RD Sharma Solutions Class 7 Fractions Exercise 2.2

RD Sharma Solutions Class 7 Chapter 2 Exercise 2.2

RD Sharma Class 7 Solutions Chapter 2 Ex 2.2 PDF Download

Exercise 2.2

Q1. Multiply

\(\frac{7}{11}\; by \; \frac{3}{5}\)

\(\frac{3}{5} \; by \; 25\)

\(3\frac{4}{15}\) by 24

\(3\frac{1}{8} \; by \; 4\frac{10}{11}\)

Solution:

We have, \(\frac{7}{11}\; by \; \frac{3}{5}\)

= \(\frac{7}{11}\; \times \; \frac{3}{5} \\ = \frac{21}{55}\)

(ii) We have, \(\frac{3}{5} \; by \; 25\)

= \(\frac{3}{5} \; \times \; 25 \\ = 15\)

(iii) We have, \(3\frac{4}{15}\) by 24

= \(3\frac{4}{15} \times 24 \\ = \frac{49}{15} \times 24 \\ = \frac{1176}{24} \\ = 78\frac{2}{5}\)

(iv) We have, \(3\frac{1}{8} \; by \; 4\frac{10}{11}\)

= \(3\frac{1}{8} \; by \; 4\frac{10}{11}\\ = \frac{25}{8} \times \frac{54}{11} \\ = \frac{25 \times 54}{88} \\ = 15\frac{15}{44}\)

 

Q2.  Find the product:

\(\frac{4}{7} \times \frac{14}{25} \\\)

\(7\frac{1}{2} \times 2\frac{4}{15}\)

\(3\frac{6}{7} \times 4\frac{2}{3}\)

\(6\frac{11}{14} \times 3\frac{1}{2}\)

Solution:

We have, \(\frac{4}{7} \times \frac{14}{25} \\ = \frac{4 \times 14}{7 \times 25} \\ = \frac{56}{175} \\ = \frac{8}{25}\)

We have, \(7\frac{1}{2} \times 2\frac{4}{15} \\ = \frac{15}{2} \times \frac{34}{15} \\ = \frac{15 \times 34}{2 \times 15} \\ = \frac{510}{30} \\ = 17\)

We have, \(3\frac{6}{7} \times 4\frac{2}{3} \\ = \frac{27}{7} \times \frac{14}{3} \\ = 3 \times \frac{14}{3} \\ = 14\)

We have, \(6\frac{11}{14} \times 3\frac{1}{2} \\ = \frac{95}{14} \times \frac{7}{2} \\ = \frac{95 \times 7}{28} \\ = \frac{665}{28} \\ = 23\frac{3}{4}\)

 

Q3. Simplify:

\(\frac{12}{25} \times \frac{15}{28} \times \frac{35}{36}\)

\(\frac{10}{27} \times \frac{39}{56} \times \frac{28}{65}\)

\(2\frac{2}{17} \times 7\frac{2}{9} \times 1\frac{33}{52}\)

Solution:

We have, \(\frac{12}{25} \times \frac{15}{28} \times \frac{35}{36} \\ = \frac{12 \times 15 \times 35}{25 \times 28 \times 36} \\ = \frac{6300}{25200} \\ = \frac{1}{4}\)

\(\frac{10}{27} \times \frac{39}{56} \times \frac{28}{65} \\ = \frac{10 \times 39 \times 28}{27 \times 56 \times 65} \\ = \frac{10920}{98280} \\ = \frac{1}{9}\)

We have, \(2\frac{2}{17} \times 7\frac{2}{9} \times 1\frac{33}{52} \\ = \frac{36}{17} \times \frac{65}{9} \times \frac{85}{52} \\ = \frac{36 \times 65 \times 85}{17 \times 9 \times 52} \\ = \frac{198900}{7956} \\ = 25\)

Q4. Find:

\(\frac{1}{2} of 4\frac{2}{9}\)

\(\frac{5}{8} \; of \; 9\frac{2}{3}\)

\(\frac{2}{3} \; of \frac{9}{16}\)

Solution:

We have, \(\frac{1}{2} of 4\frac{2}{9} \\ = \frac{1}{2} \times \frac{38}{9} \\ = \frac{38}{18}\\ = 2\frac{1}{9}\)

\(\frac{5}{8} of 9\frac{2}{3} \\ = \frac{5}{8} \times \frac{29}{3} \\ = \frac{5 \times 29}{8 \times 3} \\ = \frac{145}{24} \\ = 6\frac{1}{24}\)

We have, \(\frac{2}{3} \; of \frac{9}{16} \\ =\frac{2}{3} \times \frac{9}{16}\\=\frac{2 \times 9}{3 \times 16} \\ = \frac{18}{48} \\ = \frac{3}{8}\)

Q5. Which is greater ? \(\frac{1}{2} \; of \; \frac{6}{7}\) or \(\frac{2}{3} \; of \; \frac{3}{7}\).

Solution:

Given, \(\frac{1}{2} \; of \; \frac{6}{7}\; or \; \frac{2}{3} \; of \; \frac{3}{7} \\ = \frac{1}{2} \times \frac{6}{7} \; or \; \frac{2}{3} \times \frac{3}{7} \\ = \frac{1 \times 6}{2 \times 7} \times \frac{2 \times 3}{3 \times 7} \\ = \frac{6}{14} \; or \; \frac{6}{21}\)

While comparing two fractions, when the numerators of both the fractions are same, then the denominator having higher value shows the fraction has lower value.

So, \(\frac{6}{14}\) is greater.

Therefore, \(\frac{1}{2} \; of \; \frac{6}{7}\) is greater.

Q6. Find,

\(\frac{7}{11} \; of \; 330\)

\(\frac{5}{9} \; of \; 108 \; meters\)

\(\frac{3}{7} \; of \; 42 \; litres\)

\(\frac{1}{12}\) of an hour

\(\frac{5}{6}\) of an year

\(\frac{3}{20}\) of a Kg

\(\frac{7}{20}\) of a litres

\(\frac{5}{6}\) of a day

\(\frac{2}{7}\) of a week

Solution:

We have, \(\frac{7}{11} \; of \; 330 \\ = \frac{7}{11} \times 330 \\ = 7 \times 30 \\ = 210\)

We have, \(\frac{5}{9} \; of \; 108 \; meters \\ = \frac{5}{9} \times 108 \; meters \\ = 5 \times 12 \; meters \\ = 60 \; meters\)

We have, \(\frac{3}{7} \; of \; 42 \; litres \\ = \frac{3}{7} \times 42 \; litres \\ = 3 \times 6 \; litres \\ = 18 \; litres\)

We have, \(\frac{1}{12}\) of an hour

An hour = 60 minutes

Therefore, \(\frac{1}{12} \times 6o \; minutes \\ = 5 \; minutes\)

(v) We have, \(\frac{5}{6}\) of an year

I Year = 12 months

Therefore, \(\frac{5}{6} \times 12 \; months \\ = 5 \times 2 \; months \\ = 10 \; months\)

(vi) We have, \(\frac{3}{20}\)

1 Kg = 1000 gms

Therefore, \(\frac{3}{20} \times 1000 \; gms \\ = 3 \times 50 \; gms \\ = 150 \; gms\)

(vii) We have,  \(\frac{7}{20}\) of a litre

1 litre = 1000 ml

Therefore, \(\frac{7}{20} \times 1000 \; ml \\ = 7 \times 50 \; ml \\ = 350 \; ml\)

(viii) We have, \(\frac{5}{6}\) of a day

I day = 24 hours

Therefore, \(\frac{5}{6} \times 24 \; hours \\ = 5 \times 4 \; hours \\ = 20 \; hours\)

(ix) We have, \(\frac{2}{7}\) of a week

I week = 7 days

Therefore, \(\frac{2}{7} \times 7 \; days \\ = 2 \; days\)

 

Q7. Shikha plans 5 saplings in a row in her garden. The distance between two adjacent saplings is \(\frac{3}{4}\) m. Find the distance between first and last sapling.

Solution:

There are 4 adjacent spacing for 5 saplings.

