Chapter 6: Fractions

Exercise 6(A)

Question 1

For each expression, given below, write a fraction:

(i) 2 out of 7 = ...............

(ii) 5 out of 17 = ...............

(iii) three-fifths = ...............

Answer

As we know, 'out of' means division, so it is written as a fraction $\dfrac{\text{part}}{\text{whole}}$

(i) 2 out of 7

⇒ 2 out of 7 = $\dfrac{2}{7}$ .

Hence, 2 out of 7 = $\dfrac{2}{7}$ .

(ii) 5 out of 17

⇒ 5 out of 17 = $\dfrac{5}{17}$ .

Hence, 5 out of 17 = $\dfrac{5}{17}$ .

(iii) three-fifths

⇒ three-fifths = $\dfrac{3}{5}$ .

Hence, three-fifths = $\dfrac{3}{5}$ .

Question 2(i)

Fill in the blanks:

$\dfrac{5}{8}$ is ............... fraction.

Answer

In $\dfrac{5}{8}$ , numerator (5) < denominator (8).

Hence, $\dfrac{5}{8}$ is a proper fraction.

Question 2(ii)

Fill in the blanks:

$\dfrac{8}{5}$ is ............... fraction.

Answer

In $\dfrac{8}{5}$ , numerator (8) > denominator (5).

Hence, $\dfrac{8}{5}$ is an improper fraction.

Question 2(iii)

Fill in the blanks:

$\dfrac{15}{15}$ is ............... fraction.

Answer

In $\dfrac{15}{15}$ , numerator (15) = denominator (15), i.e. numerator is not less than denominator.

Hence, $\dfrac{15}{15}$ is an improper fraction.

Question 2(iv)

Fill in the blanks:

The value of $\dfrac{23}{23}$ = ............... .

Answer

$\dfrac{23}{23} = 23 \div 23 = 1$ .

Hence, the value of $\dfrac{23}{23}$ is 1.

Question 2(v)

Fill in the blanks:

The value of $\dfrac{5}{5}$ = ...............

Answer

$\dfrac{5}{5} = 5 \div 5 = 1$ .

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

Question 2(vi)

Fill in the blanks:

$3\dfrac{3}{10}$ is ............... fraction.

Answer

$3\dfrac{3}{10}$ consists of a natural number 3 and a proper fraction $\dfrac{3}{10}$ .

Hence, $3\dfrac{3}{10}$ is a mixed fraction.

Question 2(vii)

Fill in the blanks:

$\dfrac{2}{15} \text{ and } \dfrac{7}{15}$ are ............... fractions.

Answer

$\dfrac{2}{15} \text{ and } \dfrac{7}{15}$ have the same denominator 15.

Hence, $\dfrac{2}{15} \text{ and } \dfrac{7}{15}$ are like fractions.

Question 2(viii)

Fill in the blanks:

$\dfrac{23}{12} \text{ and } \dfrac{23}{15}$ are ............... fractions.

Answer

$\dfrac{23}{12} \text{ and } \dfrac{23}{15}$ have different denominators (12 and 15).

Hence, $\dfrac{23}{12} \text{ and } \dfrac{23}{15}$ are unlike fractions.

Question 2(ix)

Fill in the blanks:

$\dfrac{6}{15} \text{ and } \dfrac{28}{70}$ are ............... fractions.

Answer

Reducing to lowest terms, $\dfrac{6}{15} = \dfrac{2}{5} \text{ and } \dfrac{28}{70} = \dfrac{2}{5}$ .

Both are equal to $\dfrac{2}{5}$ .

Hence, $\dfrac{6}{15} \text{ and } \dfrac{28}{70}$ are equal (equivalent) fractions.

Question 2(x)

Fill in the blanks:

$\dfrac{8}{24} \text{ and } \dfrac{8}{32}$ are not ............... fractions.

Answer

$\dfrac{8}{24} \text{ and } \dfrac{8}{32}$ have different denominators (24 and 32).

Hence, $\dfrac{8}{24} \text{ and } \dfrac{8}{32}$ are not like fractions.

Question 2(xi)

Fill in the blanks:

$3\dfrac{2}{13} = \dfrac{3 \times 13 + ...............}{13}$ = ...............

Answer

$3\dfrac{2}{13}$

$\Rightarrow 3\dfrac{2}{13} = \dfrac{3 \times 13 + 2}{13} \\[1em] = \dfrac{39 + 2}{13} \\[1em] = \dfrac{41}{13}.$

Hence, $3\dfrac{2}{13} = \dfrac{3 \times 13 + 2}{13} = \dfrac{41}{13}$ .

Question 2(xii)

Fill in the blanks:

$4\dfrac{3}{5}$ = ............... = ...............

Answer

$4\dfrac{3}{5}$

$\Rightarrow 4\dfrac{3}{5} = \dfrac{4 \times 5 + 3}{5} \\[1em] = \dfrac{20 + 3}{5} \\[1em] = \dfrac{23}{5}.$

Hence, $4\dfrac{3}{5} = \dfrac{4 \times 5 + 3}{5} = \dfrac{23}{5}$ .

Question 3

From the following fractions, separate (i) proper fractions and (ii) improper fractions:

$\dfrac{2}{9}, \dfrac{4}{3}, \dfrac{7}{15}, \dfrac{11}{20}, \dfrac{20}{11}, \dfrac{18}{23}, \dfrac{27}{35}$ .

Answer

A fraction is proper if numerator < denominator, and improper if numerator ≥ denominator.

(i) Proper fractions (numerator < denominator) are:

$\dfrac{2}{9}, \dfrac{7}{15}, \dfrac{11}{20}, \dfrac{18}{23} \text{ and } \dfrac{27}{35}$ .

(ii) Improper fractions (numerator > denominator) are:

$\dfrac{4}{3} \text{ and } \dfrac{20}{11}$ .

Hence, proper fractions are $\dfrac{2}{9}, \dfrac{7}{15}, \dfrac{11}{20}, \dfrac{18}{23}, \dfrac{27}{35}$ and improper fractions are $\dfrac{4}{3}, \dfrac{20}{11}$ .

Question 4(i)

Change the following mixed fractions to improper fractions:

$2\dfrac{1}{5}$

Answer

$2\dfrac{1}{5}$

$\Rightarrow 2\dfrac{1}{5} = \dfrac{2 \times 5 + 1}{5} \\[1em] = \dfrac{10 + 1}{5} \\[1em] = \dfrac{11}{5}.$

Hence, $2\dfrac{1}{5} = \dfrac{11}{5}$ .

Question 4(ii)

Change the following mixed fractions to improper fractions:

$3\dfrac{1}{4}$

Answer

$3\dfrac{1}{4}$

$\Rightarrow 3\dfrac{1}{4} = \dfrac{3 \times 4 + 1}{4} \\[1em] = \dfrac{12 + 1}{4} \\[1em] = \dfrac{13}{4}.$

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

Question 4(iii)

Change the following mixed fractions to improper fractions:

$7\dfrac{1}{8}$

Answer

$7\dfrac{1}{8}$

$\Rightarrow 7\dfrac{1}{8} = \dfrac{7 \times 8 + 1}{8} \\[1em] = \dfrac{56 + 1}{8} \\[1em] = \dfrac{57}{8}.$

Hence, $7\dfrac{1}{8} = \dfrac{57}{8}$ .

Question 4(iv)

Change the following mixed fractions to improper fractions:

$2\dfrac{1}{11}$

Answer

$2\dfrac{1}{11}$

$\Rightarrow 2\dfrac{1}{11} = \dfrac{2 \times 11 + 1}{11} \\[1em] = \dfrac{22 + 1}{11} \\[1em] = \dfrac{23}{11}.$

Hence, $2\dfrac{1}{11} = \dfrac{23}{11}$ .

