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 a=^i+^j+2^k, b=2^i^j^k then c=2^i+^j+3^k, then the angle between 2a-c and a+b is: