RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Fractions and Decimal Numbers |

Exercise |
Additional Questions |

Number of Questions |
30 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers Additional Questions

**Multiple Choice Questions**

Question 1

Find equivalent (RBSESolutions.com) fraction of \(\frac { 3 }{ 7 }\).

Question 2

Simplest form of \(\frac { 20 }{ 25 }\) will be :

Question 3

Value of \(\frac { 20 }{ 25 }\) ÷ \(\frac { 20 }{ 25 }\) will be :

(A) 1

(B) 3

(C) 2

(D) 4

Question 4

Value of 4.105 x 4 is :

(A) 16.420

(B) 15.420

(C) 14.413

(D) 18.121

Question 5

Solution (RBSESolutions.com) of 6.25 ÷ 5 is :

(A) 2.25

(B) 1.25

(C) 0.25

(D) 1.08

Question 6

Value of \(\frac { 16 }{ 3 }\) x 9 is :

(A) 50

(B) 49

(C) 48

(D) 47

Question 7

\(\frac { 1 }{ 10 }\) th part (RBSESolutions.com) of 80 is :

(A) 18

(B) 19

(C) 17

(D) 18

Question 8

Value of \(\frac { 4 }{ 3 }\) x ( reciprocal of\(\frac { 6 }{ 8 }\)) will be :

Question 9

Value of 7.9 ÷ 1000 will be :

(A) 3.0079

(B) 0.079

(C) 0.79

(D) 7.9

Question 10

Value of 2\(\frac { 2 }{ 3 }\) ÷ 2 will be :

Answers:

1. (A), 2. (B), 3. (C), 4. (A), 5. (B), 6. (C), 7. (A), 8. (B), 9. (A), 10. (D)

**Fill in the blanks:**

(i) 2.7 x 4 = ……..

(ii) 25.5 ÷ 3 = ……

(iii) 1\(\frac { 1 }{ 4 }\) ÷ 3\(\frac { 1 }{ 4 }\) = ….

(iv) 200 gm = …….

Solutions:

(i) 10.8

(ii) 8.5

(iii) \(\frac { 5 }{ 13 }\)

(iv) 0.2 gm

**True/False :**

(i) 37.8 ÷ 1.4 = 27

(ii) 272.23 ÷ 10 = 272.33

(iii) 2 x 0.86 = 1.72

(iv) In 2.03 and 2.30, 2.30 is greater.

Answers:

(i) True,

(ii) false,

(iii) True,

(iv) True

**Very Short Questions**

Question 1

Find the (RBSESolutions.com) average of 4.2, 4.8 and 6.6.

Solution:

Question 2

Write the following extended form in decimal fraction

Solution:

Question 3

(i) Express 5 cm (RBSESolutions.com) in m and km

(ii) Express 35 mm in cm, m and km.

Solution:

Question 4

Express the (RBSESolutions.com) following in kg:

(i) 200 g

(ii) 3470 g

Solution:

**Short Answer Type Questions**

Question 1

Find:

Solution:

Question 2

Vidya and Pratap went on picnic. Their (RBSESolutions.com) mother gave a bottle of water of 5l. Vidya used \(\frac { 2 }{ 5 }\) part of water of total water.

Remaining water used by Pratap. Tell

(i) How much water vidya drunk ?

(ii) How much fraction of total Quantity of water Pratap drunk?

Solution:

Total quantity of water = 5l

∴ Quantity of water used by vidya

= \(\frac { 2 }{ 5 }\) of 5l

= (\(\frac { 2 }{ 5 }\) x 5) = 2l

(ii) fraction of (RBSESolutions.com) water used by Pratap

= 1 – \(\frac { 2 }{ 5 }\) = \(\frac { 5 }{ 5 }\) – \(\frac { 2 }{ 5 }\) = \(\frac { 5-2 }{ 5 }\) = \(\frac { 3 }{ 5 }\)

Question 3

Dinesh went from place A to place B then went to C. Distance between A and B is 7.5 km and from B to C is 12.7 km. Ayub went from A to D then went to C. A Distance of A from D is 9.3 km and from D to C is 11.8 km. who covered the greater distance and how much more that distance?

Solution:

Distance (RBSESolutions.com) covered by Dinesh = AB + BC

= 7.5 + 12.7 = 20.2 km

Distance covered by Ayub = AD + DC

= 9.3 + 11.8= 21.1 km

Clearly, 21.1 > 20.2

∴ Ayub travelled greater than Dinesh and he covered 21.1 km – 20.2 km = 0.9 km more.

Question 4

Shyama purchased 5 kg 300 gm (RBSESolutions.com) apples and 3 kg 250 gm mango. Sarla purchased 4 kg 800 gm oranges and 4 kg 150 gm banana. Who purchased more fruits?

Solution:

Shyama purchased total fruits = 5 kg 300 g + 3 kg 250 g

= 5.300 kg + 3.250 kg = 8.550 kg

Sarla purchased total fruits = 4 kg 800 g + 4 kg 150 g

= 4.800 kg + 4.150 kg = 8.950 kg

So, 8.950 > 8.550

∴ Sarla purchased more fruits.

Question 5

How much (RBSESolutions.com) less is 28 km, from 42.6 km?

Solution:

Difference = 42.6 km – 28 km = 16.6 km

**Long Answer Type Questions**

Question 1

Saile plants 4 small sapplings in a row in her garden. Distance between two consecutive sapplings is \(\frac { 3 }{ 4 }\) m. Find the distance betwen first and the last sapplings.

Solution:

Let 4 sapplings are planted in row like A, B, C and D.

so that AB = BC = CD = \(\frac { 3 }{ 4 }\) m

∴ Distance between 1st and last sapplings

Question 2

Lipika reads a book (RBSESolutions.com) daily for 1\(\frac { 3 }{ 4 }\) hrs. She reads complete book in 6 days. How many hours she took to read the complet book?

Solution:

Lipika takes = 1\(\frac { 3 }{ 4 }\) hrs in a day to read

She took total 6 days

∴ Total hours she took = 6 x 1\(\frac { 3 }{ 4 }\) hours

= 6 x \(\frac { 7 }{ 4 }\) = \(\frac { 42 }{ 4 }\)

= \(\frac { 21 }{ 2 }\) = 10\(\frac { 1 }{ 2 }\) hours.

Question 3

A car runs 16 km in 1 litre petrol. How much (RBSESolutions.com) distance will it cover in 2\(\frac { 3 }{ 4 }\) litre petrol?

Solution:

∵ Distance covers in 1 litre petrol = 16 km

∴ Distance covered in 2 \(\frac { 3 }{ 4 }\) litre petrol.

We hope the RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimal Numbers Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 2 Fractions and Decimal Numbers Additional Questions, drop a comment below and we will get back to you at the earliest.

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