Question 5(i)

Change the following improper fractions to mixed fractions:

$\dfrac{100}{17}$

Answer

$\dfrac{100}{17}$

Dividing 100 by 17, we get quotient = 5 and remainder = 15.

$\Rightarrow \dfrac{100}{17} = 5\dfrac{15}{17}.$

Hence, $\dfrac{100}{17} = 5\dfrac{15}{17}$ .

Question 5(ii)

Change the following improper fractions to mixed fractions:

$\dfrac{81}{11}$

Answer

$\dfrac{81}{11}$

Dividing 81 by 11, we get quotient = 7 and remainder = 4.

$\Rightarrow \dfrac{81}{11} = 7\dfrac{4}{11}.$

Hence, $\dfrac{81}{11} = 7\dfrac{4}{11}$ .

Question 5(iii)

Change the following improper fractions to mixed fractions:

$\dfrac{209}{7}$

Answer

$\dfrac{209}{7}$

Dividing 209 by 7, we get quotient = 29 and remainder = 6.

$\Rightarrow \dfrac{209}{7} = 29\dfrac{6}{7}.$

Hence, $\dfrac{209}{7} = 29\dfrac{6}{7}$ .

Question 5(iv)

Change the following improper fractions to mixed fractions:

$\dfrac{113}{15}$

Answer

$\dfrac{113}{15}$

Dividing 113 by 15, we get quotient = 7 and remainder = 8.

$\Rightarrow \dfrac{113}{15} = 7\dfrac{8}{15}.$

Hence, $\dfrac{113}{15} = 7\dfrac{8}{15}$ .

Question 6(i)

Change the following groups of fractions to like fractions:

$\dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4}, \dfrac{1}{6}$

Answer

$\dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4}, \dfrac{1}{6}$

LCM of 3, 5, 4 and 6 = 60.

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

Hence, the required like fractions are $\dfrac{20}{60}, \dfrac{24}{60}, \dfrac{45}{60} \text{ and } \dfrac{10}{60}$ .

Question 6(ii)

Change the following groups of fractions to like fractions:

$\dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12}, \dfrac{3}{10}$

Answer

$\dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12}, \dfrac{3}{10}$

LCM of 6, 8, 12 and 10 = 120.

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

Hence, the required like fractions are $\dfrac{100}{120}, \dfrac{105}{120}, \dfrac{110}{120} \text{ and } \dfrac{36}{120}$ .

Question 6(iii)

Change the following groups of fractions to like fractions:

$\dfrac{2}{7}, \dfrac{7}{8}, \dfrac{5}{14}, \dfrac{9}{16}$

Answer

$\dfrac{2}{7}, \dfrac{7}{8}, \dfrac{5}{14}, \dfrac{9}{16}$

LCM of 7, 8, 14 and 16 = 112.

$\Rightarrow \dfrac{2}{7} = \dfrac{2 \times 16}{7 \times 16} = \dfrac{32}{112} \\[1em] \Rightarrow \dfrac{7}{8} = \dfrac{7 \times 14}{8 \times 14} = \dfrac{98}{112} \\[1em] \Rightarrow \dfrac{5}{14} = \dfrac{5 \times 8}{14 \times 8} = \dfrac{40}{112} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{9 \times 7}{16 \times 7} = \dfrac{63}{112}$

Hence, the required like fractions are $\dfrac{32}{112}, \dfrac{98}{112}, \dfrac{40}{112} \text{ and } \dfrac{63}{112}$ .

Exercise 6(B)

Question 1(i)

Reduce the given fractions to their lowest terms:

$\dfrac{8}{10}$

Answer

$\dfrac{8}{10}$

By prime factorisation,

$\Rightarrow \dfrac{8}{10} = \dfrac{2 \times 2 \times 2}{2 \times 5} \\[1em] = \dfrac{2 \times 2}{5} \\[1em] = \dfrac{4}{5}.$

Hence, $\dfrac{8}{10}$ in lowest terms is $\dfrac{4}{5}$ .

Question 1(ii)

Reduce the given fractions to their lowest terms:

$\dfrac{50}{75}$

Answer

$\dfrac{50}{75}$

By prime factorisation,

$\Rightarrow \dfrac{50}{75} = \dfrac{2 \times 5 \times 5}{3 \times 5 \times 5} \\[1em] = \dfrac{2}{3}.$

Hence, $\dfrac{50}{75}$ in lowest terms is $\dfrac{2}{3}$ .

Question 1(iii)

Reduce the given fractions to their lowest terms:

$\dfrac{18}{81}$

Answer

$\dfrac{18}{81}$

By prime factorisation,

$\Rightarrow \dfrac{18}{81} = \dfrac{2 \times 3 \times 3}{3 \times 3 \times 3 \times 3} \\[1em] = \dfrac{2}{3 \times 3} \\[1em] = \dfrac{2}{9}.$

Hence, $\dfrac{18}{81}$ in lowest terms is $\dfrac{2}{9}$ .

Question 1(iv)

Reduce the given fractions to their lowest terms:

$\dfrac{40}{120}$

Answer

$\dfrac{40}{120}$

By prime factorisation,

$\Rightarrow \dfrac{40}{120} = \dfrac{2 \times 2 \times 2 \times 5}{2 \times 2 \times 2 \times 3 \times 5} \\[1em] = \dfrac{1}{3}.$

Hence, $\dfrac{40}{120}$ in lowest terms is $\dfrac{1}{3}$ .

Question 1(v)

Reduce the given fractions to their lowest terms:

$\dfrac{105}{70}$

Answer

$\dfrac{105}{70}$

By prime factorisation,

$\Rightarrow \dfrac{105}{70} = \dfrac{3 \times 5 \times 7}{2 \times 5 \times 7} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.$

Hence, $\dfrac{105}{70}$ in lowest terms is $\dfrac{3}{2} \text{ or } 1\dfrac{1}{2}$ .

Question 2(i)

State whether true or false:

$\dfrac{2}{5} = \dfrac{10}{15}$

Answer

$\dfrac{2}{5} = \dfrac{10}{15}$

By cross multiplication, 2 × 15 = 30 and 5 × 10 = 50.

Since 30 ≠ 50, the cross products are not equal.

Hence, the statement is False.

Question 2(ii)

State whether true or false:

$\dfrac{35}{42} = \dfrac{5}{6}$

Answer

$\dfrac{35}{42} = \dfrac{5}{6}$

By cross multiplication, 35 × 6 = 210 and 42 × 5 = 210.

Since the cross products are equal, the fractions are equal.

Hence, the statement is True.

Question 2(iii)

State whether true or false:

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

Answer

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

By cross multiplication, 5 × 5 = 25 and 4 × 4 = 16.

Since 25 ≠ 16, the cross products are not equal.

Hence, the statement is False.

Question 2(iv)

State whether true or false:

$\dfrac{7}{9} = 1\dfrac{1}{7}$

Answer

$\dfrac{7}{9} = 1\dfrac{1}{7}$

$\dfrac{7}{9}$ is a proper fraction (numerator < denominator), while $1\dfrac{1}{7} = \dfrac{8}{7}$ is an improper fraction (denominator < numerator).

Hence, the statement is False.

Question 2(v)

State whether true or false:

$\dfrac{9}{7} = 1\dfrac{1}{7}$

Answer

$\dfrac{9}{7} = 1\dfrac{1}{7}$

Converting, improper fraction into mixed fraction:

$\dfrac{9}{7} = 1\dfrac{2}{7}$ , which is not equal to $1\dfrac{1}{7}$ .

Hence, the statement is False.

Question 3(i)

Which fraction is greater?

$\dfrac{3}{5} \text{ or } \dfrac{2}{3}$

Answer

$\dfrac{3}{5} \text{ or } \dfrac{2}{3}$

LCM of 5 and 3 = 15.

