Fractions

(i) $\frac{3}{5}$ : The denominator is neither 10 nor any higher power of 10. Thus it is a vulgar fraction. Since the denominator (5) is greater than the numerator (3), it is a proper fraction.

(ii) $\frac{11}{10}$ : The denominator is 10, so it is a decimal fraction. Since the denominator (10) is less than the numerator (11), it is an improper fraction.

(iii) $\frac{13}{20}$ : The denominator is neither 10 nor any higher power of 10. Thus it is a vulgar fraction. Since the denominator (20) is greater than the numerator (13), it is a proper fraction.

(iv) $\frac{18}{7}$ : The denominator is neither 10 nor any higher power of 10, so it is a vulgar fraction. Since the denominator (7) is less than the numerator (18), it is an improper fraction.

(v) $3\frac{2}{9}$ : The denominator is neither 10 nor any higher power of 10, so it is a vulgar fraction. It consists of a natural number and a proper fraction, so it is a mixed fraction.

(i) Dividing 18 by 5 :

$$\begin{array}{l} 5\overline{\smash{\big)} \phantom{)}18\smash{\big(}} 3 \\ \phantom{a}\underline{-15} \\ \phantom{+()}3 \\ \end{array}$$

Quotient = 3 and remainder = 3.

$\frac{18}{5} = 3\frac{3}{5}$

Hence, $\frac{18}{5} = 3\frac{3}{5}$

(ii) Dividing 7 by 4 :

$$\begin{array}{l} 4\overline{\smash{\big)} \phantom{)}7\smash{\big(}} 1 \\ \phantom{a}\underline{-4} \\ \phantom{+)}3 \\ \end{array}$$

Quotient = 1 and remainder = 3.

$\frac{7}{4} = 1\frac{3}{4}$

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

(iii) Dividing 25 by 6 :

$$\begin{array}{l} 6\overline{\smash{\big)} \phantom{)}25\smash{\big(}} 4 \\ \phantom{a}\underline{-24} \\ \phantom{+))}1 \\ \end{array}$$

Quotient = 4 and remainder = 1.

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

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

(iv) Dividing 38 by 5 :

$$\begin{array}{l} 5\overline{\smash{\big)} \phantom{)}38\smash{\big(}} 7 \\ \phantom{a}\underline{-35} \\ \phantom{+((}3 \\ \end{array}$$

Quotient = 7 and remainder = 3.

$\frac{38}{5} = 7\frac{3}{5}$

Hence, $\frac{38}{5} = 7\frac{3}{5}$

(v) Dividing 22 by 5 :

$$\begin{array}{l} 5\overline{\smash{\big)} \phantom{)}22\smash{\big(}} 4 \\ \phantom{a}\underline{-20} \\ \phantom{+((}2 \\ \end{array}$$

Quotient = 4 and remainder = 2.

$\frac{22}{5} = 4\frac{2}{5}$

Hence, $\frac{22}{5} = 4\frac{2}{5}$

(i) $2\frac{4}{9} = \frac{2 \times 9 + 4}{9} = \frac{18 + 4}{9} = \frac{22}{9}$

Hence, $2\frac{4}{9} = \frac{22}{9}$

(ii) $7\frac{5}{13} = \frac{7 \times 13 + 5}{13} = \frac{91 + 5}{13} = \frac{96}{13}$

Hence, $7\frac{5}{13} = \frac{96}{13}$

(iii) $3\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{12 + 1}{4} = \frac{13}{4}$

Hence, $3\frac{1}{4} = \frac{13}{4}$

(iv) $2\frac{5}{48} = \frac{2 \times 48 + 5}{48} = \frac{96 + 5}{48} = \frac{101}{48}$

Hence, $2\frac{5}{48} = \frac{101}{48}$

(v) $12\frac{7}{11} = \frac{12 \times 11 + 7}{11} = \frac{132 + 7}{11} = \frac{139}{11}$

Hence, $12\frac{7}{11} = \frac{139}{11}$

(i) HCF of 8 and 18 = 2.

$\frac{8}{18} = \frac{8 \div 2}{18 \div 2} = \frac{4}{9}$

Hence, $\frac{8}{18} = \frac{4}{9}$

(ii) HCF of 27 and 36 = 9.

$\frac{27}{36} = \frac{27 \div 9}{36 \div 9} = \frac{3}{4}$

Hence, $\frac{27}{36} = \frac{3}{4}$

(iii) HCF of 18 and 42 = 6.

$\frac{18}{42} = \frac{18 \div 6}{42 \div 6} = \frac{3}{7}$

Hence, $\frac{18}{42} = \frac{3}{7}$

(iv) HCF of 35 and 75 = 5.

$\frac{35}{75} = \frac{35 \div 5}{75 \div 5} = \frac{7}{15}$

Hence, $\frac{35}{75} = \frac{7}{15}$

(v) HCF of 18 and 45 = 9.

$\frac{18}{45} = \frac{18 \div 9}{45 \div 9} = \frac{2}{5}$

Hence, $\frac{18}{45} = \frac{2}{5}$

(i) $\frac{30}{40} = \frac{30 \div 10}{40 \div 10} = \frac{3}{4} \text{ and } \frac{12}{16} = \frac{12 \div 4}{16 \div 4} = \frac{3}{4}$

Both reduce to $\frac{3}{4}$, so they are equivalent fractions.

Hence, the statement is True.

(ii) $\frac{10}{25} = \frac{10 \div 5}{25 \div 5} = \frac{2}{5} \text{ and } \frac{25}{10} = \frac{25 \div 5}{10 \div 5} = \frac{5}{2}$

Since $\frac{2}{5} ≠ \frac{5}{2}$, they are not equivalent fractions.

Hence, the statement is False.

(iii) $\frac{35}{49} = \frac{5}{7}, \frac{20}{28} = \frac{5}{7}, \frac{45}{63} = \frac{5}{7} \text{ and } \frac{100}{140} = \frac{5}{7}$

All four reduce to $\frac{5}{7}$, so they are equivalent fractions.

Hence, the statement is True.

A simple fraction has both numerator and denominator as whole numbers (with denominator ≠ 0); a complex fraction has a fraction in its numerator or denominator or both.

(i) $\frac{0}{8}$ : both terms are whole numbers and denominator ≠ 0, so it is a Simple fraction.

(ii) $\frac{3}{8}$ : both terms are whole numbers, so it is a Simple fraction.

(iii) $\frac{5}{7}$ : both terms are whole numbers, so it is a Simple fraction.

(iv) $\frac{3\frac{3}{5}}{18}$ : the numerator is a fraction, so it is a Complex fraction.

(v) $\frac{6}{2\frac{2}{5}}$ : the denominator is a fraction, so it is a Complex fraction.

(vi) $\frac{3\frac{1}{3}}{7\frac{2}{7}}$ : both numerator and denominator are fractions, so it is a Complex fraction.

(vii) $\frac{5\frac{2}{9}}{5}$ : the numerator is a fraction, so it is a Complex fraction.

(viii) $\frac{8}{0}$ : division by 0 is not defined, so it is Neither a simple nor a complex fraction.

(i) $\frac{5}{8} \text{ and } \frac{7}{8}$ have the same denominator (8), so they are Like fractions.

(ii) $\frac{8}{15} \text{ and } \frac{8}{21}$ have different denominators (15 and 21), so they are Unlike fractions.

(iii) $\frac{4}{9} \text{ and } \frac{9}{4}$ have different denominators (9 and 4), so they are Unlike fractions.

(i) LCM of 6 and 9 = 18.

$$\begin{aligned}\frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18} \\[1em] \frac{7}{9} = \frac{7 \times 2}{9 \times 2} = \frac{14}{18}\end{aligned}$$

Hence, the required fractions are $\frac{15}{18} \text{ and } \frac{14}{18}$.

(ii) LCM of 3, 6 and 12 = 12.

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

Hence, the required fractions are $\frac{8}{12}, \frac{10}{12} \text{ and } \frac{7}{12}$.

(iii) LCM of 5, 20, 40 and 16 = 80.

$$\begin{aligned}\frac{4}{5} = \frac{4 \times 16}{5 \times 16} = \frac{64}{80} \\[1em] \frac{17}{20} = \frac{17 \times 4}{20 \times 4} = \frac{68}{80} \\[1em] \frac{23}{40} = \frac{23 \times 2}{40 \times 2} = \frac{46}{80} \\[1em] \frac{11}{16} = \frac{11 \times 5}{16 \times 5} = \frac{55}{80}\end{aligned}$$

Hence, the required fractions are $\frac{64}{80}, \frac{68}{80}, \frac{46}{80} \text{ and } \frac{55}{80}$.

(i) LCM of the numerators 8 and 12 = 24.

$$\begin{aligned}\frac{8}{9} = \frac{8 \times 3}{9 \times 3} = \frac{24}{27} \\[1em] \frac{12}{17} = \frac{12 \times 2}{17 \times 2} = \frac{24}{34}\end{aligned}$$

Hence, the required fractions are $\frac{24}{27} \text{ and } \frac{24}{34}$.

(ii) LCM of the numerators 6, 15 and 12 = 60.

$$\begin{aligned}\frac{6}{13} = \frac{6 \times 10}{13 \times 10} = \frac{60}{130} \\[1em] \frac{15}{23} = \frac{15 \times 4}{23 \times 4} = \frac{60}{92} \\[1em] \frac{12}{17} = \frac{12 \times 5}{17 \times 5} = \frac{60}{85}\end{aligned}$$

Hence, the required fractions are $\frac{60}{130}, \frac{60}{92} \text{ and } \frac{60}{85}$.

