Chapter 3: Fractions

Exercise 3(A)

Question 1

Classify each fraction, given below, as decimal or vulgar fraction, proper or improper fraction and mixed fraction:

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

(ii) $\dfrac{11}{10}$

(iii) $\dfrac{13}{20}$

(iv) $\dfrac{18}{7}$

(v) $3\dfrac{2}{9}$

Answer

(i) $\dfrac{3}{5}$ : 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) $\dfrac{11}{10}$ : 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) $\dfrac{13}{20}$ : 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) $\dfrac{18}{7}$ : 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) $3\dfrac{2}{9}$ : 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.

Question 2

Express the following improper fractions as mixed fractions:

(i) $\dfrac{18}{5}$

(ii) $\dfrac{7}{4}$

(iii) $\dfrac{25}{6}$

(iv) $\dfrac{38}{5}$

(v) $\dfrac{22}{5}$

Answer

(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.

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

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

(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.

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

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

(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.

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

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

(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.

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

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

(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.

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

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

Question 3

Express the following mixed fractions as improper fractions:

(i) $2\dfrac{4}{9}$

(ii) $7\dfrac{5}{13}$

(iii) $3\dfrac{1}{4}$

(iv) $2\dfrac{5}{48}$

(v) $12\dfrac{7}{11}$

Answer

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

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

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

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

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

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

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

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

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

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

Question 4

Reduce the given fractions to lowest terms:

(i) $\dfrac{8}{18}$

(ii) $\dfrac{27}{36}$

(iii) $\dfrac{18}{42}$

(iv) $\dfrac{35}{75}$

(v) $\dfrac{18}{45}$

Answer

(i) HCF of 8 and 18 = 2.

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

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

(ii) HCF of 27 and 36 = 9.

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

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

(iii) HCF of 18 and 42 = 6.

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

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

(iv) HCF of 35 and 75 = 5.

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

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

(v) HCF of 18 and 45 = 9.

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

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

Question 5

State true or false:

(i) $\dfrac{30}{40} \text{ and } \dfrac{12}{16}$ are equivalent fractions.

(ii) $\dfrac{10}{25} \text{ and } \dfrac{25}{10}$ are equivalent fractions.

(iii) $\dfrac{35}{49}, \dfrac{20}{28}, \dfrac{45}{63} \text{ and } \dfrac{100}{140}$ are equivalent fractions.

Answer

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

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

Hence, the statement is True.

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

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

Hence, the statement is False.

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

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

Hence, the statement is True.

Question 6

Distinguish each of the fractions, given below, as a simple fraction or a complex fraction:

(i) $\dfrac{0}{8}$

(ii) $\dfrac{3}{8}$

(iii) $\dfrac{5}{7}$

(iv) $\dfrac{3\dfrac{3}{5}}{18}$

(v) $\dfrac{6}{2\dfrac{2}{5}}$

(vi) $\dfrac{3\dfrac{1}{3}}{7\dfrac{2}{7}}$

(vii) $\dfrac{5\dfrac{2}{9}}{5}$

(viii) $\dfrac{8}{0}$

Answer

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

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

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

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

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

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

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

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

Exercise 3(B)

Question 1

For each pair, given below, state whether it forms like fractions or unlike fractions:

(i) $\dfrac{5}{8} \text{ and } \dfrac{7}{8}$

(ii) $\dfrac{8}{15} \text{ and } \dfrac{8}{21}$

(iii) $\dfrac{4}{9} \text{ and } \dfrac{9}{4}$

Answer

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

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

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

Question 2

Convert given fractions into fractions with equal denominators:

(i) $\dfrac{5}{6} \text{ and } \dfrac{7}{9}$

(ii) $\dfrac{2}{3}, \dfrac{5}{6} \text{ and } \dfrac{7}{12}$

(iii) $\dfrac{4}{5}, \dfrac{17}{20}, \dfrac{23}{40} \text{ and } \dfrac{11}{16}$

Answer

(i) LCM of 6 and 9 = 18.

$\dfrac{5}{6} = \dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18} \\[1em] \dfrac{7}{9} = \dfrac{7 \times 2}{9 \times 2} = \dfrac{14}{18}$

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

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

$\dfrac{2}{3} = \dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12} \\[1em] \dfrac{5}{6} = \dfrac{5 \times 2}{6 \times 2} = \dfrac{10}{12} \\[1em] \dfrac{7}{12} = \dfrac{7}{12}$

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

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

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

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

Question 3

Convert given fractions into fractions with equal numerators:

(i) $\dfrac{8}{9} \text{ and } \dfrac{12}{17}$

(ii) $\dfrac{6}{13}, \dfrac{15}{23} \text{ and } \dfrac{12}{17}$

(iii) $\dfrac{15}{19}, \dfrac{25}{28}, \dfrac{9}{11} \text{ and } \dfrac{45}{47}$

Answer

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

$\dfrac{8}{9} = \dfrac{8 \times 3}{9 \times 3} = \dfrac{24}{27} \\[1em] \dfrac{12}{17} = \dfrac{12 \times 2}{17 \times 2} = \dfrac{24}{34}$

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

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

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

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

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

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

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

Question 4

Compare the given fractions by making the denominators equal:

(i) $\dfrac{2}{5} \text{ and } \dfrac{4}{9}$

(ii) $\dfrac{5}{7} \text{ and } \dfrac{8}{11}$

(iii) $\dfrac{7}{15} \text{ and } \dfrac{9}{20}$

Answer

(i) LCM of 5 and 9 = 45.

$\dfrac{2}{5} = \dfrac{2 \times 9}{5 \times 9} = \dfrac{18}{45} \\[1em] \dfrac{4}{9} = \dfrac{4 \times 5}{9 \times 5} = \dfrac{20}{45}$

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

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

(ii) LCM of 7 and 11 = 77.

$\dfrac{5}{7} = \dfrac{5 \times 11}{7 \times 11} = \dfrac{55}{77} \\[1em] \dfrac{8}{11} = \dfrac{8 \times 7}{11 \times 7} = \dfrac{56}{77}$

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

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

(iii) LCM of 15 and 20 = 60.

