Fractions and Decimals

(i) Total parts = 8

Shaded parts = 2

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{2}{8}$.

Hence, fraction = $\frac{1}{4}$.

(ii) Total parts = 10

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{10}$.

Hence, fraction = $\frac{3}{10}$.

(iii) Total parts = 12

Shaded parts = 5

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{5}{12}$.

Hence, fraction = $\frac{5}{12}$.

(iv) Total parts = 13

Shaded parts = 7

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{7}{13}$.

Hence, fraction = $\frac{7}{13}$.

We know, 1 hour = 60 minutes.

Fraction = $\frac{35 \text{ minutes}}{60 \text{ minutes}}$

$$\begin{aligned}\Rightarrow \frac{35}{60}\\[1em] \Rightarrow \frac{35 \div 5}{60 \div 5}\\[1em] \Rightarrow \frac{7}{12}\end{aligned}$$

Hence, the required fraction = $\frac{7}{12}$.

We use the rule: improper fraction = $\frac{(\text{natural number} \times \text{denominator}) + \text{numerator}}{\text{denominator}}$.

(i) $2\frac{7}{9}$

$$\begin{aligned}\Rightarrow 2\frac{7}{9} = \frac{2 \times 9 + 7}{9}\\[1em] = \frac{18 + 7}{9}\\[1em] = \frac{25}{9}.\end{aligned}$$

Hence, the required fraction is $2\frac{7}{9} = \frac{25}{9}$.

(ii) $5\frac{4}{11}$

$$\begin{aligned}\Rightarrow 5\frac{4}{11} = \frac{5 \times 11 + 4}{11}\\[1em] = \frac{55 + 4}{11}\\[1em] = \frac{59}{11}.\end{aligned}$$

Hence, the required fraction is $5\frac{4}{11} = \frac{59}{11}$.

(i) $\frac{73}{8}$

Dividing 73 by 8, we get quotient = 9 and remainder = 1.

$\Rightarrow \frac{73}{8} = 9\frac{1}{8}.$

Hence, $\frac{73}{8} = 9\frac{1}{8}$.

(ii) $\frac{94}{13}$

Dividing 94 by 13, we get quotient = 7 and remainder = 3.

$\Rightarrow \frac{94}{13} = 7\frac{3}{13}.$

Hence, $\frac{94}{13} = 7\frac{3}{13}$.

(i) $\frac{3}{7} = \frac{...}{35}$

To get 35 from 7, we have to multiply 7 by 5.

So, multiply both numerator and denominator by 5.

$\Rightarrow \frac{3}{7} = \frac{3 \times 5}{7 \times 5}$

= $\frac{15}{35}$.

Hence, the missing number is 15.

(ii) $\frac{5}{...} = \frac{30}{18}$

To get 5 from 30, we have to divide 30 by 6.

So, divide both numerator and denominator by 6.

$\Rightarrow \frac{30}{18} = \frac{30 \div 6}{18 \div 6}$

= $\frac{5}{3}.$

Hence, the missing number is 3.

(iii) $\frac{...}{9} = \frac{56}{72}$

To get 9 from 72, we have to divide 72 by 8.

So, divide both numerator and denominator by 8.

$\Rightarrow \frac{56}{72} = \frac{56 \div 8}{72 \div 8}$

= $\frac{7}{9}.$

Hence, the missing number is 7.

(i) $\frac{48}{72}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{48}{72} = \frac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 2 \times 2 \times 3 \times 3}\\[1em] = \frac{2}{3}.\end{aligned}$$

Hence, $\frac{48}{72}$ in simplest form is $\frac{2}{3}$.

(ii) $\frac{276}{115}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{276}{115} = \frac{2 \times 2 \times 3 \times 23}{5 \times 23}\\[1em] = \frac{2 \times 2 \times 3}{5}\\[1em] = \frac{12}{5}.\end{aligned}$$

Hence, $\frac{276}{115}$ in simplest form is $\frac{12}{5}$.

(iii) $\frac{72}{336}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{72}{336} = \frac{2 \times 2 \times 2 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 3 \times 7}\\[1em] = \frac{3}{2 \times 7}\\[1em] = \frac{3}{14}.\end{aligned}$$

Hence, $\frac{72}{336}$ in simplest form is $\frac{3}{14}$.

(i) $\frac{3}{4}, \frac{5}{6}, \frac{7}{8}$

L.C.M. of 4, 6 and 8 is :

$$\begin{array}{l|r} 2 & 4, 6, 8 \\ \hline 2 & 2, 3, 4 \\ \hline 2 & 1, 3, 2 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 4, 6 and 8 = 2 x 2 x 2 x 3 = 24.

$$\begin{aligned}\Rightarrow \frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}\\[1em] \Rightarrow \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}\\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}\end{aligned}$$

Hence, the equivalent like fractions are $\frac{18}{24}, \frac{20}{24}$ and $\frac{21}{24}$.

(ii) $\frac{7}{25}, \frac{9}{10}, \frac{19}{40}$

L.C.M. of 25, 10 and 40 is :

$$\begin{array}{l|r} 2 & 25, 10, 40 \\ \hline 2 & 25, 5, 20 \\ \hline 2 & 25, 5, 10 \\ \hline 5 & 25, 5, 5 \\ \hline 5 & 5, 1, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 25, 10 and 40 = 2 x 2 x 2 x 5 x 5 = 200.

$$\begin{aligned}\Rightarrow \frac{7}{25} = \frac{7 \times 8}{25 \times 8} = \frac{56}{200}\\[1em] \Rightarrow \frac{9}{10} = \frac{9 \times 20}{10 \times 20} = \frac{180}{200}\\[1em] \Rightarrow \frac{19}{40} = \frac{19 \times 5}{40 \times 5} = \frac{95}{200}\end{aligned}$$

Hence, the equivalent like fractions are $\frac{56}{200}, \frac{180}{200}$ and $\frac{95}{200}$.

(i) $\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$

LCM of 9, 3 and 21:

$$\begin{array}{l|r} 3 & 9, 3, 21 \\ \hline 3 & 3, 1, 7 \\ \hline 7 & 1, 1, 7 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 9, 3 and 21 = 3 x 3 x 7 = 63.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63}\\[1em] \Rightarrow \frac{2}{3} = \frac{2 \times 21}{3 \times 21} = \frac{42}{63}\\[1em] \Rightarrow \frac{8}{21} = \frac{8 \times 3}{21 \times 3} = \frac{24}{63}\end{aligned}$$

As 42 > 24 > 14,

$\frac{42}{63} \gt \frac{24}{63} \gt \frac{14}{63} \Rightarrow \frac{2}{3} \gt \frac{8}{21} \gt \frac{2}{9}$.

Hence, the given fractions in descending order are $\frac{2}{3}, \frac{8}{21}, \frac{2}{9}$.

(ii) $\frac{1}{5}, \frac{3}{7}, \frac{7}{10}$

LCM of 5, 7 and 10 :

$$\begin{array}{l|r} 2 & 5, 7, 10 \\ \hline 5 & 5, 7, 5 \\ \hline 7 & 1, 7, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 5, 7 and 10 = 2 x 5 x 7 = 70.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{1}{5} = \frac{1 \times 14}{5 \times 14} = \frac{14}{70}\\[1em] \Rightarrow \frac{3}{7} = \frac{3 \times 10}{7 \times 10} = \frac{30}{70}\\[1em] \Rightarrow \frac{7}{10} = \frac{7 \times 7}{10 \times 7} = \frac{49}{70}\end{aligned}$$

As 49 > 30 > 14,

$\frac{49}{70} \gt \frac{30}{70} \gt \frac{14}{70} \Rightarrow \frac{7}{10} \gt \frac{3}{7} \gt \frac{1}{5}$.

Hence, the given fractions in descending order are $\frac{7}{10}, \frac{3}{7}, \frac{1}{5}$.

(i) $\frac{5}{7}, \frac{3}{8}, \frac{9}{14}, \frac{20}{21}$

LCM of 7, 8, 14 and 21:

$$\begin{array}{l|r} 2 & 7, 8, 14, 21 \\ \hline 2 & 7, 4, 7, 21 \\ \hline 2 & 7, 2, 7, 21 \\ \hline 3 & 7, 1, 7, 21 \\ \hline 7 & 7, 1, 7, 7 \\ \hline & 1, 1, 1, 1 \end{array}$$

LCM of 7, 8, 14 and 21 = 2 x 2 x 2 x 3 x 7 = 168.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168}\\[1em] \Rightarrow \frac{3}{8} = \frac{3 \times 21}{8 \times 21} = \frac{63}{168}\\[1em] \Rightarrow \frac{9}{14} = \frac{9 \times 12}{14 \times 12} = \frac{108}{168}\\[1em] \Rightarrow \frac{20}{21} = \frac{20 \times 8}{21 \times 8} = \frac{160}{168}\end{aligned}$$

As 63 < 108 < 120 < 160,

$\frac{63}{168} \lt \frac{108}{168} \lt \frac{120}{168} \lt \frac{160}{168} \Rightarrow \frac{3}{8} \lt \frac{9}{14} \lt \frac{5}{7} \lt \frac{20}{21}$.

Hence, the given fractions in ascending order are $\frac{3}{8}, \frac{9}{14}, \frac{5}{7}, \frac{20}{21}$.

(ii) $\frac{13}{18}, \frac{8}{15}, \frac{17}{24}, \frac{7}{12}$

LCM of 18, 15, 24 and 12:

$$\begin{array}{l|r} 2 & 18, 15, 24, 12 \\ \hline 2 & 9, 15, 12, 6 \\ \hline 2 & 9, 15, 6, 3 \\ \hline 3 & 9, 15, 3, 3 \\ \hline 3 & 3, 5, 1, 1 \\ \hline 5 & 1, 5, 1, 1 \\ \hline & 1, 1, 1, 1 \end{array}$$

LCM of 18, 15, 24 and 12 = 2 x 2 x 2 x 3 x 3 x 5 = 360.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{13}{18} = \frac{13 \times 20}{18 \times 20} = \frac{260}{360}\\[1em] \Rightarrow \frac{8}{15} = \frac{8 \times 24}{15 \times 24} = \frac{192}{360}\\[1em] \Rightarrow \frac{17}{24} = \frac{17 \times 15}{24 \times 15} = \frac{255}{360}\\[1em] \Rightarrow \frac{7}{12} = \frac{7 \times 30}{12 \times 30} = \frac{210}{360}\end{aligned}$$

As 192 < 210 < 255 < 260,

$\frac{192}{360} \lt \frac{210}{360} \lt \frac{255}{360} \lt \frac{260}{360} \Rightarrow \frac{8}{15} \lt \frac{7}{12} \lt \frac{17}{24} \lt \frac{13}{18}$.

Hence, the given fractions in ascending order are $\frac{8}{15}, \frac{7}{12}, \frac{17}{24}, \frac{13}{18}$.

(i) $\frac{4}{3} + \frac{7}{8}$

LCM of 3 and 8

$$\begin{array}{l|r} 2 & 3, 8 \\ \hline 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 8 = 2 x 2 x 2 x 3 = 24.

$$\begin{aligned}\Rightarrow \frac{4 \times 8}{3 \times 8} + \frac{7 \times 3}{8 \times 3}\\[1em] \Rightarrow \frac{32}{24} + \frac{21}{24}\\[1em] \Rightarrow \frac{32 + 21}{24}\\[1em] \Rightarrow \frac{53}{24}\\[1em] \Rightarrow 2\frac{5}{24}\end{aligned}$$

Hence, $\frac{4}{3} + \frac{7}{8} = 2\frac{5}{24}$.

(ii) $8\frac{1}{2} - 3\frac{5}{8}$

LCM of 2 and 8:

$$\begin{array}{l|r} 2 & 2, 8 \\ \hline 2 & 2, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$$

LCM of 2 and 8 = 2 x 2 x 2 = 8.

$$\begin{aligned}\Rightarrow 8\frac{1}{2} - 3\frac{5}{8} \\[1em] \Rightarrow \frac{17}{2} - \frac{29}{8}\\[1em] \Rightarrow \frac{17 \times 4}{2 \times 4} - \frac{29}{8}\\[1em] \Rightarrow \frac{68}{8} - \frac{29}{8}\\[1em] \Rightarrow \frac{68 - 29}{8}\\[1em] \Rightarrow \frac{39}{8}\\[1em] \Rightarrow 4\frac{7}{8}\end{aligned}$$

Hence, $8\frac{1}{2} - 3\frac{5}{8} = 4\frac{7}{8}$.

