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.
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