NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Algebra |

Exercise |
Ex 11.5 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## 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)** \(\frac { 4 }{ 2 } \) = 2

**(d)** 7 x 3 – 19 = 8

**(e)** 5 x 4 – 8 = 2x

**(f)** x – 2 = 0

**(g)** 2m < 30

**(h)** 2n + 1 = 11

**(i)** 7 = (11 x 5) – (12 x 4)

**(j)** 7 = 11 x 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)** An equation, equation with variable (x).

**(b)** Not an equation (as have no = sign).

**(c)** An equation, equation with numbers.

**(d)** An equation, equation with numbers.

**(e)** An equation, equation with variable (x).

**(f)** An equation, equation with variable (x).

**(g)** Not an equation (as have no = sign).

**(h)** An equation, equation with variable (n).

**(i)** An equation, equation with numbers.

**(j)** An equation, equation with variable (p).

**(k)** An equation, equation with variable (y).

**(l)** Not an equation (as have no = sign).

**(m)** Not an equation (as have no = sign),

**(n)** An equation, equation with numbers.

**(o)** An equation, equation with variable (x).

Ex 11.5 Class 6 Maths Question 2.

Complete the entries in the third column of the table.

Solution:

**The table duly completed is as under :**

Ex 11.5 Class 6 Maths 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 (7, 2, 10, 14)

**(d)** \(\frac { q }{ 2 } \) = 7 (4, -2, 8, 0)

**(e)** r – 4 = 0 (4, -4, 8, 0)

**(f)** x + 4 = 2 (-2, 0, 2, 4)

Solution:

**(a)** Given equation is 5m = 60.

For m=10: 5m = 5 x 10 = 50 ≠ 60. So, m=10 does not satisfy the given equation.

For m = 5 5m = 5 x 5 = 25 ≠ 60. So, m = 5 does not satisfy the given equation.

For m=12: 5m = 5 x 12 = 60. So, m=12 does not satisfy the given equation.

Thus, n = 12 is its solution.

For m = 15 5m = 5 x 15 = 75 ≠ 60. So, m = 15 does not satisfy the given equation.

**(b)** Given equation is n +12 = 20.

For n=12: n +12 = 12 + 12 = 24 ≠ 20. So, n = 12 does not satisfy the the given equation.

For n = 8: n +12 = 8 + 12 = 20. So, n = 8 satisfies the given equation.

Thus, n = 8 is its solution.

For n = 20: n + 12 = 20 + 12 = 32 ≠ 20. So, n = 20 does not satisfy the given equation.

For n = 0: n + 12 = 0 + 12 = 12 ≠ 20. So, n = 0 does not satisfy the given equation.

**(c)** Given equation is p – 5 = 5.

For p = 0: p – 5 = 0 – 5 = -5 ≠ 5. So, p = 0 does not satisfy the given equation.

For p = 10: p – 5 = 10 – 5 = 5. So, p = 10 satisfies the given equation.

Thus, its solution is p = 10.

For p = 5: p – 5 = 5 – 5 = 0≠ 5. So, p = 5 does not satisfy the given equation.

For p = -5: p – 5 = -5 – 5 = -10 ≠ 5. So, p = -5 does not satisfy the given equation.

**(d)** Given equation is \(\frac { q }{ 2 } \) = 7.

For q = 7: \(\frac { q }{ 2 } \) = \(\frac { 7 }{ 2 } \) ≠ 7. So, q = 7 does not satisfy the given equation.

For q = 2: \(\frac { q }{ 2 } \) = \(\frac { 2 }{ 2 } \) = 1 ≠ 7. So, q = 2 does not satisfy the given equation.

For q = 10: \(\frac { q }{ 2 } \) = \(\frac { 10 }{ 2 } \) = 5 ≠ 7. So, q = 10 does not satisfy the given equation.

For q = 14: \(\frac { q }{ 2 } \) = \(\frac { 14 }{ 2 } \) = 7. So, q = 14 satisfies the given equation. Thus, q = 14 is its solution.

**(e)** Given equation is r – 4 = 0.

For r = 4: r – 4 = 4 – 4 = 0. So, r = 4 satisfies the given equation. Thus, r = 4 is its solution.

For r = -4: r – 4 = -4 – 4 = 8 ≠ 0. So, r = -4 does not satisfy the given equation.

For r = 8: r – 4 = 8 – 4 = 4 ≠ 0. So, r-8 does not satisfy the given equation.

For r = 0: r – 4 = 0 – 4 = -4 ≠ 0. So, r = 0 does hot satisfy thh given equation.

**(f)** Given equation is x + 4 = 2.

For x = -2: x + 4 = -2 + 4 = 2. So, x = -2 satisfies the given equation.

Thus, x = -2 is its solution.

For x = 0: x + 4 = 0 + 4 = 4 ≠ 2. So,- x =0 does not satisfy the given equation.

For x=2: x + 4= 2 + 4= 6 ≠ 2. So, x = 2 does not satisfy the given equation.

For x = 4: x + 4 = 4 + 4= 8 ≠ 2. So, x = 4 does not satisfy the given equation.

Ex 11.5 Class 6 Maths Question 4.

Question (a)

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

Solution:

Completing the table, we have

By inspection of the above table, we find that m = 6 satisfies the equation m+10 = 16. So, m = 6 is its solution.

Question (b)

Complete the table and by inspection of the table find the solution to the equation 51 = 35.

Solution:

By inspection of the above table, we find that t = 7 satisfies the equation 5t = 35. So, t = 7 is its solution.

Question (c)

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

Solution:

Completing the table, we have

By inspection of the above table, we find that t=12 satisfies the equation \(\frac { z }{ 3 } \) = 4. So, t = 12 is its solution.

Question (d)

Complete the table and find the solution to .the equation m – 7 = 3.

Solution:

Completing the table, we have

By inspection of the above table, we find that m = 10 satisfies the equation m – 7 = 3. So, m = 10 is its solution.

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 comer Thrice and no more!

Add the count to me

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 a pretty clue!

You will get me hack

If you take me out of twenty two!

Solution:

**(i)**Let I be denoted by x. There are 4 comers of a square.

On counting each comer thrice, we get 3 x 4 = 12

As per problem:

x + 12 = 34 ⇒ x +12 – 12 = 34 – 12

⇒ x + 0 = 22 ⇒ x = 22

Thus, I am 22.

**(ii)** According to the question:

x + 7 =23

x + 7 – 7 = 23 – 7

⇒ x + 0 =16 ⇒ x =16

**(iii)** Let the special number be x. Then, according to the problem, we have

x – 6 = 11 [ In a cricket team, no. of players =11]

⇒ x – 6 + 6 = 11 + 6

⇒ x + 0 = 17 ⇒ x = 17

Thus, the special number is 17.

**(iv)** Let I be denoted by x.

According to the question:

22 – x = x

⇒ 22 – x + x = x + x

⇒ 22 + 0 =2x

⇒ 2x = 22 ⇒ \(\frac { 2x }{ 2 } =\frac { 22 }{ 2 } \)

⇒ x =11

Thus, I am 11.

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