(iii) $\frac{5}{12} + \frac{1}{18} - \frac{2}{9}$

LCM of 12, 18 and 9 :

$$\begin{array}{l|r} 2 & 12, 18, 9 \\ \hline 2 & 6, 9, 9 \\ \hline 3 & 3, 9, 9 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 12, 18 and 9 =2 x 2 x 3 x 3 = 36.

$$\begin{aligned}\Rightarrow \frac{5 \times 3}{12 \times 3} + \frac{1 \times 2}{18 \times 2} - \frac{2 \times 4}{9 \times 4}\\[1em] \Rightarrow \frac{15}{36} + \frac{2}{36} - \frac{8}{36}\\[1em] \Rightarrow \frac{15 + 2 - 8}{36}\\[1em] \Rightarrow \frac{9}{36}\\[1em] \Rightarrow \frac{1}{4}\end{aligned}$$

Hence, $\frac{5}{12} + \frac{1}{18} - \frac{2}{9} = \frac{1}{4}$.

(i) $7\frac{3}{4} - 3\frac{5}{6} + \frac{7}{8}$

LCM of 4, 6 and 8 :

$$\begin{array}{l|r} 2 & 4, 6, 8 \\ \hline 2 & 2, 3, 4 \\ \hline 2 & 1, 3, 2 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$$

LCM of 4, 6 and 8 = 2 x 2 x 2 x 3 = 24.

$$\begin{aligned}\Rightarrow \frac{31}{4} - \frac{23}{6} + \frac{7}{8}\\[1em] \Rightarrow \frac{31 \times 6}{4 \times 6} - \frac{23 \times 4}{6 \times 4} + \frac{7 \times 3}{8 \times 3}\\[1em] \Rightarrow \frac{186}{24} - \frac{92}{24} + \frac{21}{24}\\[1em] \Rightarrow \frac{186 - 92 + 21}{24}\\[1em] \Rightarrow \frac{115}{24}\\[1em] \Rightarrow 4\frac{19}{24}\end{aligned}$$

Hence, $7\frac{3}{4} - 3\frac{5}{6} + \frac{7}{8} = 4\frac{19}{24}$.

(ii) $6\frac{1}{8} - 2\frac{1}{12} - 5\frac{1}{10} + 3\frac{7}{25}$

LCM of 8, 12, 10 and 25 :

$$\begin{array}{l|r} 2 & 8, 12, 10, 25 \\ \hline 2 & 4, 6, 5, 25 \\ \hline 2 & 2, 3, 5, 25 \\ \hline 3 & 1, 3, 5, 25 \\ \hline 5 & 1, 1, 5, 25 \\ \hline 5 & 1, 1, 1, 5 \\ \hline & 1, 1, 1, 1 \end{array}$$

LCM of 8, 12, 10 and 25 = 2 x 2 x 2 x 3 x 5 x 5 = 600.

$$\begin{aligned}\Rightarrow \frac{49}{8} - \frac{25}{12} - \frac{51}{10} + \frac{82}{25}\\[1em] \Rightarrow \frac{49 \times 75}{8 \times 75} - \frac{25 \times 50}{12 \times 50} - \frac{51 \times 60}{10 \times 60} + \frac{82 \times 24}{25 \times 24}\\[1em] \Rightarrow \frac{3675}{600} - \frac{1250}{600} - \frac{3060}{600} + \frac{1968}{600}\\[1em] \Rightarrow \frac{3675 - 1250 - 3060 + 1968}{600}\\[1em] \Rightarrow \frac{1333}{600}\\[1em] \Rightarrow 2\frac{133}{600}\end{aligned}$$

Hence, $6\frac{1}{8} - 2\frac{1}{12} - 5\frac{1}{10} + 3\frac{7}{25} = 2\frac{133}{600}$.

Total time Aliyah studies daily = $5\frac{2}{3}$ hours.

Time devoted for science and mathematics = $2\frac{4}{5}$ hours.

Time devoted for other subjects = $5\frac{2}{3} - 2\frac{4}{5}$

LCM of 3 and 5:

$$\begin{array}{l|r} 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$$

LCM of 3 and 5 = 3 x 5 = 15.

$$\begin{aligned}\Rightarrow \frac{17}{3} - \frac{14}{5}\\[1em] \Rightarrow \frac{17 \times 5}{3 \times 5} - \frac{14 \times 3}{5 \times 3}\\[1em] \Rightarrow \frac{85}{15} - \frac{42}{15}\\[1em] \Rightarrow \frac{85 - 42}{15}\\[1em] \Rightarrow \frac{43}{15}\\[1em] \Rightarrow 2\frac{13}{15}\end{aligned}$$

Hence, Aliyah devotes $2\frac{13}{15}$ hours for other subjects.

Part of exercise solved by Ram = $\frac{2}{7}$.

Part of exercise solved by Shwetha = $\frac{4}{5}$.

Compare $\frac{2}{7}$ and $\frac{4}{5}$ by cross multiplication:

2 × 5 = 10 and 7 × 4 = 28.

Since 10 < 28, therefore, $\frac{2}{7} \lt \frac{4}{5}$.

So, Ram solved the lesser part of the exercise.

Difference = $\frac{4}{5} - \frac{2}{7}$

LCM of 5 and 7 = 35.

$$\begin{aligned}\Rightarrow \frac{4 \times 7}{5 \times 7} - \frac{2 \times 5}{7 \times 5}\\[1em] \Rightarrow \frac{28}{35} - \frac{10}{35}\\[1em] \Rightarrow \frac{28 - 10}{35}\\[1em] \Rightarrow \frac{18}{35}\end{aligned}$$

Hence, Ram solved the lesser part by $\frac{18}{35}$ of the exercise.

Money Sonali had = ₹$35\frac{3}{5}$.

Money she got from her mother = ₹$16\frac{1}{15}$.

Money she spent on food = ₹$28\frac{2}{3}$.

Money left = $35\frac{3}{5} + 16\frac{1}{15} - 28\frac{2}{3}$

LCM of 5, 15 and 3 = 15.

$$\begin{aligned}\Rightarrow \frac{178}{5} + \frac{241}{15} - \frac{86}{3}\\[1em] \Rightarrow \frac{178 \times 3}{5 \times 3} + \frac{241}{15} - \frac{86 \times 5}{3 \times 5}\\[1em] \Rightarrow \frac{534}{15} + \frac{241}{15} - \frac{430}{15}\\[1em] \Rightarrow \frac{534 + 241 - 430}{15}\\[1em] \Rightarrow \frac{345}{15}\\[1em] \Rightarrow 23\end{aligned}$$

Hence, ₹23 is left with Sonali.

(i) $7 \times \frac{3}{5}$

$$\begin{aligned}\Rightarrow \frac{7 \times 3}{5}\\[1em] \Rightarrow \frac{21}{5}\\[1em] \Rightarrow 4\frac{1}{5}\end{aligned}$$

Hence, $7 \times \frac{3}{5} = 4\frac{1}{5}$.

(ii) $21 \times \frac{3}{14}$

$$\begin{aligned}\Rightarrow \frac{21 \times 3}{14}\\[1em] \Rightarrow \frac{21}{14} \times 3\\[1em] \Rightarrow \frac{3}{2} \times 3\\[1em] \Rightarrow \frac{9}{2}\\[1em] \Rightarrow 4\frac{1}{2}\end{aligned}$$

Hence, $21 \times \frac{3}{14} = 4\frac{1}{2}$.

(iii) $3\frac{2}{5} \times 8$

$$\begin{aligned}\Rightarrow \frac{17}{5} \times 8\\[1em] \Rightarrow \frac{17 \times 8}{5}\\[1em] \Rightarrow \frac{136}{5}\\[1em] \Rightarrow 27\frac{1}{5}\end{aligned}$$

Hence, $3\frac{2}{5} \times 8 = 27\frac{1}{5}$.

(iv) $5 \times 6\frac{3}{4}$

$$\begin{aligned}\Rightarrow 5 \times \frac{27}{4}\\[1em] \Rightarrow \frac{5 \times 27}{4}\\[1em] \Rightarrow \frac{135}{4}\\[1em] \Rightarrow 33\frac{3}{4}\end{aligned}$$

Hence, $5 \times 6\frac{3}{4} = 33\frac{3}{4}$.

(i) $\frac{2}{3}$ of 18

$$\begin{aligned}\Rightarrow \frac{2}{3} \times 18\\[1em] \Rightarrow \frac{2 \times 18}{3}\\[1em] \Rightarrow 2 \times 6\\[1em] \Rightarrow 12\end{aligned}$$

Hence, $\frac{2}{3}$ of 18 = 12.

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 4\frac{2}{9}\\[1em] \Rightarrow \frac{1}{2} \times \frac{38}{9}\\[1em] \Rightarrow \frac{1 \times 38}{2 \times 9}\\[1em] \Rightarrow \frac{38}{18}\\[1em] \Rightarrow \frac{19}{9}\\[1em] \Rightarrow 2\frac{1}{9}\end{aligned}$$

Hence, $\frac{1}{2}$ of $4\frac{2}{9} = 2\frac{1}{9}$.

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

$$\begin{aligned}\Rightarrow \frac{5}{8} \times 9\frac{2}{3}\\[1em] \Rightarrow \frac{5}{8} \times \frac{29}{3}\\[1em] \Rightarrow \frac{5 \times 29}{8 \times 3}\\[1em] \Rightarrow \frac{145}{24}\\[1em] \Rightarrow 6\frac{1}{24}\end{aligned}$$

Hence, $\frac{5}{8}$ of $9\frac{2}{3} = 6\frac{1}{24}$.

(i) $\frac{3}{7} \times \frac{5}{9}$

$$\begin{aligned}\Rightarrow \frac{3 \times 5}{7 \times 9}\\[1em] \Rightarrow \frac{3}{9} \times \frac{5}{7}\\[1em] \Rightarrow \frac{1}{3} \times \frac{5}{7}\\[1em] \Rightarrow \frac{5}{21}\end{aligned}$$

Hence, $\frac{3}{7} \times \frac{5}{9} = \frac{5}{21}$.

(ii) $\frac{2}{5} \times 5\frac{1}{4}$

$$\begin{aligned}\Rightarrow \frac{2}{5} \times \frac{21}{4}\\[1em] \Rightarrow \frac{2}{4} \times \frac{21}{5}\\[1em] \Rightarrow \frac{1}{2} \times \frac{21}{5}\\[1em] \Rightarrow \frac{21}{10}\\[1em] \Rightarrow 2\frac{1}{10}\end{aligned}$$

Hence, $\frac{2}{5} \times 5\frac{1}{4} = 2\frac{1}{10}$.

(iii) $2\frac{1}{3} \times 5\frac{4}{21}$

$$\begin{aligned}\Rightarrow \frac{7}{3} \times \frac{109}{21}\\[1em] \Rightarrow \frac{7}{21} \times \frac{109}{3}\\[1em] \Rightarrow \frac{1}{3} \times \frac{109}{3}\\[1em] \Rightarrow \frac{109}{9}\\[1em] \Rightarrow 12\frac{1}{9}\end{aligned}$$

Hence, $2\frac{1}{3} \times 5\frac{4}{21} = 12\frac{1}{9}$.

(iv) $3\frac{1}{6} \times 7\frac{4}{23}$

$$\begin{aligned}\Rightarrow \frac{19}{6} \times \frac{165}{23}\\[1em] \Rightarrow \frac{19}{23} \times \frac{165}{6}\\[1em] \Rightarrow \frac{19}{23} \times \frac{55}{2}\\[1em] \Rightarrow \frac{19 \times 55}{23 \times 2}\\[1em] \Rightarrow \frac{1045}{46}\\[1em] \Rightarrow 22\frac{33}{46}\end{aligned}$$

Hence, $3\frac{1}{6} \times 7\frac{4}{23} = 22\frac{33}{46}$.

(i) $\frac{1}{3}$ of ₹42

$$\begin{aligned}\Rightarrow \frac{1}{3} \times 42\\[1em] \Rightarrow \frac{42}{3}\\[1em] \Rightarrow ₹14\end{aligned}$$

Hence, $\frac{1}{3}$ of ₹42 = ₹14.

(ii) $\frac{3}{7}$ of $4\frac{2}{3}$ kg

$$\begin{aligned}\Rightarrow \frac{3}{7} \times 4\frac{2}{3}\\[1em] \Rightarrow \frac{3}{7} \times \frac{14}{3}\\[1em] \Rightarrow \frac{3}{3} \times \frac{14}{7}\\[1em] \Rightarrow 1 \times 2\\[1em] \Rightarrow 2 \text{ kg}\end{aligned}$$

Hence, $\frac{3}{7}$ of $4\frac{2}{3}$ kg = 2 kg.

(iii) $4\frac{1}{2}$ times of $5\frac{1}{2}$ metres

$$\begin{aligned}\Rightarrow 4\frac{1}{2} \times 5\frac{1}{2}\\[1em] \Rightarrow \frac{9}{2} \times \frac{11}{2}\\[1em] \Rightarrow \frac{9 \times 11}{2 \times 2}\\[1em] \Rightarrow \frac{99}{4}\\[1em] \Rightarrow 24\frac{3}{4} \text{ metres}\end{aligned}$$

Hence, $4\frac{1}{2}$ times of $5\frac{1}{2}$ metres = $24\frac{3}{4}$ metres.

