# HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.5

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.5 Textbook Exercise Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.5

Question 1.
State whether the following are equations or not. Give reason for your answer. Further identify which are equations with only numbers and which are equations with variables. Identify the variables from the equations with variables. Give reason for your answer.
(a) 17 = x+7
(b) (t-7) > 5
(c) $$\frac{4}{2}$$ = 8
(ci) 7 x 3 – 19 = 8
(e) 5 x 4 – 8 = 2x
(f) x-2 = 0
(g) 2m < 30
(h) 2n+11 = 1
(j) 7 = 11 x 5 – 12 x 4
(j) 7 = 11 x 2 + p
(k) 20= 5y
(l) $$\frac{3 q}{2}$$< 5 (m) z + 12 > 24
(n) 20- (10-5) = 3×5
(o) 7 – x = 5.
Solution:
(6) It is not an equation because there is no sign of (=).
(c) It is an equation with numbers only.
(d) It is an equation with numbers only
(e) It is an equation with variable ‘r’
(f) It is an equation with variable ‘x’.
(g) It is not an equation because there is no sign of (=).
(h) It is an equation with variable ‘n’
(i) It is an equation with numbers only
(j) It is an equation with variable ‘p’
(k) It is an equation with variable ‘y’
(l) It is not an equation because there is no sign of (=).
(m) It is not an equation because there is no sign of (=).
(n) It is an equation with numbers only
(o) It is an equation with variable ‘x’.

Question 2.
Complete the entries in the third column of the table :
Solution:

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 (O, 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)
Solution:

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

m = 6 is the solution of the equation m + 10 = 16.

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

Solution:
t = 7 is the solution of the equation 5t = 35.

(c) Complete the table and find the table the solution of the equation $$\frac{z}{3}$$ = 4

z – 12 is the solution of the equation $$\frac{z}{3}$$ = 4.

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

m = 10 is the solution of the equation m – 7 = 3.

Question 5.
Solve:
(a) x + 5 = 12
(b) y – 2 = 10
(c) 7p = 210
(d) $$\frac{q}{2}$$ = 5
(e) t + 100 = 125
(f) l – 20 = 30
(g) 9u = 81
(h) $$\frac{k}{8}$$ = 20
(i) 3y = 33
(j) x-3 = 0
(k) $$\frac{k}{8}$$ = 8
(l) 13y = 65
Solution:
(a) x + 5 = 12
x = 12 – 5 = 7
Thus, x = 7 is the required solution of the given equation.

(b) y- 2 = 10
⇒ y = 10 + 2 = 12
Thus, y = 12 is the required solution of the given equation.

(c) 7p = 210
⇒ P = $$\frac{210}{7}$$ = 30
Thus, p = 30 is the required solution of the given equation.

(d) $$\frac{q}{2}$$ = 5
⇒ q = 2 x 5 = 10
Thus, q = 10 is the required solution of the given equation.

(e) t + 100 = 125
⇒ t = 125 – 100 = 25
Thus, t = 25 is the required solution of the given equation.

(f) 1 – 20 = 30
⇒ l = 30 + 20 = 50
Thus, 1 = 50 is the required solution of the given equation.

(g) 9 u = 81
⇒ u = $$\frac{81}{9}$$ = 9
Thus, u = 9 is the required solution of the given equation.

(h) $$\frac{k}{8}$$ = 20
⇒ k = 8 x 20= 160
Thus, ft = 160 is the required solution of the given equation.

(i) 3y = 33
⇒ y = $$\frac{33}{3}$$ = 11
Thus, y = 11 is the required solution of the given equation.

(j) x- 3 = 0
⇒ x = 3
Thus, x = 3 is the required solution of the given equation.

(k) $$\frac{k}{8}$$ = 8
⇒ k = 8 x 8 = 64
Thus, ft = 64 is the required solution of the given equation.

(l) 13y = 65
y = $$\frac{65}{13}$$ = 5
Thus, y = 5 is the required solution of the given equation.

Question 6.
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) I am a special number. Take away from me a six ! A whole cricket team, you will still be able to fix !
(iii) For each day of the week. Make an up cohnt from me A. If you make no mistake, you will get twenty three !
(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 !
Solution:
(i) Let I be V. Since a square has four corners.
Counting every corner thrice, we get 4 x 3 = 12
Now x + 12 = 34
⇒ x = 34 – 12 = 22
Thus, I am 22.

(ii) Let I be ‘x’. By taking away 6 from me, you get x – 6.
∵ There are 11 members in a cricket team.
∴ x – 6 = 11
⇒ x – 11 + 6 = 17.
Thus, I am 17.

(iii) There are 7 days in a week.
∴ A + 7 = 23
⇒ A = 23 – 7 = 16
Thus A = 16.

(iv) Let I be x.
Now, 22 – x = x
⇒ 2x = 22
⇒ x = $$\frac{22}{2}$$ = 11
Thus, I am 11.