These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.5

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 × 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) \(\frac{3 \mathrm{q}}{2}\) < 5

(m) z + 12 > 24

(n) 20 – (10 – 5) = 3 × 5

(o) 7 – x = 5

Answer:

(a) It is an equation of variable as both the sides are equal. The variable is x.

(b) It is not an equation as L.H.S. is greater than R.H.S.

(c) It is an equation with no variable. But it is a false equation.

(d) It is an equation with no variable. But it is a false equation.

(e) It is an equation of variable as both the sides are equal. The variable is x.

(f) It is an equation of variable x.

(g) It is not an equation as L.H.S. is less than R.H.S.

(h) It is an equation of variable as both the sides are equal. The variable is n.

(i) It is an equation with no variable as its both sides are equal.

(j) It is an equation of variable p.

(k) It is an equation of variable y.

(l) It is not an equation as L.H.S. is less than R.H.S.

(m) It is not an equation as L.H.S. is greater than R.H.S.

(n) It is an equation with no variable.

(o) It is an equation of variable x

Question 2.

Complete the entries in the third column of the table.

Equation | Value of variable | Equation 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 | – |

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) \(\frac{\mathrm{q}}{2}\) = 7 (7, 2, 10, 14)

(e) r – 4 = 0 (4,-4, 8,0)

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

Answer:

(a) 5m = 60

Putting the given values in L.H.S.,

5 × 10 = 50

∵ L.H.S. ≠ R.H.S

∴ m = 10 is not the solution

5 × 5 = 25

∵ L.H.S. ≠ R.H.S

∴ m = 5 is not the solution

5 × 12 = 60

∵ L.H.S. = R.H.S

∴ m = 12 is a solution

5 × 15 = 75

∵ L.H.S. ≠ R.H.S

∴ m = 15 is not the solution

(b) n + 12 = 20

Putting the given values in L.H.S.,

12 + 12 = 24

∵ L.H.S. ≠ R.H.S

∴ n = 12 is not the solution

8 + 12 = 20

∵ L.H.S. = R.H.S

∴ n = 8 is a solution

20 + 12 = 32

∵ L.H.S. ≠ R.H.S

∴ n = 20 is not the solution

0 + 12 = 12

∵ L.H.S. ≠ R.H.S

∴ n = 0 is not the solution

(c) p – 5 = 5

Putting the given values in L.H.S.,

0 – 5 = -5

∵ L.H.S. ≠ R.H.S

∴ p = 0 is not the solution

10 – 5 = 5

∵ L.H.S. = R.H.S

∴ p = 10 is a solution

5 – 5 = 0

∵ L.H.S. ≠ R.H.S

∴ p = 5 is not the solution

-5 -5 = -10

∵ L.H.S. ≠ R.H.S

∴ p = -5 is not the solution

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

Putting the given values in L.H.S.,

\(\frac{\mathrm{7}}{2}\)

∵ L.H.S. ≠ R.H.S

∴ q = 7 is not the solution

\(\frac{\mathrm{2}}{2}\) = 1

∵ L.H.S. ≠ R.H.S

∴ q = 2 is not the solution

\(\frac{\mathrm{10}}{2}\) = 5

∵ L.H.S. ≠ R.H.S

∴ q = 10 is not the solution

\(\frac{\mathrm{14}}{2}\) = 7

∵ L.H.S. = R.H.S

∴ q = 14 is a solution

(e) r – 4 = 0

Putting the given values in L.H.S.,

4 – 4 = 0

∵ L.H.S. = R.H.S

∴ r = 4 is a solution

– 4 -4 = -8

∵ L.H.S. ≠ R.H.S

∴ r = – 4 is not the solution

8 – 4 = 4

∵ L.H.S. ≠ R.H.S

∴ r = 8 is not the solution

0 – 4 = – 4

∵ L.H.S. ≠ R.H.S

∴ r = 0 is not the solution

(f) x + 4 = 2

Putting the given values in L.H.S.,

-2 + 4 = 2

∵ L.H.S. = R.H.S

∴ x = -2 is a solution

0 + 4 = 4

∵ L.H.S. ≠ R.H.S

∴ x = 0 is not the solution

2 + 4 = 6

∵ L.H.S. ≠ R.H.S

∴ x = 0 is not the solution

4 + 4 = 8

∵ L.H.S. ≠ R.H.S

∴ x = 4 is not the 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 of the equation z/3 = 4 using the table.

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

Answer:

(a)

∵ At m = 6, m + 10 = 16 ∴ m = 6 is the solution.

(b)

∵ At t = 7, 5t = 35 ∴ t = 7 is the solution.

(c)

∵ At z = 12, z/3 = 4 ∴ z = 12 is the solution.

(d)

∵ At m = 10, m- 7= 3 ∴ m = 10 is the solution.

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 a pretty clue!

You will get me back

If you take me out of twenty two!

Answer:

(i) According to the condition,

1 + 12 = 34 or × + 12 = 34

∴ By inspection, we have

22 + 12 = 34

So, I am 22.

(ii) Let I am ‘×’.

We know that there are 7 days in a week.

∴ up-counting from × for 7, the sum = 23

By inspections, we have

16 + 7 = 23

∴ × = 16

Thus I am 16.

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

∴ Special Number -6 = 11

× – 6 = 11

By inspection, we get

17 – 6 = 11

∴ × = 17

Thus I am 17.

(iv) Suppose I am ‘×’

∴ 22 – 1 = 1

or 22 – × = ×

By inspection, we have

22 -11 = 11

∴ × = 11

Thus I am 11.