(iii) LCM of the numerators 15, 25, 9 and 45 = 225.

$$\begin{aligned}\frac{15}{19} = \frac{15 \times 15}{19 \times 15} = \frac{225}{285} \\[1em] \frac{25}{28} = \frac{25 \times 9}{28 \times 9} = \frac{225}{252} \\[1em] \frac{9}{11} = \frac{9 \times 25}{11 \times 25} = \frac{225}{275} \\[1em] \frac{45}{47} = \frac{45 \times 5}{47 \times 5} = \frac{225}{235}\end{aligned}$$

Hence, the required fractions are $\frac{225}{285}, \frac{225}{252}, \frac{225}{275} \text{ and } \frac{225}{235}$.

(i) LCM of 5 and 9 = 45.

$$\begin{aligned}\frac{2}{5} = \frac{2 \times 9}{5 \times 9} = \frac{18}{45} \\[1em] \frac{4}{9} = \frac{4 \times 5}{9 \times 5} = \frac{20}{45}\end{aligned}$$

Since 18 < 20, $\frac{18}{45} \lt \frac{20}{45}$

Hence, $\frac{2}{5} \lt \frac{4}{9}$

(ii) LCM of 7 and 11 = 77.

$$\begin{aligned}\frac{5}{7} = \frac{5 \times 11}{7 \times 11} = \frac{55}{77} \\[1em] \frac{8}{11} = \frac{8 \times 7}{11 \times 7} = \frac{56}{77}\end{aligned}$$

Since 55 < 56, $\frac{55}{77} \lt \frac{56}{77}$

Hence, $\frac{5}{7} \lt \frac{8}{11}$

(iii) LCM of 15 and 20 = 60.

$$\begin{aligned}\frac{7}{15} = \frac{7 \times 4}{15 \times 4} = \frac{28}{60} \\[1em] \frac{9}{20} = \frac{9 \times 3}{20 \times 3} = \frac{27}{60}\end{aligned}$$

Since 28 > 27, $\frac{28}{60} \gt \frac{27}{60}$

Hence, $\frac{7}{15} \gt \frac{9}{20}$

(i) LCM of the numerators 4 and 2 = 4.

$$\begin{aligned}\frac{4}{9} = \frac{4}{9} \\[1em] \frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}\end{aligned}$$

Since the numerators are equal and 9 < 10, the fraction with the smaller denominator is greater.

Hence, $\frac{4}{9} \gt \frac{2}{5}$

(ii) LCM of the numerators 5 and 8 = 40.

$$\begin{aligned}\frac{5}{12} = \frac{5 \times 8}{12 \times 8} = \frac{40}{96} \\[1em] \frac{8}{19} = \frac{8 \times 5}{19 \times 5} = \frac{40}{95}\end{aligned}$$

Since the numerators are equal and 96 > 95, $\frac{40}{96} \lt \frac{40}{95}$

Hence, $\frac{5}{12} \lt \frac{8}{19}$

(iii) LCM of the numerators 5 and 9 = 45.

$$\begin{aligned}\frac{5}{7} = \frac{5 \times 9}{7 \times 9} = \frac{45}{63} \\[1em] \frac{9}{14} = \frac{9 \times 5}{14 \times 5} = \frac{45}{70}\end{aligned}$$

Since the numerators are equal and 63 < 70, $\frac{45}{63} \gt \frac{45}{70}$

Hence, $\frac{5}{7} \gt \frac{9}{14}$

(i) For $\frac{2}{5} \text{ and } \frac{4}{9}$, cross-multiplying gives 2 × 9 = 18 and 5 × 4 = 20.

Since 18 < 20,

Hence, $\frac{2}{5} \lt \frac{4}{9}$

(ii) For $\frac{3}{8} \text{ and } \frac{6}{11}$, cross-multiplying gives 3 × 11 = 33 and 8 × 6 = 48.

Since 33 < 48,

Hence, $\frac{3}{8} \lt \frac{6}{11}$

(iii) For $\frac{5}{18} \text{ and } \frac{11}{21}$, cross-multiplying gives 5 × 21 = 105 and 18 × 11 = 198.

Since 105 < 198,

Hence, $\frac{5}{18} \lt \frac{11}{21}$

(i) LCM of 3, 5, 4 and 6 = 60.

$$\begin{aligned}\frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} \\[1em] \frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60} \\[1em] \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \\[1em] \frac{1}{6} = \frac{1 \times 10}{6 \times 10} = \frac{10}{60}\end{aligned}$$

Arranging the numerators in ascending order: 10 < 20 < 24 < 45

Hence, the ascending order is $\frac{1}{6}, \frac{1}{3}, \frac{2}{5}, \frac{3}{4}$.

(ii) LCM of 6, 8, 12 and 10 = 120.

$$\begin{aligned}\frac{5}{6} = \frac{5 \times 20}{6 \times 20} = \frac{100}{120} \\[1em] \frac{7}{8} = \frac{7 \times 15}{8 \times 15} = \frac{105}{120} \\[1em] \frac{11}{12} = \frac{11 \times 10}{12 \times 10} = \frac{110}{120} \\[1em] \frac{3}{10} = \frac{3 \times 12}{10 \times 12} = \frac{36}{120}\end{aligned}$$

Arranging the numerators in ascending order: 36 < 100 < 105 < 110

Hence, the ascending order is $\frac{3}{10}, \frac{5}{6}, \frac{7}{8}, \frac{11}{12}$.

(iii) LCM of 7, 8, 14 and 21 = 168.

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

Arranging the numerators in ascending order: 63 < 108 < 120 < 160

Hence, the ascending order is $\frac{3}{8}, \frac{9}{14}, \frac{5}{7}, \frac{20}{21}$.

(i) LCM of the numerators 5, 4, 8 and 1 = 40.

$$\begin{aligned}\frac{5}{6} = \frac{5 \times 8}{6 \times 8} = \frac{40}{48} \\[1em] \frac{4}{15} = \frac{4 \times 10}{15 \times 10} = \frac{40}{150} \\[1em] \frac{8}{9} = \frac{8 \times 5}{9 \times 5} = \frac{40}{45} \\[1em] \frac{1}{3} = \frac{1 \times 40}{3 \times 40} = \frac{40}{120}\end{aligned}$$

With equal numerators, the fraction with the smaller denominator is greater. Arranging the denominators in ascending order: 45 < 48 < 120 < 150

Hence, the descending order is $\frac{8}{9}, \frac{5}{6}, \frac{1}{3}, \frac{4}{15}$.

(ii) LCM of the numerators 3, 4, 5 and 8 = 120.

$$\begin{aligned}\frac{3}{7} = \frac{3 \times 40}{7 \times 40} = \frac{120}{280} \\[1em] \frac{4}{9} = \frac{4 \times 30}{9 \times 30} = \frac{120}{270} \\[1em] \frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168} \\[1em] \frac{8}{11} = \frac{8 \times 15}{11 \times 15} = \frac{120}{165}\end{aligned}$$

With equal numerators, the fraction with the smaller denominator is greater. Arranging the denominators in ascending order: 165 < 168 < 270 < 280

Hence, the descending order is $\frac{8}{11}, \frac{5}{7}, \frac{4}{9}, \frac{3}{7}$.

(iii) LCM of the numerators 1, 6, 8 and 3 = 24.

$$\begin{aligned}\frac{1}{10} = \frac{1 \times 24}{10 \times 24} = \frac{24}{240} \\[1em] \frac{6}{11} = \frac{6 \times 4}{11 \times 4} = \frac{24}{44} \\[1em] \frac{8}{11} = \frac{8 \times 3}{11 \times 3} = \frac{24}{33} \\[1em] \frac{3}{5} = \frac{3 \times 8}{5 \times 8} = \frac{24}{40}\end{aligned}$$

With equal numerators, the fraction with the smaller denominator is greater. Arranging the denominators in ascending order: 33 < 40 < 44 < 240

Hence, the descending order is $\frac{8}{11}, \frac{3}{5}, \frac{6}{11}, \frac{1}{10}$.

LCM of 2 and 3 = 6.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{2} + \frac{2}{3} \\[1em] = \frac{1 \times 3}{6} + \frac{2 \times 2}{6} \\[1em] = \frac{3 + 4}{6} = \frac{7}{6} \\[1em] = 1\frac{1}{6}\end{aligned}$$

Hence, the required fraction is $1\frac{1}{6}$.

LCM of 5 and 10 = 10.

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

Hence, the required fraction is $\frac{1}{2}$.

LCM of 3 and 6 = 6.

$$\begin{aligned}\Rightarrow \frac{2}{3} - \frac{1}{6} = \frac{2 \times 2}{6} - \frac{1}{6}\\[1em] = \frac{4 - 1}{6} = \frac{3}{6}\\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the required fraction is $\frac{1}{2}$.

$1\frac{1}{3} + 2\frac{1}{4} = \frac{4}{3} + \frac{9}{4}$

LCM of 3 and 4 = 12.

$$\begin{aligned}= \frac{4 \times 4}{12} + \frac{9 \times 3}{12}\\[1em] = \frac{16 + 27}{12} \\[1em] = \frac{43}{12}\\[1em] = 3\frac{7}{12}\end{aligned}$$

Hence, the required fraction is $3\frac{7}{12}$.

LCM of 4, 6 and 12 = 12.

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

Hence, the required fraction is 1.

LCM of 1, 3 and 5 = 15.

$$\begin{aligned}\Rightarrow \frac{2}{3} - \frac{3}{5} + 3 - \frac{1}{5} \\[1em] = \frac{2 \times 5}{15} - \frac{3 \times 3}{15} + \frac{3 \times 15}{15} - \frac{1 \times 3}{15} \\[1em] = \frac{10 - 9 + 45 - 3}{15} = \frac{43}{15} = 2\frac{13}{15}\end{aligned}$$

Hence, the required fraction is $2\frac{13}{15}$.