(i) $\frac{2}{7}$ of $\frac{3}{4}$ or $\frac{3}{5}$ of $\frac{5}{8}$

First, find $\frac{2}{7}$ of $\frac{3}{4}$:

$$\begin{aligned}\Rightarrow \frac{2}{7} \times \frac{3}{4}\\[1em] \Rightarrow \frac{2}{4} \times \frac{3}{7}\\[1em] \Rightarrow \frac{1}{2} \times \frac{3}{7}\\[1em] \Rightarrow \frac{3}{14}\end{aligned}$$

Next, find $\frac{3}{5}$ of $\frac{5}{8}$:

$$\begin{aligned}\Rightarrow \frac{3}{5} \times \frac{5}{8}\\[1em] \Rightarrow \frac{3}{8} \times \frac{5}{5}\\[1em] \Rightarrow \frac{3}{8} \times 1\\[1em] \Rightarrow \frac{3}{8}\end{aligned}$$

Now compare $\frac{3}{14}$ and $\frac{3}{8}$.

These are fractions with same numerator 3.

In fractions with same numerator, the fraction with smaller denominator is greater.

Since 8 < 14, therefore, $\frac{3}{8} \gt \frac{3}{14}$.

Hence, $\frac{3}{5}$ of $\frac{5}{8}$ is greater.

(ii) $\frac{1}{2}$ of $\frac{6}{7}$ or $\frac{2}{3}$ of $\frac{3}{7}$

First, find $\frac{1}{2}$ of $\frac{6}{7}$:

$$\begin{aligned}\Rightarrow \frac{1}{2} \times \frac{6}{7}\\[1em] \Rightarrow \frac{1 \times 6}{2 \times 7}\\[1em] \Rightarrow \frac{6}{14}\\[1em] \Rightarrow \frac{3}{7}\end{aligned}$$

Next, find $\frac{2}{3}$ of $\frac{3}{7}$:

$$\begin{aligned}\Rightarrow \frac{2}{3} \times \frac{3}{7}\\[1em] \Rightarrow \frac{2}{7} \times \frac{3}{3}\\[1em] \Rightarrow \frac{2}{7} \times 1\\[1em] \Rightarrow \frac{2}{7}\end{aligned}$$

Now compare $\frac{3}{7}$ and $\frac{2}{7}$.

These are like fractions (same denominator 7).

In like fractions, the fraction with greater numerator is greater.

Since 3 > 2, therefore, $\frac{3}{7} \gt \frac{2}{7}$.

Hence, $\frac{1}{2}$ of $\frac{6}{7}$ is greater.

Cost of 1 metre cloth = ₹$131\frac{3}{4}$.

Cost of $5\frac{1}{2}$ metres cloth = $131\frac{3}{4} \times 5\frac{1}{2}$

$$\begin{aligned}\Rightarrow \frac{527}{4} \times \frac{11}{2}\\[1em] \Rightarrow \frac{527 \times 11}{4 \times 2}\\[1em] \Rightarrow \frac{5797}{8}\\[1em] \Rightarrow 724\frac{5}{8}\end{aligned}$$

Hence, the cost of $5\frac{1}{2}$ metres cloth = ₹$724\frac{5}{8}$.

Speed of the car = $105\frac{1}{5}$ km/h.

Time taken = $3\frac{3}{5}$ hours.

Distance covered = Speed × Time = $105\frac{1}{5} \times 3\frac{3}{5}$

$$\begin{aligned}\Rightarrow \frac{526}{5} \times \frac{18}{5}\\[1em] \Rightarrow \frac{526 \times 18}{5 \times 5}\\[1em] \Rightarrow \frac{9468}{25}\\[1em] \Rightarrow 378\frac{18}{25}\end{aligned}$$

Hence, the distance covered by the car = $378\frac{18}{25}$ km.

Distance covered in 1 litre of petrol = 16 km.

Distance covered in $2\frac{3}{4}$ litres of petrol = $16 \times 2\frac{3}{4}$

$$\begin{aligned}\Rightarrow 16 \times \frac{11}{4}\\[1em] \Rightarrow \frac{16 \times 11}{4}\\[1em] \Rightarrow 4 \times 11\\[1em] \Rightarrow 44 \text{ km}\end{aligned}$$

Hence, the car will cover 44 km in $2\frac{3}{4}$ litres of petrol.

Part of the book read in 1 hour = $\frac{1}{3}$.

Part of the book read in $2\frac{1}{5}$ hours = $\frac{1}{3} \times 2\frac{1}{5}$

$$\begin{aligned}\Rightarrow \frac{1}{3} \times \frac{11}{5}\\[1em] \Rightarrow \frac{1 \times 11}{3 \times 5}\\[1em] \Rightarrow \frac{11}{15}\end{aligned}$$

Hence, Sushant will read $\frac{11}{15}$ part of the book in $2\frac{1}{5}$ hours.

Total weight of the ornament = 52 grams.

Fraction of copper = $\frac{2}{13}$.

Weight of copper = $\frac{2}{13} \times 52$

$$\begin{aligned}\Rightarrow \frac{2 \times 52}{13}\\[1em] \Rightarrow 2 \times 4\\[1em] \Rightarrow 8 \text{ grams}\end{aligned}$$

Weight of pure gold = Total weight − Weight of copper

⇒ 52 - 8

⇒ 44 grams

Hence, the weight of pure gold in the ornament = 44 grams.

Total number of students = 40.

(i) Number of students who like to study English = $\frac{1}{5} \times 40$

$\Rightarrow \frac{40}{5}$

⇒ 8

Hence, 8 students like to study English.

(ii) Number of students who like to study Mathematics = $\frac{2}{5} \times 40$

$$\begin{aligned}\Rightarrow \frac{2 \times 40}{5}\\[1em] \Rightarrow 2 \times 8\\[1em] \Rightarrow 16\end{aligned}$$

Hence, 16 students like to study Mathematics.

(iii) Fraction of students who like English and Mathematics = $\frac{1}{5} + \frac{2}{5} = \frac{1 + 2}{5} = \frac{3}{5}$.

Fraction of students who like to study Science = $1 - \frac{3}{5}$

$$\begin{aligned}\Rightarrow \frac{5}{5} - \frac{3}{5}\\[1em] \Rightarrow \frac{5 - 3}{5}\\[1em] \Rightarrow \frac{2}{5}\end{aligned}$$

Hence, $\frac{2}{5}$ of the total number of students like to study Science.

Length of the sheet = $12\frac{1}{2}$ cm = $\frac{25}{2}$ cm.

Width of the sheet = $10\frac{2}{3}$ cm = $\frac{32}{3}$ cm.

(i) Perimeter = 2 × (Length + Width)

$$\begin{aligned}\Rightarrow 2 \times \left(\frac{25}{2} + \frac{32}{3}\right)\\[1em] \Rightarrow 2 \times \left(\frac{25 \times 3}{2 \times 3} + \frac{32 \times 2}{3 \times 2}\right)\\[1em] \Rightarrow 2 \times \left(\frac{75}{6} + \frac{64}{6}\right)\\[1em] \Rightarrow 2 \times \frac{139}{6}\\[1em] \Rightarrow \frac{139}{3}\\[1em] \Rightarrow 46\frac{1}{3} \text{ cm}\end{aligned}$$

Hence, the perimeter of the sheet = $46\frac{1}{3}$ cm.

(ii) Area = Length × Width

$$\begin{aligned}\Rightarrow \frac{25}{2} \times \frac{32}{3}\\[1em] \Rightarrow \frac{25 \times 32}{2 \times 3}\\[1em] \Rightarrow \frac{25 \times 16}{3}\\[1em] \Rightarrow \frac{400}{3}\\[1em] \Rightarrow 133\frac{1}{3} \text{ sq. cm}\end{aligned}$$

Hence, the area of the sheet = $133\frac{1}{3}$ sq. cm.

Fraction of girls = $\frac{25}{54}$.

Fraction of boys = $1 - \frac{25}{54}$

$$\begin{aligned}\Rightarrow \frac{54}{54} - \frac{25}{54}\\[1em] \Rightarrow \frac{54 - 25}{54}\\[1em] \Rightarrow \frac{29}{54}\end{aligned}$$

Let the total number of students be x.

Given, number of boys = 2030.

$$\begin{aligned}\Rightarrow \frac{29}{54} \times x = 2030\\[1em] \Rightarrow x = 2030 \times \frac{54}{29}\\[1em] \Rightarrow x = 70 \times 54\\[1em] \Rightarrow x = 3780\end{aligned}$$

Number of girls = $\frac{25}{54} \times 3780$

⇒ 25 × 70

⇒ 1750

Hence, the number of girls in the school = 1750.

Fraction of orange trees = $\frac{1}{5}$.

Fraction of mango trees = $\frac{3}{13}$.

Fraction of banana trees = $1 - \left(\frac{1}{5} + \frac{3}{13}\right)$

LCM of 5 and 13 = 65.

$$\begin{aligned}\Rightarrow 1 - \left(\frac{1 \times 13}{5 \times 13} + \frac{3 \times 5}{13 \times 5}\right)\\[1em] \Rightarrow 1 - \left(\frac{13}{65} + \frac{15}{65}\right)\\[1em] \Rightarrow 1 - \frac{28}{65}\\[1em] \Rightarrow \frac{65}{65} - \frac{28}{65}\\[1em] \Rightarrow \frac{37}{65}\end{aligned}$$

Let the total number of trees be x.

Given, number of banana trees = 148.

$$\begin{aligned}\Rightarrow \frac{37}{65} \times x = 148\\[1em] \Rightarrow x = 148 \times \frac{65}{37}\\[1em] \Rightarrow x = 4 \times 65\\[1em] \Rightarrow x = 260\end{aligned}$$

Hence, the total number of trees in the orchard = 260.

The reciprocal of a fraction $\frac{a}{b}$ is $\frac{b}{a}$.

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

The reciprocal of $\frac{3}{7}$ is $\frac{7}{3}$.

Hence, the reciprocal of $\frac{3}{7}$ is $\frac{7}{3}$.

(ii) $\frac{13}{9}$

The reciprocal of $\frac{13}{9}$ is $\frac{9}{13}$.

Hence, the reciprocal of $\frac{13}{9}$ is $\frac{9}{13}$.

(iii) 8

We can write 8 as $\frac{8}{1}$.

The reciprocal of $\frac{8}{1}$ is $\frac{1}{8}$.

Hence, the reciprocal of 8 is $\frac{1}{8}$.

(i) $14 \div \frac{5}{6}$

$$\begin{aligned}\Rightarrow \frac{14}{1} \times \frac{6}{5}\\[1em] \Rightarrow \frac{14 \times 6}{1 \times 5}\\[1em] \Rightarrow \frac{84}{5}\\[1em] \Rightarrow 16\frac{4}{5}\end{aligned}$$

Hence, $14 \div \frac{5}{6} = 16\frac{4}{5}$.

(ii) $5 \div 3\frac{4}{7}$

$$\begin{aligned}\Rightarrow 5 \div \frac{25}{7}\\[1em] \Rightarrow \frac{5}{1} \times \frac{7}{25}\\[1em] \Rightarrow \frac{5}{25} \times \frac{7}{1}\\[1em] \Rightarrow \frac{1}{5} \times 7\\[1em] \Rightarrow \frac{7}{5}\\[1em] \Rightarrow 1\frac{2}{5}\end{aligned}$$

Hence, $5 \div 3\frac{4}{7} = 1\frac{2}{5}$.

(iii) $3\frac{1}{5} \div 1\frac{2}{3}$

$$\begin{aligned}\Rightarrow \frac{16}{5} \div \frac{5}{3}\\[1em] \Rightarrow \frac{16}{5} \times \frac{3}{5}\\[1em] \Rightarrow \frac{16 \times 3}{5 \times 5}\\[1em] \Rightarrow \frac{48}{25}\\[1em] \Rightarrow 1\frac{23}{25}\end{aligned}$$

Hence, $3\frac{1}{5} \div 1\frac{2}{3} = 1\frac{23}{25}$.

(iv) $2\frac{5}{8} \div 1\frac{1}{6}$

$$\begin{aligned}\Rightarrow \frac{21}{8} \div \frac{7}{6}\\[1em] \Rightarrow \frac{21}{8} \times \frac{6}{7}\\[1em] \Rightarrow \frac{21}{7} \times \frac{6}{8}\\[1em] \Rightarrow 3 \times \frac{3}{4}\\[1em] \Rightarrow \frac{9}{4}\\[1em] \Rightarrow 2\frac{1}{4}\end{aligned}$$

Hence, $2\frac{5}{8} \div 1\frac{1}{6} = 2\frac{1}{4}$.

Total length of cloth = $77\frac{1}{2}$ metres.

Length of each piece = $5\frac{1}{6}$ metres.