$\dfrac{7}{15} = \dfrac{7 \times 4}{15 \times 4} = \dfrac{28}{60} \\[1em] \dfrac{9}{20} = \dfrac{9 \times 3}{20 \times 3} = \dfrac{27}{60}$

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

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

Question 5

Compare the given fractions by making the numerators equal:

(i) $\dfrac{4}{9} \text{ and } \dfrac{2}{5}$

(ii) $\dfrac{5}{12} \text{ and } \dfrac{8}{19}$

(iii) $\dfrac{5}{7} \text{ and } \dfrac{9}{14}$

Answer

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

$\dfrac{4}{9} = \dfrac{4}{9} \\[1em] \dfrac{2}{5} = \dfrac{2 \times 2}{5 \times 2} = \dfrac{4}{10}$

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

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

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

$\dfrac{5}{12} = \dfrac{5 \times 8}{12 \times 8} = \dfrac{40}{96} \\[1em] \dfrac{8}{19} = \dfrac{8 \times 5}{19 \times 5} = \dfrac{40}{95}$

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

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

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

$\dfrac{5}{7} = \dfrac{5 \times 9}{7 \times 9} = \dfrac{45}{63} \\[1em] \dfrac{9}{14} = \dfrac{9 \times 5}{14 \times 5} = \dfrac{45}{70}$

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

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

Question 6

Compare the given fractions by cross-multiplication method:

(i) $\dfrac{2}{5} \text{ and } \dfrac{4}{9}$

(ii) $\dfrac{3}{8} \text{ and } \dfrac{6}{11}$

(iii) $\dfrac{5}{18} \text{ and } \dfrac{11}{21}$

Answer

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

Since 18 < 20,

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

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

Since 33 < 48,

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

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

Since 105 < 198,

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

Question 7

Arrange the given fractions in ascending order by making the denominators equal:

(i) $\dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4} \text{ and } \dfrac{1}{6}$

(ii) $\dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12} \text{ and } \dfrac{3}{10}$

(iii) $\dfrac{5}{7}, \dfrac{3}{8}, \dfrac{9}{14} \text{ and } \dfrac{20}{21}$

Answer

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

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

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

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

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

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

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

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

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

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

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

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

Question 8

Arrange the given fractions in descending order by making the numerators equal:

(i) $\dfrac{5}{6}, \dfrac{4}{15}, \dfrac{8}{9} \text{ and } \dfrac{1}{3}$

(ii) $\dfrac{3}{7}, \dfrac{4}{9}, \dfrac{5}{7} \text{ and } \dfrac{8}{11}$

(iii) $\dfrac{1}{10}, \dfrac{6}{11}, \dfrac{8}{11} \text{ and } \dfrac{3}{5}$

Answer

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

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

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 $\dfrac{8}{9}, \dfrac{5}{6}, \dfrac{1}{3}, \dfrac{4}{15}$ .

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

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

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 $\dfrac{8}{11}, \dfrac{5}{7}, \dfrac{4}{9}, \dfrac{3}{7}$ .

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

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

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 $\dfrac{8}{11}, \dfrac{3}{5}, \dfrac{6}{11}, \dfrac{1}{10}$ .

Exercise 3(C)

Question 1(i)

Reduce to a single fraction:

$\dfrac{1}{2} + \dfrac{2}{3}$

Answer

LCM of 2 and 3 = 6.

Solving,

$\Rightarrow \dfrac{1}{2} + \dfrac{2}{3} \\[1em] = \dfrac{1 \times 3}{6} + \dfrac{2 \times 2}{6} \\[1em] = \dfrac{3 + 4}{6} = \dfrac{7}{6} \\[1em] = 1\dfrac{1}{6}$

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

Question 1(ii)

Reduce to a single fraction:

$\dfrac{3}{5} - \dfrac{1}{10}$

Answer

LCM of 5 and 10 = 10.

$\Rightarrow \dfrac{3}{5} - \dfrac{1}{10} = \dfrac{3 \times 2}{10} - \dfrac{1}{10}\\[1em] = \dfrac{6 - 1}{10} = \dfrac{5}{10}\\[1em] = \dfrac{1}{2}$

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

Question 1(iii)

Reduce to a single fraction:

$\dfrac{2}{3} - \dfrac{1}{6}$

Answer

LCM of 3 and 6 = 6.

$\Rightarrow \dfrac{2}{3} - \dfrac{1}{6} = \dfrac{2 \times 2}{6} - \dfrac{1}{6}\\[1em] = \dfrac{4 - 1}{6} = \dfrac{3}{6}\\[1em] = \dfrac{1}{2}$

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

Question 1(iv)

Reduce to a single fraction:

$1\dfrac{1}{3} + 2\dfrac{1}{4}$

Answer

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

LCM of 3 and 4 = 12.

$= \dfrac{4 \times 4}{12} + \dfrac{9 \times 3}{12}\\[1em] = \dfrac{16 + 27}{12} \\[1em] = \dfrac{43}{12}\\[1em] = 3\dfrac{7}{12}$

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

Question 1(v)

Reduce to a single fraction:

$\dfrac{1}{4} + \dfrac{5}{6} - \dfrac{1}{12}$

Answer

LCM of 4, 6 and 12 = 12.

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

Hence, the required fraction is 1.

Question 1(vi)

Reduce to a single fraction:

$\dfrac{2}{3} - \dfrac{3}{5} + 3 - \dfrac{1}{5}$

Answer

LCM of 1, 3 and 5 = 15.

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

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

Question 1(vii)

Reduce to a single fraction:

$\dfrac{2}{3} - \dfrac{1}{5} + \dfrac{1}{10}$

Answer

LCM of 3, 5 and 10 = 30.

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

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

Question 1(viii)

Reduce to a single fraction:

$2\dfrac{1}{2} + 2\dfrac{1}{3} - 1\dfrac{1}{4}$

Answer

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

LCM of 2, 3 and 4 = 12.

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

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

Question 1(ix)

Reduce to a single fraction:

$2\dfrac{5}{8} - 2\dfrac{1}{6} + 4\dfrac{3}{4}$

Answer

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

LCM of 8, 6 and 4 = 24.

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

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

Question 2(i)

Simplify:

$\dfrac{3}{4} \times 6$

Answer

Solving,

$\Rightarrow \dfrac{3}{4} \times 6 \\[1em] = \dfrac{3 \times 6}{4} \\[1em] = \dfrac{18}{4} \\[1em] = \dfrac{9}{2} \\[1em] = 4\dfrac{1}{2}$

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

Question 2(ii)

Simplify:

$\dfrac{2}{3} \times 15$

Answer

Solving,

$\Rightarrow \dfrac{2}{3} \times 15 \\[1em] = \dfrac{2 \times 15}{3} \\[1em] = \dfrac{30}{3} \\[1em] = 10$

Hence, the required value is 10.