LCM of 3, 5 and 10 = 30.

$$\begin{aligned}\Rightarrow \frac{2}{3} - \frac{1}{5} + \frac{1}{10}\\[1em] = \frac{2 \times 10}{30} - \frac{1 \times 6}{30} + \frac{1 \times 3}{30} \\[1em] = \frac{20 - 6 + 3}{30} \\[1em] = \frac{17}{30}\end{aligned}$$

Hence, the required fraction is $\frac{17}{30}$.

$2\frac{1}{2} + 2\frac{1}{3} - 1\frac{1}{4} = \frac{5}{2} + \frac{7}{3} - \frac{5}{4}$

LCM of 2, 3 and 4 = 12.

$$\begin{aligned}= \frac{5 \times 6}{12} + \frac{7 \times 4}{12} - \frac{5 \times 3}{12} \\[1em] = \frac{30 + 28 - 15}{12} \\[1em] = \frac{43}{12}\\[1em] = 3\frac{7}{12}\end{aligned}$$

Hence, the required fraction is $3\frac{7}{12}$.

$2\frac{5}{8} - 2\frac{1}{6} + 4\frac{3}{4} = \frac{21}{8} - \frac{13}{6} + \frac{19}{4}$

LCM of 8, 6 and 4 = 24.

$$\begin{aligned}= \frac{21 \times 3}{24} - \frac{13 \times 4}{24} + \frac{19 \times 6}{24} \\[1em] = \frac{63 - 52 + 114}{24} = \frac{125}{24} = 5\frac{5}{24}\end{aligned}$$

Hence, the required fraction is $5\frac{5}{24}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{4} \times 6 \\[1em] = \frac{3 \times 6}{4} \\[1em] = \frac{18}{4} \\[1em] = \frac{9}{2} \\[1em] = 4\frac{1}{2}\end{aligned}$$

Hence, the required value is $4\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2}{3} \times 15 \\[1em] = \frac{2 \times 15}{3} \\[1em] = \frac{30}{3} \\[1em] = 10\end{aligned}$$

Hence, the required value is 10.

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{4} \times \frac{1}{2} \\[1em] = \frac{3 \times 1}{4 \times 2} \\[1em] = \frac{3}{8}\end{aligned}$$

Hence, the required value is $\frac{3}{8}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{9}{12} \times \frac{4}{7} \\[1em] = \frac{9 \times 4}{12 \times 7} \\[1em] = \frac{36}{84} \\[1em] = \frac{3}{7}\end{aligned}$$

Hence, the required value is $\frac{3}{7}$.

Solving,

$$\begin{aligned}\Rightarrow 45 \times 2\frac{1}{3} \\[1em] = 45 \times \frac{7}{3} \\[1em] = \frac{45 \times 7}{3} \\[1em] = \frac{315}{3} \\[1em] = 105\end{aligned}$$

Hence, the required value is 105.

Solving,

$$\begin{aligned}\Rightarrow 36 \times 3\frac{1}{4} \\[1em] = 36 \times \frac{13}{4} \\[1em] = \frac{36 \times 13}{4} \\[1em] = \frac{468}{4} \\[1em] = 117\end{aligned}$$

Hence, the required value is 117.

Solving,

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

Hence, the required value is 6.

Solving,

$$\begin{aligned}\Rightarrow 3 \div \frac{2}{5} \\[1em] = 3 \times \frac{5}{2} \\[1em] = \frac{15}{2} \\[1em] = 7\frac{1}{2}\end{aligned}$$

Hence, the required value is $7\frac{1}{2}$.

Solving,

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

Hence, the required value is $1\frac{2}{3}$.

Solving,

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

Hence, the required value is $1\frac{1}{3}$.

Solving,

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

Hence, the required value is $-\frac{5}{6}$.

Solving,

$$\begin{aligned}\Rightarrow 3\frac{3}{7} \div 1\frac{1}{14} \\[1em] = \frac{24}{7} \div \frac{15}{14} \\[1em] = \frac{24}{7} \times \frac{14}{15} \\[1em] = \frac{24 \times 14}{7 \times 15} \\[1em] = \frac{336}{105} \\[1em] = \frac{16}{5} \\[1em] = 3\frac{1}{5}\end{aligned}$$

Hence, the required value is $3\frac{1}{5}$.

Solving,

$$\begin{aligned}\Rightarrow 3\frac{3}{4} \times 1\frac{1}{5} \times \frac{20}{21} \\[1em] = \frac{15}{4} \times \frac{6}{5} \times \frac{20}{21} \\[1em] = \frac{15 \times 6 \times 20}{4 \times 5 \times 21} \\[1em] = \frac{1800}{420} \\[1em] = \frac{30}{7} \\[1em] = 4\frac{2}{7}\end{aligned}$$

Hence, the required value is $4\frac{2}{7}$.

Solving,

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

Hence, the required value is $-1\frac{1}{3}$.

Solving,

$$\begin{aligned}\frac{5}{8} - \frac{1}{8} = \frac{5 - 1}{8} \\[1em] = \frac{4}{8} = \frac{1}{2}\end{aligned}$$

Hence, the required value is $\frac{1}{2}$.

Solving,

$$\begin{aligned}\frac{2}{5} - \left(-\frac{2}{5}\right) = \frac{2}{5} + \frac{2}{5} \\[1em] = \frac{2 + 2}{5} = \frac{4}{5}\end{aligned}$$

Hence, the required value is $\frac{4}{5}$.

Solving,

$$\begin{aligned}\frac{3}{7} - \left(-\frac{3}{7}\right) = \frac{3}{7} + \frac{3}{7} \\[1em] = \frac{3 + 3}{7} = \frac{6}{7}\end{aligned}$$

Hence, the required value is $\frac{6}{7}$.

Solving,

$-\frac{4}{5} - 0 = -\frac{4}{5}$

Hence, the required value is $-\frac{4}{5}$.

$\frac{4}{5} - \frac{2}{9}$

LCM of 5 and 9 = 45.

$$\begin{aligned}= \frac{4 \times 9}{45} - \frac{2 \times 5}{45} \\[1em] = \frac{36 - 10}{45} = \frac{26}{45}\end{aligned}$$

Hence, the required value is $\frac{26}{45}$.

$-\frac{6}{11} - \left(-\frac{4}{7}\right) = -\frac{6}{11} + \frac{4}{7}$

LCM of 11 and 7 = 77.

$$\begin{aligned}= -\frac{6 \times 7}{77} + \frac{4 \times 11}{77} \\[1em] = \frac{-42 + 44}{77} = \frac{2}{77}\end{aligned}$$

Hence, the required value is $\frac{2}{77}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{2} \text{ of } 10 \text{ kg} \\[1em] \Rightarrow \frac{1}{2} \times 10 \\[1em] \Rightarrow 5 \text{ kg}\end{aligned}$$

Hence, the required value is 5 kg.

1 hour = 60 minutes.

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{5} \text{ of } 1 \text{ hour} \\[1em] \Rightarrow \frac{3}{5} \times 60 \\[1em] \Rightarrow 36 \text{ minutes}\end{aligned}$$

Hence, the required value is 36 minutes.

Solving,

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

Hence, the required value is $1\frac{1}{3}$ kg.

Solving,

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

Hence, the required value is 7 metres.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{2} \text{ of } 2\frac{2}{3} \\[1em] \Rightarrow \frac{1}{2} \times \frac{8}{3} \\[1em] \Rightarrow \frac{8}{6} \\[1em] \Rightarrow \frac{4}{3} \\[1em] \Rightarrow 1\frac{1}{3}\end{aligned}$$

Hence, the required value is $1\frac{1}{3}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{5}{11} \text{ of } \frac{4}{5} \text{ of } 22 \text{ kg} \\[1em] \Rightarrow \frac{5}{11} \times \frac{4}{5} \times 22 \\[1em] \Rightarrow \frac{5 \times 4 \times 22}{11 \times 5} \\[1em] \Rightarrow \frac{440}{55} \\[1em] \Rightarrow 8 \text{ kg}\end{aligned}$$

Hence, the required value is 8 kg.

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{3\frac{3}{4}} \\[1em] \Rightarrow 3 \div \frac{15}{4} \\[1em] \Rightarrow 3 \times \frac{4}{15} \\[1em] \Rightarrow \frac{12}{15} \\[1em] \Rightarrow \frac{4}{5}\end{aligned}$$

Hence, the required value is $\frac{4}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\frac{3}{5}}{7} \\[1em] \Rightarrow \frac{3}{5} \div 7 \\[1em] \Rightarrow \frac{3}{5} \times \frac{1}{7} \\[1em] \Rightarrow \frac{3}{35}\end{aligned}$$

Hence, the required value is $\frac{3}{35}$.

Solving,

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

Hence, the required value is $4\frac{1}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2\frac{1}{5}}{1\frac{1}{10}} \\[1em] \Rightarrow \frac{11}{5} \div \frac{11}{10} \\[1em] \Rightarrow \frac{11}{5} \times \frac{10}{11} \\[1em] \Rightarrow \frac{110}{55} \\[1em] \Rightarrow 2\end{aligned}$$

Hence, the required value is 2.

Solving,

$$\begin{aligned}\Rightarrow \frac{2}{5} \text{ of } \frac{6}{11} \times 1\frac{1}{4} \\[1em] \Rightarrow \frac{2}{5} \times \frac{6}{11} \times \frac{5}{4} \\[1em] \Rightarrow \frac{2 \times 6 \times 5}{5 \times 11 \times 4} \\[1em] \Rightarrow \frac{60}{220} \\[1em] \Rightarrow \frac{3}{11}\end{aligned}$$

Hence, the required value is $\frac{3}{11}$.