$\Rightarrow \dfrac{3}{5} = \dfrac{3 \times 3}{5 \times 3} = \dfrac{9}{15} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

Since 9 < 10, $\dfrac{9}{15} \lt \dfrac{10}{15}$ .

Hence, $\dfrac{2}{3}$ is greater.

Question 3(ii)

Which fraction is greater?

$\dfrac{5}{9} \text{ or } \dfrac{3}{4}$

Answer

$\dfrac{5}{9} \text{ or } \dfrac{3}{4}$

LCM of 9 and 4 = 36.

$\Rightarrow \dfrac{5}{9} = \dfrac{5 \times 4}{9 \times 4} = \dfrac{20}{36} \\[1em] \Rightarrow \dfrac{3}{4} = \dfrac{3 \times 9}{4 \times 9} = \dfrac{27}{36}$

Since 20 < 27, $\dfrac{20}{36} \lt \dfrac{27}{36}$ .

Hence, $\dfrac{3}{4}$ is greater.

Question 3(iii)

Which fraction is greater?

$\dfrac{11}{14} \text{ or } \dfrac{26}{35}$

Answer

$\dfrac{11}{14} \text{ or } \dfrac{26}{35}$

LCM of 14 and 35 = 70.

$\Rightarrow \dfrac{11}{14} = \dfrac{11 \times 5}{14 \times 5} = \dfrac{55}{70} \\[1em] \Rightarrow \dfrac{26}{35} = \dfrac{26 \times 2}{35 \times 2} = \dfrac{52}{70}$

Since 55 > 52, $\dfrac{55}{70} \gt \dfrac{52}{70}$ .

Hence, $\dfrac{11}{14}$ is greater.

Question 4(i)

Which fraction is smaller?

$\dfrac{3}{8} \text{ or } \dfrac{4}{5}$

Answer

$\dfrac{3}{8} \text{ or } \dfrac{4}{5}$

LCM of 8 and 5 = 40.

$\Rightarrow \dfrac{3}{8} = \dfrac{3 \times 5}{8 \times 5} = \dfrac{15}{40} \\[1em] \Rightarrow \dfrac{4}{5} = \dfrac{4 \times 8}{5 \times 8} = \dfrac{32}{40}$

Since 15 < 32, $\dfrac{15}{40} \lt \dfrac{32}{40}$ .

Hence, $\dfrac{3}{8}$ is smaller.

Question 4(ii)

Which fraction is smaller?

$\dfrac{8}{15} \text{ or } \dfrac{4}{7}$

Answer

$\dfrac{8}{15} \text{ or } \dfrac{4}{7}$

LCM of 15 and 7 = 105.

$\Rightarrow \dfrac{8}{15} = \dfrac{8 \times 7}{15 \times 7} = \dfrac{56}{105} \\[1em] \Rightarrow \dfrac{4}{7} = \dfrac{4 \times 15}{7 \times 15} = \dfrac{60}{105}$

Since 56 < 60, $\dfrac{56}{105} \lt \dfrac{60}{105}$ .

Hence, $\dfrac{8}{15}$ is smaller.

Question 4(iii)

Which fraction is smaller?

$\dfrac{7}{26} \text{ or } \dfrac{10}{39}$

Answer

$\dfrac{7}{26} \text{ or } \dfrac{10}{39}$

LCM of 26 and 39 = 78.

$\Rightarrow \dfrac{7}{26} = \dfrac{7 \times 3}{26 \times 3} = \dfrac{21}{78} \\[1em] \Rightarrow \dfrac{10}{39} = \dfrac{10 \times 2}{39 \times 2} = \dfrac{20}{78}$

Since 20 < 21, $\dfrac{20}{78} \lt \dfrac{21}{78}$ .

Hence, $\dfrac{10}{39}$ is smaller.

Question 5(i)

Arrange the given fractions in descending order of magnitude:

$\dfrac{5}{16}, \dfrac{13}{24} \text{ and } \dfrac{7}{8}$

Answer

$\dfrac{5}{16}, \dfrac{13}{24} \text{ and } \dfrac{7}{8}$

LCM of 16, 24 and 8 = 48.

$\Rightarrow \dfrac{5}{16} = \dfrac{5 \times 3}{16 \times 3} = \dfrac{15}{48} \\[1em] \Rightarrow \dfrac{13}{24} = \dfrac{13 \times 2}{24 \times 2} = \dfrac{26}{48} \\[1em] \Rightarrow \dfrac{7}{8} = \dfrac{7 \times 6}{8 \times 6} = \dfrac{42}{48}$

Since 42 > 26 > 15, we have $\dfrac{42}{48} \gt \dfrac{26}{48} \gt \dfrac{15}{48}$ .

Hence, the descending order is $\dfrac{7}{8} \gt \dfrac{13}{24} \gt \dfrac{5}{16}$ .

Question 5(ii)

Arrange the given fractions in descending order of magnitude:

$\dfrac{4}{5}, \dfrac{7}{15}, \dfrac{11}{20} \text{ and } \dfrac{3}{4}$

Answer

$\dfrac{4}{5}, \dfrac{7}{15}, \dfrac{11}{20} \text{ and } \dfrac{3}{4}$

LCM of 5, 15, 20 and 4 = 60.

$\Rightarrow \dfrac{4}{5} = \dfrac{4 \times 12}{5 \times 12} = \dfrac{48}{60} \\[1em] \Rightarrow \dfrac{7}{15} = \dfrac{7 \times 4}{15 \times 4} = \dfrac{28}{60} \\[1em] \Rightarrow \dfrac{11}{20} = \dfrac{11 \times 3}{20 \times 3} = \dfrac{33}{60} \\[1em] \Rightarrow \dfrac{3}{4} = \dfrac{3 \times 15}{4 \times 15} = \dfrac{45}{60}$

Since 48 > 45 > 33 > 28, we have $\dfrac{48}{60} \gt \dfrac{45}{60} \gt \dfrac{33}{60} \gt \dfrac{28}{60}$ .

Hence, the descending order is $\dfrac{4}{5} \gt \dfrac{3}{4} \gt \dfrac{11}{20} \gt \dfrac{7}{15}$ .

Question 5(iii)

Arrange the given fractions in descending order of magnitude:

$\dfrac{5}{7}, \dfrac{3}{8} \text{ and } \dfrac{9}{11}$

Answer

$\dfrac{5}{7}, \dfrac{3}{8} \text{ and } \dfrac{9}{11}$

LCM of 7, 8 and 11 = 616.

$\Rightarrow \dfrac{5}{7} = \dfrac{5 \times 88}{7 \times 88} = \dfrac{440}{616} \\[1em] \Rightarrow \dfrac{3}{8} = \dfrac{3 \times 77}{8 \times 77} = \dfrac{231}{616} \\[1em] \Rightarrow \dfrac{9}{11} = \dfrac{9 \times 56}{11 \times 56} = \dfrac{504}{616}$

Since 504 > 440 > 231, we have $\dfrac{504}{616} \gt \dfrac{440}{616} \gt \dfrac{231}{616}$ .

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

Question 6(i)

Arrange the given fractions in ascending order of magnitude:

$\dfrac{9}{16}, \dfrac{7}{12} \text{ and } \dfrac{1}{4}$

Answer

$\dfrac{9}{16}, \dfrac{7}{12} \text{ and } \dfrac{1}{4}$

LCM of 16, 12 and 4 = 48.

$\Rightarrow \dfrac{9}{16} = \dfrac{9 \times 3}{16 \times 3} = \dfrac{27}{48} \\[1em] \Rightarrow \dfrac{7}{12} = \dfrac{7 \times 4}{12 \times 4} = \dfrac{28}{48} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{1 \times 12}{4 \times 12} = \dfrac{12}{48}$

Since 12 < 27 < 28, we have $\dfrac{12}{48} \lt \dfrac{27}{48} \lt \dfrac{28}{48}$ .