Number of pieces = $77\frac{1}{2} \div 5\frac{1}{6}$

$$\begin{aligned}\Rightarrow \frac{155}{2} \div \frac{31}{6}\\[1em] \Rightarrow \frac{155}{2} \times \frac{6}{31}\\[1em] \Rightarrow \frac{155}{31} \times \frac{6}{2}\\[1em] \Rightarrow 5 \times 3\\[1em] \Rightarrow 15\end{aligned}$$

Hence, 15 pieces can be cut from the cloth.

Let the required number be x.

$$\begin{aligned}\Rightarrow 4\frac{7}{8} \times x = 87\frac{3}{4}\\[1em] \Rightarrow x = 87\frac{3}{4} \div 4\frac{7}{8}\\[1em] \Rightarrow x = \frac{351}{4} \div \frac{39}{8}\\[1em] \Rightarrow x = \frac{351}{4} \times \frac{8}{39}\\[1em] \Rightarrow x = \frac{351}{39} \times \frac{8}{4}\\[1em] \Rightarrow x = 9 \times 2\\[1em] \Rightarrow x = 18\end{aligned}$$

Hence, the required number is 18.

Milk consumed by each student = $\frac{1}{3}$ litre.

Total consumption of milk = $57\frac{2}{3}$ litres.

Number of students = $57\frac{2}{3} \div \frac{1}{3}$

$$\begin{aligned}\Rightarrow \frac{173}{3} \div \frac{1}{3}\\[1em] \Rightarrow \frac{173}{3} \times \frac{3}{1}\\[1em] \Rightarrow \frac{173 \times 3}{3}\\[1em] \Rightarrow 173\end{aligned}$$

Hence, there are 173 students in the hostel.

Cost of $5\frac{1}{4}$ kg oranges = ₹336.

Rate of oranges per kg = $336 \div 5\frac{1}{4}$

$$\begin{aligned}\Rightarrow 336 \div \frac{21}{4}\\[1em] \Rightarrow \frac{336}{1} \times \frac{4}{21}\\[1em] \Rightarrow \frac{336}{21} \times 4\\[1em] \Rightarrow 16 \times 4\\[1em] \Rightarrow ₹64\end{aligned}$$

Hence, the rate of oranges is ₹64 per kg.

Area of the rectangular plot = $68\frac{3}{4}$ sq. m.

Length of the plot = $12\frac{1}{2}$ m.

Width = Area ÷ Length = $68\frac{3}{4} \div 12\frac{1}{2}$

$$\begin{aligned}\Rightarrow \frac{275}{4} \div \frac{25}{2}\\[1em] \Rightarrow \frac{275}{4} \times \frac{2}{25}\\[1em] \Rightarrow \frac{275}{25} \times \frac{2}{4}\\[1em] \Rightarrow 11 \times \frac{1}{2}\\[1em] \Rightarrow \frac{11}{2}\\[1em] \Rightarrow 5\frac{1}{2} \text{ m}\end{aligned}$$

Hence, the width of the plot = $5\frac{1}{2}$ m.

Cost of $5\frac{1}{2}$ kg of sugar = ₹$206\frac{1}{4}$.

Cost of 1 kg of sugar = $206\frac{1}{4} \div 5\frac{1}{2}$

$$\begin{aligned}\Rightarrow \frac{825}{4} \div \frac{11}{2}\\[1em] \Rightarrow \frac{825}{4} \times \frac{2}{11}\\[1em] \Rightarrow \frac{825}{11} \times \frac{2}{4}\\[1em] \Rightarrow 75 \times \frac{1}{2}\\[1em] \Rightarrow \frac{75}{2}\end{aligned}$$

Cost of $8\frac{1}{4}$ kg of sugar = $\frac{75}{2} \times 8\frac{1}{4}$

$$\begin{aligned}\Rightarrow \frac{75}{2} \times \frac{33}{4}\\[1em] \Rightarrow \frac{75 \times 33}{2 \times 4}\\[1em] \Rightarrow \frac{2475}{8}\\[1em] \Rightarrow 309\frac{3}{8}\end{aligned}$$

Hence, the cost of $8\frac{1}{4}$ kg of sugar = ₹$309\frac{3}{8}$.

Part of homework completed in 2 hours = $\frac{2}{3}$.

Part of homework completed in 1 hour = $\frac{2}{3} \div 2$

$$\begin{aligned}\Rightarrow \frac{2}{3} \times \frac{1}{2}\\[1em] \Rightarrow \frac{2}{2} \times \frac{1}{3}\\[1em] \Rightarrow \frac{1}{3}\end{aligned}$$

Part of homework completed in $1\frac{1}{4}$ hours = $\frac{1}{3} \times 1\frac{1}{4}$

$$\begin{aligned}\Rightarrow \frac{1}{3} \times \frac{5}{4}\\[1em] \Rightarrow \frac{1 \times 5}{3 \times 4}\\[1em] \Rightarrow \frac{5}{12}\end{aligned}$$

Hence, Vamika completed $\frac{5}{12}$ part of her homework in $1\frac{1}{4}$ hours.

(i) 2.56

The digit 2 is in the ones (units) place.

Hence, the place value of 2 in 2.56 is 2.

(ii) 21.37

The digit 2 is in the tens place.

Hence, the place value of 2 in 21.37 is 20.

(iii) 10.25

The digit 2 is in the tenths place.

Hence, the place value of 2 in 10.25 is $\frac{2}{10}$.

(iv) 63.352

The digit 2 is in the thousandths place.

Hence, the place value of 2 in 63.352 is $\frac{2}{1000}$.

(i) 0.8

$$\begin{aligned}\Rightarrow \frac{8}{10}\\[1em] \Rightarrow \frac{8 \div 2}{10 \div 2}\\[1em] \Rightarrow \frac{4}{5}\end{aligned}$$

Hence, 0.8 = $\frac{4}{5}$.

(ii) 0.225

$$\begin{aligned}\Rightarrow \frac{225}{1000}\\[1em] \Rightarrow \frac{225 \div 25}{1000 \div 25}\\[1em] \Rightarrow \frac{9}{40}\end{aligned}$$

Hence, 0.225 = $\frac{9}{40}$.

(iii) 0.0092

$$\begin{aligned}\Rightarrow \frac{92}{10000}\\[1em] \Rightarrow \frac{92 \div 4}{10000 \div 4}\\[1em] \Rightarrow \frac{23}{2500}\end{aligned}$$

Hence, 0.0092 = $\frac{23}{2500}$.

(iv) 3.025

$$\begin{aligned}\Rightarrow \frac{3025}{1000}\\[1em] \Rightarrow \frac{3025 \div 25}{1000 \div 25}\\[1em] \Rightarrow \frac{121}{40}\end{aligned}$$

Hence, 3.025 = $\frac{121}{40}$.

(i) 5.05

$$\begin{aligned}\Rightarrow 5 + \frac{5}{100}\\[1em] \Rightarrow 5 + \frac{1}{20}\\[1em] \Rightarrow 5\frac{1}{20}\end{aligned}$$

Hence, 5.05 = $5\frac{1}{20}$.

(ii) 63.125

$$\begin{aligned}\Rightarrow 63 + \frac{125}{1000}\\[1em] \Rightarrow 63 + \frac{1}{8}\\[1em] \Rightarrow 63\frac{1}{8}\end{aligned}$$

Hence, 63.125 = $63\frac{1}{8}$.

(iii) 17.075

$$\begin{aligned}\Rightarrow 17 + \frac{75}{1000}\\[1em] \Rightarrow 17 + \frac{3}{40}\\[1em] \Rightarrow 17\frac{3}{40}\end{aligned}$$

Hence, 17.075 = $17\frac{3}{40}$.

(iv) 317.0006

$$\begin{aligned}\Rightarrow 317 + \frac{6}{10000}\\[1em] \Rightarrow 317 + \frac{3}{5000}\\[1em] \Rightarrow 317\frac{3}{5000}\end{aligned}$$

Hence, 317.0006 = $317\frac{3}{5000}$.

(i) $\frac{3}{5}$

$$\begin{aligned}\Rightarrow \frac{3 \times 2}{5 \times 2}\\[1em] \Rightarrow \frac{6}{10}\\[1em] \Rightarrow 0.6\end{aligned}$$

Hence, $\frac{3}{5}$ = 0.6.

(ii) $\frac{7}{8}$

$$\begin{aligned}\Rightarrow \frac{7 \times 125}{8 \times 125}\\[1em] \Rightarrow \frac{875}{1000}\\[1em] \Rightarrow 0.875\end{aligned}$$

Hence, $\frac{7}{8}$ = 0.875.

(iii) $3\frac{5}{16}$

$$\begin{aligned}\Rightarrow 3 + \frac{5}{16}\\[1em] \Rightarrow 3 + \frac{5 \times 625}{16 \times 625}\\[1em] \Rightarrow 3 + \frac{3125}{10000}\\[1em] \Rightarrow 3 + 0.3125\\[1em] \Rightarrow 3.3125\end{aligned}$$

Hence, $3\frac{5}{16}$ = 3.3125.

(iv) $137\frac{13}{625}$

$$\begin{aligned}\Rightarrow 137 + \frac{13}{625}\\[1em] \Rightarrow 137 + \frac{13 \times 16}{625 \times 16}\\[1em] \Rightarrow 137 + \frac{208}{10000}\\[1em] \Rightarrow 137 + 0.0208\\[1em] \Rightarrow 137.0208\end{aligned}$$

Hence, $137\frac{13}{625}$ = 137.0208.

(i) 0.5 or 0.05

Expressing as like decimals: 0.50 and 0.05.

Comparing, 50 > 5, so 0.50 > 0.05.

Hence, 0.5 is greater.

(ii) 7 or 0.7

Expressing as like decimals: 7.0 and 0.7.

Comparing the whole number parts, 7 > 0, so 7 > 0.7.

Hence, 7 is greater.

(iii) 2.03 or 2.30

The whole number parts are equal (2 = 2).

Comparing the tenths digits, 3 > 0, so 2.30 > 2.03.

Hence, 2.30 is greater.

(iv) 0.8 or 0.88

Expressing as like decimals: 0.80 and 0.88.

Comparing, 88 > 80, so 0.88 > 0.80.

Hence, 0.88 is greater.

(i) 38.02, 38.021, 3.802, 83.02, 38.002

Expressing as like decimals: 38.020, 38.021, 3.802, 83.020, 38.002.

Comparing the numbers, we get :

3.802 < 38.002 < 38.020 < 38.021 < 83.020.

Hence, the numbers in ascending order are 3.802, 38.002, 38.02, 38.021, 83.02 .

(ii) 46.542, 46.452, 46.254, 46.05, 64.542, 46.0542

Expressing as like decimals: 46.5420, 46.4520, 46.2540, 46.0500, 64.5420, 46.0542.

Comparing the numbers, we get :

46.0500 < 46.0542 < 46.2540 < 46.4520 < 46.5420 < 64.5420.

Hence, the numbers in ascending order are 46.05, 46.0542, 46.254, 46.452, 46.542, 64.542.

(i) 5.6, 0.93, 1.87, 1.9, 1.78, 0.39

Expressing as like decimals: 5.60, 0.93, 1.87, 1.90, 1.78, 0.39.

Comparing the numbers, we get :

5.60 > 1.90 > 1.87 > 1.78 > 0.93 > 0.39.

Hence, the numbers in descending order are 5.6, 1.9, 1.87, 1.78, 0.93, 0.39.

(ii) 71.201, 20.1, 2.01, 3.1, 2.14, 0.652

Expressing as like decimals: 71.201, 20.100, 2.010, 3.100, 2.140, 0.652.

Comparing the numbers, we get :

71.201 > 20.100 > 3.100 > 2.140 > 2.010 > 0.652.

Hence, the numbers in descending order are 71.201, 20.1, 3.1, 2.14, 2.01, 0.652.

We know, 100 paise = ₹1, so 1 paise = ₹$\frac{1}{100}$ = ₹0.01.

(i) 7 paise

$\Rightarrow ₹\frac{7}{100}$

⇒ ₹0.07

Hence, 7 paise = ₹0.07.

(ii) 77 rupees 77 paise

$$\begin{aligned}\Rightarrow ₹77 + ₹\frac{77}{100}\\[1em] \Rightarrow ₹77 + ₹0.77\\[1em] \Rightarrow ₹77.77\end{aligned}$$

Hence, 77 rupees 77 paise = ₹77.77.

(iii) 235 paise

$\Rightarrow ₹\frac{235}{100}$

⇒ ₹2.35

Hence, 235 paise = ₹2.35.

We know, 100 cm = 1 m and 1000 m = 1 km.

In metre:

$\Rightarrow 5 \text{ cm} = \frac{5}{100} \text{ m}$

$\Rightarrow$ 0.05 m

In kilometre:

$\Rightarrow 0.05 \text{ m} = \frac{0.05}{1000} \text{ km}$

$\Rightarrow$ 0.00005 km

Hence, 5 cm = 0.05 m = 0.00005 km.

We know, 1000 g = 1 kg, so 1 g = $\frac{1}{1000}$ kg = 0.001 kg.