Solving,

$$\begin{aligned}\Rightarrow 2\frac{1}{4} \div \frac{1}{7} \times \frac{1}{3} \\[1em] \Rightarrow \frac{9}{4} \div \frac{1}{7} \times \frac{1}{3} \\[1em] \Rightarrow \frac{9}{4} \times \frac{7}{1} \times \frac{1}{3} \\[1em] \Rightarrow \frac{9 \times 7 \times 1}{4 \times 1 \times 3} \\[1em] \Rightarrow \frac{63}{12} \\[1em] \Rightarrow \frac{21}{4} \\[1em] \Rightarrow 5\frac{1}{4}\end{aligned}$$

Hence, the required value is $5\frac{1}{4}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{3} \times 4\frac{2}{3} \div 3\frac{1}{2} \times \frac{1}{2} \\[1em] \Rightarrow \frac{1}{3} \times \frac{14}{3} \div \frac{7}{2} \times \frac{1}{2} \\[1em] \Rightarrow \frac{14}{9} \times \frac{2}{7} \times \frac{1}{2} \\[1em] \Rightarrow \frac{14 \times 2 \times 1}{9 \times 7 \times 2} \\[1em] \Rightarrow \frac{28}{126} \\[1em] \Rightarrow \frac{2}{9}\end{aligned}$$

Hence, the required value is $\frac{2}{9}$.

Solving,

Simplifying 'of' first: $\frac{3}{7} \text{ of } 2\frac{5}{8} = \frac{3}{7} \times \frac{21}{8} = \frac{63}{56} = \frac{9}{8}$

$$\begin{aligned}\Rightarrow \frac{2}{3} \times 1\frac{1}{4} \div \frac{3}{7} \text{ of } 2\frac{5}{8} \\[1em] \Rightarrow \frac{2}{3} \times \frac{5}{4} \div \frac{9}{8} \\[1em] \Rightarrow \frac{2}{3} \times \frac{5}{4} \times \frac{8}{9} \\[1em] \Rightarrow \frac{2 \times 5 \times 8}{3 \times 4 \times 9} \\[1em] \Rightarrow \frac{80}{108} \\[1em] \Rightarrow \frac{20}{27}\end{aligned}$$

Hence, the required value is $\frac{20}{27}$.

Solving,

$$\begin{aligned}\Rightarrow 0 \div \frac{8}{11} \\[1em] \Rightarrow 0 \times \frac{11}{8} \\[1em] \Rightarrow 0\end{aligned}$$

Hence, the required value is 0.

Solving,

Simplifying 'of' first: $\frac{7}{15} \text{ of } \frac{8}{9} = \frac{7}{15} \times \frac{8}{9} = \frac{56}{135}$

$$\begin{aligned}\Rightarrow \frac{4}{5} \div \frac{7}{15} \text{ of } \frac{8}{9} \\[1em] \Rightarrow \frac{4}{5} \div \frac{56}{135} \\[1em] \Rightarrow \frac{4}{5} \times \frac{135}{56} \\[1em] \Rightarrow \frac{4 \times 135}{5 \times 56} \\[1em] \Rightarrow \frac{540}{280} \\[1em] \Rightarrow \frac{27}{14} \\[1em] \Rightarrow 1\frac{13}{14}\end{aligned}$$

Hence, the required value is $1\frac{13}{14}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{4}{5} \div \frac{7}{15} \times \frac{8}{9} \\[1em] \Rightarrow \frac{4}{5} \times \frac{15}{7} \times \frac{8}{9} \\[1em] \Rightarrow \frac{4 \times 15 \times 8}{5 \times 7 \times 9} \\[1em] \Rightarrow \frac{480}{315} \\[1em] \Rightarrow \frac{32}{21} \\[1em] \Rightarrow 1\frac{11}{21}\end{aligned}$$

Hence, the required value is $1\frac{11}{21}$.

Solving,

Simplifying 'of' first: $\frac{4}{5} \text{ of } \frac{7}{15} = \frac{4}{5} \times \frac{7}{15} = \frac{28}{75}$

$$\begin{aligned}\Rightarrow \frac{4}{5} \text{ of } \frac{7}{15} \div \frac{8}{9} \\[1em] \Rightarrow \frac{28}{75} \div \frac{8}{9} \\[1em] \Rightarrow \frac{28}{75} \times \frac{9}{8} \\[1em] \Rightarrow \frac{28 \times 9}{75 \times 8} \\[1em] \Rightarrow \frac{252}{600} \\[1em] \Rightarrow \frac{21}{50}\end{aligned}$$

Hence, the required value is $\frac{21}{50}$.

Solving,

Simplifying 'of' first: $\frac{1}{2} \text{ of } \frac{3}{4} = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$

$$\begin{aligned}\Rightarrow \frac{1}{2} \text{ of } \frac{3}{4} \times \frac{1}{2} \div \frac{2}{3} \\[1em] \Rightarrow \frac{3}{8} \times \frac{1}{2} \div \frac{2}{3} \\[1em] \Rightarrow \frac{3}{16} \div \frac{2}{3} \\[1em] \Rightarrow \frac{3}{16} \times \frac{3}{2} \\[1em] \Rightarrow \frac{9}{32}\end{aligned}$$

Hence, the required value is $\frac{9}{32}$.

Given,

Weight of wheat bought = $3\frac{3}{4}$ kg

Weight of rice bought = $2\frac{1}{2}$ kg

Total weight = Weight of wheat + Weight of rice

Total weight $= 3\frac{3}{4} + 2\frac{1}{2} = \frac{15}{4} + \frac{5}{2}$

LCM of 4 and 2 = 4.

$$\begin{aligned}= \frac{15 \times 1}{4 \times 1} + \frac{5 \times 2}{2 \times 2} \\[1em] = \frac{15}{4} + \frac{10}{4} \\[1em] = \frac{15 + 10}{4} \\[1em] = \frac{25}{4} \\[1em] = 6\frac{1}{4} \text{ kg}\end{aligned}$$

Hence, the total weight of wheat and rice is $6\frac{1}{4}$ kg.

LCM of 5 and 10 = 10.

$\frac{3}{5} = \frac{3 \times 2}{10} = \frac{6}{10} \text{ and } \frac{7}{10} = \frac{7}{10}$

Since 7 > 6, $\frac{7}{10} \gt \frac{3}{5}$

Difference = $\frac{7}{10} - \frac{6}{10} = \frac{1}{10}$

Hence, $\frac{7}{10}$ is greater than $\frac{3}{5}\text { by } \frac{1}{10}$.

Let the required number be x.

Then, $8\frac{2}{3} + x = 12\frac{5}{6}$

$$\begin{aligned}\Rightarrow x = 12\frac{5}{6} - 8\frac{2}{3} \\[1em] = \frac{77}{6} - \frac{26}{3} \\[1em] = \frac{77}{6} - \frac{26 \times 2}{6} \\[1em] = \frac{77 - 52}{6} = \frac{25}{6} = 4\frac{1}{6}\end{aligned}$$

Hence, the required number is $4\frac{1}{6}$.

Let the required number be x.

Then, $8\frac{3}{4} - x = 2\frac{2}{3}$

$$\begin{aligned}\Rightarrow x = 8\frac{3}{4} - 2\frac{2}{3} \\[1em] = \frac{35}{4} - \frac{8}{3} \\[1em] = \frac{35 \times 3}{12} - \frac{8 \times 4}{12} \\[1em] = \frac{105 - 32}{12} = \frac{73}{12} = 6\frac{1}{12}\end{aligned}$$

Hence, the required number is $6\frac{1}{12}$.

Given,

Length of the field = $16\frac{1}{2}$ m

Breadth of the field = $12\frac{2}{5}$ m

Length = $16\frac{1}{2} \text { m }= \frac{33}{2}$ m and breadth $= 12\frac{2}{5}\text { m }= \frac{62}{5}$ m

Perimeter = 2 x (length + breadth)

$$\begin{aligned}= 2 \times \left(\frac{33}{2} + \frac{62}{5}\right) \\[1em] = 2 \times \left(\frac{33 \times 5}{2 \times 5} + \frac{62 \times 2}{5 \times 2}\right) \\[1em] = 2 \times \left(\frac{165}{10} + \frac{124}{10}\right) \\[1em] = 2 \times \frac{165 + 124}{10} \\[1em] = 2 \times \frac{289}{10} \\[1em] = \frac{289}{5} = 57\frac{4}{5} \text{ m}\end{aligned}$$

Hence, the perimeter of the field is $57\frac{4}{5} \text { m }.$

Given,

Cost of sugar per kg = ₹ $37\frac{1}{2}$

Quantity of sugar = $8\frac{3}{4}$ kg

Cost of $8\frac{3}{4}$ kg sugar = Cost per kg × Quantity

$$\begin{aligned}= 37\frac{1}{2} \times 8\frac{3}{4} \\[1em] = \frac{75}{2} \times \frac{35}{4} \\[1em] = \frac{75 \times 35}{2 \times 4} = \frac{2625}{8} \\[1em] = 328\frac{1}{8}\end{aligned}$$

Hence, the cost of $8\frac{3}{4}$ kg sugar is ₹ $328\frac{1}{8}$.

Given,

Distance run on 1 litre of petrol = $31\frac{1}{4}$ km

Quantity of petrol = $1\frac{3}{5}$ litres

Distance run on $1\frac{3}{5}$ litres = Distance per litre × Quantity of petrol

$$\begin{aligned}= 31\frac{1}{4} \times 1\frac{3}{5} \\[1em] = \frac{125}{4} \times \frac{8}{5} \\[1em] = \frac{125 \times 8}{4 \times 5} = \frac{1000}{20} \\[1em] = 50 \text{ km}\end{aligned}$$

Hence, the motor cycle will run 50 km.

Given,

Length of the park = $23\frac{2}{5}$ m

Breadth of the park = $16\frac{2}{3}$ m

Area = length × breadth

$$\begin{aligned}\text{Area } = 23\frac{2}{5} \times 16\frac{2}{3} \\[1em] = \frac{117}{5} \times \frac{50}{3} \\[1em] = 39 \times 10 \\[1em] = 390 \text{ m}^2.\end{aligned}$$

Hence, the area of the park is 390 m².

