RD Sharma solution class 7 chapter 2 Fractions Exercise 2.1

Exercise 2.1

Page-2.12

Question 1:

Compare the following fractions by using the symbol > or < or =:
(i) $\frac{7}{9} \mathrm{and} \frac{8}{13}$
(ii) $\frac{11}{9} \mathrm{and} \frac{5}{9}$
(iii) $\frac{37}{41} \mathrm{and} \frac{19}{30}$
(iv) $\frac{17}{15} \mathrm{and} \frac{119}{105}$

Answer 1:

First, we need to find the LCM of denominators in each case. After that, we will equate the denominators in order to compare the two fractions. 

(i) 
     LCM of 9 and 13 is 117.

     $$\begin{gathered} \mathrm{Now} \mathrm{make} \mathrm{both} \mathrm{fraction} \mathrm{equivalent} \mathrm{with} \mathrm{denominator} \mathrm{as} 117 \\ \frac{7}{9}=\frac{7}{9}\times \frac{13}{13} \\ \Rightarrow \frac{7}{9}=\frac{91}{117} \\ \frac{8}{13}=\frac{8}{13}\times \frac{9}{9} \\ \Rightarrow \frac{8}{13}=\frac{72}{117} \\ \mathrm{we} \mathrm{know} \\ 91>72 \\ \Rightarrow \frac{91}{117}>\frac{72}{117} \\ \Rightarrow \frac{7}{9}>\frac{8}{13} \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{both} \mathrm{fraction} \mathrm{have} \mathrm{same} \mathrm{denominator} \mathrm{as} 9 \\ \mathrm{we} \mathrm{know} \\ 11>5 \\ \Rightarrow \frac{11}{9}>\frac{5}{9} \end{gathered}$$
(iii)

$$\begin{gathered} \mathrm{LCM} \mathrm{of} 41 \mathrm{and} 30 \mathrm{is} 1230 \\ \mathrm{Now} \mathrm{convert} \mathrm{both} \mathrm{fraction} \mathrm{to} \mathrm{their} \mathrm{equivalent} \mathrm{fractions} \mathrm{with} \mathrm{denominator} \mathrm{as} 1230 \\ \frac{37}{41}=\frac{37}{41}\times \frac{30}{30} \\ \Rightarrow \frac{37}{41}=\frac{1110}{1230} \\ \frac{19}{30}=\frac{19}{30}\times \frac{41}{41} \\ \Rightarrow \frac{19}{30}=\frac{779}{1230} \\ \mathrm{we} \mathrm{know} \\ 1110>779 \\ \Rightarrow \frac{1110}{1230}>\frac{779}{1230} \\ \Rightarrow \frac{37}{41}>\frac{19}{30} \end{gathered}$$

(iv)

LCM of 15 and 105 is 105.

$$\begin{gathered} \mathrm{Now} \mathrm{convert} \mathrm{fraction} \mathrm{to} \mathrm{its} \mathrm{equivalent} \mathrm{fractions} \mathrm{with} \mathrm{denominator} \mathrm{as} 105 \\ \frac{17}{15}=\frac{17}{15}\times \frac{7}{7} \\ \Rightarrow \frac{17}{15}=\frac{119}{105} \end{gathered}$$

Question 2:

Arrange the following fractions in ascending order:
(i) $\frac{3}{8},\frac{5}{6},\frac{6}{8}\frac{2}{4},\frac{1}{3}$
(ii) $\frac{4}{3},\frac{3}{8},\frac{6}{12},\frac{5}{16}$

Answer 2:

(i)
$$\begin{gathered} \mathrm{LCM} \mathrm{of} \mathrm{the} \mathrm{denominators} 8, 6, 4 \mathrm{and} 3 \mathrm{is} 24. \\ \mathrm{Now}, \mathrm{convert} \mathrm{all} \mathrm{fractions} \mathrm{into} \mathrm{their} \mathrm{equivalent} \mathrm{fractions} \mathrm{with} \mathrm{denominator} 24. \\ \frac{3}{8}=\frac{3}{8}\times \frac{3}{3} \\ \Rightarrow \frac{3}{8}=\frac{9}{24} \\ \frac{5}{6}=\frac{5}{6}\times \frac{4}{4} \\ \Rightarrow \frac{5}{6}=\frac{20}{24} \\ \frac{6}{8}=\frac{6}{8}\times \frac{3}{3} \\ \Rightarrow \frac{6}{8}=\frac{18}{24} \\ \frac{2}{4}=\frac{2}{4}\times \frac{6}{6} \\ \Rightarrow \frac{2}{4}=\frac{12}{24} \\ \frac{1}{3}=\frac{1}{3}\times \frac{8}{8} \\ \Rightarrow \frac{1}{3}=\frac{8}{24} \\ \mathrm{We} \mathrm{know}: \\ 8<9<12<18<20 \\ \Rightarrow \frac{8}{24}<\frac{9}{24}<\frac{12}{24}<\frac{18}{24}<\frac{20}{24} \\ \Rightarrow \frac{1}{3}<\frac{3}{8}<\frac{2}{4}<\frac{6}{8}<\frac{5}{6} \end{gathered}$$
(ii)
$$\begin{gathered} \mathrm{LCM} \mathrm{of} \mathrm{the} \mathrm{denominators} 8, 12, 16 \mathrm{and} 3 \mathrm{is} 48. \\ \mathrm{Now}, \mathrm{convert} \mathrm{all} \mathrm{fractions} \mathrm{into} \mathrm{their} \mathrm{equivalent} \mathrm{fractions} \mathrm{with} \mathrm{denominator} 48. \\ \frac{4}{3}=\frac{4}{3}\times \frac{16}{16} \\ \Rightarrow \frac{4}{3}=\frac{64}{48} \\ \frac{3}{8}=\frac{3}{8}\times \frac{6}{6} \\ \Rightarrow \frac{3}{8}=\frac{18}{48} \\ \frac{6}{12}=\frac{6}{12}\times \frac{4}{4} \\ \Rightarrow \frac{6}{12}=\frac{24}{48} \\ \frac{5}{16}=\frac{5}{16}\times \frac{3}{3} \\ \Rightarrow \frac{5}{16}=\frac{15}{48} \\ \mathrm{we} \mathrm{know} \\ 15<18<24<64 \\ \Rightarrow \frac{15}{48}<\frac{18}{48}<\frac{24}{48}<\frac{64}{48} \\ \Rightarrow \frac{5}{16}<\frac{3}{8}<\frac{6}{12}<\frac{4}{3} \end{gathered}$$

Question 3:

Arrange the following fractions in descending order:
(i) $\frac{4}{5},\frac{7}{10},\frac{11}{15},\frac{17}{20}$
(ii) $\frac{2}{7},\frac{11}{35},\frac{9}{14},\frac{13}{28}$

Answer 3:

(i)
$$\begin{gathered} \mathrm{LCM} \mathrm{of} \mathrm{the} \mathrm{denominators} 5, 10, 15 \mathrm{and} 20 \mathrm{is} 60. \\ \mathrm{Now}, \mathrm{convert} \mathrm{all} \mathrm{fractions} \mathrm{to} \mathrm{their} \mathrm{equivalent} \mathrm{fractions} \mathrm{with} \mathrm{denominator} 60. \\ \frac{4}{5}=\frac{4}{5}\times \frac{12}{12}=\frac{48}{60} \\ \frac{7}{10}=\frac{7}{10}\times \frac{6}{6}=\frac{42}{60} \\ \frac{11}{15}=\frac{11}{15}\times \frac{4}{4}=\frac{44}{60} \\ \frac{17}{20}=\frac{17}{20}\times \frac{3}{3}=\frac{51}{60} \\ \mathrm{We} \mathrm{know}: \\ 51>48>44>42 \\ \Rightarrow \frac{51}{60}>\frac{48}{60}>\frac{44}{60}>\frac{42}{60} \\ \Rightarrow \frac{17}{20}>\frac{4}{5}>\frac{11}{15}>\frac{7}{10} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{LCM} \mathrm{of} \mathrm{the} \mathrm{denominators} 7, 35, 14 \mathrm{and} 28 \mathrm{is} 140. \\ \mathrm{Now}, \mathrm{convert} \mathrm{all} \mathrm{fractions} \mathrm{to} \mathrm{their} \mathrm{equivalent} \mathrm{fractions} \mathrm{with} \mathrm{denominator} 140. \\ \frac{2}{7}=\frac{2}{7}\times \frac{20}{20}=\frac{40}{140} \\ \frac{11}{35}=\frac{11}{35}\times \frac{4}{4}=\frac{44}{140} \\ \frac{9}{14}=\frac{9}{14}\times \frac{10}{10}=\frac{90}{140} \\ \frac{13}{28}=\frac{13}{28}\times \frac{5}{5}=\frac{65}{140} \\ \mathrm{We} \mathrm{know}: \\ 90>65>44>40 \\ \Rightarrow \frac{90}{140}>\frac{65}{140}>\frac{44}{140}>\frac{40}{140} \\ \Rightarrow \frac{9}{14}>\frac{13}{28}>\frac{11}{35}>\frac{2}{7} \end{gathered}$$