(i) 200 g

$\Rightarrow \frac{200}{1000} \text{ kg}$

⇒ 0.2 kg

Hence, 200 g = 0.2 kg.

(ii) 3470 g

$\Rightarrow \frac{3470}{1000} \text{ kg}$

⇒ 3.470 kg

Hence, 3470 g = 3.470 kg.

(iii) 4 kg 8 g

$$\begin{aligned}\Rightarrow 4 \text{ kg} + \frac{8}{1000} \text{ kg}\\[1em] \Rightarrow 4 \text{ kg} + 0.008 \text{ kg}\\[1em] \Rightarrow 4.008 \text{ kg}\end{aligned}$$

Hence, 4 kg 8 g = 4.008 kg.

(i) 5.765, 9.2, 3.08

Writing the numbers as like decimals and adding:

⇒ 5.765 + 9.200 + 3.080

⇒ 18.045

Hence, the sum = 18.045.

(ii) 15.49, 8.3572, 0.903, 7.8

Writing the numbers as like decimals and adding:

⇒ 15.4900 + 8.3572 + 0.9030 + 7.8000

⇒ 32.5502

Hence, the sum = 32.5502.

(i) 72.53 - 46.782

Writing as like decimals and subtracting:

⇒ 72.530 - 46.782

⇒ 25.748

Hence, 72.53 - 46.782 = 25.748.

(ii) 18.376 - 5.43 - 8.8976

First add the numbers to be subtracted:

⇒ 5.4300 + 8.8976

⇒ 14.3276

Now subtract:

⇒ 18.3760 - 14.3276

⇒ 4.0484

Hence, 18.376 - 5.43 - 8.8976 = 4.0484.

(iii) 28.5 - 9.708 - 6.234

First add the numbers to be subtracted:

⇒ 9.708 + 6.234

⇒ 15.942

Now subtract:

⇒ 28.500 - 15.942

⇒ 12.558

Hence, 28.5 - 9.708 - 6.234 = 12.558.

(iv) 8.2 - 4.56 - 0.7912 + 2.67

Group the positive and negative numbers separately.

Sum of positive numbers = 8.2 + 2.67 = 10.87.

Sum of numbers to be subtracted = 4.56 + 0.7912 = 5.3512.

⇒ 10.8700 - 5.3512

⇒ 5.5188

Hence, 8.2 - 4.56 - 0.7912 + 2.67 = 5.5188.

(i) What number added to 3.56 gives 13.016?

Required number = 13.016 - 3.56

⇒ 13.016 - 3.560

⇒ 9.456

Hence, the required number is 9.456.

(ii) What number should be subtracted from 30 to get 23.709?

Required number = 30 - 23.709

⇒ 30.000 - 23.709

⇒ 6.291

Hence, the required number is 6.291.

(iii) What is the excess of 20.4 over 9.7403?

Excess = 20.4 - 9.7403

⇒ 20.4000 - 9.7403

⇒ 10.6597

Hence, the excess of 20.4 over 9.7403 is 10.6597.

(i) 2.7 × 4

Multiplying without the decimal point, 27 × 4 = 108.

The number of decimal places in 2.7 is 1, so the product has 1 decimal place.

⇒ 2.7 × 4

⇒ 10.8

Hence, 2.7 × 4 = 10.8.

(ii) 2.71 × 5

Multiplying without the decimal point, 271 × 5 = 1355.

The number of decimal places in 2.71 is 2, so the product has 2 decimal places.

⇒ 2.71 × 5

⇒ 13.55

Hence, 2.71 × 5 = 13.55.

(iii) 2.5 × 0.3

Multiplying without the decimal points, 25 × 3 = 75.

The number of decimal places in 2.5 and 0.3 together is 1 + 1 = 2, so the product has 2 decimal places.

⇒ 2.5 × 0.3

⇒ 0.75

Hence, 2.5 × 0.3 = 0.75.

(iv) 2.3 × 4.35

Multiplying without the decimal points, 23 × 435 = 10005.

The number of decimal places in 2.3 and 4.35 together is 1 + 2 = 3, so the product has 3 decimal places.

⇒ 2.3 × 4.35

⇒ 10.005

Hence, 2.3 × 4.35 = 10.005.

(v) 238.06 × 7.5

Multiplying without the decimal points, 23806 × 75 = 1785450.

The number of decimal places in 238.06 and 7.5 together is 2 + 1 = 3, so the product has 3 decimal places.

⇒ 238.06 × 7.5

⇒ 1785.45

Hence, 238.06 × 7.5 = 1785.45.

(vi) 0.79 × 32.4

Multiplying without the decimal points, 79 × 324 = 25596.

The number of decimal places in 0.79 and 32.4 together is 2 + 1 = 3, so the product has 3 decimal places.

⇒ 0.79 × 32.4

⇒ 25.596

Hence, 0.79 × 32.4 = 25.596.

(vii) 1.07 × 0.02

Multiplying without the decimal points, 107 × 2 = 214.

The number of decimal places in 1.07 and 0.02 together is 2 + 2 = 4, so the product has 4 decimal places.

⇒ 1.07 × 0.02

⇒ 0.0214

Hence, 1.07 × 0.02 = 0.0214.

(viii) 10.05 × 1.05

Multiplying without the decimal points, 1005 × 105 = 105525.

The number of decimal places in 10.05 and 1.05 together is 2 + 2 = 4, so the product has 4 decimal places.

⇒ 10.05 × 1.05

⇒ 10.5525

Hence, 10.05 × 1.05 = 10.5525.

(i) 10.8 ÷ 4

$\Rightarrow \frac{10.8}{4}$

⇒ 2.7

Hence, 10.8 ÷ 4 = 2.7.

(ii) 126.35 ÷ 7

$\Rightarrow \frac{126.35}{7}$

⇒ 18.05

Hence, 126.35 ÷ 7 = 18.05.

(iii) 22.5 ÷ 1.5

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{22.5}{1.5}\\[1em] \Rightarrow \frac{22.5 \times 10}{1.5 \times 10}\\[1em] \Rightarrow \frac{225}{15}\\[1em] \Rightarrow 15\end{aligned}$$

Hence, 22.5 ÷ 1.5 = 15.

(iv) 4.28 ÷ 0.02

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{4.28}{0.02}\\[1em] \Rightarrow \frac{4.28 \times 100}{0.02 \times 100}\\[1em] \Rightarrow \frac{428}{2}\\[1em] \Rightarrow 214\end{aligned}$$

Hence, 4.28 ÷ 0.02 = 214.

(v) 3.645 ÷ 1.35

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{3.645}{1.35}\\[1em] \Rightarrow \frac{3.645 \times 100}{1.35 \times 100}\\[1em] \Rightarrow \frac{364.5}{135}\\[1em] \Rightarrow 2.7\end{aligned}$$

Hence, 3.645 ÷ 1.35 = 2.7.

(vi) 0.728 ÷ 0.04

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{0.728}{0.04}\\[1em] \Rightarrow \frac{0.728 \times 100}{0.04 \times 100}\\[1em] \Rightarrow \frac{72.8}{4}\\[1em] \Rightarrow 18.2\end{aligned}$$

Hence, 0.728 ÷ 0.04 = 18.2.

(vii) 13.06 ÷ 0.08

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{13.06}{0.08}\\[1em] \Rightarrow \frac{13.06 \times 100}{0.08 \times 100}\\[1em] \Rightarrow \frac{1306}{8}\\[1em] \Rightarrow 163.25\end{aligned}$$

Hence, 13.06 ÷ 0.08 = 163.25.

(viii) 58.635 ÷ 4.5

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{58.635}{4.5}\\[1em] \Rightarrow \frac{58.635 \times 10}{4.5 \times 10}\\[1em] \Rightarrow \frac{586.35}{45}\\[1em] \Rightarrow 13.03\end{aligned}$$

Hence, 58.635 ÷ 4.5 = 13.03.

While multiplying a decimal by 10, 100 or 1000, the decimal point is shifted to the right by as many places as there are zeros.

(i) 5.9

$$\begin{aligned}\Rightarrow 5.9 \times 10 = 59\\[1em] \Rightarrow 5.9 \times 100 = 590\\[1em] \Rightarrow 5.9 \times 1000 = 5900\end{aligned}$$

Hence, 5.9 × 10 = 59, 5.9 × 100 = 590 and 5.9 × 1000 = 5900.

(ii) 3.76

$$\begin{aligned}\Rightarrow 3.76 \times 10 = 37.6\\[1em] \Rightarrow 3.76 \times 100 = 376\\[1em] \Rightarrow 3.76 \times 1000 = 3760\end{aligned}$$

Hence, 3.76 × 10 = 37.6, 3.76 × 100 = 376 and 3.76 × 1000 = 3760.

(iii) 0.549

$$\begin{aligned}\Rightarrow 0.549 \times 10 = 5.49\\[1em] \Rightarrow 0.549 \times 100 = 54.9\\[1em] \Rightarrow 0.549 \times 1000 = 549\end{aligned}$$

Hence, 0.549 × 10 = 5.49, 0.549 × 100 = 54.9 and 0.549 × 1000 = 549.

While dividing a decimal by 10, 100 or 1000, the decimal point is shifted to the left by as many places as there are zeros.

(i) 4.8

$$\begin{aligned}\Rightarrow 4.8 \div 10 = 0.48\\[1em] \Rightarrow 4.8 \div 100 = 0.048\\[1em] \Rightarrow 4.8 \div 1000 = 0.0048\end{aligned}$$

Hence, 4.8 ÷ 10 = 0.48, 4.8 ÷ 100 = 0.048 and 4.8 ÷ 1000 = 0.0048.

(ii) 38.53

$$\begin{aligned}\Rightarrow 38.53 \div 10 = 3.853\\[1em] \Rightarrow 38.53 \div 100 = 0.3853\\[1em] \Rightarrow 38.53 \div 1000 = 0.03853\end{aligned}$$

Hence, 38.53 ÷ 10 = 3.853, 38.53 ÷ 100 = 0.3853 and 38.53 ÷ 1000 = 0.03853.

(iii) 128.9

$$\begin{aligned}\Rightarrow 128.9 \div 10 = 12.89\\[1em] \Rightarrow 128.9 \div 100 = 1.289\\[1em] \Rightarrow 128.9 \div 1000 = 0.1289\end{aligned}$$

Hence, 128.9 ÷ 10 = 12.89, 128.9 ÷ 100 = 1.289 and 128.9 ÷ 1000 = 0.1289.

Length of rectangle = 5.7 cm.

Breadth of rectangle = 3.5 cm.

Area of rectangle = length × breadth = 5.7 × 3.5

Multiplying without the decimal points, 57 × 35 = 1995.

The total number of decimal places is 1 + 1 = 2.

⇒ 5.7 × 3.5

⇒ 19.95 sq. cm

Hence, the area of the rectangle = 19.95 sq. cm.

Cost of 1 metre of cloth = ₹38.50.

Cost of 3.6 m of cloth = 38.50 × 3.6

Multiplying without the decimal points, 3850 × 36 = 138600.

The total number of decimal places is 2 + 1 = 3.

⇒ 38.50 × 3.6

⇒ 138.60

Hence, the cost of 3.6 m cloth = ₹138.60.

Distance covered in 1 litre of petrol = 45.3 km.

Distance covered in 5.9 litres of petrol = 45.3 × 5.9

Multiplying without the decimal points, 453 × 59 = 26727.

The total number of decimal places is 1 + 1 = 2.

⇒ 45.3 × 5.9

⇒ 267.27 km

Hence, the two-wheeler will cover 267.27 km in 5.9 litres of petrol.

Fat in 1 kg of milk = 0.245 kg.

Fat in 12.8 kg of milk = 0.245 × 12.8

Multiplying without the decimal points, 245 × 128 = 31360.

The total number of decimal places is 3 + 1 = 4.

⇒ 0.245 × 12.8

⇒ 3.136 kg

Hence, there is 3.136 kg of fat in 12.8 kg of milk.

Total amount = ₹242.46.

Number of children = 6.

Share of each child = $\frac{242.46}{6}$

$\Rightarrow \frac{242.46}{6}$

⇒ 40.41

Hence, the share of each child = ₹40.41.

Distance covered in 2.4 litres of petrol = 43.2 km.

Distance covered in 1 litre of petrol = $\frac{43.2}{2.4}$

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{43.2}{2.4}\\[1em] \Rightarrow \frac{43.2 \times 10}{2.4 \times 10}\\[1em] \Rightarrow \frac{432}{24}\\[1em] \Rightarrow 18 \text{ km}\end{aligned}$$

Hence, the vehicle will cover 18 km in one litre of petrol.

Total ice cream = 8.4 litres = 8.4 × 1000 = 8400 mL. (Since 1 litre = 1000 mL)

Ice cream filled in one cone = 35 mL.

Number of cones = $\frac{8400}{35}$

$\Rightarrow \frac{8400}{35}$

⇒ 240

Hence, 240 ice cream cones can be filled.

Product of two numbers = 38.745.