Hence, the ascending order is $\dfrac{1}{4} \lt \dfrac{9}{16} \lt \dfrac{7}{12}$ .

Question 6(ii)

Arrange the given fractions in ascending order of magnitude:

$\dfrac{5}{6}, \dfrac{2}{7}, \dfrac{8}{9} \text{ and } \dfrac{1}{3}$

Answer

$\dfrac{5}{6}, \dfrac{2}{7}, \dfrac{8}{9} \text{ and } \dfrac{1}{3}$

LCM of 6, 7, 9 and 3 = 126.

$\Rightarrow \dfrac{5}{6} = \dfrac{5 \times 21}{6 \times 21} = \dfrac{105}{126} \\[1em] \Rightarrow \dfrac{2}{7} = \dfrac{2 \times 18}{7 \times 18} = \dfrac{36}{126} \\[1em] \Rightarrow \dfrac{8}{9} = \dfrac{8 \times 14}{9 \times 14} = \dfrac{112}{126} \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{1 \times 42}{3 \times 42} = \dfrac{42}{126}$

Since 36 < 42 < 105 < 112, we have $\dfrac{36}{126} \lt \dfrac{42}{126} \lt \dfrac{105}{126} \lt \dfrac{112}{126}$ .

Hence, the ascending order is $\dfrac{2}{7} \lt \dfrac{1}{3} \lt \dfrac{5}{6} \lt \dfrac{8}{9}$ .

Question 6(iii)

Arrange the given fractions in ascending order of magnitude:

$\dfrac{2}{3}, \dfrac{5}{9}, \dfrac{5}{6} \text{ and } \dfrac{3}{8}$

Answer

$\dfrac{2}{3}, \dfrac{5}{9}, \dfrac{5}{6} \text{ and } \dfrac{3}{8}$

LCM of 3, 9, 6 and 8 = 72.

$\Rightarrow \dfrac{2}{3} = \dfrac{2 \times 24}{3 \times 24} = \dfrac{48}{72} \\[1em] \Rightarrow \dfrac{5}{9} = \dfrac{5 \times 8}{9 \times 8} = \dfrac{40}{72} \\[1em] \Rightarrow \dfrac{5}{6} = \dfrac{5 \times 12}{6 \times 12} = \dfrac{60}{72} \\[1em] \Rightarrow \dfrac{3}{8} = \dfrac{3 \times 9}{8 \times 9} = \dfrac{27}{72}$

Since 27 < 40 < 48 < 60, we have $\dfrac{27}{72} \lt \dfrac{40}{72} \lt \dfrac{48}{72} \lt \dfrac{60}{72}$ .

Hence, the ascending order is $\dfrac{3}{8} \lt \dfrac{5}{9} \lt \dfrac{2}{3} \lt \dfrac{5}{6}$ .

Question 7(i)

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{6}{11} \text{...............} \dfrac{5}{9}$

Answer

$\dfrac{6}{11} \text{ and } \dfrac{5}{9}$

By cross multiplication, 6 × 9 = 54 and 11 × 5 = 55.

Since 54 < 55, therefore $\dfrac{6}{11} \lt \dfrac{5}{9}$ .

Hence, $\dfrac{6}{11} \lt \dfrac{5}{9}$ .

Question 7(ii)

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{3}{7} \text{...............} \dfrac{9}{13}$

Answer

$\dfrac{3}{7} \text{ and } \dfrac{9}{13}$

By cross multiplication, 3 × 13 = 39 and 7 × 9 = 63.

Since 39 < 63, therefore $\dfrac{3}{7} \lt \dfrac{9}{13}$ .

Hence, $\dfrac{3}{7} \lt \dfrac{9}{13}$ .

Question 7(iii)

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{56}{64} \text{...............} \dfrac{7}{8}$

Answer

$\dfrac{56}{64} \text{ and } \dfrac{7}{8}$

Reducing $\dfrac{56}{64}$ to lowest terms, $\dfrac{56}{64} = \dfrac{7}{8}$ .

Hence, $\dfrac{56}{64} = \dfrac{7}{8}$ .

Question 7(iv)

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{5}{12} \text{...............} \dfrac{8}{33}$

Answer

$\dfrac{5}{12} \text{ and } \dfrac{8}{33}$

By cross multiplication, 5 × 33 = 165 and 12 × 8 = 96.

Since 165 > 96, therefore $\dfrac{5}{12} \gt \dfrac{8}{33}$ .

Hence, $\dfrac{5}{12} \gt \dfrac{8}{33}$ .

Exercise 6(C)

Question 1(i)

Add the following fractions:

$1\dfrac{3}{4} \text{ and } \dfrac{3}{8}$

Answer

$1\dfrac{3}{4} \text{ and } \dfrac{3}{8}$

$\Rightarrow 1\dfrac{3}{4} + \dfrac{3}{8} = \dfrac{7}{4} + \dfrac{3}{8} \\[1em] \text{ L.C.M. of 4 and 8 = 8 }\\[1em] = \dfrac{7 \times 2}{4 \times 2} + \dfrac{3}{8} \\[1em] =\dfrac{14}{8} + \dfrac{3}{8} \\[1em] = \dfrac{17}{8} \\[1em] = 2\dfrac{1}{8}.$

Hence, the required sum = $2\dfrac{1}{8}$ .

Question 1(ii)

Add the following fractions:

$\dfrac{2}{5}, 2\dfrac{3}{15} \text{ and } \dfrac{7}{10}$

Answer

$\dfrac{2}{5}, 2\dfrac{3}{15} \text{ and } \dfrac{7}{10}$

LCM of 5, 15 and 10 = 30.

$\Rightarrow \dfrac{2}{5} + 2\dfrac{3}{15} + \dfrac{7}{10} = \dfrac{2}{5} + \dfrac{33}{15} + \dfrac{7}{10} \\[1em] = \dfrac{12}{30} + \dfrac{66}{30} + \dfrac{21}{30} \\[1em] = \dfrac{99}{30}\\[1em] = \dfrac{33}{10}\\[1em] = 3\dfrac{3}{10}.$

Hence, the required sum = $3\dfrac{3}{10}$ .

Question 1(iii)

Add the following fractions:

$1\dfrac{7}{8}, 1\dfrac{1}{2} \text{ and } 1\dfrac{3}{4}$

Answer

$1\dfrac{7}{8}, 1\dfrac{1}{2} \text{ and } 1\dfrac{3}{4}$

LCM of 8, 2 and 4 = 8.

$\Rightarrow 1\dfrac{7}{8} + 1\dfrac{1}{2} + 1\dfrac{3}{4} = \dfrac{15}{8} + \dfrac{3}{2} + \dfrac{7}{4} \\[1em] = \dfrac{15}{8} + \dfrac{12}{8} + \dfrac{14}{8} \\[1em] = \dfrac{41}{8} \\[1em] = 5\dfrac{1}{8}.$

Hence, the required sum = $5\dfrac{1}{8}$ .

Question 1(iv)

Add the following fractions:

$3\dfrac{3}{4}, 2\dfrac{1}{6} \text{ and } 1\dfrac{5}{8}$

Answer

$3\dfrac{3}{4}, 2\dfrac{1}{6} \text{ and } 1\dfrac{5}{8}$

LCM of 4, 6 and 8 = 24.

$\Rightarrow 3\dfrac{3}{4} + 2\dfrac{1}{6} + 1\dfrac{5}{8} = \dfrac{15}{4} + \dfrac{13}{6} + \dfrac{13}{8} \\[1em] = \dfrac{90}{24} + \dfrac{52}{24} + \dfrac{39}{24} \\[1em] = \dfrac{181}{24} \\[1em] = 7\dfrac{13}{24}.$

Hence, the required sum = $7\dfrac{13}{24}$ .