Question 4:

Write five equivalent fractions of $\frac{3}{5}$.

Answer 4:

Five equivalent fractions of $\frac{3}{5}$ are:

$$\begin{gathered} \text{(i)} \frac{3}{5}=\frac{3}{5}\times \frac{2}{2} \\ \Rightarrow \frac{3}{5}=\frac{6}{10} \\ \text{(ii)} \frac{3}{5}=\frac{3}{5}\times \frac{3}{3} \\ \Rightarrow \frac{3}{5}=\frac{9}{15} \\ \text{(iii)} \frac{3}{5}=\frac{3}{5}\times \frac{4}{4} \\ \Rightarrow \frac{3}{5}=\frac{12}{20} \\ \text{(iv)} \frac{3}{5}=\frac{3}{5}\times \frac{5}{5} \\ \Rightarrow \frac{3}{5}=\frac{15}{25} \\ \text{(v)} \frac{3}{5}=\frac{3}{5}\times \frac{6}{6} \\ \Rightarrow \frac{3}{5}=\frac{18}{30} \end{gathered}$$

Question 5:

Find the sum:
(i) $\frac{5}{8}+\frac{3}{10}$
(ii) $4\frac{3}{4}+9\frac{2}{5}$
(iii) $\frac{5}{6}+3+\frac{3}{4}$
(iv) $2\frac{3}{5}+4\frac{7}{10}+2\frac{4}{15}$

Answer 5:

(i)

$$\begin{gathered} \frac{5}{8}+\frac{3}{10} \\ LCM of 8,10 is 40. \\ \Rightarrow \frac{(5\times 5)+(3\times 4)}{40}=\frac{37}{40} \end{gathered}$$
(ii)

$$\begin{gathered} 4\frac{3}{4}+9\frac{2}{5} or \frac{(4\times 4)+3}{4}+\frac{(9\times 5)+2}{5} \\ \Rightarrow \frac{19}{4}+\frac{47}{5} \\ LCM of 5,4 is 20. \\ \frac{(19\times 5)+(47\times 4)}{20}=\frac{283}{20} \end{gathered}$$
(iii)

$$\begin{gathered} \frac{5}{6}+3+\frac{3}{4} \\ LCM of 6,4 is 24. \\ \frac{(5\times 4)+(3\times 24)+(3\times 6)}{24}=\frac{110}{24} \end{gathered}$$
(iv)

$$\begin{gathered} 2\frac{3}{5}+4\frac{7}{10}+2\frac{4}{15} or \frac{(2\times 5)+3}{5}+\frac{(4\times 10)+7}{10}+\frac{(2\times 15)+4}{15} \\ \Rightarrow \frac{13}{5}+\frac{47}{10}+\frac{34}{15} \\ LCM of 15,10 and 5 is 30. \\ \frac{(13\times 6)+(47\times 3)+(34\times 2)}{30}=\frac{287}{30} \end{gathered}$$

Question 6:

Find the difference of
(i) $\frac{13}{24} \mathrm{and} \frac{7}{16}$
(ii) $6 \mathrm{and} \frac{23}{3}$
(iii) $\frac{21}{25} \mathrm{and} \frac{18}{20}$
(iv) $3\frac{3}{10} \mathrm{and} 2\frac{7}{15}$