One of the numbers = 2.7.

Other number = $\frac{38.745}{2.7}$

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{38.745}{2.7}\\[1em] \Rightarrow \frac{38.745 \times 10}{2.7 \times 10}\\[1em] \Rightarrow \frac{387.45}{27}\\[1em] \Rightarrow 14.35\end{aligned}$$

Hence, the other number is 14.35.

Let the number be $x$.

$$\begin{aligned}\Rightarrow \frac{2}{3} \times x = 10\\[1em] \Rightarrow x = 10 \times \frac{3}{2}\\[1em] \Rightarrow x = 15\end{aligned}$$

Now, 1.75 times of the number = 1.75 × 15

$\Rightarrow 1.75 \times 15$

$\Rightarrow$ 26.25

Hence, 1.75 times of the number = 26.25.

(i) $\frac{3}{5}$ of $1\frac{1}{9} + 3\frac{1}{2}$

Solving 'of' first,

$$\begin{aligned}\Rightarrow \frac{3}{5} \text{ of } \frac{10}{9} + \frac{7}{2}\\[1em] \Rightarrow \frac{3}{5} \times \frac{10}{9} + \frac{7}{2}\\[1em] \Rightarrow \frac{2}{3} + \frac{7}{2}\\[1em] \Rightarrow \frac{2 \times 2}{3 \times 2} + \frac{7 \times 3}{2 \times 3}\\[1em] \Rightarrow \frac{4}{6} + \frac{21}{6}\\[1em] \Rightarrow \frac{4 + 21}{6}\\[1em] \Rightarrow \frac{25}{6}\\[1em] \Rightarrow 4\frac{1}{6}\end{aligned}$$

Hence, $\frac{3}{5}$ of $1\frac{1}{9} + 3\frac{1}{2} = 4\frac{1}{6}$.

(ii) $\frac{4}{5} \times 2\frac{3}{8} - 2 \times \frac{3}{5}$

Solving multiplication first,

$$\begin{aligned}\Rightarrow \frac{4}{5} \times \frac{19}{8} - 2 \times \frac{3}{5}\\[1em] \Rightarrow \frac{19}{10} - \frac{6}{5}\\[1em] \Rightarrow \frac{19}{10} - \frac{6 \times 2}{5 \times 2}\\[1em] \Rightarrow \frac{19}{10} - \frac{12}{10}\\[1em] \Rightarrow \frac{19 - 12}{10}\\[1em] \Rightarrow \frac{7}{10}\end{aligned}$$

Hence, $\frac{4}{5} \times 2\frac{3}{8} - 2 \times \frac{3}{5} = \frac{7}{10}$.

(iii) $\left(\frac{4}{5} + 2\right)\left(3 - \frac{2}{3}\right)$

Solving the brackets first,

$$\begin{aligned}\Rightarrow \left(\frac{4}{5} + \frac{10}{5}\right)\left(\frac{9}{3} - \frac{2}{3}\right)\\[1em] \Rightarrow \left(\frac{4 + 10}{5}\right)\left(\frac{9 - 2}{3}\right)\\[1em] \Rightarrow \frac{14}{5} \times \frac{7}{3}\\[1em] \Rightarrow \frac{98}{15}\\[1em] \Rightarrow 6\frac{8}{15}\end{aligned}$$

Hence, $\left(\frac{4}{5} + 2\right)\left(3 - \frac{2}{3}\right) = 6\frac{8}{15}$.

(i) $\left(\frac{1}{4} \text{ of } 2 \frac{2}{7}\right) \div \frac{3}{5}$

Solving 'of' inside the bracket first,

$$\begin{aligned}\Rightarrow \left(\frac{1}{4} \times \frac{16}{7}\right) \div \frac{3}{5}\\[1em] \Rightarrow \frac{4}{7} \div \frac{3}{5}\\[1em] \Rightarrow \frac{4}{7} \times \frac{5}{3}\\[1em] \Rightarrow \frac{20}{21}\end{aligned}$$

Hence, $\left(\frac{1}{4} \text{ of } 2 \frac{2}{7}\right) \div \frac{3}{5} = \frac{20}{21}$.

(ii) $\left(\frac{3}{7} \div \frac{1}{2}\right) \div \frac{7}{8}$

Solving the bracket first,

$$\begin{aligned}\Rightarrow \left(\frac{3}{7} \times \frac{2}{1}\right) \div \frac{7}{8}\\[1em] \Rightarrow \frac{6}{7} \div \frac{7}{8}\\[1em] \Rightarrow \frac{6}{7} \times \frac{8}{7}\\[1em] \Rightarrow \frac{48}{49}\end{aligned}$$

Hence, $\left(\frac{3}{7} \div \frac{1}{2}\right) \div \frac{7}{8} = \frac{48}{49}$.

(iii) $\frac{5}{8} \div \frac{3}{4} + \frac{2}{5}$

Solving division first,

$$\begin{aligned}\Rightarrow \frac{5}{8} \times \frac{4}{3} + \frac{2}{5}\\[1em] \Rightarrow \frac{5}{6} + \frac{2}{5}\\[1em] \Rightarrow \frac{5 \times 5}{6 \times 5} + \frac{2 \times 6}{5 \times 6}\\[1em] \Rightarrow \frac{25}{30} + \frac{12}{30}\\[1em] \Rightarrow \frac{25 + 12}{30}\\[1em] \Rightarrow \frac{37}{30}\\[1em] \Rightarrow 1\frac{7}{30}\end{aligned}$$

Hence, $\frac{5}{8} \div \frac{3}{4} + \frac{2}{5} = 1\frac{7}{30}$.

(i) $\left(4\frac{1}{2} - 2\frac{2}{3}\right) \div \frac{7}{12} + 5\frac{1}{2} \text{ of } 3 \frac{5}{6}$

Solving the bracket and 'of' first,

$$\begin{aligned}\Rightarrow \left(\frac{9}{2} - \frac{8}{3}\right) \div \frac{7}{12} + \frac{11}{2} \times \frac{23}{6}\\[1em] \Rightarrow \left(\frac{27}{6} - \frac{16}{6}\right) \div \frac{7}{12} + \frac{253}{12}\\[1em] \Rightarrow \frac{11}{6} \div \frac{7}{12} + \frac{253}{12}\\[1em] \Rightarrow \frac{11}{6} \times \frac{12}{7} + \frac{253}{12}\\[1em] \Rightarrow \frac{22}{7} + \frac{253}{12}\end{aligned}$$

LCM of 7 and 12 = 84.

$$\begin{aligned}\Rightarrow \frac{22 \times 12}{7 \times 12} + \frac{253 \times 7}{12 \times 7}\\[1em] \Rightarrow \frac{264}{84} + \frac{1771}{84}\\[1em] \Rightarrow \frac{264 + 1771}{84}\\[1em] \Rightarrow \frac{2035}{84}\\[1em] \Rightarrow 24\frac{19}{84}\end{aligned}$$

Hence, $\left(4\frac{1}{2} - 2\frac{2}{3}\right) \div \frac{7}{12} + 5\frac{1}{2} \text{ of } 3 \frac{5}{6} = 24\frac{19}{84}$.

(ii) $\left(\frac{1}{2} + \frac{1}{3}\right) \div \left(\frac{1}{4} - \frac{1}{6}\right) - \left[8 - \Big\lbrace5\frac{1}{3} - \left(3 - 2\frac{1}{2}\right)\Big\rbrace\right]$

Solving the brackets first,

$$\begin{aligned}\Rightarrow \left(\frac{3 + 2}{6}\right) \div \left(\frac{3 - 2}{12}\right) - \left[8 - \Big\lbrace\frac{16}{3} - \left(3 - \frac{5}{2}\right)\Big\rbrace\right]\\[1em] \Rightarrow \frac{5}{6} \div \frac{1}{12} - \left[8 - \Big\lbrace\frac{16}{3} - \frac{1}{2}\Big\rbrace\right]\\[1em] \Rightarrow \frac{5}{6} \times \frac{12}{1} - \left[8 - \Big\lbrace\frac{32 - 3}{6}\Big\rbrace\right]\\[1em] \Rightarrow 10 - \left[8 - \frac{29}{6}\right]\\[1em] \Rightarrow 10 - \left[\frac{48 - 29}{6}\right]\\[1em] \Rightarrow 10 - \frac{19}{6}\\[1em] \Rightarrow \frac{60 - 19}{6}\\[1em] \Rightarrow \frac{41}{6}\\[1em] \Rightarrow 6\frac{5}{6}\end{aligned}$$

Hence, $\left(\frac{1}{2} + \frac{1}{3}\right) \div \left(\frac{1}{4} - \frac{1}{6}\right) - \left[8 - \Big\lbrace5\frac{1}{3} - \left(3 - 2\frac{1}{2}\right)\Big\rbrace\right] = 6\frac{5}{6}$.

(i) $2.3 - [1.89 - {3.6 - (2.7 - \overline{0.8 - 0.03})}]$

Solving the bar (vinculum) first,

$$\begin{aligned}\Rightarrow 2.3 - [1.89 - {3.6 - (2.7 - 0.77)}]\\[1em] \Rightarrow 2.3 - [1.89 - {3.6 - 1.93}]\\[1em] \Rightarrow 2.3 - [1.89 - 1.67]\\[1em] \Rightarrow 2.3 - 0.22\\[1em] \Rightarrow 2.08\end{aligned}$$

Hence, $2.3 - [1.89 - {3.6 - (2.7 - \overline{0.8 - 0.03})}] = 2.08$.

(ii) $4.5 - \frac{1}{2}$ of (7.6 - 3.5) + 2.3 × 4.05

Solving the bracket first,

$$\begin{aligned}\Rightarrow 4.5 - \frac{1}{2} \text{ of } 4.1 + 2.3 \times 4.05\\[1em] \Rightarrow 4.5 - \frac{1}{2} \times 4.1 + 2.3 \times 4.05\\[1em] \Rightarrow 4.5 - 2.05 + 9.315\\[1em] \Rightarrow 2.45 + 9.315\\[1em] \Rightarrow 11.765\end{aligned}$$

Hence, $4.5 - \frac{1}{2}$ of (7.6 - 3.5) + 2.3 × 4.05 = 11.765.

(i) $\frac{2\frac{1}{2} + \frac{1}{5}}{2\frac{1}{2} \div \frac{1}{5}}$

Solving the numerator and denominator separately.

Numerator = $2\frac{1}{2} + \frac{1}{5} = \frac{5}{2} + \frac{1}{5} = \frac{25 + 2}{10} = \frac{27}{10}$.

Denominator = $2\frac{1}{2} \div \frac{1}{5} = \frac{5}{2} \times \frac{5}{1} = \frac{25}{2}$.

$$\begin{aligned}\Rightarrow \frac{27}{10} \div \frac{25}{2}\\[1em] \Rightarrow \frac{27}{10} \times \frac{2}{25}\\[1em] \Rightarrow \frac{27}{125}\end{aligned}$$

Hence, $\frac{2\frac{1}{2} + \frac{1}{5}}{2\frac{1}{2} \div \frac{1}{5}} = \frac{27}{125}$.

(ii) $\frac{3.5 \times 0.24}{0.21} - 0.037$

Solving the numerator first,

$$\begin{aligned}\Rightarrow \frac{0.84}{0.21} - 0.037\\[1em] \Rightarrow \frac{0.84 \times 100}{0.21 \times 100} - 0.037\\[1em] \Rightarrow \frac{84}{21} - 0.037\\[1em] \Rightarrow 4 - 0.037\\[1em] \Rightarrow 3.963\end{aligned}$$

Hence, $\frac{3.5 \times 0.24}{0.21} - 0.037 = 3.963$.

(i) In fractions with same numerator, the fraction with greater denominator is smaller.

(ii) $\frac{114}{138}$

By prime factorisation,

$\Rightarrow \frac{114}{138} = \frac{2 \times 3 \times 19}{2 \times 3 \times 23}$

= $\frac{19}{23}.$

So, $\frac{114}{138}$ reduced to simplest form is $\frac{19}{23}$.

(iii) $\frac{154}{286} = \frac{...}{13}$

To get 13 from 286, we have to divide 286 by 22. So, divide 154 by 22.

$\frac{154}{286} = \frac{154 \div 22}{286 \div 22} = \frac{7}{13}$.

So, $\frac{154}{286} = \frac{7}{13}$.

(iv) The reciprocal of $2\frac{3}{8} = \frac{19}{8}$ is obtained by interchanging the numerator and denominator.

So, the reciprocal of the fraction $2\frac{3}{8}$ is $\frac{8}{19}$.

(v) 2 hours = 2 × 60 = 120 minutes.

$\frac{2}{5}$ of 120 = $\frac{2}{5} \times 120 = 48$ minutes.

So, there are 48 minutes in $\frac{2}{5}$ of 2 hours.

(vi) $2\frac{3}{7} \times 4\frac{2}{3} = \frac{17}{7} \times \frac{14}{3} = \frac{17 \times 2}{3} = \frac{34}{3} = 11\frac{1}{3}$.

