NCERT Solutions For Class 6 Maths Algebra Exercise 11.5

### NCERT Solutions For Class 6 Maths Algebra Exercise 11.5

**NCERT Solutions For Class 6 Maths Chapter 11 Algebra Ex 11.5**

**Exercise 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 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) 3q/2 < 5 (m) z + 12 > 24

(n) 20 – (10 – 5) = 3 x 5

(o) 7 – x = 5

Solution:

(a) 17 = x + 7 is an equation with a variable x.

(b) (t – 7) > 5 is not an equation because it does not have ‘=’ sign.

(c) 4/2 = 2 is not an equation because it has no variable.

(d) (7 x 3) – 19 = 8 is not an equation because it has no variable.

(e) 5 x 4 – 8 = 2x is an equation with a variable x.

(f) x – 2 = 0 is an equation with a variable x.

(g) 2m < 30 is not an equation because it does not have ‘=’ sign.

(h) 2n + 1 = 11 is an equation with a variable n.

(i) 7 = (11 x 5) – (12 x 4) is not an equation because it does not have a variable.

(j) 7 = (11 x 2) + p is an equation with a variable p.

(k) 20 = 5y is an equation with a variable y.

(l) 3q/2 < 5 is not an equation because it does not have ‘=’ sign. (m) z + 12 > 24 is not an equation because it does not have ‘=’ sign.

(n) 20 – (10 – 5) = 3 x 5 is not an equation because it has no variable.

(o) 7 – x = 5 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.

(а) 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 x 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 x 12 = 60, RHS = 60

Here, LHS = RHS

∴ m = 12 is the solution of the equation

For m = 15 LHS = 5 x 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 = 0, RHS = 5

Here, LHS ≠ RHS

∴ p = -5 is not 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

For 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 = 6

(b) Complete the table and by inspection of the table find the solution to the equation 51 – 35

(c) Complete the table and find the solution of the equation g = 4 using the table.

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

Solution:

(a) By inspections, we have

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

(b) Given that 5t = 35

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

(c) Given that z/3 = 35

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

(d) 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!

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 you a pretty clue!

you will get me back

If you take me out of twenty two!

Solution:

(i) According to the condition,

I + 12 = 34 or x + 12 = 34

∴ By inspection, 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

By inspections, we have

16 + 7 = 23

∴ x = 16

Thus I am 16.

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

∴ Special Number -6 = 11

∴ x – 6 = 11

By inspection, we get

17 – 6 = 11

∴ x = 17

Thus I am 17.

(iv) Suppose I am ‘x’.

∴ 22 – I = I

or 22 – x = x

By inspection, we have

22 – 11 = 11

∴ x = 11

Thus I am 11.