Question 1(v)

Add the following fractions:

$2\dfrac{8}{9}, \dfrac{11}{18} \text{ and } 3\dfrac{5}{6}$

Answer

$2\dfrac{8}{9}, \dfrac{11}{18} \text{ and } 3\dfrac{5}{6}$

LCM of 9, 18 and 6 = 18.

$\Rightarrow 2\dfrac{8}{9} + \dfrac{11}{18} + 3\dfrac{5}{6} = \dfrac{26}{9} + \dfrac{11}{18} + \dfrac{23}{6} \\[1em] = \dfrac{52}{18} + \dfrac{11}{18} + \dfrac{69}{18} \\[1em] = \dfrac{132}{18} \\[1em] = \dfrac{22}{3} \\[1em] = 7\dfrac{1}{3}.$

Hence, the required sum = $7\dfrac{1}{3}$ .

Question 2(i)

Simplify:

$1\dfrac{11}{12} - \dfrac{13}{16}$

Answer

$1\dfrac{11}{12} - \dfrac{13}{16}$

LCM of 12 and 16 = 48.

$\Rightarrow 1\dfrac{11}{12} - \dfrac{13}{16} = \dfrac{23}{12} - \dfrac{13}{16} \\[1em] = \dfrac{92}{48} - \dfrac{39}{48} \\[1em] = \dfrac{53}{48} \\[1em] = 1\dfrac{5}{48}.$

Hence, $1\dfrac{11}{12} - \dfrac{13}{16} = 1\dfrac{5}{48}$ .

Question 2(ii)

Simplify:

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

Answer

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

LCM of 4 and 6 = 12.

$\Rightarrow 2\dfrac{3}{4} - 1\dfrac{5}{6} = \dfrac{11}{4} - \dfrac{11}{6} \\[1em] = \dfrac{33}{12} - \dfrac{22}{12} \\[1em] = \dfrac{11}{12}.$

Hence, $2\dfrac{3}{4} - 1\dfrac{5}{6} = \dfrac{11}{12}$ .

Question 2(iii)

Simplify:

$2\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21}$

Answer

$2\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21}$

LCM of 7, 14 and 21 = 42.

$\Rightarrow 2\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21} = \dfrac{19}{7} + \dfrac{3}{14} - \dfrac{13}{21} \\[1em] = \dfrac{114}{42} + \dfrac{9}{42} - \dfrac{26}{42} \\[1em] = \dfrac{97}{42} \\[1em] = 2\dfrac{13}{42}.$

Hence, $2\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21} = 2\dfrac{13}{42}$ .

Question 2(iv)

Simplify:

$3\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12}$

Answer

$3\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12}$

LCM of 6 and 12 = 12.

$\Rightarrow 3\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12} = \dfrac{23}{6} - \dfrac{1}{6} - \dfrac{13}{12} \\[1em] = \dfrac{46}{12} - \dfrac{2}{12} - \dfrac{13}{12} \\[1em] = \dfrac{31}{12} \\[1em] = 2\dfrac{7}{12}.$

Hence, $3\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12} = 2\dfrac{7}{12}$ .

Question 2(v)

Simplify:

$6 + \dfrac{3}{10} - 1\dfrac{8}{15}$

Answer

$6 + \dfrac{3}{10} - 1\dfrac{8}{15}$

LCM of 1, 10 and 15 = 30.

$\Rightarrow 6 + \dfrac{3}{10} - 1\dfrac{8}{15} = 6 + \dfrac{3}{10} - \dfrac{23}{15} \\[1em] = \dfrac{180}{30} + \dfrac{9}{30} - \dfrac{46}{30} \\[1em] = \dfrac{143}{30} \\[1em] = 4\dfrac{23}{30}.$

Hence, $6 + \dfrac{3}{10} - 1\dfrac{8}{15} = 4\dfrac{23}{30}$ .

Question 2(vi)

Simplify:

$1\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14}$

Answer

$1\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14}$

LCM of 4, 7 and 14 = 28.

$\Rightarrow 1\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14} = \dfrac{7}{4} + \dfrac{19}{7} - \dfrac{17}{14} \\[1em] = \dfrac{49}{28} + \dfrac{76}{28} - \dfrac{34}{28} \\[1em] = \dfrac{91}{28} \\[1em] = \dfrac{13}{4} \\[1em] = 3\dfrac{1}{4}.$

Hence, $1\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14} = 3\dfrac{1}{4}$ .

Question 2(vii)

Simplify:

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

Answer

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

LCM of 1, 8 and 6 = 24.

$\Rightarrow 4 + 3\dfrac{1}{8} - 3\dfrac{1}{6} = 4 + \dfrac{25}{8} - \dfrac{19}{6} \\[1em] = \dfrac{96}{24} + \dfrac{75}{24} - \dfrac{76}{24} \\[1em] = \dfrac{95}{24} \\[1em] = 3\dfrac{23}{24}.$

Hence, $4 + 3\dfrac{1}{8} - 3\dfrac{1}{6} = 3\dfrac{23}{24}$ .

Question 2(viii)

Simplify:

$6 - 3\dfrac{1}{2} - 2\dfrac{1}{5}$

Answer

$6 - 3\dfrac{1}{2} - 2\dfrac{1}{5}$

LCM of 1, 2 and 5 = 10.

$\Rightarrow 6 - 3\dfrac{1}{2} - 2\dfrac{1}{5} = 6 - \dfrac{7}{2} - \dfrac{11}{5} \\[1em] = \dfrac{60}{10} - \dfrac{35}{10} - \dfrac{22}{10} \\[1em] = \dfrac{3}{10}.$

Hence, $6 - 3\dfrac{1}{2} - 2\dfrac{1}{5} = \dfrac{3}{10}$ .

Question 2(ix)

Simplify:

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

Answer

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

LCM of 8, 6 and 4 = 24.

$\Rightarrow 1\dfrac{5}{8} - 2\dfrac{1}{6} + 3\dfrac{3}{4} = \dfrac{13}{8} - \dfrac{13}{6} + \dfrac{15}{4} \\[1em] = \dfrac{39}{24} - \dfrac{52}{24} + \dfrac{90}{24} \\[1em] = \dfrac{77}{24} \\[1em] = 3\dfrac{5}{24}.$

Hence, $1\dfrac{5}{8} - 2\dfrac{1}{6} + 3\dfrac{3}{4} = 3\dfrac{5}{24}$ .

Question 2(x)

Simplify:

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

Answer

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

LCM of 2, 3 and 4 = 12.

$\Rightarrow 3\dfrac{1}{2} + 1\dfrac{2}{3} - 2\dfrac{1}{4} = \dfrac{7}{2} + \dfrac{5}{3} - \dfrac{9}{4} \\[1em] = \dfrac{42}{12} + \dfrac{20}{12} - \dfrac{27}{12} \\[1em] = \dfrac{35}{12} \\[1em] = 2\dfrac{11}{12}.$

Hence, $3\dfrac{1}{2} + 1\dfrac{2}{3} - 2\dfrac{1}{4} = 2\dfrac{11}{12}$ .

Question 2(xi)

Simplify:

$4\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45}$

Answer

$4\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45}$

LCM of 5, 9, 15 and 45 = 45.

$\Rightarrow 4\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45} = \dfrac{23}{5} - \dfrac{25}{9} - \dfrac{17}{15} - \dfrac{2}{45} \\[1em] = \dfrac{207}{45} - \dfrac{125}{45} - \dfrac{51}{45} - \dfrac{2}{45} \\[1em] = \dfrac{29}{45}.$

Hence, $4\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45} = \dfrac{29}{45}$ .

Exercise 6(D)

Question 1(i)

Simplify:

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

Answer

Solving,

$\Rightarrow \dfrac{3}{7} \times \dfrac{2}{5} \\[1em] = \dfrac{3 \times 2}{7 \times 5} \\[1em] = \dfrac{6}{35}$

Hence, $\dfrac{3}{7} \times \dfrac{2}{5} = \dfrac{6}{35}$ .