So, $2\frac{3}{7} \times 4\frac{2}{3} = 11\frac{1}{3}$.

(vii) $1\frac{2}{3} \div 2\frac{1}{5} = \frac{5}{3} \div \frac{11}{5} = \frac{5}{3} \times \frac{5}{11} = \frac{25}{33}$.

So, $1\frac{2}{3} \div 2\frac{1}{5} = \frac{25}{33}$.

(viii) Price of each pen = $\frac{37.80}{7} = 5.40$.

So, the price of each pen is ₹5.40.

(ix) Multiplying without the decimal points, 54 × 235 = 12690. The total number of decimal places is 1 + 2 = 3.

So, 5.4 × 2.35 = 12.69.

(x) 0.32 ÷ 8 = 0.04.

So, 0.32 ÷ 8 = 0.04.

(xi) Since 1 mg = $\frac{1}{1000}$ g, 45 mg = $\frac{45}{1000}$ = 0.045 g.

So, 45 mg = 0.045 g.

(xii) 5.06 kg = 5 kg + 0.06 kg = 5 kg + (0.06 × 1000) g = 5 kg + 60 g.

So, 5.06 kg = 5 kg 60 g.

(xiii) 7.035 m = 7 m + 0.035 m = 7 m + 3.5 cm = 7 m + 3 cm + 5 mm.

So, 7.035 m = 7 m 3 cm 5 mm.

(xiv) The product of a proper fraction and an improper fraction is less than the improper fraction.

(xv) $2\frac{3}{7} \times \frac{7}{9} = \frac{17}{7} \times \frac{7}{9} = \frac{17}{9}$.

So, the lowest form of the product is $\frac{17}{9}$.

(xvi) Part of cake eaten by Ravi = $\frac{2}{7}$.

Remaining part = $1 - \frac{2}{7} = \frac{5}{7}$.

Part eaten by Rani = $\frac{4}{5}$ of $\frac{5}{7} = \frac{4}{5} \times \frac{5}{7} = \frac{4}{7}$.

Total part eaten = $\frac{2}{7} + \frac{4}{7} = \frac{6}{7}$.

Part of cake left = $1 - \frac{6}{7} = \frac{1}{7}$.

So, part of the cake left is $\frac{1}{7}$.

(i) False.

Reason: The reciprocal of 1 is 1 itself, since 1 × 1 = 1.

(ii) False.

Reason: The reciprocal of a proper fraction is an improper fraction. For example, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$, which is improper.

(iii) False.

Reason: The reciprocal of an improper fraction is a proper fraction. For example, the reciprocal of $\frac{5}{2}$ is $\frac{2}{5}$, which is proper.

(iv) False.

Reason: Product of two fractions = $\frac{\text{product of their numerators}}{\text{product of their denominators}}$.

(v) True.

Reason: $\frac{3}{20}$ of 2 kg = $\frac{3}{20} \times 2 = \frac{6}{20} = 0.3$ kg = 300 g.

(vi) True.

Reason: $3\frac{5}{7} = \frac{26}{7}$, and its multiplicative inverse is $\frac{7}{26}$.

(vii) True.

Reason: LCM of 15 and 20 = 60. So, $\frac{11}{15} - \frac{7}{20} = \frac{44}{60} - \frac{21}{60} = \frac{23}{60}$.

(viii) False.

Reason: $\frac{2}{3}$ of 8 = $\frac{2}{3} \times 8 = \frac{16}{3}$, whereas $\frac{2}{3} \div 8 = \frac{2}{3} \times \frac{1}{8} = \frac{1}{12}$. They are not the same.

(ix) False.

Reason: The product of two proper fractions is smaller than each of the two fractions.

(x) False.

Reason: To multiply a decimal number by 10, move the decimal point to the right by one place.

(xi) True.

Reason: To divide a decimal number by 100, the decimal point is moved to the left by two places.

$\frac{5}{6}$ of 480 = $\frac{5}{6} \times 480 = 5 \times 80 = 400$.

Hence, option 1 is the correct option.

$5\frac{2}{3} = \frac{17}{3}$.

The reciprocal of $\frac{17}{3}$ is $\frac{3}{17}$.

Hence, option 4 is the correct option.

$\frac{11}{7} = 1\frac{4}{7}$, which is greater than 1 and less than 2.

So, $\frac{11}{7}$ lies between 1 and 2.

Hence, option 2 is the correct option.

Cost of 1 kg almonds = ₹460.

Cost of $\frac{2}{5}$ kg almonds = $\frac{2}{5} \times 460 = 2 \times 92 = ₹184$.

Hence, option 2 is the correct option.

$$\begin{aligned}\Rightarrow 2\frac{1}{5} \div 1\frac{1}{5}\\[1em] \Rightarrow \frac{11}{5} \div \frac{6}{5}\\[1em] \Rightarrow \frac{11}{5} \times \frac{5}{6}\\[1em] \Rightarrow \frac{11}{6}\\[1em] \Rightarrow 1\frac{5}{6}\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}\Rightarrow 5\frac{1}{6} \div 4\frac{1}{2}\\[1em] \Rightarrow \frac{31}{6} \div \frac{9}{2}\\[1em] \Rightarrow \frac{31}{6} \times \frac{2}{9}\\[1em] \Rightarrow \frac{31}{27}\\[1em] \Rightarrow 1\frac{4}{27}\end{aligned}$$

Hence, option 4 is the correct option.

Let the number be x.

$$\begin{aligned}\Rightarrow \frac{3}{4} \times x = 12\\[1em] \Rightarrow x = 12 \times \frac{4}{3}\\[1em] \Rightarrow x = 16\end{aligned}$$

Hence, option 2 is the correct option.

Total eggs = 3 dozen = 3 × 12 = 36.

Number of rotten eggs = $\frac{1}{9}$ of 36 = $\frac{1}{9} \times 36 = 4$.

Hence, option 1 is the correct option.

Time taken daily = $1\frac{3}{4}$ hours.

Total time = $1\frac{3}{4} \times 6 = \frac{7}{4} \times 6 = \frac{42}{4} = \frac{21}{2} = 10\frac{1}{2}$ hours.

Hence, option 3 is the correct option.

In 35.0471, the digit 7 is at the thousandths place.

So, its place value = $7 \times \frac{1}{1000} = \frac{7}{1000}$.

Hence, option 4 is the correct option.

Multiplying without the decimal points, 2 × 3 = 6.

The total number of decimal places is 3 + 1 = 4.

So, 0.002 × 0.3 = 0.0006.

Hence, option 4 is the correct option.

$5\frac{3}{8} = 5 + \frac{3}{8} = 5 + 0.375 = 5.375$.

Hence, option 2 is the correct option.

$$\begin{aligned}\Rightarrow \frac{0.35}{0.7}\\[1em] \Rightarrow \frac{0.35 \times 10}{0.7 \times 10}\\[1em] \Rightarrow \frac{3.5}{7}\\[1em] \Rightarrow 0.5\end{aligned}$$

Hence, option 3 is the correct option.

Since 1 cm = $\frac{1}{100}$ m, 5 cm = 0.05 m.

So, 30 m 5 cm = 30 m + 0.05 m = 30.05 m.

Hence, option 3 is the correct option.

While multiplying by 1000, the decimal point shifts 3 places to the right.

So, 0.05309 × 1000 = 53.09.

Hence, option 2 is the correct option.

While dividing by 1000, the decimal point shifts 3 places to the left.

So, 2.305 ÷ 1000 = 0.002305.

Hence, option 3 is the correct option.

A regular hexagon has 6 equal sides.

Perimeter = 6 × side = 6 × 3.5 = 21 cm.

Hence, option 2 is the correct option.

Evaluating each option:

Option 1: 0.0002

Option 2: $\frac{2}{1000} = 0.002$

Option 3: 0.02 × 0.001 = 0.00002

Option 4: $\frac{2}{1000} \div 0.01 = 0.002 \div 0.01 = 0.2$

Comparing all the values, 0.00002 is the smallest.

Hence, option 3 is the correct option.

According to Statement I, consider the proper fraction $\frac{1}{2}$ and the improper fraction $\frac{3}{2}$.

Their product = $\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$, which is a proper fraction.

So, the product of a proper fraction and an improper fraction is not always an improper fraction.

∴ Statement I is false.

According to Statement II, the product of two proper fractions is always smaller than each of them, and is therefore a proper fraction. For example, $\frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$, which is proper.

∴ Statement II is true.

Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

According to Statement I, any improper fraction can be written as a mixed fraction by dividing the numerator by the denominator. For example, $\frac{11}{4} = 2\frac{3}{4}$.

∴ Statement I is true.

According to Statement II, a mixed fraction is made up of a natural number and a proper fraction, so it is always greater than 1. For example, 2\dfrac{3}{4} = 2 + \dfrac{3}{4} &ampgt 1.

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

According to Statement I, $\frac{2}{5} \div \frac{5}{2} = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$, which is not equal to 1.

∴ Statement I is false.

According to Statement II, $\frac{2}{5} \times \frac{5}{2} = 1$, so $\frac{2}{5}$ and $\frac{5}{2}$ are reciprocals of each other.

∴ Statement II is true.

Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

According to Statement I, comparing the whole number parts, 2 > 1, so 2.01 > 1.99.

∴ Statement I is true.

According to Statement II, decimal numbers having an equal number of decimal places are called like decimals. For example, 2.01 and 1.99 each have 2 decimal places, so they are like decimals.

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

According to Statement I, evaluate each number:

$$\begin{aligned}\Rightarrow 1.03 + 2.6 = 3.63\\[1em] \Rightarrow 6.75 - 4.879 = 1.871\\[1em] \Rightarrow 3.67 \times 0.4 = 1.468\\[1em] \Rightarrow 10.326 \div 0.3 = 34.42\end{aligned}$$

Arranging in descending order: 34.42, 3.63, 1.871, 1.468, i.e. 10.326 ÷ 0.3, 1.03 + 2.6, 6.75 − 4.879, 3.67 × 0.4.

The number on the extreme right is 3.67 × 0.4 (= 1.468), not 6.75 − 4.879.

∴ Statement I is false.

According to Statement II, to add two decimal numbers we write them one below the other so that the decimal points are aligned in the same column, and then add like whole numbers, placing the decimal point of the sum directly below the other decimal points. The rule described (counting and summing the number of decimal places) is the rule for multiplication, not addition.

∴ Statement II is false.

Both Statement I and Statement II are false.

Hence, option 4 is the correct option.

3 kilogram = 3 × 1000 = 3000 gram. (Since 1 kg = 1000 g)

Required fraction = $\frac{270}{3000}$

$\Rightarrow \frac{270}{3000}$

$\Rightarrow \frac{9}{100}$

Hence, 270 gram is $\frac{9}{100}$ of 3 kilogram.

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

$$\begin{aligned}\Rightarrow \frac{15}{2} \times \frac{34}{15}\\[1em] \Rightarrow \frac{34}{2}\\[1em] \Rightarrow 17\end{aligned}$$

Hence, $7\frac{1}{2} \times 2\frac{4}{15} = 17$.

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

$$\begin{aligned}\Rightarrow \frac{27}{7} \times \frac{14}{3}\\[1em] \Rightarrow \frac{27 \times 2}{3}\\[1em] \Rightarrow \frac{54}{3}\\[1em] \Rightarrow 18\end{aligned}$$

Hence, $3\frac{6}{7} \times 4\frac{2}{3} = 18$.

(iii) $3\frac{3}{7} \div \frac{16}{21}$

$$\begin{aligned}\Rightarrow \frac{24}{7} \div \frac{16}{21}\\[1em] \Rightarrow \frac{24}{7} \times \frac{21}{16}\\[1em] \Rightarrow \frac{3 \times 3}{2}\\[1em] \Rightarrow \frac{9}{2}\\[1em] \Rightarrow 4\frac{1}{2}\end{aligned}$$

Hence, $3\frac{3}{7} \div \frac{16}{21} = 4\frac{1}{2}$.

(iv) $15\frac{3}{7} \div 1\frac{23}{49}$

$$\begin{aligned}\Rightarrow \frac{108}{7} \div \frac{72}{49}\\[1em] \Rightarrow \frac{108}{7} \times \frac{49}{72}\\[1em] \Rightarrow \frac{108 \times 7}{72}\\[1em] \Rightarrow \frac{756}{72}\\[1em] \Rightarrow \frac{21}{2}\\[1em] \Rightarrow 10\frac{1}{2}\end{aligned}$$

Hence, $15\frac{3}{7} \div 1\frac{23}{49} = 10\frac{1}{2}$.

Marked price of shirt = ₹540.

Sale price = $\frac{3}{4}$ of 540 = $\frac{3}{4} \times 540$

$$\begin{aligned}\Rightarrow \frac{3}{4} \times 540\\[1em] \Rightarrow 3 \times 135\\[1em] \Rightarrow 405\end{aligned}$$

Hence, the sale price of the shirt = ₹405.