Given,

Number of identical boxes = 40

Weight of each box = $4\frac{4}{5}$ kg

Total weight = Number of boxes × Weight of each box

Total weight $= 40 \times 4\frac{4}{5} = 40 \times \frac{24}{5}$

= $\frac{40 \times 24}{5} = \frac{960}{5} = 192 \text{ kg}$

Hence, the total weight of all the boxes is 192 kg.

Given,

Total quantity of wheat = 24 kg

Fraction of wheat consumed = $\frac{5}{6}$

Fraction of wheat left $= 1 - \frac{5}{6} = \frac{6 - 5}{6} = \frac{1}{6}$

Wheat left = Fraction left × Total quantity

Wheat left $= \frac{1}{6} \times 24 = 4$ kg

Hence, 4 kg of wheat is still left.

Given,

Total length of the rod = $2\frac{2}{5}$ metre

Number of equal parts = 5

Length of each part = Total length ÷ Number of parts

Length of each part $= 2\frac{2}{5} \div 5 = \frac{12}{5} \div 5$

= $\frac{12}{5} \times \frac{1}{5} = \frac{12}{25} \text{ m}$

Hence, the length of each part is $\frac{12}{25}$ m.

A $= 3\frac{3}{8} = \frac{27}{8}$ and B $= 6\frac{5}{8} = \frac{53}{8}$

(i) A ÷ B $= \frac{27}{8} \div \frac{53}{8} = \frac{27}{8} \times \frac{8}{53} = \frac{27}{53}$

Hence, A ÷ B = $\frac{27}{53}$.

(ii) B ÷ A

$$\begin{aligned}= \frac{53}{8} \div \frac{27}{8} \\[1em] = \frac{53}{8} \times \frac{8}{27}\\[1em] = \frac{53}{27} = 1\frac{26}{27}\end{aligned}$$

Hence, B ÷ A = $1\frac{26}{27}$.

Given,

Quantity of oil = $3\frac{5}{7}$ litres

Cost of $3\frac{5}{7}$ litres of oil = ₹ $83\frac{1}{2}$

Cost of one litre = Total cost ÷ Quantity of oil

Cost of one litre $= 83\frac{1}{2} \div 3\frac{5}{7}$

$$\begin{aligned}= \frac{167}{2} \div \frac{26}{7} \\[1em] = \frac{167}{2} \times \frac{7}{26} \\[1em] = \frac{167 \times 7}{2 \times 26} = \frac{1169}{52} \\[1em] = 22\frac{25}{52}\end{aligned}$$

Hence, the cost of one litre of oil is ₹ $22\frac{25}{52}$.

Let the other number be x.

Then, $6\frac{2}{3} \times x = 20\frac{5}{7}$

$$\begin{aligned}\Rightarrow x = 20\frac{5}{7} \div 6\frac{2}{3} \\[1em] = \frac{145}{7} \div \frac{20}{3} \\[1em] = \frac{145}{7} \times \frac{3}{20} \\[1em] = \frac{145 \times 3}{7 \times 20} = \frac{435}{140} \\[1em] = \frac{87}{28} = 3\frac{3}{28}\end{aligned}$$

Hence, the other number is $3\frac{3}{28}$.

Let the required number be x.

Then, $5\frac{5}{6} \times x = 3\frac{1}{3}$

$$\begin{aligned}\Rightarrow x = 3\frac{1}{3} \div 5\frac{5}{6} \\[1em] = \frac{10}{3} \div \frac{35}{6} \\[1em] = \frac{10}{3} \times \frac{6}{35} \\[1em] = \frac{10 \times 6}{3 \times 35} = \frac{60}{105} \\[1em] = \frac{4}{7}\end{aligned}$$

Hence, the required number is $\frac{4}{7}$.

Simplifying,

$$\begin{aligned}6 + \Bigg\lbrace\frac{4}{3} + \left(\frac{3}{4} - \frac{1}{3}\right)\Bigg\rbrace \\[1em] = 6 + \Bigg\lbrace\frac{4}{3} + \left(\frac{9 - 4}{12}\right)\Bigg\rbrace \\[1em] = 6 + \Bigg\lbrace\frac{4}{3} + \frac{5}{12}\Bigg\rbrace \\[1em] = 6 + \Bigg\lbrace\frac{16}{12} + \frac{5}{12}\Bigg\rbrace \\[1em] = 6 + \frac{21}{12} \\[1em] = 6 + \frac{7}{4} \\[1em] = \frac{24 + 7}{4} = \frac{31}{4} = 7\frac{3}{4}\end{aligned}$$

Hence, the required value is $7\frac{3}{4}$.

Simplifying,

$$\begin{aligned}8 - \Bigg\lbrace\frac{3}{2} + \left(\frac{3}{5} - \frac{1}{2}\right)\Bigg\rbrace \\[1em] = 8 - \Bigg\lbrace\frac{3}{2} + \left(\frac{6 - 5}{10}\right)\Bigg\rbrace \\[1em] = 8 - \Bigg\lbrace\frac{3}{2} + \frac{1}{10}\Bigg\rbrace \\[1em] = 8 - \Bigg\lbrace\frac{15}{10} + \frac{1}{10}\Bigg\rbrace \\[1em] = 8 - \frac{16}{10} \\[1em] = 8 - \frac{8}{5} \\[1em] = \frac{40 - 8}{5} \\[1em] = \frac{32}{5} = 6\frac{2}{5}\end{aligned}$$

Hence, the required value is $6\frac{2}{5}$.

Simplifying,

$$\begin{aligned}\frac{1}{4}\left(\frac{1}{4} + \frac{1}{3}\right) - \frac{2}{5} \\[1em] = \frac{1}{4}\left(\frac{3 + 4}{12}\right) - \frac{2}{5} \\[1em] = \frac{1}{4} \times \frac{7}{12} - \frac{2}{5} \\[1em] = \frac{7}{48} - \frac{2}{5} \\[1em] = \frac{7 \times 5}{240} - \frac{2 \times 48}{240} \\[1em] = \frac{35 - 96}{240} = -\frac{61}{240}\end{aligned}$$

Hence, the required value is $-\frac{61}{240}$.

Simplifying,

$$\begin{aligned}2\frac{3}{4} - \left[3\frac{1}{8} \div \Bigg\lbrace 5 - \left(4\frac{2}{3} - \frac{11}{12}\right)\Bigg\rbrace\right] \\[1em] = 2\frac{3}{4} - \left[3\frac{1}{8} \div \Bigg\lbrace 5 - \left(\frac{14}{3} - \frac{11}{12}\right)\Bigg\rbrace\right] \\[1em] = 2\frac{3}{4} - \left[3\frac{1}{8} \div \Bigg\lbrace 5 - \left(\frac{56 - 11}{12}\right)\Bigg\rbrace\right] \\[1em] = 2\frac{3}{4} - \left[3\frac{1}{8} \div \Bigg\lbrace 5 - \frac{45}{12}\Bigg\rbrace\right] \\[1em] = 2\frac{3}{4} - \left[3\frac{1}{8} \div \Bigg\lbrace 5 - \frac{15}{4}\Bigg\rbrace\right] \\[1em] = 2\frac{3}{4} - \left[3\frac{1}{8} \div \left(\frac{20 - 15}{4}\right)\right] \\[1em] = 2\frac{3}{4} - \left[3\frac{1}{8} \div \frac{5}{4}\right] \\[1em] = 2\frac{3}{4} - \left[\frac{25}{8} \times \frac{4}{5}\right] \\[1em] = 2\frac{3}{4} - \frac{5}{2} \\[1em] = \frac{11}{4} - \frac{10}{4} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the required value is $\frac{1}{4}$.

Simplifying,

$$\begin{aligned}12\frac{1}{2} - \left[8\frac{1}{2} + \Bigg\lbrace 9 - \left(5 - \overline{3 - 2}\right)\Bigg\rbrace\right] \\[1em] = 12\frac{1}{2} - \left[8\frac{1}{2} + \Bigg\lbrace 9 - \left(5 - 1\right)\Bigg\rbrace\right] \\[1em] = 12\frac{1}{2} - \left[8\frac{1}{2} + \Bigg\lbrace 9 - 4\Bigg\rbrace\right] \\[1em] = 12\frac{1}{2} - \left[8\frac{1}{2} + 5\right] \\[1em] = 12\frac{1}{2} - 13\frac{1}{2} \\[1em] = \frac{25}{2} - \frac{27}{2} \\[1em] = -\frac{2}{2} = -1\end{aligned}$$

Hence, the required value is -1.

Simplifying,

$$\begin{aligned}1\frac{1}{5} \div \Bigg\lbrace 2\frac{1}{3} - \left(5 + \overline{2 - 3}\right)\Bigg\rbrace - 3\frac{1}{2} \\[1em] = 1\frac{1}{5} \div \Bigg\lbrace 2\frac{1}{3} - \left(5 + (-1)\right)\Bigg\rbrace - 3\frac{1}{2} \\[1em] = 1\frac{1}{5} \div \Bigg\lbrace 2\frac{1}{3} - 4\Bigg\rbrace - 3\frac{1}{2} \\[1em] = 1\frac{1}{5} \div \left(\frac{7}{3} - \frac{12}{3}\right) - 3\frac{1}{2} \\[1em] = 1\frac{1}{5} \div \left(-\frac{5}{3}\right) - 3\frac{1}{2} \\[1em] = \frac{6}{5} \times \left(-\frac{3}{5}\right) - 3\frac{1}{2} \\[1em] = -\frac{18}{25} - \frac{7}{2} \\[1em] = -\frac{36}{50} - \frac{175}{50} \\[1em] = -\frac{211}{50} = -4\frac{11}{50}\end{aligned}$$

Hence, the required value is $-4\frac{11}{50}$.