Question 2(iii)

Simplify:

$\dfrac{3}{4} \times \dfrac{1}{2}$

Answer

Solving,

$\Rightarrow \dfrac{3}{4} \times \dfrac{1}{2} \\[1em] = \dfrac{3 \times 1}{4 \times 2} \\[1em] = \dfrac{3}{8}$

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

Question 2(iv)

Simplify:

$\dfrac{9}{12} \times \dfrac{4}{7}$

Answer

Solving,

$\Rightarrow \dfrac{9}{12} \times \dfrac{4}{7} \\[1em] = \dfrac{9 \times 4}{12 \times 7} \\[1em] = \dfrac{36}{84} \\[1em] = \dfrac{3}{7}$

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

Question 2(v)

Simplify:

$45 \times 2\dfrac{1}{3}$

Answer

Solving,

$\Rightarrow 45 \times 2\dfrac{1}{3} \\[1em] = 45 \times \dfrac{7}{3} \\[1em] = \dfrac{45 \times 7}{3} \\[1em] = \dfrac{315}{3} \\[1em] = 105$

Hence, the required value is 105.

Question 2(vi)

Simplify:

$36 \times 3\dfrac{1}{4}$

Answer

Solving,

$\Rightarrow 36 \times 3\dfrac{1}{4} \\[1em] = 36 \times \dfrac{13}{4} \\[1em] = \dfrac{36 \times 13}{4} \\[1em] = \dfrac{468}{4} \\[1em] = 117$

Hence, the required value is 117.

Question 2(vii)

Simplify:

$2 \div \dfrac{1}{3}$

Answer

Solving,

$\Rightarrow 2 \div \dfrac{1}{3} \\[1em] = 2 \times \dfrac{3}{1} \\[1em] = 6$

Hence, the required value is 6.

Question 2(viii)

Simplify:

$3 \div \dfrac{2}{5}$

Answer

Solving,

$\Rightarrow 3 \div \dfrac{2}{5} \\[1em] = 3 \times \dfrac{5}{2} \\[1em] = \dfrac{15}{2} \\[1em] = 7\dfrac{1}{2}$

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

Question 2(ix)

Simplify:

$1 \div \dfrac{3}{5}$

Answer

Solving,

$\Rightarrow 1 \div \dfrac{3}{5} \\[1em] = 1 \times \dfrac{5}{3} \\[1em] = \dfrac{5}{3} \\[1em] = 1\dfrac{2}{3}$

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

Question 2(x)

Simplify:

$\dfrac{1}{3} \div \dfrac{1}{4}$

Answer

Solving,

$\Rightarrow \dfrac{1}{3} \div \dfrac{1}{4} \\[1em] = \dfrac{1}{3} \times \dfrac{4}{1} \\[1em] = \dfrac{4}{3} \\[1em] = 1\dfrac{1}{3}$

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

Question 2(xi)

Simplify:

$-\dfrac{5}{8} \div \dfrac{3}{4}$

Answer

Solving,

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

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

Question 2(xii)

Simplify:

$3\dfrac{3}{7} \div 1\dfrac{1}{14}$

Answer

Solving,

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

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

Question 2(xiii)

Simplify:

$3\dfrac{3}{4} \times 1\dfrac{1}{5} \times \dfrac{20}{21}$

Answer

Solving,

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

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

Question 3(i)

Subtract:

2 from $\dfrac{2}{3}$

Answer

Solving,

$\Rightarrow \dfrac{2}{3} - 2 \\[1em] \Rightarrow \dfrac{2}{3} - \dfrac{6}{3} \\[1em] \Rightarrow \dfrac{2 - 6}{3} \\[1em] \Rightarrow -\dfrac{4}{3} \\[1em] \Rightarrow -1\dfrac{1}{3}$

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

Question 3(ii)

Subtract:

$\dfrac{1}{8} \text{ from } \dfrac{5}{8}$

Answer

Solving,

$\dfrac{5}{8} - \dfrac{1}{8} = \dfrac{5 - 1}{8} \\[1em] = \dfrac{4}{8} = \dfrac{1}{2}$

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

Question 3(iii)

Subtract:

$-\dfrac{2}{5} \text{ from } \dfrac{2}{5}$

Answer

Solving,

$\dfrac{2}{5} - \left(-\dfrac{2}{5}\right) = \dfrac{2}{5} + \dfrac{2}{5} \\[1em] = \dfrac{2 + 2}{5} = \dfrac{4}{5}$

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

Question 3(iv)

Subtract:

$-\dfrac{3}{7} \text{ from } \dfrac{3}{7}$

Answer

Solving,

$\dfrac{3}{7} - \left(-\dfrac{3}{7}\right) = \dfrac{3}{7} + \dfrac{3}{7} \\[1em] = \dfrac{3 + 3}{7} = \dfrac{6}{7}$

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

Question 3(v)

Subtract:

0 from $-\dfrac{4}{5}$

Answer

Solving,

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

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

Question 3(vi)

Subtract:

$\dfrac{2}{9} \text{ from } \dfrac{4}{5}$

Answer

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

LCM of 5 and 9 = 45.

$= \dfrac{4 \times 9}{45} - \dfrac{2 \times 5}{45} \\[1em] = \dfrac{36 - 10}{45} = \dfrac{26}{45}$

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

Question 3(vii)

Subtract:

$-\dfrac{4}{7} \text{ from } -\dfrac{6}{11}$

Answer

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

LCM of 11 and 7 = 77.

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

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

Question 4(i)

Find the value of:

$\dfrac{1}{2}$ of 10 kg

Answer

Solving,

$\Rightarrow \dfrac{1}{2} \text{ of } 10 \text{ kg} \\[1em] \Rightarrow \dfrac{1}{2} \times 10 \\[1em] \Rightarrow 5 \text{ kg}$

Hence, the required value is 5 kg.

Question 4(ii)

Find the value of:

$\dfrac{3}{5}$ of 1 hour

Answer

1 hour = 60 minutes.

Solving,

$\Rightarrow \dfrac{3}{5} \text{ of } 1 \text{ hour} \\[1em] \Rightarrow \dfrac{3}{5} \times 60 \\[1em] \Rightarrow 36 \text{ minutes}$

Hence, the required value is 36 minutes.