Given, the distance between two adjacent saplings is \(\frac{3}{4}\) m.

4 adjacent spacing for 5 saplings = \(\frac{3}{4} \times 4\) = 3 m

Therefore, the distance between first and last sapling is 3 m.

Q8. Ravish reads \(\frac{1}{3}\) part of a book in one hour. How much part of the book will he read in \(2\frac{1}{5}\) hours?

Solution:

Let x be the full part of book.

Given, Ravish reads \(\frac{1}{3}\) part of a book in one hour

1 hour = \(\frac{1}{3}\) x

Part of the book will he read in \(2\frac{1}{5}\) hours

\(2\frac{1}{5} = \frac{11}{5}\) hours = \(\frac{1}{3} \times x \times \frac{11}{5}\)

\(\frac{11}{15}\) x = \(\frac{11}{15}\) part of book

 

Q9. Lipika reads a book for \(1\frac{3}{4}\) hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Solution:

Given,

Time taken by Lipika to read a book per day = \(1\frac{3}{4} = \frac{7}{4}\) hours

Time taken by Lipika to read a book for 6 days  = \(\frac{7}{4} \times 6\) = \(\frac{42}{4} = 10\frac{1}{2}\) hours.

 

Q10. Find the area of a rectangular park which is \(41\frac{2}{3} \; m\) long and \(18\frac{3}{5} \; m\) broad.

Solution:

Given, \(41\frac{2}{3} \; m = \frac{145}{3} \; m \\ And, \; 18\frac{3}{5} \; m = \frac{93}{5} \; m \\\)

Area of a rectangular park = (length x breadth) = \((\frac{125}{3} \; m \times \frac{93}{5} \; m) \\ = (\frac{125 \times 93}{15}) m^{2} \\ = (\frac{11625}{15}) m^{2} \\ = 775 m^{2}\)

 

Q11. If milk is available at Rs \(17\frac{3}{4}\) per litre, find the cost of \(7\frac{2}{5}\) litres of milk.

Solution:

Given,

\(Rs 17\frac{3}{4} = Rs \frac{71}{4} \\ And, 7\frac{2}{5} \; litres = \frac{37}{5} \; litres \\\)

The cost of milk per litre = \(Rs \frac{71}{4}\)

The cost of milk per \(\frac{37}{5} \; litres\) = Rs \(\frac{37}{5} \times \frac{71}{4} \\ = Rs \frac{2327}{20} \\ = Rs 131\frac{7}{20}\)

 

Q12. Sharda can walk \(8\frac{1}{3}\) km in one hour. How much distance will she cover in \(2\frac{2}{5}\) hours.

Solution:

Given,

\(8\frac{1}{3} km = \frac{25}{3} km \\ 2\frac{2}{5} hours = \frac{12}{5} hours\)

Distance covered by Sharda in one hour = \(\frac{25}{3} km\)

Distance covered by Sharda in \(\frac{12}{5} hours\) = \(2\frac{2}{5} \times \frac{25}{3}\) = 20 km

 

Q13. A sugar bag contains 30 kg of sugar. After consuming \(\frac{2}{3}\) of it, how much sugar is left in the bag

Solution:

Given, A sugar bag contains 30 kg of sugar.

After consuming \(\frac{2}{3}\) of it, the amount of sugar left in the bag = \(30 kg – \frac{2}{3} \times 30 kg \\ = 30 kg – 20 kg \\ = 10 kg\)

 

Q14. Each side of a square is \(6\frac{2}{3}\) m long. Find its area.

Solution:

Given,

Each side = \(6\frac{2}{3} m = \frac{20}{3} m\)

Area = side2 = \((\frac{20}{3})^{2} m^{2}\) = \(\frac{400}{9} m^{2} \\ = 44\frac{4}{9} m^{2}\)

 

Q15. There are 45 students in a class and \(\frac{3}{5}\) of them are boys. How many girls are there in the class?

Solution:

Given,

There are 45 students in a class,

And \(\frac{3}{5}\) of them are boys.

Therefore, no of girls in the class = 45 – \(\frac{3}{5}\times 45\)

= 45 – 27

= 18