Simplifying,

$$\begin{aligned}\left(\frac{1}{2} + \frac{2}{3}\right) \div \left(\frac{3}{4} - \frac{2}{9}\right) \\[1em] = \left(\frac{3 + 4}{6}\right) \div \left(\frac{27 - 8}{36}\right) \\[1em] = \frac{7}{6} \div \frac{19}{36} \\[1em] = \frac{7}{6} \times \frac{36}{19} \\[1em] = \frac{7 \times 36}{6 \times 19} = \frac{252}{114} \\[1em] = \frac{42}{19} = 2\frac{4}{19}\end{aligned}$$

Hence, the required value is $2\frac{4}{19}$.

Simplifying,

$$\begin{aligned}\frac{6}{5} \text{ of } \left(3\frac{1}{3} - 2\frac{1}{2}\right) \div \left(2\frac{5}{21} - 2\right) \\[1em] = \frac{6}{5} \text{ of } \left(\frac{10}{3} - \frac{5}{2}\right) \div \left(\frac{47}{21} - 2\right) \\[1em] = \frac{6}{5} \text{ of } \left(\frac{20 - 15}{6}\right) \div \left(\frac{47 - 42}{21}\right) \\[1em] = \frac{6}{5} \text{ of } \frac{5}{6} \div \frac{5}{21} \\[1em] = \frac{6}{5} \times \frac{5}{6} \div \frac{5}{21} \\[1em] = 1 \div \frac{5}{21} \\[1em] = 1 \times \frac{21}{5} = \frac{21}{5} = 4\frac{1}{5}\end{aligned}$$

Hence, the required value is $4\frac{1}{5}$.

Simplifying,

$$\begin{aligned}10\frac{1}{8} \text{ of } \frac{4}{5} \div \frac{35}{36} \text{ of } \frac{20}{49} \\[1em] = \left(\frac{81}{8} \times \frac{4}{5}\right) \div \left(\frac{35}{36} \times \frac{20}{49}\right) \\[1em] = \frac{81}{10} \div \frac{25}{63} \\[1em] = \frac{81}{10} \times \frac{63}{25} \\[1em] = \frac{81 \times 63}{10 \times 25} = \frac{5103}{250} \\[1em] = 20\frac{103}{250}\end{aligned}$$

Hence, the required value is $20\frac{103}{250}$.

Simplifying,

$$\begin{aligned}5\frac{3}{4} - \frac{3}{7} \times 15\frac{3}{4} + 2\frac{2}{35} \div 1\frac{11}{25} \\[1em] = 5\frac{3}{4} - \frac{3}{7} \times \frac{63}{4} + \frac{72}{35} \div \frac{36}{25} \\[1em] = 5\frac{3}{4} - \frac{3}{7} \times \frac{63}{4} + \frac{72}{35} \times \frac{25}{36} \\[1em] = 5\frac{3}{4} - \frac{27}{4} + \frac{10}{7} \\[1em] = \frac{23}{4} - \frac{27}{4} + \frac{10}{7} \\[1em] = -\frac{4}{4} + \frac{10}{7} \\[1em] = -1 + \frac{10}{7} \\[1em] = \frac{-7 + 10}{7} = \frac{3}{7}\end{aligned}$$

Hence, the required value is $\frac{3}{7}$.

Simplifying,

$$\begin{aligned}\frac{3}{4} \text{ of } 7\frac{3}{7} - 5\frac{3}{5} \div 3\frac{4}{15} \\[1em] = \frac{3}{4} \times \frac{52}{7} - \frac{28}{5} \div \frac{49}{15} \\[1em] = \frac{3}{4} \times \frac{52}{7} - \frac{28}{5} \times \frac{15}{49} \\[1em] = \frac{39}{7} - \frac{12}{7} \\[1em] = \frac{27}{7} = 3\frac{6}{7}\end{aligned}$$

Hence, the required value is $3\frac{6}{7}$.

Given,

Length of line AB = 6 cm

Length of line CD = 15 cm

(i) Required fraction $= \frac{AB}{CD} = \frac{6}{15} = \frac{2}{5}$

Hence, the required fraction is $\frac{2}{5}$.

(ii) Solving,

$$\begin{aligned}\frac{1}{2} \text{ of } AB = \frac{1}{2} \times 6 = 3 \text { cm and }\\[1em] \frac{1}{3} \text{ of CD }= \frac{1}{3} \times 15 = 5 \text{cm}\end{aligned}$$

Required fraction $= \frac{3}{5}$

Hence, the required fraction is $\frac{3}{5}$.

(iii) $\frac{1}{5}\text{ of CD }= \frac{1}{5} \times 15 = 3$ cm

Required fraction = $\frac{3}{6} = \frac{1}{2}$

Hence, the required fraction is $\frac{1}{2}$.

Solving, $\left(\frac{2}{7} - \frac{5}{21}\right)$ :

$\frac{2}{7} - \frac{5}{21} = \frac{6 - 5}{21} = \frac{1}{21}$

Sum of $\frac{3}{4}, \frac{5}{7} \text{ and } \frac{7}{12}$ (LCM of 4, 7 and 12 = 84):

$$\begin{aligned}\Rightarrow \frac{3}{4} + \frac{5}{7} + \frac{7}{12}\\[1em] = \frac{63}{84} + \frac{60}{84} + \frac{49}{84} \\[1em] = \frac{172}{84}\\[1em] = \frac{43}{21}\end{aligned}$$

Now subtracting $\left(\frac{2}{7} - \frac{5}{21}\right)$ from the sum of $\frac{3}{4}, \frac{5}{7} \text{ and } \frac{7}{12}$ :

$\frac{43}{21} - \frac{1}{21} = \frac{42}{21} = 2$

Hence, the required value is 2.

Given,

Total weight of the sack of potatoes = 120 kg

Portions sold = 6 kg, $5\frac{1}{4}$ kg, $9\frac{1}{2}$ kg and $9\frac{3}{4}$ kg

(i) Total sold = sum of all the portions sold

$$\begin{aligned}\Rightarrow 6 + 5\frac{1}{4} + 9\frac{1}{2} + 9\frac{3}{4} \\[1em] = 6 + \frac{21}{4} + \frac{19}{2} + \frac{39}{4}\end{aligned}$$

LCM of 4 and 2 = 4.

$$\begin{aligned}= \frac{24}{4} + \frac{21}{4} + \frac{38}{4} + \frac{39}{4} \\[1em] = \frac{122}{4} = \frac{61}{2} \\[1em] = 30\frac{1}{2} \text{ kg}\end{aligned}$$

Hence, he sold $30\frac{1}{2}$ kg.

(ii) Potatoes left

$$\begin{aligned}= 120 - 30\frac{1}{2} \\[1em] = \frac{240}{2} - \frac{61}{2} \\[1em] = \frac{179}{2} = 89\frac{1}{2} \text{ kg}\end{aligned}$$

Hence, $89\frac{1}{2}$ kg are still left in the sack.

Given,

Hours worked on six consecutive days = 8 hours, $7\frac{1}{2}$ hours, $8\frac{1}{4}$ hours, $6\frac{1}{4}$ hours, $6\frac{3}{4}$ hours and 7 hours

Rate of pay = ₹ 36 per hour

Total hours worked = sum of the hours worked on the six days

$$\begin{aligned}= 8 + 7\frac{1}{2} + 8\frac{1}{4} + 6\frac{1}{4} + 6\frac{3}{4} + 7 \\[1em] = 8 + \frac{15}{2} + \frac{33}{4} + \frac{25}{4} + \frac{27}{4} + 7\end{aligned}$$

LCM of 2 and 4 = 4.

$$\begin{aligned}= \frac{32}{4} + \frac{30}{4} + \frac{33}{4} + \frac{25}{4} + \frac{27}{4} + \frac{28}{4} \\[1em] = \frac{175}{4} = 43\frac{3}{4} \text{ hours}\end{aligned}$$

Money earned = No. of hours worked × Rate per hour

$$\begin{aligned}= 43\frac{3}{4} \times 36 = \frac{175}{4} \times 36 \\[1em] = \frac{175 \times 36}{4} = \frac{6300}{4} = 1575\end{aligned}$$

Hence, the boy will earn ₹ 1,575.

Given,

Yellow ribbon bought = $4\frac{1}{3}$ m

Red ribbon bought = $6\frac{1}{6}$ m

Blue ribbon bought = $3\frac{2}{9}$ m

Total ribbon bought = Yellow ribbon + Red ribbon + Blue ribbon

Total ribbon $= 4\frac{1}{3} + 6\frac{1}{6} + 3\frac{2}{9} = \frac{13}{3} + \frac{37}{6} + \frac{29}{9}$

LCM of 3, 6 and 9 = 18.

$$\begin{aligned}= \frac{78}{18} + \frac{111}{18} + \frac{58}{18} \\[1em] = \frac{247}{18} = 13\frac{13}{18} \text{ metres}\end{aligned}$$

Hence, the student bought $13\frac{13}{18}$ metres of ribbon.

Given,

Total investment = ₹ 40,000

Ram's share of the investment = $\frac{3}{5}$ of the total

Deepak's share of the investment = $\frac{2}{5}$ of the total

Ram's investment = $\frac{3}{5}$ × Total investment

Ram's investment $= \frac{3}{5} \times 40000 = \frac{3 \times 40000}{5} = 24000$

Deepak's investment = $\frac{2}{5}$ × Total investment

Deepak's investment $= \frac{2}{5} \times 40000 = \frac{2 \times 40000}{5} = 16000$

Hence, Ram's investment is ₹ 24,000 and Deepak's investment is ₹ 16,000.

Given,

Total problems for homework = 30

Problems worked out = $\frac{2}{3}$ of the total

Problems worked out = $\frac{2}{3}$ × Total problems

Problems worked out $= \frac{2}{3} \times 30 = 20$

Problems still left = Total problems - Problems worked out

Problems still left = 30 - 20 = 10

Hence, 10 problems were still left to be worked out.