Question 4(iii)

Find the value of:

$\dfrac{4}{7} \text{ of } 2\dfrac{1}{3}$ kg

Answer

Solving,

$\Rightarrow \dfrac{4}{7} \text{ of } 2\dfrac{1}{3} \text{ kg} \\[1em] \Rightarrow \dfrac{4}{7} \times \dfrac{7}{3} \\[1em] \Rightarrow \dfrac{4 \times 7}{7 \times 3} \\[1em] \Rightarrow \dfrac{4}{3} \\[1em] \Rightarrow 1\dfrac{1}{3} \text{ kg}$

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

Question 4(iv)

Find the value of:

$3\dfrac{1}{2}$ times of 2 metre

Answer

Solving,

$\Rightarrow 3\dfrac{1}{2} \text{ times of } 2 \text{ metre} \\[1em] \Rightarrow \dfrac{7}{2} \times 2 \\[1em] \Rightarrow 7 \text{ metres}$

Hence, the required value is 7 metres.

Question 4(v)

Find the value of:

$\dfrac{1}{2} \text{ of } 2\dfrac{2}{3}$

Answer

Solving,

$\Rightarrow \dfrac{1}{2} \text{ of } 2\dfrac{2}{3} \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{8}{6} \\[1em] \Rightarrow \dfrac{4}{3} \\[1em] \Rightarrow 1\dfrac{1}{3}$

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

Question 4(vi)

Find the value of:

$\dfrac{5}{11} \text{ of } \dfrac{4}{5}$ of 22 kg

Answer

Solving,

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

Hence, the required value is 8 kg.

Question 5(i)

Simplify and reduce to a simple fraction:

$\dfrac{3}{3\dfrac{3}{4}}$

Answer

Solving,

$\Rightarrow \dfrac{3}{3\dfrac{3}{4}} \\[1em] \Rightarrow 3 \div \dfrac{15}{4} \\[1em] \Rightarrow 3 \times \dfrac{4}{15} \\[1em] \Rightarrow \dfrac{12}{15} \\[1em] \Rightarrow \dfrac{4}{5}$

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

Question 5(ii)

Simplify and reduce to a simple fraction:

$\dfrac{\dfrac{3}{5}}{7}$

Answer

Solving,

$\Rightarrow \dfrac{\dfrac{3}{5}}{7} \\[1em] \Rightarrow \dfrac{3}{5} \div 7 \\[1em] \Rightarrow \dfrac{3}{5} \times \dfrac{1}{7} \\[1em] \Rightarrow \dfrac{3}{35}$

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

Question 5(iii)

Simplify and reduce to a simple fraction:

$\dfrac{3}{\dfrac{5}{7}}$

Answer

Solving,

$\Rightarrow \dfrac{3}{\dfrac{5}{7}} \\[1em] \Rightarrow 3 \div \dfrac{5}{7} \\[1em] \Rightarrow 3 \times \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{21}{5} \\[1em] \Rightarrow 4\dfrac{1}{5}$

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

Question 5(iv)

Simplify and reduce to a simple fraction:

$\dfrac{2\dfrac{1}{5}}{1\dfrac{1}{10}}$

Answer

Solving,

$\Rightarrow \dfrac{2\dfrac{1}{5}}{1\dfrac{1}{10}} \\[1em] \Rightarrow \dfrac{11}{5} \div \dfrac{11}{10} \\[1em] \Rightarrow \dfrac{11}{5} \times \dfrac{10}{11} \\[1em] \Rightarrow \dfrac{110}{55} \\[1em] \Rightarrow 2$

Hence, the required value is 2.

Question 5(v)

Simplify and reduce to a simple fraction:

$\dfrac{2}{5} \text{ of } \dfrac{6}{11} \times 1\dfrac{1}{4}$

Answer

Solving,

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

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

Question 5(vi)

Simplify and reduce to a simple fraction:

$2\dfrac{1}{4} \div \dfrac{1}{7} \times \dfrac{1}{3}$

Answer

Solving,

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

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

Question 5(vii)

Simplify and reduce to a simple fraction:

$\dfrac{1}{3} \times 4\dfrac{2}{3} \div 3\dfrac{1}{2} \times \dfrac{1}{2}$

Answer

Solving,

$\Rightarrow \dfrac{1}{3} \times 4\dfrac{2}{3} \div 3\dfrac{1}{2} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{14}{3} \div \dfrac{7}{2} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{14}{9} \times \dfrac{2}{7} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{14 \times 2 \times 1}{9 \times 7 \times 2} \\[1em] \Rightarrow \dfrac{28}{126} \\[1em] \Rightarrow \dfrac{2}{9}$

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

Question 5(viii)

Simplify and reduce to a simple fraction:

$\dfrac{2}{3} \times 1\dfrac{1}{4} \div \dfrac{3}{7} \text{ of } 2\dfrac{5}{8}$

Answer

Solving,

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

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

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

Question 5(ix)

Simplify and reduce to a simple fraction:

$0 \div \dfrac{8}{11}$

Answer

Solving,

$\Rightarrow 0 \div \dfrac{8}{11} \\[1em] \Rightarrow 0 \times \dfrac{11}{8} \\[1em] \Rightarrow 0$

Hence, the required value is 0.

Question 5(x)

Simplify and reduce to a simple fraction:

$\dfrac{4}{5} \div \dfrac{7}{15} \text{ of } \dfrac{8}{9}$

Answer

Solving,

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

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

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

Question 5(xi)

Simplify and reduce to a simple fraction:

$\dfrac{4}{5} \div \dfrac{7}{15} \times \dfrac{8}{9}$

Answer

Solving,

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

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

Question 5(xii)

Simplify and reduce to a simple fraction:

$\dfrac{4}{5} \text{ of } \dfrac{7}{15} \div \dfrac{8}{9}$

Answer

Solving,

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

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

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

Question 5(xiii)

Simplify and reduce to a simple fraction:

$\dfrac{1}{2} \text{ of } \dfrac{3}{4} \times \dfrac{1}{2} \div \dfrac{2}{3}$

Answer

Solving,

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

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

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

Question 6

A bought $3\dfrac{3}{4}$ kg of wheat and $2\dfrac{1}{2}$ kg of rice. Find the total weight of wheat and rice bought by A.

Answer

Given,

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

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

Total weight = Weight of wheat + Weight of rice

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

LCM of 4 and 2 = 4.

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

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

Question 7

Which is greater, $\dfrac{3}{5} \text{ or } \dfrac{7}{10}$ and by how much?

Answer

LCM of 5 and 10 = 10.

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

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

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

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

Question 8

What number should be added to $8\dfrac{2}{3}$ to get $12\dfrac{5}{6}$ ?

Answer

Let the required number be x.

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

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

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

Question 9

What should be subtracted from $8\dfrac{3}{4}$ to get $2\dfrac{2}{3}$ ?

Answer

Let the required number be x.

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

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

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

Question 10

A rectangular field is $16\dfrac{1}{2}$ m long and $12\dfrac{2}{5}$ m wide. Find the perimeter of the field.