Total number of students = 56.

Number of students in blue house = $\frac{1}{4} \times 56 = 14$.

Number of students in yellow house = $\frac{3}{14} \times 56 = 3 \times 4 = 12$.

Remaining students = 56 - 14 - 12 = 30.

Number of students in green house = $\frac{1}{3} \times 30 = 10$.

Number of students in red house = 30 - 10 = 20.

Hence, blue house = 14, yellow house = 12, green house = 10 and red house = 20 students.

Total price of motor cycle = ₹36000.

Amount paid in cash = $\frac{1}{6}$ of 36000 = $\frac{1}{6} \times 36000 = ₹6000$.

Remaining amount = 36000 - 6000 = ₹30000.

Amount paid every month = $\frac{30000}{12}$

$\Rightarrow \frac{30000}{12}$

⇒ 2500

Hence, Rohit had to pay ₹2500 every month.

Let the total number of pages in the book be 1 (whole).

Part read on first day = $\frac{4}{9}$.

Remaining part = $1 - \frac{4}{9} = \frac{5}{9}$.

Part read on second day = $\frac{3}{5}$ of $\frac{5}{9} = \frac{3}{5} \times \frac{5}{9} = \frac{1}{3}$.

Total part read = $\frac{4}{9} + \frac{1}{3} = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}$.

Part left unread = $1 - \frac{7}{9} = \frac{2}{9}$.

Given, $\frac{2}{9}$ of the book = 100 pages.

$$\begin{aligned}\Rightarrow \frac{2}{9} \times \text{Total pages} = 100\\[1em] \Rightarrow \text{Total pages} = 100 \times \frac{9}{2}\\[1em] \Rightarrow \text{Total pages} = 450\end{aligned}$$

Hence, the book contained 450 pages.

Converting the mixed fractions in the denominators to improper fractions,

$4\frac{2}{7} = \frac{30}{7}$, $1\frac{11}{13} = \frac{24}{13}$.

$$\begin{aligned}\Rightarrow \frac{1}{\frac{30}{7}} + \frac{1}{\frac{24}{13}} + \frac{1}{\frac{5}{9}}\\[1em] \Rightarrow \frac{7}{30} + \frac{13}{24} + \frac{9}{5}\end{aligned}$$

LCM of 30, 24 and 5 = 120.

$$\begin{aligned}\Rightarrow \frac{7 \times 4}{30 \times 4} + \frac{13 \times 5}{24 \times 5} + \frac{9 \times 24}{5 \times 24}\\[1em] \Rightarrow \frac{28}{120} + \frac{65}{120} + \frac{216}{120}\\[1em] \Rightarrow \frac{28 + 65 + 216}{120}\\[1em] \Rightarrow \frac{309}{120}\\[1em] \Rightarrow \frac{103}{40}\\[1em] \Rightarrow 2\frac{23}{40}\end{aligned}$$

Hence, the value = $2\frac{23}{40}$.

(i) 0.025

$\Rightarrow \frac{25}{1000}$

$\Rightarrow \frac{1}{40}$

Hence, 0.025 = $\frac{1}{40}$.

(ii) 0.876

$\Rightarrow \frac{876}{1000}$

$\Rightarrow \frac{219}{250}$

Hence, 0.876 = $\frac{219}{250}$.

(iii) 4.3125

$$\begin{aligned}\Rightarrow \frac{43125}{10000}\\[1em] \Rightarrow \frac{69}{16}\\[1em] \Rightarrow 4\frac{5}{16}\end{aligned}$$

Hence, 4.3125 = $4\frac{5}{16}$.

(i) $1\frac{3}{8}$

$$\begin{aligned}\Rightarrow \frac{11}{8}\\[1em] \Rightarrow \frac{11 \times 125}{8 \times 125}\\[1em] \Rightarrow \frac{1375}{1000}\\[1em] \Rightarrow 1.375\end{aligned}$$

Hence, $1\frac{3}{8} = 1.375$.

(ii) $\frac{47}{125}$

$$\begin{aligned}\Rightarrow \frac{47 \times 8}{125 \times 8}\\[1em] \Rightarrow \frac{376}{1000}\\[1em] \Rightarrow 0.376\end{aligned}$$

Hence, $\frac{47}{125} = 0.376$.

(iii) $2\frac{9}{40}$

$$\begin{aligned}\Rightarrow \frac{89}{40}\\[1em] \Rightarrow \frac{89 \times 25}{40 \times 25}\\[1em] \Rightarrow \frac{2225}{1000}\\[1em] \Rightarrow 2.225\end{aligned}$$

Hence, $2\frac{9}{40} = 2.225$.

Sum of 17.443 and 29.657 = 17.443 + 29.657 = 47.1.

Sum of 13.687 and 18.548 = 13.687 + 18.548 = 32.235.

Required excess = 47.1 - 32.235

⇒ 47.100 - 32.235

⇒ 14.865

Hence, the first sum exceeds the second by 14.865.

(i) 4.27 × 0.036

Multiplying without the decimal points, 427 × 36 = 15372. The total number of decimal places is 2 + 3 = 5.

⇒ 4.27 × 0.036

⇒ 0.15372

Hence, 4.27 × 0.036 = 0.15372.

(ii) 0.09 × 1.04

Multiplying without the decimal points, 9 × 104 = 936. The total number of decimal places is 2 + 2 = 4.

⇒ 0.09 × 1.04

⇒ 0.0936

Hence, 0.09 × 1.04 = 0.0936.

(iii) 1.32 ÷ 0.8

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{1.32}{0.8}\\[1em] \Rightarrow \frac{1.32 \times 10}{0.8 \times 10}\\[1em] \Rightarrow \frac{13.2}{8}\\[1em] \Rightarrow 1.65\end{aligned}$$

Hence, 1.32 ÷ 0.8 = 1.65.

(iv) 0.7038 ÷ 0.34

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{0.7038}{0.34}\\[1em] \Rightarrow \frac{0.7038 \times 100}{0.34 \times 100}\\[1em] \Rightarrow \frac{70.38}{34}\\[1em] \Rightarrow 2.07\end{aligned}$$

Hence, 0.7038 ÷ 0.34 = 2.07.

Cost of 1 kg rice = ₹52.70.

Cost of 12.5 kg rice = 52.70 × 12.5

Multiplying without the decimal points, 5270 × 125 = 658750. The total number of decimal places is 2 + 1 = 3.

⇒ 52.70 × 12.5

⇒ 658.75

Hence, the cost of 12.5 kg rice = ₹658.75.

Total length of cloth = 24.5 m.

Length of each piece = 1.75 m.

Number of pieces = $\frac{24.5}{1.75}$

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{24.5}{1.75}\\[1em] \Rightarrow \frac{24.5 \times 100}{1.75 \times 100}\\[1em] \Rightarrow \frac{2450}{175}\\[1em] \Rightarrow 14\end{aligned}$$

Hence, 14 pieces can be cut from the cloth.

Product of two numbers = 1.599.

One of the numbers = 0.65.

Other number = $\frac{1.599}{0.65}$

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$$\begin{aligned}\Rightarrow \frac{1.599}{0.65}\\[1em] \Rightarrow \frac{1.599 \times 100}{0.65 \times 100}\\[1em] \Rightarrow \frac{159.9}{65}\\[1em] \Rightarrow 2.46\end{aligned}$$

Hence, the other number is 2.46.

(i) $\left(\frac{4}{5} - \frac{1}{3}\right) \div 4\frac{1}{5} + \frac{2}{3}$ of $\left(5\frac{1}{6} - 4\frac{3}{8}\right)$

Solving the brackets first,

$$\begin{aligned}\Rightarrow \left(\frac{12 - 5}{15}\right) \div \frac{21}{5} + \frac{2}{3} \text{ of } \left(\frac{31}{6} - \frac{35}{8}\right)\\[1em] \Rightarrow \frac{7}{15} \div \frac{21}{5} + \frac{2}{3} \text{ of } \left(\frac{124 - 105}{24}\right)\\[1em] \Rightarrow \frac{7}{15} \times \frac{5}{21} + \frac{2}{3} \times \frac{19}{24}\\[1em] \Rightarrow \frac{1}{9} + \frac{19}{36}\\[1em] \Rightarrow \frac{4}{36} + \frac{19}{36}\\[1em] \Rightarrow \frac{4 + 19}{36}\\[1em] \Rightarrow \frac{23}{36}\end{aligned}$$

Hence, $\left(\frac{4}{5} - \frac{1}{3}\right) \div 4\frac{1}{5} + \frac{2}{3}$ of $\left(5\frac{1}{6} - 4\frac{3}{8}\right) = \frac{23}{36}$.

(ii) $1\frac{2}{3}$ of $\left(\frac{3}{8} - \frac{1}{12}\right) - \left[4\frac{2}{3} - \Big\lbrace6 - \left(2\frac{2}{3} - \overline{4\frac{1}{2} - 3\frac{1}{3}}\right)\Big\rbrace\right]$

Solving the bar (vinculum) first,

$$\begin{aligned}\Rightarrow 4\frac{1}{2} - 3\frac{1}{3}\\[1em] = \frac{9}{2} - \frac{10}{3}\\[1em] = \frac{27 - 20}{6} = \frac{7}{6}\end{aligned}$$

Now substituting and solving the innermost bracket,

$$\begin{aligned}\Rightarrow 1\frac{2}{3} \text{ of } \left(\frac{3}{8} - \frac{1}{12}\right) - \left[4\frac{2}{3} - \Big\lbrace6 - \left(\frac{8}{3} - \frac{7}{6}\right)\Big\rbrace\right]\\[1em] \Rightarrow 1\frac{2}{3} \text{ of } \left(\frac{9 - 2}{24}\right) - \left[\frac{14}{3} - \Big\lbrace6 - \left(\frac{16 - 7}{6}\right)\Big\rbrace\right]\\[1em] \Rightarrow \frac{5}{3} \text{ of } \frac{7}{24} - \left[\frac{14}{3} - \Big\lbrace6 - \frac{9}{6}\Big\rbrace\right]\\[1em] \Rightarrow \frac{5}{3} \times \frac{7}{24} - \left[\frac{14}{3} - \Big\lbrace\frac{36 - 9}{6}\Big\rbrace\right]\\[1em] \Rightarrow \frac{35}{72} - \left[\frac{14}{3} - \frac{9}{2}\right]\\[1em] \Rightarrow \frac{35}{72} - \left[\frac{28 - 27}{6}\right]\\[1em] \Rightarrow \frac{35}{72} - \frac{1}{6}\\[1em] \Rightarrow \frac{35 - 12}{72}\\[1em] \Rightarrow \frac{23}{72}\end{aligned}$$

Hence, $1\frac{2}{3}$ of $\left(\frac{3}{8} - \frac{1}{12}\right) - \left[4\frac{2}{3} - \Big\lbrace6 - \left(2\frac{2}{3} - \overline{4\frac{1}{2} - 3\frac{1}{3}}\right)\Big\rbrace\right] = \frac{23}{72}$.

First, simplify $\left(1\frac{1}{3} \div 1\frac{1}{2}\right) \div 1\frac{1}{9}$.

$$\begin{aligned}\Rightarrow \left(\frac{4}{3} \div \frac{3}{2}\right) \div \frac{10}{9}\\[1em] \Rightarrow \left(\frac{4}{3} \times \frac{2}{3}\right) \div \frac{10}{9}\\[1em] \Rightarrow \frac{8}{9} \div \frac{10}{9}\\[1em] \Rightarrow \frac{8}{9} \times \frac{9}{10}\\[1em] \Rightarrow \frac{8}{10}\\[1em] \Rightarrow \frac{4}{5}\end{aligned}$$

The result is $\frac{4}{5}$, which is a proper fraction. The smallest natural number greater than $\frac{4}{5}$ is 1.

Least fraction to be added = $1 - \frac{4}{5} = \frac{5 - 4}{5} = \frac{1}{5}$.

Hence, the least fraction that must be added is $\frac{1}{5}$.

Let the capacity of the drum be 1 (whole).

Part of drum that is full at first = $\frac{3}{4}$.

Part of drum that is full after drawing oil = $\frac{7}{12}$.

Part of drum corresponding to 30 litres = $\frac{3}{4} - \frac{7}{12}$

$$\begin{aligned}\Rightarrow \frac{3}{4} - \frac{7}{12}\\[1em] \Rightarrow \frac{9 - 7}{12}\\[1em] \Rightarrow \frac{2}{12}\\[1em] \Rightarrow \frac{1}{6}\end{aligned}$$

So, $\frac{1}{6}$ of the capacity = 30 litres.

$$\begin{aligned}\Rightarrow \frac{1}{6} \times \text{Capacity} = 30\\[1em] \Rightarrow \text{Capacity} = 30 \times 6\\[1em] \Rightarrow \text{Capacity} = 180 \text{ litres}\end{aligned}$$

Hence, the capacity of the drum is 180 litres.