Question 1(ii)

Simplify:

$\dfrac{4}{9} \times \dfrac{3}{5}$

Answer

$\dfrac{4}{9} \times \dfrac{3}{5}$

$\Rightarrow \dfrac{4}{9} \times \dfrac{3}{5} = \dfrac{4 \times 3}{9 \times 5} \\[1em] = \dfrac{12}{45} \\[1em] = \dfrac{4}{15}.$

Hence, $\dfrac{4}{9} \times \dfrac{3}{5} = \dfrac{4}{15}$ .

Question 1(iii)

Simplify:

$\dfrac{5}{12}$ × 8

Answer

$\dfrac{5}{12} \times 8$

$\Rightarrow \dfrac{5}{12} \times 8 = \dfrac{5 \times 8}{12} \\[1em] = \dfrac{40}{12} \\[1em] = \dfrac{10}{3} \\[1em] = 3\dfrac{1}{3}.$

Hence, $\dfrac{5}{12} \times 8 = 3\dfrac{1}{3}$ .

Question 1(iv)

Simplify:

$\dfrac{7}{6} \text{ of } \dfrac{3}{14}$

Answer

$\dfrac{7}{6} \text{ of } \dfrac{3}{14}$

$\Rightarrow \dfrac{7}{6} \text{ of } \dfrac{3}{14} = \dfrac{7}{6} \times \dfrac{3}{14} \\[1em] = \dfrac{7 \times 3}{6 \times 14} \\[1em] = \dfrac{21}{84} \\[1em] = \dfrac{1}{4}.$

Hence, $\dfrac{7}{6} \text{ of } \dfrac{3}{14} = \dfrac{1}{4}$ .

Question 1(v)

Simplify:

$3\dfrac{3}{8} \times 3\dfrac{6}{7}$

Answer

$3\dfrac{3}{8} \times 3\dfrac{6}{7}$

$\Rightarrow 3\dfrac{3}{8} \times 3\dfrac{6}{7} = \dfrac{27}{8} \times \dfrac{27}{7} \\[1em] = \dfrac{27 \times 27}{8 \times 7} \\[1em] = \dfrac{729}{56} \\[1em] = 13\dfrac{1}{56}.$

Hence, $3\dfrac{3}{8} \times 3\dfrac{6}{7} = 13\dfrac{1}{56}$ .

Question 1(vi)

Simplify:

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

Answer

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

$\Rightarrow \dfrac{1}{2} \text{ of } \dfrac{1}{3} \times \dfrac{3}{4} = \dfrac{1}{6} \times \dfrac{3}{4} \\[1em] = \dfrac{3}{24} \\[1em] = \dfrac{1}{8}.$

Hence, $\dfrac{1}{2} \text{ of } \dfrac{1}{3} \times \dfrac{3}{4} = \dfrac{1}{8}$ .

Question 1(vii)

Simplify:

$\dfrac{3}{7} \times \dfrac{5}{9} \times 4\dfrac{1}{5}$

Answer

$\dfrac{3}{7} \times \dfrac{5}{9} \times 4\dfrac{1}{5}$

$\Rightarrow \dfrac{3}{7} \times \dfrac{5}{9} \times 4\dfrac{1}{5} = \dfrac{3}{7} \times \dfrac{5}{9} \times \dfrac{21}{5} \\[1em] = \dfrac{3 \times 5 \times 21}{7 \times 9 \times 5} \\[1em] = \dfrac{315}{315} \\[1em] = 1.$

Hence, $\dfrac{3}{7} \times \dfrac{5}{9} \times 4\dfrac{1}{5} = 1$ .

Question 1(viii)

Simplify:

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

Answer

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

$\Rightarrow 1\dfrac{1}{3} \times 1\dfrac{2}{7} \text{ of } 1\dfrac{1}{4} = \dfrac{4}{3} \times \dfrac{9}{7} \times \dfrac{5}{4} \\[1em] = \dfrac{4 \times 9 \times 5}{3 \times 7 \times 4} \\[1em] = \dfrac{180}{84} \\[1em] = \dfrac{15}{7} \\[1em] = 2\dfrac{1}{7}.$

Hence, $1\dfrac{1}{3} \times 1\dfrac{2}{7} \text{ of } 1\dfrac{1}{4} = 2\dfrac{1}{7}$ .

Question 2(i)

Simplify:

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

Answer

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

$\Rightarrow \dfrac{2}{3} \div 1\dfrac{1}{5} = \dfrac{2}{3} \div \dfrac{6}{5} \\[1em] = \dfrac{2}{3} \times \dfrac{5}{6} \\[1em] = \dfrac{10}{18} \\[1em] = \dfrac{5}{9}.$

Hence, $\dfrac{2}{3} \div 1\dfrac{1}{5} = \dfrac{5}{9}$ .

Question 2(ii)

Simplify:

$4\dfrac{1}{2} \div \dfrac{4}{9}$

Answer

$4\dfrac{1}{2} \div \dfrac{4}{9}$

$\Rightarrow 4\dfrac{1}{2} \div \dfrac{4}{9} = \dfrac{9}{2} \div \dfrac{4}{9} \\[1em] = \dfrac{9}{2} \times \dfrac{9}{4} \\[1em] = \dfrac{81}{8} \\[1em] = 10\dfrac{1}{8}.$

Hence, $4\dfrac{1}{2} \div \dfrac{4}{9} = 10\dfrac{1}{8}$ .

Question 2(iii)

Simplify:

1 ÷ $\dfrac{2}{5}$

Answer

1 ÷ $\dfrac{2}{5}$

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

Hence, $1 \div \dfrac{2}{5} = 2\dfrac{1}{2}$ .

Question 2(iv)

Simplify:

$\dfrac{4}{9} \div \dfrac{4}{9}$

Answer

$\dfrac{4}{9} \div \dfrac{4}{9}$

$\Rightarrow \dfrac{4}{9} \div \dfrac{4}{9} = \dfrac{4}{9} \times \dfrac{9}{4} \\[1em] = 1.$

Hence, $\dfrac{4}{9} \div \dfrac{4}{9} = 1$ .

Question 2(v)

Simplify:

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

Answer

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

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

Hence, $2\dfrac{1}{3} \div 1\dfrac{3}{4} = 1\dfrac{1}{3}$ .

Question 3(i)

Simplify:

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

Answer

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

$\Rightarrow \dfrac{1}{4} \text{ of } 2\dfrac{2}{7} \div \dfrac{3}{5} = \dfrac{1}{4} \times \dfrac{16}{7} \div \dfrac{3}{5} \\[1em] = \dfrac{4}{7} \div \dfrac{3}{5} \\[1em] = \dfrac{4}{7} \times \dfrac{5}{3} \\[1em] = \dfrac{20}{21}.$

Hence, $\dfrac{1}{4} \text{ of } 2\dfrac{2}{7} \div \dfrac{3}{5} = \dfrac{20}{21}$ .

Question 3(ii)

Simplify:

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

Answer

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

$\Rightarrow 1\dfrac{1}{4} \times \dfrac{1}{2} \div 1\dfrac{1}{3} = \dfrac{5}{4} \times \dfrac{1}{2} \div \dfrac{4}{3} \\[1em] = \dfrac{5}{8} \div \dfrac{4}{3} \\[1em] = \dfrac{5}{8} \times \dfrac{3}{4} \\[1em] = \dfrac{15}{32}.$

Hence, $1\dfrac{1}{4} \times \dfrac{1}{2} \div 1\dfrac{1}{3} = \dfrac{15}{32}$ .

Question 3(iii)

Simplify:

$6\dfrac{1}{7} \times 0 \times 5\dfrac{3}{8}$

Answer

$6\dfrac{1}{7} \times 0 \times 5\dfrac{3}{8}$

Any number multiplied by 0 is 0.