Answer

Given,

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

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

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

Perimeter = 2 x (length + breadth)

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

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

Question 11

Sugar costs ₹ $37\dfrac{1}{2}$ per kg. Find the cost of $8\dfrac{3}{4}$ kg sugar.

Answer

Given,

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

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

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

$= 37\dfrac{1}{2} \times 8\dfrac{3}{4} \\[1em] = \dfrac{75}{2} \times \dfrac{35}{4} \\[1em] = \dfrac{75 \times 35}{2 \times 4} = \dfrac{2625}{8} \\[1em] = 328\dfrac{1}{8}$

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

Question 12

A motor cycle runs $31\dfrac{1}{4}$ km consuming 1 litre of petrol. How much distance will it run consuming $1\dfrac{3}{5}$ litre of petrol?

Answer

Given,

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

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

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

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

Hence, the motor cycle will run 50 km.

Question 13

A rectangular park has length = $23\dfrac{2}{5}$ m and breadth = $16\dfrac{2}{3}$ m. Find the area of the park.

Answer

Given,

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

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

Area = length × breadth

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

Hence, the area of the park is 390 m².

Question 14

Each of 40 identical boxes weighs $4\dfrac{4}{5}$ kg. Find the total weight of all the boxes.

Answer

Given,

Number of identical boxes = 40

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

Total weight = Number of boxes × Weight of each box

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

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

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

Question 15

Out of 24 kg of wheat, $\dfrac{5}{6}$ th of wheat is consumed. Find how much wheat is still left?

Answer

Given,

Total quantity of wheat = 24 kg

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

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

Wheat left = Fraction left × Total quantity

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

Hence, 4 kg of wheat is still left.

Question 16

A rod of length $2\dfrac{2}{5}$ metre is divided into five equal parts. Find the length of each part so obtained.

Answer

Given,

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

Number of equal parts = 5

Length of each part = Total length ÷ Number of parts

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

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

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

Question 17

If A = $3\dfrac{3}{8}$ and B = $6\dfrac{5}{8}$ , find:

(i) A ÷ B

(ii) B ÷ A.

Answer

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

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

Hence, A ÷ B = $\dfrac{27}{53}$ .

(ii) B ÷ A

$= \dfrac{53}{8} \div \dfrac{27}{8} \\[1em] = \dfrac{53}{8} \times \dfrac{8}{27}\\[1em] = \dfrac{53}{27} = 1\dfrac{26}{27}$

Hence, B ÷ A = $1\dfrac{26}{27}$ .

Question 18

Cost of $3\dfrac{5}{7}$ litres of oil is ₹ $83\dfrac{1}{2}$ . Find the cost of one litre oil.

Answer

Given,

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

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

Cost of one litre = Total cost ÷ Quantity of oil

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

$= \dfrac{167}{2} \div \dfrac{26}{7} \\[1em] = \dfrac{167}{2} \times \dfrac{7}{26} \\[1em] = \dfrac{167 \times 7}{2 \times 26} = \dfrac{1169}{52} \\[1em] = 22\dfrac{25}{52}$

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

Question 19

The product of two numbers is $20\dfrac{5}{7}$ . If one of these numbers is $6\dfrac{2}{3}$ , find the other.

Answer

Let the other number be x.

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

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

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

Question 20

By what number should $5\dfrac{5}{6}$ be multiplied to get $3\dfrac{1}{3}$ ?

Answer

Let the required number be x.

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

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

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

Exercise 3(D)

Question 1

Simplify:

$6 + \Bigg\lbrace\dfrac{4}{3} + \left(\dfrac{3}{4} - \dfrac{1}{3}\right)\Bigg\rbrace$

Answer

Simplifying,

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

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

Question 2

Simplify:

$8 - \Bigg\lbrace\dfrac{3}{2} + \left(\dfrac{3}{5} - \dfrac{1}{2}\right)\Bigg\rbrace$

Answer

Simplifying,

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

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

Question 3

Simplify:

$\dfrac{1}{4}\left(\dfrac{1}{4} + \dfrac{1}{3}\right) - \dfrac{2}{5}$

Answer

Simplifying,

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

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

Question 4

Simplify:

$2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \left(4\dfrac{2}{3} - \dfrac{11}{12}\right)\Bigg\rbrace\right]$

Answer

Simplifying,

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

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

Question 5

Simplify:

$12\dfrac{1}{2} - \left[8\dfrac{1}{2} + \Bigg\lbrace 9 - \left(5 - \overline{3 - 2}\right)\Bigg\rbrace\right]$

Answer

Simplifying,

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

Hence, the required value is -1.

Question 6

Simplify:

$1\dfrac{1}{5} \div \Bigg\lbrace 2\dfrac{1}{3} - \left(5 + \overline{2 - 3}\right)\Bigg\rbrace - 3\dfrac{1}{2}$

Answer

Simplifying,

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

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

Question 7

Simplify:

$\left(\dfrac{1}{2} + \dfrac{2}{3}\right) \div \left(\dfrac{3}{4} - \dfrac{2}{9}\right)$

Answer

Simplifying,

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

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

Question 8

Simplify:

$\dfrac{6}{5} \text{ of } \left(3\dfrac{1}{3} - 2\dfrac{1}{2}\right) \div \left(2\dfrac{5}{21} - 2\right)$

Answer

Simplifying,

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

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

Question 9

Simplify:

$10\dfrac{1}{8} \text{ of } \dfrac{4}{5} \div \dfrac{35}{36} \text{ of } \dfrac{20}{49}$

Answer

Simplifying,

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

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

Question 10

Simplify:

$5\dfrac{3}{4} - \dfrac{3}{7} \times 15\dfrac{3}{4} + 2\dfrac{2}{35} \div 1\dfrac{11}{25}$

Answer

Simplifying,

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

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

Question 11

Simplify:

$\dfrac{3}{4} \text{ of } 7\dfrac{3}{7} - 5\dfrac{3}{5} \div 3\dfrac{4}{15}$

Answer

Simplifying,

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

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

Exercise 3(E)

Question 1

A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is:

(i) the length of AB to that of CD?

(ii) $\dfrac{1}{2}$ the length of AB to that of $\dfrac{1}{3}$ of CD?

(iii) $\dfrac{1}{5}$ of CD to that of AB?

Answer

Given,

Length of line AB = 6 cm

Length of line CD = 15 cm

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

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

(ii) Solving,

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

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

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

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

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

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

Question 2

Subtract $\left(\dfrac{2}{7} - \dfrac{5}{21}\right)$ from the sum of $\dfrac{3}{4}, \dfrac{5}{7} \text{ and } \dfrac{7}{12}$ .

