# RD Sharma Solutions Class 7 Fractions Exercise 2.3

## RD Sharma Solutions Class 7 Chapter 2 Exercise 2.3

### RD Sharma Class 7 Solutions Chapter 2 Ex 2.3 PDF Free Download

#### Exercise 2.3

Q1. Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers

(i) $\frac{3}{7}$

(ii) $\frac{5}{8}$

(iii) $\frac{9}{7}$

(iv) $\frac{6}{5}$

(v) $\frac{12}{7}$

(vi) $\frac{1}{8}$

Solution:

(i) $\frac{3}{7}$

$\frac{7}{3}$ = improper number

(ii) $\frac{5}{8}$

$\frac{8}{5}$ = improper number

(iii) $\frac{9}{7}$

$\frac{7}{9}$ = proper number

(iv) $\frac{6}{5}$

$\frac{5}{6}$ = proper number

(v) $\frac{12}{7}$

$\frac{7}{12}$ = proper number

(vi) $\frac{1}{8}$

8 = whole number

Q2. Divide:

(i) $\frac{3}{8} \; by \; \frac{5}{9}$

(ii) $3\frac{1}{4} \; by \; \frac{2}{3}$

(iii) $\frac{7}{8} \; by \; 4\frac{1}{2}$

(iv) $6\frac{1}{4} by 2\frac{3}{5}$

Solution:

(i) $\frac{3}{8} \; by \; \frac{5}{9}$

= $\frac{\frac{3}{8}}{\frac{5}{9}} \\ = \frac{3 \times 9}{8 \times 5} \\ = \frac{27}{40}$

(ii) $3\frac{1}{4} \; by \; \frac{2}{3}$

= $\frac{3\frac{1}{4}}{\frac{2}{3}} \\ = \frac{\frac{13}{4}}{\frac{2}{3}} \\ = \frac{13 \times 3}{4 \times 2} \\ = \frac{39}{8} \\ = 4\frac{7}{8}$

(iii) $\frac{7}{8} \; by \; 4\frac{1}{2}$

= $\frac{\frac{7}{8}}{\frac{9}{2}} \\ = \frac{7 \times 2}{9 \times 8} \\ = \frac{14}{72} \\ = \frac{7}{36}$

(iv) $6\frac{1}{4} by 2\frac{3}{5}$

= $\frac{6\frac{1}{4}}{2\frac{3}{5}} \\ = \frac{\frac{25}{4}}{\frac{13}{5}} \\ = \frac{25 \times 5}{4 \times 13} \\ = \frac{75}{52} \\ = 2\frac{21}{52}$

Q3. Divide:

(i) $\frac{3}{8}$ by 4

(ii) $\frac{9}{16}$ by 6

(iii) 9 by $\frac{3}{16}$

(iv) 10 by $\frac{100}{3}$

Solution:

(i) $\frac{3}{8}$ by 4

= $\frac{\frac{3}{8}}{4} \\ = \frac{3}{8 \times 4} \\ = \frac{3}{32}$

(ii) $\frac{9}{16}$ by 6

= $\frac{\frac{9}{16}}{6} \\ = \frac{9}{16 \times 6} \\ = \frac{9}{96} \\ = \frac{3}{32}$

(iii) 9 by $\frac{3}{16}$

= $\frac{9}{\frac{3}{16}} \\ = \frac{9 \times 16}{3} \\ = 3 \times 16 \\ = 48$

(iv) 10 by $\frac{100}{3}$

= $\frac{10}{\frac{100}{3}} \\ = \frac{10 \times 3}{100} \\ = \frac{3}{10}$

Q4. Simplify:

(i) $\frac{3}{10} \div \frac{10}{3}$

(ii) $4\frac{3}{5} \div \frac{4}{5}$

(iii) $5\frac{4}{7} \div 1\frac{3}{10}$

(iv) $4 \div 2\frac{2}{5}\\$

Solution:

(i) $\frac{3}{10} \div \frac{10}{3} \\ = \frac{3 \times 3}{10 \times 10} \\ = \frac{9}{100}$

(ii) $4\frac{3}{5} \div \frac{4}{5} \\ = \frac{23}{5} \div \frac{4}{5} \\ = \frac{23 \times 5}{5 \times 4} \\ = \frac{23}{4} \\ = 5\frac{3}{4}$

(iii) $5\frac{4}{7} \div 1\frac{3}{10} \\ = \frac{39}{7} \div \frac{13}{10} \\ = \frac{39 \times 10}{7 \times 13} \\ = \frac{390}{91} \\ = 4\frac{2}{7}$

(iv) $4 \div 2\frac{2}{5}\\ = 4 \div \frac{12}{5} \\ = \frac{4}{\frac{12}{5}} \\ = \frac{20}{12} \\ = 1\frac{2}{3}$

Q5.  A wire of length $12\frac{1}{2}$ m is cut into 10 pieces of equal length . Find the length of each piece.