$\Rightarrow 6\dfrac{1}{7} \times 0 \times 5\dfrac{3}{8} = 0.$

Hence, $6\dfrac{1}{7} \times 0 \times 5\dfrac{3}{8} = 0$ .

Question 3(iv)

Simplify:

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

Answer

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

Solving 'of' first,

$\Rightarrow \dfrac{3}{7} \text{ of } 2\dfrac{5}{8} = \dfrac{3}{7} \times \dfrac{21}{8} = \dfrac{9}{8}\\[1em] \Rightarrow \dfrac{3}{4} \times 1\dfrac{1}{3} \div \dfrac{9}{8} = \dfrac{3}{4} \times \dfrac{4}{3} \div \dfrac{9}{8} \\[1em] = 1 \div \dfrac{9}{8} \\[1em] = 1 \times \dfrac{8}{9} \\[1em] = \dfrac{8}{9}.$

Hence, $\dfrac{3}{4} \times 1\dfrac{1}{3} \div \dfrac{3}{7} \text{ of } 2\dfrac{5}{8} = \dfrac{8}{9}$ .

Question 3(v)

Simplify:

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

Answer

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

Solving 'of' first,

$\Rightarrow \dfrac{2}{7} \text{ of } 1\dfrac{1}{3} = \dfrac{2}{7} \times \dfrac{4}{3} = \dfrac{8}{21}\\[1em] \Rightarrow 2\dfrac{1}{4} \div \dfrac{8}{21} \times \dfrac{2}{3} = \dfrac{9}{4} \div \dfrac{8}{21} \times \dfrac{2}{3} \\[1em] = \dfrac{9}{4} \times \dfrac{21}{8} \times \dfrac{2}{3} \\[1em] = \dfrac{9 \times 21 \times 2}{4 \times 8 \times 3} \\[1em] = \dfrac{378}{96} \\[1em] = \dfrac{63}{16} \\[1em] = 3\dfrac{15}{16}.$

Hence, $2\dfrac{1}{4} \div \dfrac{2}{7} \text{ of } 1\dfrac{1}{3} \times \dfrac{2}{3} = 3\dfrac{15}{16}$ .

Question 4(i)

Simplify:

$5 - \left(\dfrac{8}{11} - 3\dfrac{3}{11}\right)$

Answer

Using BODMAS we simplify bracket first.

$5 - \left(\dfrac{8}{11} - 3\dfrac{3}{11}\right)$

$\Rightarrow 5 - \left(\dfrac{8}{11} - \dfrac{36}{11}\right) = 5 - \left(\dfrac{8 - 36}{11}\right) \\[1em] = 5 - \left(\dfrac{-28}{11}\right) \\[1em] = 5 + \dfrac{28}{11} \\[1em] = \dfrac{55}{11} + \dfrac{28}{11} \\[1em] = \dfrac{83}{11} \\[1em] = 7\dfrac{6}{11}.$

Hence, $5 - \left(\dfrac{8}{11} - 3\dfrac{3}{11}\right) = 7\dfrac{6}{11}$ .

Question 4(ii)

Simplify:

$\dfrac{1}{2} \div \left(\dfrac{7}{8} - \dfrac{3}{5}\right)$

Answer

Using BODMAS we simplify bracket first.

$\dfrac{1}{2} \div \left(\dfrac{7}{8} - \dfrac{3}{5}\right)$

$\Rightarrow \dfrac{1}{2} \div \left(\dfrac{7}{8} - \dfrac{3}{5}\right) = \dfrac{1}{2} \div \left(\dfrac{35}{40} - \dfrac{24}{40}\right) \\[1em] = \dfrac{1}{2} \div \dfrac{11}{40} \\[1em] = \dfrac{1}{2} \times \dfrac{40}{11} \\[1em] = \dfrac{20}{11} \\[1em] = 1\dfrac{9}{11}.$

Hence, $\dfrac{1}{2} \div \left(\dfrac{7}{8} - \dfrac{3}{5}\right) = 1\dfrac{9}{11}$ .

Question 4(iii)

Simplify:

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

Answer

Using BODMAS we simplify bracket first.

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

$\Rightarrow 2\dfrac{1}{3} \div \left(\dfrac{11}{2} + \dfrac{15}{4}\right) = \dfrac{7}{3} \div \left(\dfrac{22}{4} + \dfrac{15}{4}\right) \\[1em] = \dfrac{7}{3} \div \dfrac{37}{4} \\[1em] = \dfrac{7}{3} \times \dfrac{4}{37} \\[1em] = \dfrac{28}{111}.$

Hence, $2\dfrac{1}{3} \div \left(5\dfrac{1}{2} + 3\dfrac{3}{4}\right) = \dfrac{28}{111}$ .

Question 4(iv)

Simplify:

$\left(3\dfrac{7}{8} - 3\dfrac{3}{5}\right) \div \dfrac{1}{2}$

Answer

Using BODMAS we simplify bracket first.

$\left(3\dfrac{7}{8} - 3\dfrac{3}{5}\right) \div \dfrac{1}{2}$

$\Rightarrow \left(\dfrac{31}{8} - \dfrac{18}{5}\right) \div \dfrac{1}{2} = \left(\dfrac{155}{40} - \dfrac{144}{40}\right) \div \dfrac{1}{2} \\[1em] = \dfrac{11}{40} \div \dfrac{1}{2} \\[1em] = \dfrac{11}{40} \times 2 \\[1em] = \dfrac{22}{40} \\[1em] = \dfrac{11}{20}.$

Hence, $\left(3\dfrac{7}{8} - 3\dfrac{3}{5}\right) \div \dfrac{1}{2} = \dfrac{11}{20}$ .

Question 4(v)

Simplify:

$\dfrac{4}{7} \div \left(\dfrac{1}{3} \times 2\dfrac{4}{5}\right)$

Answer

Using BODMAS we simplify bracket first.

$\dfrac{4}{7} \div \left(\dfrac{1}{3} \times 2\dfrac{4}{5}\right)$

$\Rightarrow \dfrac{4}{7} \div \left(\dfrac{1}{3} \times \dfrac{14}{5}\right) = \dfrac{4}{7} \div \dfrac{14}{15} \\[1em] = \dfrac{4}{7} \times \dfrac{15}{14} \\[1em] = \dfrac{60}{98} \\[1em] = \dfrac{30}{49}.$

Hence, $\dfrac{4}{7} \div \left(\dfrac{1}{3} \times 2\dfrac{4}{5}\right) = \dfrac{30}{49}$ .

Question 5(i)

Simplify:

$\left(\dfrac{1}{2} + \dfrac{1}{3}\right) \div \left(\dfrac{1}{4} - \dfrac{1}{6}\right)$

Answer

$\left(\dfrac{1}{2} + \dfrac{1}{3}\right) \div \left(\dfrac{1}{4} - \dfrac{1}{6}\right)$

Using BODMAS we simplify bracket first.

$\Rightarrow \left(\dfrac{3}{6} + \dfrac{2}{6}\right) \div \left(\dfrac{3}{12} - \dfrac{2}{12}\right) = \dfrac{5}{6} \div \dfrac{1}{12} \\[1em] = \dfrac{5}{6} \times 12 \\[1em] = 10.$

Hence, $\left(\dfrac{1}{2} + \dfrac{1}{3}\right) \div \left(\dfrac{1}{4} - \dfrac{1}{6}\right) = 10$ .