Answer

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

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

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

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

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

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

Hence, the required value is 2.

Question 3

From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, $5\dfrac{1}{4} \text { kg }, 9\dfrac{1}{2} \text { kg and }9\dfrac{3}{4}$ kg respectively.

(i) How many kg did he sell?

(ii) How many kg are still left in the sack?

Answer

Given,

Total weight of the sack of potatoes = 120 kg

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

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

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

LCM of 4 and 2 = 4.

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

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

(ii) Potatoes left

$= 120 - 30\dfrac{1}{2} \\[1em] = \dfrac{240}{2} - \dfrac{61}{2} \\[1em] = \dfrac{179}{2} = 89\dfrac{1}{2} \text{ kg}$

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

Question 4

If a boy works for six consecutive days for 8 hours, $7\dfrac{1}{2}$ hours, $8\dfrac{1}{4}$ hours, $6\dfrac{1}{4}$ hours, $6\dfrac{3}{4}$ hours and 7 hours respectively, how much money will he earn at the rate of ₹ 36 per hour?

Answer

Given,

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

Rate of pay = ₹ 36 per hour

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

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

LCM of 2 and 4 = 4.

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

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

$= 43\dfrac{3}{4} \times 36 = \dfrac{175}{4} \times 36 \\[1em] = \dfrac{175 \times 36}{4} = \dfrac{6300}{4} = 1575$

Hence, the boy will earn ₹ 1,575.

Question 5

A student bought $4\dfrac{1}{3}$ m of yellow ribbon, $6\dfrac{1}{6}$ m of red ribbon and $3\dfrac{2}{9}$ m of blue ribbon for decorating a room. How many metres of ribbon did he buy?

Answer

Given,

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

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

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

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

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

LCM of 3, 6 and 9 = 18.

$= \dfrac{78}{18} + \dfrac{111}{18} + \dfrac{58}{18} \\[1em] = \dfrac{247}{18} = 13\dfrac{13}{18} \text{ metres}$

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

Question 6

In a business, Ram and Deepak invest $\dfrac{3}{5} \text{ and } \dfrac{2}{5}$ of the total investment. If ₹ 40,000 is the total investment, calculate the amount invested by each.

Answer

Given,

Total investment = ₹ 40,000

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

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

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

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

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

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

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

Question 7

Geeta had 30 problems for home work. She worked out $\dfrac{2}{3}$ of them. How many problems were still left to be worked out by her?

Answer

Given,

Total problems for homework = 30

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

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

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

Problems still left = Total problems - Problems worked out

Problems still left = 30 - 20 = 10

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

Question 8

A picture was marked at ₹ 90. It was sold at $\dfrac{3}{4}$ of its marked price. What was the sale price?

Answer

Given,

Marked price of the picture = ₹ 90

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

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

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

Hence, the sale price was ₹ 67.50.

Question 9

Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed $10\dfrac{1}{2}$ kg?

Answer

Given,

Number of parcels = 15

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

Total weight = Number of parcels × Weight of each parcel

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

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

Question 10

A rope is $25\dfrac{1}{2}$ m long. How many pieces each of $1\dfrac{1}{2}$ m length can be cut out from it?

Answer

Given,

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

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

Number of pieces = Total length ÷ Length of each piece

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

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

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

Question 11

The heights of two vertical poles, above the earth's surface, are $14\dfrac{1}{4} \text { m and }22\dfrac{1}{3}$ m respectively. How much higher is the second pole as compared with the height of the first pole?

Answer

Given,

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

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

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

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

LCM of 3 and 4 = 12.

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

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

Question 12

Vijay weighed $65\dfrac{1}{2}$ kg. He gained $1\dfrac{2}{5}$ kg during the first week, $1\dfrac{1}{4}$ kg during the second week, but lost $\dfrac{5}{16}$ kg during the third week. What was his weight after the third week?

Answer

Given,

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

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

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

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

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

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

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

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

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

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

Question 13

A man spends $\dfrac{2}{5}$ of his salary on food and $\dfrac{3}{10}$ on house rent, electricity, etc. What fraction of his salary is still left with him?

Answer

Given,

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

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

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

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

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

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

Question 14

A man spends $\dfrac{2}{5}$ of his salary on food and $\dfrac{3}{10}$ of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him?

Answer

Given,

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

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

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

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

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

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

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

Question 15

Shyam bought a refrigerator for ₹ 5,000. He paid $\dfrac{1}{10}$ of the price in cash and the rest in 12 equal monthly instalments. How much did he have to pay each month?

Answer

Given,

Price of the refrigerator = ₹ 5,000

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

Number of equal monthly instalments for the rest = 12

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

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

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

Amount paid each month = Remaining amount ÷ Number of instalments

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

Hence, he had to pay ₹ 375 each month.

Question 16

A lamp post has half of its length in mud and $\dfrac{1}{3}$ of its length in water.

(i) What fraction of its length is above the water?

(ii) If $3\dfrac{1}{3}$ m of the lamp post is above the water, find the whole length of the lamp post.

Answer

Given,

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

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

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

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

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

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

$\Rightarrow \dfrac{1}{6} \times x = 3\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{x}{6} = \dfrac{10}{3} \\[1em] \Rightarrow x = \dfrac{10}{3} \times 6 = 20$

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

Question 17

I spent $\dfrac{3}{5}$ of my savings and still have ₹ 2,000 left. What were my savings?

Answer

Given,

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

Amount still left = ₹ 2,000

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

Let the savings be $x$ .

$\dfrac{2}{5} \times x = 2000 \\[1em] \Rightarrow x = 2000 \times \dfrac{5}{2} = 5000$

Hence, my savings were ₹ 5,000.

Question 18

In a school $\dfrac{4}{5}$ of the children are boys. If the number of girls is 200, find the number of boys.

Answer

Given,

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

Number of girls = 200

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

Let the total number of children be $x$ .

$\dfrac{1}{5} \times x = 200 \\[1em] \Rightarrow x = 200 \times 5 = 1000$

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

Hence, the number of boys is 800.

Question 19

If $\dfrac{4}{5}$ of an estate is worth ₹ 42,000, find the worth of the whole estate. Also, find the value of $\dfrac{3}{7}$ of it.

Answer

Given,

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

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

$\dfrac{4}{5} \times x = 42000 \\[1em] \Rightarrow x = 42000 \times \dfrac{5}{4} = 52500$

Worth of the whole estate = ₹ 52,500

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

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

Question 20

After going $\dfrac{3}{4}$ of my journey, I find that I have covered 16 km. How much journey is still left?