Answer 6:

(i)

$$\begin{gathered} \frac{13}{24}-\frac{7}{16} \\ LCM of 24 and 16 is 48. \\ \frac{(13\times 2)-(7\times 3)}{48}=\frac{5}{48} \end{gathered}$$

(ii)

$$\begin{gathered} \frac{23}{3}-6 \\ LCM of 3 and 1 is 3. \\ \frac{(23\times 1)-(6\times 3)}{3}=\frac{5}{3} \end{gathered}$$

(iii)

$$\begin{gathered} \frac{18}{20}-\frac{21}{25} \\ LCM of 20,25 is 100. \\ \Rightarrow \frac{(18\times 5)-(21\times 4)}{100}=\frac{6}{100}=\frac{3}{50} \end{gathered}$$

(iv)

$$\begin{gathered} 3\frac{3}{10}-2\frac{7}{15}\iff \frac{(3\times 10)+3}{10}-\frac{(2\times 15)+7}{15}\iff \frac{33}{10}-\frac{37}{15} \\ LCM of 10 and 15 is 30. \\ \frac{(33\times 3)-(37\times 2)}{30}=\frac{25}{30}\iff \frac{5}{6} \end{gathered}$$

Page-2.13

Question 7:

Find the difference:
(i) $\frac{6}{7}-\frac{9}{11}$
(ii) $8-\frac{5}{9}$
(iii) $9-5\frac{2}{3}$
(iv) $4\frac{3}{10}-1\frac{2}{15}$

Answer 7:

(i)
$$\begin{gathered} \frac{6}{7}-\frac{9}{11} \\ LCM of 7 and 11 is 77. \\ \Rightarrow \frac{6}{7}-\frac{9}{11}\iff \frac{(6\times 11)-(9\times 7)}{77}\iff \frac{3}{77} \end{gathered}$$

(ii)
$$\begin{gathered} 8-\frac{5}{9} \\ LCM of 1 and 9 is 9. \\ \Rightarrow \frac{8}{1}-\frac{5}{9}\iff \frac{(8\times 9)-(5\times 1)}{9}\iff \frac{67}{9} \end{gathered}$$

(iii)
$$\begin{gathered} 9-5\frac{2}{3}\iff \frac{9}{1}-\frac{(5\times 3)+2}{3} \\ LCM of 1 and 3 is 3. \\ \Rightarrow \frac{9}{1}-\frac{17}{3}\iff \frac{(9\times 3)-(17\times 1)}{3}\iff \frac{10}{3} \end{gathered}$$

(iv)
$$\begin{gathered} 4\frac{3}{10}-1\frac{2}{15}\iff \frac{(4\times 10)+3}{10}-\frac{(15\times 1)+2}{15} \\ LCM of 10 and 15 is 30. \\ \Rightarrow \frac{43}{10}-\frac{17}{15}\iff \frac{(43\times 3)-(17\times 2)}{30}\iff \frac{95}{30}\iff \frac{19}{6} \end{gathered}$$

Question 8:

Simplify:
(i) $\frac{2}{3}+\frac{1}{6}-\frac{2}{9}$
(ii) $12-3\frac{1}{2}$
(iii) $7\frac{5}{6}-4\frac{3}{8}+2\frac{7}{12}$

Answer 8:

(i)

$$\begin{gathered} \frac{2}{3}+\frac{1}{6}-\frac{2}{9} \\ \mathrm{LCM} \mathrm{of} 3.6 \mathrm{and} 9 \mathrm{is} 18. \\ =\frac{2}{3}+\frac{1}{6}-\frac{2}{9} \\ =\frac{(2\times 6)+(1\times 3)-(2\times 2)}{18} \\ =\frac{11}{18} \end{gathered}$$
(ii)

$$\begin{gathered} 12-3\frac{1}{2}=\frac{12}{1}-\frac{(3\times 2)+1}{2} \\ \mathrm{LCM} \mathrm{of} 2 \mathrm{and} 1 \mathrm{is} 2. \\ =\frac{12}{1}-\frac{7}{2} \\ =\frac{(12\times 2)-(7\times 1)}{2} \\ =\frac{17}{2} \end{gathered}$$

