If you make no mistake
you will get twentythree!
(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 twentytwo!
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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