Given,

Marked price of the picture = ₹ 90

The picture was sold at $\frac{3}{4}$ of its marked price.

Sale price = $\frac{3}{4}$ × Marked price

Sale price = $\frac{3}{4} \times 90 = \frac{3 \times 90}{4} = \frac{270}{4} = 67.50$

Hence, the sale price was ₹ 67.50.

Given,

Number of parcels = 15

Weight of each parcel = $10\frac{1}{2}$ kg

Total weight = Number of parcels × Weight of each parcel

Total weight $= 15 \times 10\frac{1}{2} = 15 \times \frac{21}{2} = \frac{315}{2} = 157.5$ kg

Hence, the total weight of the parcels is 157.5 kg.

Given,

Total length of the rope = $25\frac{1}{2}$ m

Length of each piece = $1\frac{1}{2}$ m

Number of pieces = Total length ÷ Length of each piece

Number of pieces $= 25\frac{1}{2} \div 1\frac{1}{2} = \frac{51}{2} \div \frac{3}{2}$

= $\frac{51}{2} \times \frac{2}{3} = \frac{51}{3} = 17$

Hence, 17 pieces can be cut out from the rope.

Given,

Height of the first pole = $14\frac{1}{4}$ m

Height of the second pole = $22\frac{1}{3}$ m

Difference in heights = Height of second pole - Height of first pole

Difference in heights $= 22\frac{1}{3} - 14\frac{1}{4} = \frac{67}{3} - \frac{57}{4}$

LCM of 3 and 4 = 12.

$= \frac{268}{12} - \frac{171}{12} = \frac{97}{12} = 8\frac{1}{12} \text{ m}$

Hence, the second pole is $8\frac{1}{12}$ m higher than the first pole.

Given,

Vijay's initial weight = $65\frac{1}{2}$ kg

Weight gained in the first week = $1\frac{2}{5}$ kg

Weight gained in the second week = $1\frac{1}{4}$ kg

Weight lost in the third week = $\frac{5}{16}$ kg

Weight after third week = Initial weight + Gain in first week + Gain in second week - Loss in third week

Weight after third week $= 65\frac{1}{2} + 1\frac{2}{5} + 1\frac{1}{4} - \frac{5}{16}$

= $\frac{131}{2} + \frac{7}{5} + \frac{5}{4} - \frac{5}{16}$

LCM of 2, 5, 4 and 16 = 80.

$$\begin{aligned}= \frac{5240}{80} + \frac{112}{80} + \frac{100}{80} - \frac{25}{80} \\[1em] = \frac{5240 + 112 + 100 - 25}{80} = \frac{5427}{80} = 67\frac{67}{80} \text{ kg}\end{aligned}$$

Hence, Vijay's weight after the third week was $67\frac{67}{80}$ kg.

Given,

Fraction of salary spent on food = $\frac{2}{5}$

Fraction of salary spent on house rent, electricity, etc. = $\frac{3}{10}$

Total fraction spent = Fraction on food + Fraction on house rent, electricity, etc.

Total fraction spent $= \frac{2}{5} + \frac{3}{10} = \frac{4}{10} + \frac{3}{10} = \frac{7}{10}$

Fraction left $= 1 - \frac{7}{10} = \frac{10 - 7}{10} = \frac{3}{10}$

Hence, $\frac{3}{10}$ of his salary is still left with him.

Given,

Fraction of salary spent on food = $\frac{2}{5}$

Fraction of the remaining salary spent on house rent, electricity, etc. = $\frac{3}{10}$

Fraction spent on food = $\frac{2}{5}$

Remaining = 1 - $\frac{2}{5} = \frac{3}{5}$

Fraction spent on house rent, etc. $= \frac{3}{10} \text{ of } \frac{3}{5} = \frac{3}{10} \times \frac{3}{5} = \frac{9}{50}$

Fraction left = $\frac{3}{5} - \frac{9}{50} = \frac{30}{50} - \frac{9}{50} = \frac{21}{50}$

Hence, $\frac{21}{50}$ of his salary is still left with him.

Given,

Price of the refrigerator = ₹ 5,000

Fraction of the price paid in cash = $\frac{1}{10}$

Number of equal monthly instalments for the rest = 12

Amount paid in cash = $\frac{1}{10}$ × Price

Amount paid in cash = $\frac{1}{10} \times 5000 = 500$

Remaining amount = ₹ 5,000 - 500 = ₹ 4,500

Amount paid each month = Remaining amount ÷ Number of instalments

Amount paid each month = $\frac{4500}{12} = 375$

Hence, he had to pay ₹ 375 each month.

Given,

Fraction of the length in mud = $\frac{1}{2}$

Fraction of the length in water = $\frac{1}{3}$

(i) Fraction in mud and water $= \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6}$

Fraction above water $= 1 - \frac{5}{6} = \frac{1}{6}$

Hence, $\frac{1}{6}$ of its length is above the water.

(ii) Let the whole length be $x$ m.

$$\begin{aligned}\Rightarrow \frac{1}{6} \times x = 3\frac{1}{3} \\[1em] \Rightarrow \frac{x}{6} = \frac{10}{3} \\[1em] \Rightarrow x = \frac{10}{3} \times 6 = 20\end{aligned}$$

Hence, the whole length of the lamp post is 20 m.

Given,

Fraction of savings spent = $\frac{3}{5}$

Amount still left = ₹ 2,000

Fraction of savings left $= 1 - \frac{3}{5} = \frac{2}{5}$

Let the savings be $x$.

$$\begin{aligned}\frac{2}{5} \times x = 2000 \\[1em] \Rightarrow x = 2000 \times \frac{5}{2} = 5000\end{aligned}$$

Hence, my savings were ₹ 5,000.

Given,

Fraction of children who are boys = $\frac{4}{5}$

Number of girls = 200

Fraction of girls $= 1 - \frac{4}{5} = \frac{1}{5}$

Let the total number of children be $x$.

$$\begin{aligned}\frac{1}{5} \times x = 200 \\[1em] \Rightarrow x = 200 \times 5 = 1000\end{aligned}$$

Number of boys $= \frac{4}{5} \times 1000 = 800$

Hence, the number of boys is 800.

Given,

Worth of $\frac{4}{5}$ of the estate = ₹ 42,000

Let the worth of the whole estate be $x$.

$$\begin{aligned}\frac{4}{5} \times x = 42000 \\[1em] \Rightarrow x = 42000 \times \frac{5}{4} = 52500\end{aligned}$$

Worth of the whole estate = ₹ 52,500

Value of $\frac{3}{7}\text { of it }= \frac{3}{7} \times 52500 = 22500$

Hence, the worth of the whole estate is ₹ 52,500 and the value of $\frac{3}{7}$ of it is ₹ 22,500.

Given,

Fraction of the journey covered = $\frac{3}{4}$

Distance covered in that fraction = 16 km

Let the whole journey be $x$ km.

$$\begin{aligned}\frac{3}{4} \times x = 16 \\[1em] \Rightarrow x = 16 \times \frac{4}{3} = \frac{64}{3}\end{aligned}$$

Journey still left $= x - 16 = \frac{64}{3} - 16 = \frac{64 - 48}{3} = \frac{16}{3} = 5\frac{1}{3}$ km

Hence, $5\frac{1}{3}$ km of journey is still left.

Given,

Distance travelled = 25 km

Fraction of the journey still left = $\frac{3}{5}$

Fraction of journey travelled $= 1 - \frac{3}{5} = \frac{2}{5}$

Let the whole journey be $x$ km.

$$\begin{aligned}\frac{2}{5} \times x = 25 \\[1em] \Rightarrow x = 25 \times \frac{5}{2} = \frac{125}{2} = 62\frac{1}{2}\end{aligned}$$

Hence, the length of the whole journey is $62\frac{1}{2}$ km.

Given,

Fraction of the land bought by Rajesh = $\frac{1}{3}$ of the total

Fraction of the land bought by Manoj = $\frac{1}{3}$ of the remaining

Land still left unsold = 600 m²

Let the total area be $x$ m².

Area bought by Rajesh $= \frac{1}{3}x$

Remaining $= x - \frac{1}{3}x = \frac{2}{3}x$

Area bought by Manoj $= \frac{1}{3} \text{ of } \frac{2}{3}x = \frac{1}{3} \times \frac{2}{3}x = \frac{2}{9}x$

Land left unsold

$$\begin{aligned}= x - \frac{1}{3}x - \frac{2}{9}x \\[1em] = \frac{9x - 3x - 2x}{9} \\[1em] = \frac{4x}{9}\\[1em] \frac{4x}{9} = 600 \\[1em] \Rightarrow x = 600 \times \frac{9}{4} = 1350\end{aligned}$$

Hence, the total area of the piece of land is 1350 m².

Given,

Money the boy initially had = ₹ 2,400

Fraction of money spent on clothes = $\frac{3}{5}$

Fraction of the remaining money spent on shoes = $\frac{1}{4}$

Money spent on clothes = $\frac{3}{5}$ × Total money

Money spent on clothes $= \frac{3}{5} \times 2400 = 1440$

Remaining money = 2400 - 1440 = 960

Money spent on shoes = $\frac{1}{4}$ × Remaining money

Money spent on shoes $= \frac{1}{4} \times 960 = 240$

Hence, he spent ₹ 240 on shoes.

Given,

Money the boy initially had = ₹ 2,400

Fraction of money spent on clothes = $\frac{3}{5}$

Fraction of money spent on shoes = $\frac{1}{4}$

Money spent on shoes $= \frac{1}{4}$ of his money $= \frac{1}{4} \times 2400 = 600$

Hence, he spent ₹ 600 on shoes.

Comparing $\frac{2}{7} \text{ and } \frac{4}{5}$ by cross-multiplication:

$2 \times 5 = 10 \text{ and } 7 \times 4 = 28$

Since $10 \lt 28$, we have $\frac{2}{7} \lt \frac{4}{5}$.