Answer

Given,

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

Distance covered in that fraction = 16 km

Let the whole journey be $x$ km.

$\dfrac{3}{4} \times x = 16 \\[1em] \Rightarrow x = 16 \times \dfrac{4}{3} = \dfrac{64}{3}$

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

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

Question 21

When Krishna travelled 25 km, he found that $\dfrac{3}{5}$ of his journey was still left. What was the length of the whole journey?

Answer

Given,

Distance travelled = 25 km

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

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

Let the whole journey be $x$ km.

$\dfrac{2}{5} \times x = 25 \\[1em] \Rightarrow x = 25 \times \dfrac{5}{2} = \dfrac{125}{2} = 62\dfrac{1}{2}$

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

Question 22

From a piece of land, one-third is bought by Rajesh and one-third of remaining is bought by Manoj. If 600 m² land is still left unsold, find the total area of the piece of land.

Answer

Given,

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

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

Land still left unsold = 600 m²

Let the total area be $x$ m².

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

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

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

Land left unsold

$= x - \dfrac{1}{3}x - \dfrac{2}{9}x \\[1em] = \dfrac{9x - 3x - 2x}{9} \\[1em] = \dfrac{4x}{9}\\[1em] \dfrac{4x}{9} = 600 \\[1em] \Rightarrow x = 600 \times \dfrac{9}{4} = 1350$

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

Question 23

A boy spent $\dfrac{3}{5}$ of his money on buying clothes and $\dfrac{1}{4}$ of the remaining money on buying shoes. If he initially had ₹ 2,400 how much did he spend on shoes?

Answer

Given,

Money the boy initially had = ₹ 2,400

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

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

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

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

Remaining money = 2400 - 1440 = 960

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

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

Hence, he spent ₹ 240 on shoes.

Question 24

A boy spent $\dfrac{3}{5}$ of his money on buying clothes and $\dfrac{1}{4}$ of his money on buying shoes. If he initially had ₹ 2,400, how much did he spend on shoes?

Answer

Given,

Money the boy initially had = ₹ 2,400

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

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

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

Hence, he spent ₹ 600 on shoes.

Multiple Choice Questions

Question 1

Mercy solved $\dfrac{2}{7}$ part of an exercise while John solved $\dfrac{4}{5}$ of it, who solved lesser part?

Mercy
John
none of these

Answer

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

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

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

∴ Mercy solved the lesser part.

Hence, Option 1 is the correct option.

Question 2

Fractions $\dfrac{7}{8}, \dfrac{17}{12} \text{ and } \dfrac{41}{48}$ in the form of their equivalent fractions are:

$\dfrac{315}{360}, \dfrac{510}{360} \text{ and } \dfrac{126}{360}$
$\dfrac{315}{180}, \dfrac{510}{180} \text{ and } \dfrac{64}{180}$
$\dfrac{7}{360}, \dfrac{17}{360} \text{ and } \dfrac{41}{360}$
none of these

Answer

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

Converting each fraction to denominator 48:

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

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

Since 48 does not divide 360 a whole number of times for $\dfrac{41}{48}$ (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.

Question 3

The difference between the fractions $\dfrac{3}{5} \text{ and } \dfrac{7}{10}$ is:

$-\dfrac{1}{10}$
$\dfrac{1}{10}$
$\dfrac{4}{5}$
$-\dfrac{4}{5}$

Answer

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

LCM of 5 and 10 = 10.

$\Rightarrow \dfrac{7}{10} - \dfrac{3}{5} \\[1em] = \dfrac{7}{10} - \dfrac{6}{10} \\[1em] = \dfrac{7 - 6}{10} = \dfrac{1}{10}$

Hence, Option 2 is the correct option.

Question 4

$2\dfrac{1}{3} + 4\dfrac{1}{3} - 3\dfrac{1}{3}$ is equal to:

$3\dfrac{1}{3}$
3
$2\dfrac{2}{3}$
$3\dfrac{2}{3}$

Answer

$\Rightarrow 2\dfrac{1}{3} + 4\dfrac{1}{3} - 3\dfrac{1}{3}\\[1em] = \dfrac{7}{3} + \dfrac{13}{3} - \dfrac{10}{3} \\[1em] = \dfrac{7 + 13 - 10}{3} = \dfrac{10}{3} = 3\dfrac{1}{3}$

Hence, Option 1 is the correct option.

Question 5

Which of the following is greater? $\dfrac{2}{7} \text{ of } \dfrac{3}{4} \text{ or } \dfrac{3}{5} \text{ of } \dfrac{5}{8}$

$\dfrac{2}{7} \text{ of } \dfrac{3}{4}$
$\dfrac{3}{5} \text{ of } \dfrac{5}{8}$
none of these

Answer

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

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

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

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

Hence, Option 2 is the correct option.

Question 6

By what number should $7\dfrac{1}{3}$ be multiplied to get $5\dfrac{2}{3}$ ?

$\dfrac{22}{17}$
$\dfrac{17}{11}$
$\dfrac{17}{22}$
none of these

Answer

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

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

Hence, Option 3 is the correct option.

Question 7

The cost of $3\dfrac{3}{4}$ kg apples is ₹ 600; the rate of apple per kg is:

₹ 40
₹ 80
₹ 160
₹ $\dfrac{1}{600}$

Answer

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

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

Hence, Option 3 is the correct option.

Question 8

$\dfrac{1}{3}$ part of a work is done in one hour. The part of work done in $2\dfrac{1}{5}$ hours is:

$\dfrac{1}{3} \div 2\dfrac{1}{5}$
$2\dfrac{1}{5} \div \dfrac{1}{3}$
$2\dfrac{1}{5} \div 3$
$\dfrac{1}{3} \times 2\dfrac{1}{5}$

Answer

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

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

Hence, Option 4 is the correct option.

Question 9

A rectangular sheet of paper is $11\dfrac{1}{2}$ cm long and $8\dfrac{1}{2}$ cm wide. Its perimeter is:

20 cm
40 cm
$\dfrac{391}{4}$ cm
38 cm

Answer

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

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

Hence, Option 2 is the correct option.

Question 10

Peter can read a novel in 17 days. If he devotes $2\dfrac{1}{7}$ hours per day, in how many hours (in all) will he read the whole novel:

$17 \div 2\dfrac{1}{7}$ hours
$2\dfrac{1}{7} \div 17$ hours
$\left(2\dfrac{1}{7} \times 17\right)$ hours
none of these

Answer

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

Hence, Option 3 is the correct option.