Solution:

Given, $12\frac{1}{2} m = \frac{25}{2} m$

10 pieces of wire = $\frac{25}{2} m$

1 piece of wire = $\frac{ \frac{25}{2}}{10} \\ = \frac{25}{20} \\ = \frac{5}{4} \\ = 1\frac{1}{4}$

Q6. The length of a rectangular plot of area $65\frac{1}{3} m^{2} \; is \; 12\frac{1}{4} m$. What is the width of the plot?

Solution:

Given,

The length of a rectangular plot of area $65\frac{1}{3} m^{2} \; is \; 12\frac{1}{4} m$.

Area = $65\frac{1}{3} \; m^{2} = \frac{196}{3} \; m^{2}$

Length = $12\frac{1}{4}$ m

Now, Area = length x breadth

• $\frac{196}{3} \; m^{2} = \frac{49}{4} \; m \times breadth$

$Breadth = \frac{4}{49} \; m \times \frac{196}{3} \; m^{2} \\ Breadth = \frac{196\times 4}{49 \times 3} \\ Breadth = \frac{184}{147} \\ Breadth = 5\frac{3}{4}$

Q7. By what number $6\frac{2}{9}$ be multiplied to get                                    $4\frac{4}{9}$?

Solution:

Given,

$6\frac{2}{9} = \frac{56}{9}, \\ And, 4\frac{4}{9} =\frac{40}{9}$

Let x be the number which needs to be multiplied by $\frac{56}{9},$

Now,

$x \times \frac{56}{9} = \frac{40}{9}$

$x = \frac{40}{9} \times \frac{9}{56} \\ x = \frac{40}{56} = \frac{5}{7}$

Q8. The product of two numbers is $25\frac{5}{6}$. If one of the numbers is $6\frac{2}{3}$, find the other?

Solution:

Given,

The product of two numbers is $25\frac{5}{6}$. If one of the numbers is $6\frac{2}{3}$

$6\frac{2}{3} = \frac{20}{3} \\ And, 25\frac{5}{6} = \frac{155}{6}$

Let the other number be x.

$\frac{20}{3} \times x = \frac{155}{6} \\ x = \frac{3}{20} \times \frac{155}{6} \\ x = \frac{3 \times 155}{20 \times 6} \\ x = \frac{31}{8} = 3\frac{7}{8}$

Q9. The cost of $6\frac{1}{4}$ kg of apples is Rs 400. At what rate per kg are the apples being sold?

Solution:

Given,

The cost of $6\frac{1}{4}$ kg of apples is Rs 400

$6\frac{1}{4} = \frac{25}{4} \\$

Cost of $\frac{25}{4}$ kg of apple = Rs 400

Cost of 1 kg of apple = Rs $\frac{4}{25} \times 400$ = Rs 64

Q10. By selling oranges at the rate of Rs $5\frac{1}{4}$ per orange, a fruit seller get Rs 630. How many dozens of oranges does he sell?

Solution:

Given,

Oranges at the rate of Rs $5\frac{1}{4}$ per orange, a fruit seller get Rs 630

$5\frac{1}{4} = \frac{21}{4}$

Number of oranges for Rs $\frac{21}{4}$ = 1

Number of oranges for Re 1 = $\frac{4}{21}$

Number of oranges for Rs 630 = $\frac{4}{21} \times 630$ = 120 apples

12 apples = 1 dozen

Therefore, 120 apples = 10 dozen

Q11. In mid-day meal scheme $\frac{3}{10}$ litre of milk is given to each student of a primary school. If 30 litres of milk is distributed everyday in the school, how many students are there in the school?

Solution:

Given,

$\frac{3}{10}$ litre of milk is given to each student of a primary school.

30 litres of milk is distributed everyday in the school

Number of students given $\frac{3}{10}$ litres of milk = 1

Number of students given 1 litre of milk = $\frac{10}{3}$

Number of students given 30 litres of milk = $\frac{10}{3} \times 30$ = 100 Students

Q12. In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs $50\frac{3}{4}$, how many tickets were sold?

Solution:

Given,

Rs 6496 were collected by selling some tickets.

The price of each ticket was Rs $50\frac{3}{4}$

$50\frac{3}{4} = \frac{203}{4} \\$

Number of tickets bought at Rs $\frac{203}{4}$ = 1

Number of tickets bought at Re 1 = $\frac{4}{203}$

Number of tickets bought at Rs 6496 = $\frac{4}{203} \times 6496$ = $4 \times 32$ = 128

#### Practise This Question

If S1,S2,S3Sp are the sums of infinite geometric series whose first terms are 1,2,3,... p and whose common
ratios are 12,13,14,1p+1 respectively,
Then S1+S2+S3++Sp=