(iii)

$$\begin{gathered} 7\frac{5}{6}-4\frac{3}{8}+2\frac{7}{12}=\frac{(7\times 6)+5}{6}-\frac{(4\times 8)+3}{8}+\frac{(2\times 12)+7}{12} \\ \mathrm{LCM} \mathrm{of} 6,8 \mathrm{and} 12 \mathrm{is} 24. \\ =\frac{47}{6}-\frac{35}{8}+\frac{31}{12} \\ =\frac{(47\times 4)-(35\times 3)+(31\times 2)}{24} \\ =\frac{145}{24} \end{gathered}$$

Question 9:

What should be added to $5\frac{3}{7}$ to get 12?

Answer 9:

Let x be the required fraction.

According to the question:

$$\begin{gathered} x+5\frac{3}{7}=12 \\ \Rightarrow x+\frac{(5\times 7)+3}{7}=12 \\ \Rightarrow x=12-\frac{38}{7} \\ \Rightarrow x=\frac{(12\times 7)-(38\times 1)}{7}\iff \frac{46}{7} \end{gathered}$$

Question 10:

What should be added to $5\frac{4}{15}$ to get $12\frac{3}{5}?$

Answer 10:

Let x be the required fraction.

According to the question:

$$\begin{gathered} x+5\frac{4}{15}=12\frac{3}{5} \\ \Rightarrow x+\frac{(15\times 5)+4}{15}=\frac{(12\times 5)+3}{5} \\ \Rightarrow x=\frac{63}{5}-\frac{79}{15} \\ LCM of 5 and 15 is 15. \\ \Rightarrow x=\frac{(63\times 3)-(79\times 1)}{15}\iff \frac{110}{15} \\ \Rightarrow x=\frac{110}{15}\iff \frac{22}{3} \end{gathered}$$

Question 11:

Suman studies for $5\frac{2}{3}$ hours daily. She devotes $2\frac{4}{5}$ hours of her time for Science and Mathematics. How much time does she devote for other subjects?

Answer 11:

Suman studies for  $5\frac{2}{3}\mathrm{hours}$ daily. Therefore, we have
$5\frac{2}{3}\mathrm{hours} =\frac{(5\times 3)+2}{3}=\frac{17}{3}\mathrm{hours}$
She studies science and mathematics for  $2\frac{4}{5}\mathrm{hours}$. Therefore, we have
$2\frac{4}{5}\mathrm{hours}=\frac{(2\times 5)+4}{5}=\frac{14}{5}\mathrm{hours}$
Time devoted to other subjects = Total study time $-$ Time devoted to science and mathematics 

= $$\begin{gathered} \frac{17}{3}-\frac{14}{5}=\frac{(17\times 5)-(14\times 3)}{15} \\ =\frac{43}{15}\mathrm{hours} \end{gathered}$$

Question 12:

A piece of wire is of length $12\frac{3}{4}\mathrm{m}$. If it is cut into two pieces in such a way that the length of one piece is $5\frac{1}{4}\mathrm{m},$ what is the length of the other piece?

Answer 12:

Let the length of second piece be x.

Total length of wire = Length of one piece + Length of second piece

$$\begin{gathered} 12\frac{3}{4}=5\frac{1}{4}+x \\ \Rightarrow \frac{(12\times 4)+3}{4}=\frac{(5\times 4)+1}{4}+x \\ \Rightarrow x=\frac{(12\times 4)+3}{4}-\frac{(5\times 4)+1}{4} \\ \Rightarrow x=\frac{51}{4}-\frac{21}{4}\iff \frac{30}{4} \\ \Rightarrow x=\frac{30}{4}\iff \frac{15}{2} \end{gathered}$$

Question 13:

A rectangular sheet of paper is $12\frac{1}{2} \mathrm{cm}$ long and $10\frac{2}{3} \mathrm{cm}$ wide. Find its perimeter.

