Gujarat BoardĀ GSEB Textbook Solutions Class 6 Maths Chapter 11 Algebra Ex 11.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 1.

State which of the following are equations (with a variable). Give a 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 Ć 3) – 19 = 8

(e) 5 Ć 4 – 8 = 2x

(f) x – 2 = 0

(g) 2m < 30

(h) 2n + 1 = 11

(i) 7 = (11 Ć 5) – (12 Ć 4)

(j) 7 = (11 Ć 2) + p

(k) 20 = 5y

(l) \(\frac { 3q }{ 2 }\) < 5

(m) z + 12 > 24

(n) 20 – (10 – 5) = 3 x 5

(o) 7 – x = 5

Solution:

(a) It is an equation. It has a variable x.

(b) It is not an equation. There is no sign of equality (=).

(c) It is not an equation. It has no variable.

(d) It is not an equation. It has no variable.

(e) It is an equation. It has a variable x.

(f) It is an equation. It has a variable x.

(g) It is not an equation. It has no sign of equality (=).

(h) It is an equation. It has a variable n.

(i) It is not an equation. It has no variable.

(j) It is an equation. It has a variable p.

(k) It is an equation. It has a variable y.

(l) It is not an equation.ā It has no sign of equality (=).

(m) It is not an equation. It has no sign of equality (=).

(n) It is not an equation. It has no variable.

(o) It is an equation. It has a variable x.

Question 2.

Complete the entries in the 4th column of the table.

Answer:

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)

Solution:

(a) 5m = 60:

For m = 10, we have: LHS = 5 Ć 10 = 50 and

RHS 60

Since, LHS ā RHS

ā“ m = 10 is not a solution.

For m = 5, we have: LHS = 5 Ć 5 = 25 and

RHS = 60

Since, LHS ā RHS

ā“ m = 5 is not a solution.

For m = 12, we have: LHS = 5 Ć 12 = 60 and

RHS = 60

Since, LHS ā RHS.

ā“ m = 12 is a solution.

For m = 15, we have: LHS = 5 Ć 15 = 75 and

RHS = 60

Since, LHS ā RHS

ā“ m = 15 is not a solution

(b) n + 12 = 20:

For n = 12, LHS = 12 + 12 = 24 and RHS = 20

Since, LHS ā RHS

ā“ n = 12 is not a solution.

For n = 8, we have: LHS = 8 +12 = 20 and

RHS = 20

Since, LHS = RHS

ā“ n = 8 is a solution.

For n = 20, we have: LHS = 20 + 12 = 32 and

RHS = 20

Since LHS ā RHS

ā“ n = 20 is not a solution.

For n = 0, we have: LHS =0 + 12 = 12 and

RHS = 20

Since, LHS ā RHS

ā“ n = 0 is not a solution.

(c) p – 5 = 5

For p = 0, LHS = 0 – 5 = -5 and RHS = 5

Since, LHS ā RHS

ā“ p = 0 is not a solution.

For p = 10, LHS = 10 – 5 = 5 and RHS = 5

Since, LHS = RHS

ā“ p = 10 is a solution.

For p = 5, LHS = 5 – 5 = 0 and RHS = 5

Since, LHS ā RHS

ā“ p = 5 is not a solution.

For p = -5, LHS = – 5 – 5 = -10 and RHS = 5

Since, LHS ā RHS

ā“ p = -5 is not a solution.

(d) \(\frac { q }{ 2 }\) =7

For q = 7, LHS = \(\frac { 7 }{ 2 }\) and RHS = 7

Since, LHS ā RHS

ā“ q = 7 is not a solution.

For q = 2, LHS = \(\frac { 2 }{ 2 }\) = 1 and RHS = 7

āµ LHS ā RHS

ā“ q = 2 is not a solution.

For q = 10, LHS = \(\frac { 10 }{ 2 }\) = 5 and RHS = 7

āµ LHS ā RHS

ā“ q = 10 is not a solution.

For q = 14, LHS = \(\frac { 14 }{ 2 }\) = 7 and RHS = 7

āµ LHS = RHS

ā“ q = 14 is a solution.

(e) r – 4 = 0

For r = 4, LHS = 4 – 4 = 0 and RHS = 0

āµ LHS = RHS

ā“ r = 4 is a solution.

For r = – 4, LHS = – 4 – 4 = -8 and RHS = 0

āµ LHS ā RHS

ā“ r = – 4 is not a solution.

For r = 8, LHS = 8 – 4 = 4 and RHS = 0

āµ LHS ā RHS

ā“ r = 8 is not a solution.

For r = 0, LHS = 0 – 4 = – 4 and RHS = 0

āµ LHS ā RHS

ā“ r = 0 is not a solution.

(f) x + 4 = 2

For x = -2 , LHS = -2 + 4 = 2 and RHS = 2

āµ LHS = RHS

ā“ x = -2 is a solution.

For x = 0 , LHS = 0 + 4 = 4 and RHS = 2

āµ LHS ā RHS

ā“ x = 0 is not a solution.

For x = 2 , LHS = 2 + 4 = 6 and RHS = 2

āµ LHS ā RHS

ā“ x = 2 is not a solution.

For x = 4 , LHS = 4 + 4 = 8 and RHS = 2

Since, LHS ā RHS

ā“ x = 4 is not a solution.

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 to 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 inspection, we have:

āµ For m = 6, m + 10 = 16

ā“ m = 6 is the solution to m + 10 = 16

(b) By inspection, we have:

āµ For t = 7, 5t = 35

ā“ t = 7 is the solution to 5t = 35.

(c) By inspection, we have:

āµ For z = 12, \(\frac { z }{ 2 }\) = 4

ā“ z = 12 is the solution to \(\frac { z }{ 3 }\) = 4

(d) By inspection, we have:

āµ For m = 10, m – 7 = 3

ā“ m = 10 is the solution to m – 7 = 3

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! V Add the count to me To get exactly thirty-four!

(ii) For each day of the week Make an account 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!

Solution:

(i) Suppose I am āxā

Since there are 4 comers of a square and each corner is counted thrice

ā“ 4 Ć 3 = 12

According to the condition,

[myself] + 12 = 34

or x + 12 = 34

By inspection, we have

22 + 12 = 34

x = 22, Thus, I am 22.

(ii) There are 7 days in a week and let I am āxā.

āµ Accounting from JC for 7, the sum = 23

i.e. x + 7 = 23

ā“ By inspection, we have 16 + 7 = 23

ā“ x = 16, Thus, I am 16.

(iii) Let the special number be x and there are 11 members in a cricket team.

āµ [Special number] – 6 = [A cricket team] or x – 6 = 11

ā“ By inspection, we have 17 – 6 = 11,

x = 17, Thus, I am 17.

(iv) Suppose I am āxā.

According to the problem,

22 – (myself) = (myself) or 22 – r = x

By inspection, we have: 22 – 11 = 11

ā“ x = 11, Thus, I am 11.