RBSE Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.5 is part of RBSE Solutions for Class 6 Maths. Here we have given Rajasthan Board RBSE Class 6 Maths Chapter 5 Fractions Exercise 5.5.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Fractions |

Exercise |
Ex 5.5 |

Number of Questions |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 6 Maths Chapter 5 Fractions Ex 5.5

Question 1.

Solve the (RBSESolutions.com) following.

Solution.

Question 2.

Heera gave \(\frac { 1 }{ 4 } \) litre milk to (RBSESolutions.com) Bhavna out of her \(\frac { 3 }{ 7 } \) litre milk. How much of milk is now left with her?

Solution.

Heera have milk = \(\frac { 3 }{ 7 } \) litre

Heera gave milk to Bhavna = \(\frac { 1 }{ 4 } \) litre

∴ Remaining milk to Heera = \(\frac { 3 }{ 7 } \) – \(\frac { 1 }{ 4 } \)

L.C.M. of 7 and 4 = 28

∴ \(\frac { 3 }{ 7 } \) – \(\frac { 1 }{ 4 } \) = \(\frac { 3\times 4-1\times 7 }{ 28 } \)

= \(\frac { 12-7 }{ 28 } \) = \(\frac { 5 }{ 28 } \) litre

Question 3.

A wooden piece is \(\frac { 9 }{ 10 } \) m long (RBSESolutions.com) and a \(\frac { 2 }{ 5 } \) m long piece has been cut from it. 5 What is the length of the remaining piece?

Solution.

Length of wooden piece = \(\frac { 9 }{ 10 } \) m

Length of piece cut from it = \(\frac { 2 }{ 5 } \) m.

∴ Length of remaining piece = \(\frac { 9 }{ 10 } \) – \(\frac { 2 }{ 5 } \)

Question 4.

Anshul drink \(\frac { 2 }{ 3 } \) of one (RBSESolutions.com) glass water. Find out how much water is left in glass?

Solution.

Total water = 1 glass

Anshul drink water = \(\frac { 2 }{ 3 } \) part

∴ Remaining water = 1 – \(\frac { 2 }{ 3 } \)

= \(\frac { 1\times 3-2\times 1 }{ 3 } \)

= \(\frac { 3-2 }{ 3 } \) = \(\frac { 1 }{ 3 } \) Part

Question 5.

Sunil purchased 5 \(\frac { 1 }{ 2 } \) kg and (RBSESolutions.com) Vijay purchased 3 \(\frac { 4 }{ 5 } \) kg mangoes. Find how much more mangoes did Sunil purchase.

Solution.

Sunil purchased mangoes = 5 \(\frac { 1 }{ 2 } \) kg or \(\frac { 11 }{ 2 } \) kg

Vijay purchased mangoes = 3 \(\frac { 4 }{ 5 } \) kg or \(\frac { 19 }{ 5 } \) kg

∴ Sunil purchased more mangoes = \(\frac { 11 }{ 2 } \) – \(\frac { 19 }{ 5 } \)

∴ = \(\frac { 11 }{ 2 } \) – \(\frac { 19 }{ 5 } \) = \(\frac { 11\times 5-19\times 2 }{ 10 }\)

= \(\frac { 55-38 }{ 10 } \) = \(\frac { 17 }{ 10 } \) kg

(L.C.M. of 2 and 5 = 10)

Question 6.

Neha finished one (RBSESolutions.com) race in 3 \(\frac { 1 }{ 2 } \) minute 13 and Geeta in \(\frac { 13 }{ 4 } \) minutes. Find out who 4 finished the race in lesser time and how much time?

Solution.

Neha finished race = 3 \(\frac { 1 }{ 2 } \) minutes or \(\frac { 7 }{ 2 } \) minutes

Geeta finished race = \(\frac { 13 }{ 4 } \) minutes

L.C.M. of 2 and 4 = 4

On making fraction \(\frac { 7 }{ 2 } \) and \(\frac { 13 }{ 4 } \) like fractions

\(\frac { 7 }{ 2 } \) x \(\frac { 2 }{ 2 } \) = \(\frac { 14 }{ 4 } \)

\(\frac { 13 }{ 4 } \) x \(\frac { 1 }{ 1 } \) = \(\frac { 13 }{ 4 } \)

Thus, \(\frac { 14 }{ 4 } \) minutes or Neha took more time and geeta took less time.

∴ Geeta took less time = \(\frac { 14 }{ 4 } \) – \(\frac { 13 }{ 4 } \)

= \(\frac { 14-13 }{ 4 } \) = \(\frac { 1 }{ 4 } \) min.

Question 7.

Complete the following addition (RBSESolutions.com) and subtraction table

Solution.

We hope the RBSE Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.5 will help you. If you have any query regarding Rajasthan Board RBSE Class 6 Maths Chapter 5 Fractions Exercise 5.5, drop a comment below and we will get back to you at the earliest.

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