Answer 13:

Perimeter of rectangle = 2(length + width)

$$\begin{gathered} 2\times \left[12\frac{1}{2}+10\frac{2}{3}\right] \\ =2\times \left[\frac{(12\times 2)+1}{2}+\frac{(10\times 3)+2}{3}\right] \\ =2\times \left[\frac{(25\times 3)+(32\times 2)}{6}\right] \\ =2\times \left[\frac{139}{6}\right] \\ =\frac{139}{3} \mathrm{cm} \end{gathered}$$

Question 14:

In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic  square?

$\frac{4}{11}$	$\frac{9}{11}$	$\frac{2}{11}$
$\frac{3}{11}$	$\frac{5}{11}$	$\frac{7}{11}$
$\frac{8}{11}$	$\frac{1}{11}$	$\frac{6}{11}$

Answer 14:

$$\begin{gathered} \mathrm{Sum} \mathrm{along} \mathrm{col}\mathrm{umns} \mathrm{and} \mathrm{rows}: \\ \frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11} \\ \frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{15}{11} \\ \frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{15}{11} \\ \frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{15}{11} \\ \frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{15}{11} \\ \frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{15}{11} \\ \text{S}\mathrm{um} \mathrm{along} \mathrm{diagonals} : \\ \frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{15}{11} \\ \frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{15}{11} \\ \mathrm{Since}, \mathrm{all} \mathrm{the} \mathrm{sums} \text{in the square} \mathrm{are} \mathrm{equal} \mathrm{along} \mathrm{rows},\mathrm{columns} \mathrm{and} \mathrm{diagonals}, \\ \text{it} \mathrm{is} \mathrm{a} \mathrm{magic} \mathrm{square}. \end{gathered}$$

Question 15:

The cost of Mathematics book is Rs $25\frac{3}{4}$ and that of Science book is Rs $20\frac{1}{2}$. Which costs more and by how much?

Answer 15:

$$\begin{gathered} \mathrm{Cost} \mathrm{of} \mathrm{mathematics} \mathrm{book} =\mathrm{Rs} 25\frac{3}{4}=\frac{(25\times 4)+3}{4}=\mathrm{Rs}\frac{103}{4} \\ \mathrm{Cost} \mathrm{of} \mathrm{Science} \mathrm{book} =\mathrm{Rs} 20\frac{1}{2}=\frac{(20\times 2)+1}{2} = \frac{41}{2}=\frac{41}{2}\times \frac{2}{2} =\mathrm{Rs}\frac{82}{4} \\ \mathrm{We} \mathrm{know} \\ 82<103 \\ =>\frac{82}{4}<\frac{103}{4} \\ \mathrm{Thus}, \mathrm{Mathematics} \mathrm{book} \mathrm{costs} \mathrm{more}. \\ \mathrm{Difference} \mathrm{in} \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{Mathematics} \mathrm{and} \mathrm{Science} \mathrm{book} =\mathrm{cost} \mathrm{of} \mathrm{Mathematics} \mathrm{book}-\mathrm{Cost} \mathrm{of} \mathrm{Science} \mathrm{book} \\ =\frac{103}{4}-\frac{82}{4} \\ = \frac{21}{4} = \mathrm{Rs} 5\frac{1}{4} \\ \mathrm{So}, \mathrm{Mathematics} \mathrm{book} \mathrm{costs} \mathrm{more} \mathrm{by} \mathrm{Rs} 5\frac{1}{4} \end{gathered}$$

Question 16:

(i) Provide the number in box $\boxed{ }$ and also give its simplest from in each of the following:
(i) $\frac{2}{3}\times \boxed{ }=\frac{10}{30}$
(ii) $\frac{3}{5}\times \boxed{ }=\frac{24}{75}$.

Answer 16:

(i) $$\begin{gathered} \frac{2}{3}\times x=\frac{10}{30} \\ x=\frac{10}{30}\times \frac{3}{2}=\frac{1}{2} \end{gathered}$$

(ii) $$\begin{gathered} \frac{3}{5}\times x=\frac{24}{75} \\ x=\frac{24}{75}\times \frac{5}{3}=\frac{8}{15} \end{gathered}$$