∴ Mercy solved the lesser part.

Hence, Option 1 is the correct option.

LCM of the denominators 8, 12 and 48 = 48.

Converting each fraction to denominator 48:

$$\begin{aligned}\frac{7}{8} = \frac{7 \times 6}{8 \times 6} = \frac{42}{48} \\[1em] \frac{17}{12} = \frac{17 \times 4}{12 \times 4} = \frac{68}{48} \\[1em] \frac{41}{48} = \frac{41}{48}\end{aligned}$$

The equivalent fractions are $\frac{42}{48}, \frac{68}{48} \text{ and } \frac{41}{48}$.

Since 48 does not divide 360 a whole number of times for $\frac{41}{48}$ (as $360 \div 48 = 7.5$), the fractions cannot all be written with denominator 360, so none of the given options is correct.

Hence, Option 4 is the correct option.

The difference between two fractions is found by subtracting the smaller from the larger.

LCM of 5 and 10 = 10.

$$\begin{aligned}\Rightarrow \frac{7}{10} - \frac{3}{5} \\[1em] = \frac{7}{10} - \frac{6}{10} \\[1em] = \frac{7 - 6}{10} = \frac{1}{10}\end{aligned}$$

Hence, Option 2 is the correct option.

$$\begin{aligned}\Rightarrow 2\frac{1}{3} + 4\frac{1}{3} - 3\frac{1}{3}\\[1em] = \frac{7}{3} + \frac{13}{3} - \frac{10}{3} \\[1em] = \frac{7 + 13 - 10}{3} = \frac{10}{3} = 3\frac{1}{3}\end{aligned}$$

Hence, Option 1 is the correct option.

$$\begin{aligned}\frac{2}{7} \text{ of } \frac{3}{4} = \frac{2}{7} \times \frac{3}{4} = \frac{6}{28} = \frac{3}{14} \\[1em] \frac{3}{5} \text{ of } \frac{5}{8} = \frac{3}{5} \times \frac{5}{8} = \frac{15}{40} = \frac{3}{8}\end{aligned}$$

Comparing $\frac{3}{14} \text{ and } \frac{3}{8}$ (equal numerators), the fraction with the smaller denominator is greater.

Since $8 \lt 14$, we have $\frac{3}{8} \gt \frac{3}{14}$.

∴ $\frac{3}{5} \text{ of } \frac{5}{8}$ is greater.

Hence, Option 2 is the correct option.

Required number $= 5\frac{2}{3} \div 7\frac{1}{3}$

$= \frac{17}{3} \div \frac{22}{3} = \frac{17}{3} \times \frac{3}{22} = \frac{17}{22}$

Hence, Option 3 is the correct option.

Rate per kg $= 600 \div 3\frac{3}{4}$

$= 600 \div \frac{15}{4} = 600 \times \frac{4}{15} = \frac{2400}{15} = 160$

Hence, Option 3 is the correct option.

Work done in 1 hour $= \frac{1}{3}$

∴ Work done in $2\frac{1}{5} \text { hours }= \frac{1}{3} \times 2\frac{1}{5}$

Hence, Option 4 is the correct option.

Perimeter $= 2 \times (\text{length} + \text{breadth})$

$$\begin{aligned}= 2 \times \left(11\frac{1}{2} + 8\frac{1}{2}\right) \\[1em] = 2 \times \left(\frac{23}{2} + \frac{17}{2}\right) \\[1em] = 2 \times \frac{40}{2}\\[1em] = 2 \times 20 = 40 \text{ cm}\end{aligned}$$

Hence, Option 2 is the correct option.

Total hours = hours per day x number of days $= \left(2\frac{1}{7} \times 17\right)$ hours

Hence, Option 3 is the correct option.

Removing brackets in order — vinculum first, then parentheses, curly brackets and square brackets:

$$\begin{aligned}\frac{1}{3} - \left[\frac{1}{3} - \Bigg\lbrace \frac{1}{3} - \left(\frac{1}{3} - \overline{\frac{1}{3} - \frac{1}{3}}\right)\Bigg\rbrace\right] \\[1em] = \frac{1}{3} - \left[\frac{1}{3} - \Bigg\lbrace \frac{1}{3} - \left(\frac{1}{3} - 0\right)\Bigg\rbrace\right] \\[1em] = \frac{1}{3} - \left[\frac{1}{3} - \Bigg\lbrace \frac{1}{3} - \frac{1}{3}\Bigg\rbrace\right] \\[1em] = \frac{1}{3} - \left[\frac{1}{3} - 0\right] \\[1em] = \frac{1}{3} - \frac{1}{3} = 0\end{aligned}$$

Hence, Option 3 is the correct option.

To divide by $\frac{3}{7}$, we multiply by its reciprocal $\frac{7}{3}$:

$\frac{1}{5} \div \frac{3}{7} \times \frac{5}{4} = \frac{1}{5} \times \frac{7}{3} \times \frac{5}{4}$

Hence, Option 2 is the correct option.

$$\begin{aligned}\Rightarrow \Bigg\lbrace \frac{3}{5} \div \left(\frac{3}{5} + \frac{2}{5}\right)\Bigg\rbrace \div \frac{3}{5} \\[1em] = \Bigg\lbrace \frac{3}{5} \div \frac{5}{5}\Bigg\rbrace \div \frac{3}{5} \\[1em] = \Bigg\lbrace \frac{3}{5} \div 1\Bigg\rbrace \div \frac{3}{5} = \frac{3}{5} \div \frac{3}{5} \\[1em] = \frac{3}{5} \times \frac{5}{3} = 1\end{aligned}$$

Hence, Option 4 is the correct option.

1 hour = 60 minutes.

∴ Required fraction $= \frac{40}{60}$

Hence, Option 2 is the correct option.

Let Shyam's salary be ₹ $x$.

$$\begin{aligned}\frac{3}{11} \times x = 7260 \\[1em] \Rightarrow x = 7260 \times \frac{11}{3} = 2420 \times 11 = 26620\end{aligned}$$

Hence, Option 4 is the correct option.

Let the required number be $x$.

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

Hence, Option 3 is the correct option.

For Statement 1, the fraction between $\frac{1}{2} \text{ and } \frac{2}{3} \text{ is } \frac{1+2}{2+3} = \frac{3}{5}$.

Converting to a common denominator 30:

$\frac{1}{2} = \frac{15}{30}, \quad \frac{3}{5} = \frac{18}{30}, \quad \frac{2}{3} = \frac{20}{30}$

Since $\frac{15}{30} \lt \frac{18}{30} \lt \frac{20}{30}, \frac{3}{5}$ lies between $\frac{1}{2} \text{ and } \frac{2}{3}$. Statement 1 is true.

Statement 2 gives the general rule used above, which is true.

∴ Both the statements are true.

Hence, Option 1 is the correct option.

LCM of denominators 5, 15, 2 and 3 = 30. Converting to like fractions:

$\frac{7}{5} = \frac{42}{30}, \quad \frac{2}{15} = \frac{4}{30}, \quad \frac{5}{2} = \frac{75}{30}, \quad \frac{1}{3} = \frac{10}{30}$

The correct descending order is $\frac{75}{30} \gt \frac{42}{30} \gt \frac{10}{30} \gt \frac{4}{30}$, i.e. $\frac{5}{2} \gt \frac{7}{5} \gt \frac{1}{3} \gt \frac{2}{15}$.

The order stated in Statement 1 is $\frac{7}{5} \gt \frac{5}{2} \gt \frac{2}{15} \gt \frac{1}{3}$, which is incorrect. Statement 1 is false.

Statement 2 correctly describes comparison using like fractions, so it is true.

∴ Statement 1 is false, and statement 2 is true.

Hence, Option 4 is the correct option.

The product of $\frac{3}{4} \text{ and } 1\frac{1}{5}$ is :

$$\begin{aligned}\frac{3}{4} \times 1\frac{1}{5} \\[1em] = \frac{3}{4} \times \frac{6}{5} \\[1em] = \frac{18}{20} \\[1em] = \frac{9}{10}\end{aligned}$$

Checking the order:

$\frac{3}{4} = 0.75$, $\frac{9}{10} = 0.9 \text{ and } 1\frac{1}{5} = 1.2,$

$\text{ so }\frac{3}{4} \lt \frac{9}{10} \lt 1\frac{1}{5}$.

Thus, Assertion (A) is true.

The Reason states the product is less than the proper fraction but greater than the improper fraction. In fact the product is greater than the proper fraction but less than the improper (mixed) fraction, so the Reason is false.

Thus, Reason (R) is false.

∴ A is true, R is false.

Hence, Option 1 is the correct option.

The product of $1\frac{1}{3} \text{ and } 2\frac{1}{2}$ is :

$$\begin{aligned}\Rightarrow 1\frac{1}{3} \times 2\frac{1}{2} \\[1em] = \frac{4}{3} \times \frac{5}{2}\\[1em] = \frac{20}{6} \\[1em] = \frac{10}{3}\\[1em] = 3\frac{1}{3}\end{aligned}$$

Since $3\frac{1}{3}$ is greater than both $1\frac{1}{3} \text{ and } 2\frac{1}{2}$,

Thus, Assertion (A) is true.

We know that,

The product of two improper positive fractions is greater than each of the improper fractions multiplied together.

Thus, Reason (R) is true.

∴ Both A and R are true.

Hence, Option 3 is the correct option.

For the Assertion, HCF of 32 and 56 is 8.

$\frac{32}{56} = \frac{32 \div 8}{56 \div 8} = \frac{4}{7}$

Thus, Assertion (A) is true.

The Reason states the correct condition for two fractions to be equivalent, namely ps = rq.

Checking: $32 \times 7 = 224 \text{ and } 4 \times 56 = 224$, so the condition holds.

Thus, Reason (R) is true.

∴ Both A and R are true.

Hence, Option 3 is the correct option.