Question 11

$\dfrac{1}{3} - \left[\dfrac{1}{3} - \Bigg\lbrace \dfrac{1}{3} - \left(\dfrac{1}{3} - \overline{\dfrac{1}{3} - \dfrac{1}{3}}\right)\Bigg\rbrace\right]$ is equal to:

$\dfrac{1}{3}$
$\dfrac{2}{3}$
0
1

Answer

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

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

Hence, Option 3 is the correct option.

Question 12

$\dfrac{1}{5} \div \dfrac{3}{7} \times \dfrac{5}{4}$ is equal to:

$\dfrac{1}{5} \times \dfrac{7 \times 4}{3 \times 5}$
$\dfrac{1}{5} \times \dfrac{7}{3} \times \dfrac{5}{4}$
$5 \times \dfrac{3 \times 5}{7 \times 4}$
none of these

Answer

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

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

Hence, Option 2 is the correct option.

Question 13

$\Bigg\lbrace \dfrac{3}{5} \div \left(\dfrac{3}{5} + \dfrac{2}{5}\right)\Bigg\rbrace \div \dfrac{3}{5}$ is equal to:

$\dfrac{9}{25}$
$\dfrac{25}{9}$
0
1

Answer

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

Hence, Option 4 is the correct option.

Question 14

What fraction of an hour is 40 minutes?

$\dfrac{1}{40}$
$\dfrac{40}{60}$
$\dfrac{60}{40}$
2400

Answer

1 hour = 60 minutes.

∴ Required fraction $= \dfrac{40}{60}$

Hence, Option 2 is the correct option.

Question 15

$\dfrac{3}{11}$ of Shyam's salary is ₹ 7,260, then his salary is:

₹ 13300
₹ 26600
₹ 13320
₹ 26620

Answer

Let Shyam's salary be ₹ $x$ .

$\dfrac{3}{11} \times x = 7260 \\[1em] \Rightarrow x = 7260 \times \dfrac{11}{3} = 2420 \times 11 = 26620$

Hence, Option 4 is the correct option.

Question 16

The number that should divide $2\dfrac{1}{3}$ to get 1 is:

$\dfrac{3}{7}$
$\dfrac{3}{5}$
$2\dfrac{1}{3}$
$1\dfrac{2}{3}$

Answer

Let the required number be $x$ .

$\Rightarrow 2\dfrac{1}{3} \div x = 1 \\[1em] \Rightarrow \dfrac{7}{3} \div x = 1 \\[1em] \Rightarrow x = \dfrac{7}{3} = 2\dfrac{1}{3}$

Hence, Option 3 is the correct option.

Statement I-II Type Questions

Question 17

Statement 1: $\dfrac{3}{5}$ is a fraction between $\dfrac{1}{2} \text{ and } \dfrac{2}{3}$ .

Statement 2: If $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ are two fractions, then $\dfrac{a+c}{b+d}$ lies between $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ .

Which of the following options is correct?

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

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

Converting to a common denominator 30:

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

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

Question 18

Statement 1: $\dfrac{7}{5}, \dfrac{2}{15}, \dfrac{5}{2}, \dfrac{1}{3}$ in descending order of magnitude is $\dfrac{7}{5} \gt \dfrac{5}{2} \gt \dfrac{2}{15} \gt \dfrac{1}{3}$ .

Statement 2: After converting the given fractions into like fractions, the fraction with greater numerator is greater.

Which of the following options is correct?

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

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

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

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

The order stated in Statement 1 is $\dfrac{7}{5} \gt \dfrac{5}{2} \gt \dfrac{2}{15} \gt \dfrac{1}{3}$ , 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.

Assertion-Reason Type Questions

Question 19

Assertion (A): Let us consider the product of a proper fraction $\dfrac{3}{4}$ and a mixed fraction $1\dfrac{1}{5}$ . The product is $\dfrac{9}{10} \text{ and } \dfrac{3}{4} \lt \dfrac{9}{10} \lt 1\dfrac{1}{5}$ .

Reason (R): The product of a proper fraction and a mixed fraction is less than the proper fraction but greater than the improper fraction.

A is true, R is false.
A is false, R is true.
Both A and R are true.
Both A and R are false.

Answer

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

$\dfrac{3}{4} \times 1\dfrac{1}{5} \\[1em] = \dfrac{3}{4} \times \dfrac{6}{5} \\[1em] = \dfrac{18}{20} \\[1em] = \dfrac{9}{10}$

Checking the order:

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

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

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.

Question 20

Assertion (A): The product of two improper fractions $1\dfrac{1}{3} \text{ and } 2\dfrac{1}{2}$ is greater than both $1\dfrac{1}{3} \text{ and } 2\dfrac{1}{2}$ .

Reason (R): The product of two improper positive fractions is greater than each of the improper fractions multiplied together.

A is true, R is false.
A is false, R is true.
Both A and R are true.
Both A and R are false.

Answer

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

$\Rightarrow 1\dfrac{1}{3} \times 2\dfrac{1}{2} \\[1em] = \dfrac{4}{3} \times \dfrac{5}{2}\\[1em] = \dfrac{20}{6} \\[1em] = \dfrac{10}{3}\\[1em] = 3\dfrac{1}{3}$

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

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.

Question 21

Assertion (A): The fraction $\dfrac{32}{56}$ can be reduced to $\dfrac{4}{7}$ .

Reason (R): Two fractions $\dfrac{p}{q} \text{ and } \dfrac{r}{s}$ are said to be equivalent only when ps = rq.

A is true, R is false.
A is false, R is true.
Both A and R are true.
Both A and R are false.

Answer

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

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

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.

Chapter 3

Fractions

Class - 7 Concise Mathematics Selina

Exercise 3(A)

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Exercise 3(B)

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Exercise 3(C)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 1(vi)

Question 1(vii)

Question 1(viii)

Question 1(ix)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 2(vii)

Question 2(viii)

Question 2(ix)

Question 2(x)

Question 2(xi)

Question 2(xii)

Question 2(xiii)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 3(iv)

Question 3(v)

Question 3(vi)

Question 3(vii)

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 4(iv)

Question 4(v)

Question 4(vi)

Question 5(i)

Question 5(ii)

Question 5(iii)

Question 5(iv)

Question 5(v)

Question 5(vi)

Question 5(vii)

Question 5(viii)

Question 5(ix)

Question 5(x)

Question 5(xi)

Question 5(xii)

Question 5(xiii)

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17