Question 5(ii)

Simplify:

$\left(\dfrac{24}{35} \div \dfrac{6}{7} + \dfrac{5}{9}\right) \times \dfrac{3}{4}$

Answer

$\left(\dfrac{24}{35} \div \dfrac{6}{7} + \dfrac{5}{9}\right) \times \dfrac{3}{4}$

Using BODMAS, we solve the division inside the bracket first,

$\Rightarrow \dfrac{24}{35} \div \dfrac{6}{7} = \dfrac{24}{35} \times \dfrac{7}{6} = \dfrac{4}{5}.\\[1em] \Rightarrow \left(\dfrac{4}{5} + \dfrac{5}{9}\right) \times \dfrac{3}{4} = \left(\dfrac{36}{45} + \dfrac{25}{45}\right) \times \dfrac{3}{4} \\[1em] = \dfrac{61}{45} \times \dfrac{3}{4} \\[1em] = \dfrac{183}{180} \\[1em] = \dfrac{61}{60} \\[1em] = 1\dfrac{1}{60}.$

Hence, $\left(\dfrac{24}{35} \div \dfrac{6}{7} + \dfrac{5}{9}\right) \times \dfrac{3}{4} = 1\dfrac{1}{60}$ .

Question 5(iii)

Simplify:

$\dfrac{3}{4} \text{ of } 6\dfrac{1}{8} - \dfrac{2}{3} \text{ of } 2\dfrac{1}{4}$

Answer

$\dfrac{3}{4} \text{ of } 6\dfrac{1}{8} - \dfrac{2}{3} \text{ of } 2\dfrac{1}{4}$

Using BODMAS, we solve 'of' first,

$\Rightarrow \dfrac{3}{4} \text{ of } 6\dfrac{1}{8} = \dfrac{3}{4} \times \dfrac{49}{8} = \dfrac{147}{32}.\\[1em] \Rightarrow \dfrac{2}{3} \text{ of } 2\dfrac{1}{4} = \dfrac{2}{3} \times \dfrac{9}{4} = \dfrac{3}{2}.\\[1em] \Rightarrow \dfrac{147}{32} - \dfrac{3}{2} = \dfrac{147}{32} - \dfrac{48}{32} \\[1em] = \dfrac{99}{32} \\[1em] = 3\dfrac{3}{32}.$

Hence, $\dfrac{3}{4} \text{ of } 6\dfrac{1}{8} - \dfrac{2}{3} \text{ of } 2\dfrac{1}{4} = 3\dfrac{3}{32}$ .

Question 5(iv)

Simplify:

$\dfrac{7}{30} \text{ of } \left(\dfrac{1}{3} + \dfrac{7}{15}\right) \div \left(\dfrac{5}{6} - \dfrac{3}{5}\right)$

Answer

$\dfrac{7}{30} \text{ of } \left(\dfrac{1}{3} + \dfrac{7}{15}\right) \div \left(\dfrac{5}{6} - \dfrac{3}{5}\right)$

Using BODMAS, we simplify the brackets first,

$\Rightarrow \dfrac{1}{3} + \dfrac{7}{15} = \dfrac{5}{15} + \dfrac{7}{15} = \dfrac{12}{15} = \dfrac{4}{5}.\\[1em] \Rightarrow \dfrac{5}{6} - \dfrac{3}{5} = \dfrac{25}{30} - \dfrac{18}{30} = \dfrac{7}{30}.\\[1em] \Rightarrow \dfrac{7}{30} \text{ of } \dfrac{4}{5} \div \dfrac{7}{30} = \dfrac{7}{30} \times \dfrac{4}{5} \div \dfrac{7}{30} \\[1em] = \dfrac{28}{150} \div \dfrac{7}{30} \\[1em] = \dfrac{28}{150} \times \dfrac{30}{7} \\[1em] = \dfrac{4}{5}.$

Hence, $\dfrac{7}{30} \text{ of } \left(\dfrac{1}{3} + \dfrac{7}{15}\right) \div \left(\dfrac{5}{6} - \dfrac{3}{5}\right) = \dfrac{4}{5}$ .

Question 5(v)

Simplify:

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

Answer

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

Using BODMAS, we solve the multiplication first,

$\Rightarrow 3\dfrac{1}{2} \times 1\dfrac{3}{4} = \dfrac{7}{2} \times \dfrac{7}{4} = \dfrac{49}{8}.\\[1em] \Rightarrow 2\dfrac{1}{2} - \dfrac{49}{8} + 2\dfrac{1}{2} = \dfrac{5}{2} - \dfrac{49}{8} + \dfrac{5}{2} \\[1em] = \dfrac{20}{8} - \dfrac{49}{8} + \dfrac{20}{8} \\[1em] = \dfrac{-9}{8} \\[1em] = -1\dfrac{1}{8}.$

Hence, $2\dfrac{1}{2} - 3\dfrac{1}{2} \times 1\dfrac{3}{4} + 2\dfrac{1}{2} = -1\dfrac{1}{8}$ .

Question 5(vi)

Simplify:

$4\dfrac{5}{7}\left(3\dfrac{1}{8} \div \dfrac{11}{12}\right)$

Answer

$4\dfrac{5}{7}\left(3\dfrac{1}{8} \div \dfrac{11}{12}\right)$

Using BODMAS, we simplify the bracket first,

$\Rightarrow 3\dfrac{1}{8} \div \dfrac{11}{12} = \dfrac{25}{8} \times \dfrac{12}{11} = \dfrac{300}{88} = \dfrac{75}{22}.\\[1em] \Rightarrow 4\dfrac{5}{7} \times \dfrac{75}{22} = \dfrac{33}{7} \times \dfrac{75}{22} \\[1em] = \dfrac{33 \times 75}{7 \times 22} \\[1em] = \dfrac{2475}{154} \\[1em] = \dfrac{225}{14} \\[1em] = 16\dfrac{1}{14}.$

Hence, $4\dfrac{5}{7}\left(3\dfrac{1}{8} \div \dfrac{11}{12}\right) = 16\dfrac{1}{14}$ .

Question 5(vii)

Simplify:

$\dfrac{2}{5} \text{ of } \left(\dfrac{1}{7} - \dfrac{1}{12}\right) \text{ of } 1\dfrac{2}{5}$

Answer

$\dfrac{2}{5} \text{ of } \left(\dfrac{1}{7} - \dfrac{1}{12}\right) \text{ of } 1\dfrac{2}{5}$

Using BODMAS, we simplify the bracket first,

$\Rightarrow \dfrac{1}{7} - \dfrac{1}{12} = \dfrac{12}{84} - \dfrac{7}{84} = \dfrac{5}{84}.\\[1em] \Rightarrow \dfrac{2}{5} \text{ of } \dfrac{5}{84} \text{ of } 1\dfrac{2}{5} = \dfrac{2}{5} \times \dfrac{5}{84} \times \dfrac{7}{5} \\[1em] = \dfrac{2 \times 5 \times 7}{5 \times 84 \times 5} \\[1em] = \dfrac{70}{2100} \\[1em] = \dfrac{1}{30}.$

Hence, $\dfrac{2}{5} \text{ of } \left(\dfrac{1}{7} - \dfrac{1}{12}\right) \text{ of } 1\dfrac{2}{5} = \dfrac{1}{30}$ .

Question 5(viii)

Simplify:

$\left(\dfrac{1}{2} - \dfrac{1}{3}\right)\left(\dfrac{3}{4} - \dfrac{4}{5}\right) \div \left(\dfrac{1}{2} - \dfrac{2}{5} + \dfrac{1}{7}\right)$

Answer

$\left(\dfrac{1}{2} - \dfrac{1}{3}\right)\left(\dfrac{3}{4} - \dfrac{4}{5}\right) \div \left(\dfrac{1}{2} - \dfrac{2}{5} + \dfrac{1}{7}\right)$

Using BODMAS, we simplify the brackets first,

$\Rightarrow \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{3}{6} - \dfrac{2}{6} = \dfrac{1}{6}.\\[1em] \Rightarrow \dfrac{3}{4} - \dfrac{4}{5} = \dfrac{15}{20} - \dfrac{16}{20} = \dfrac{-1}{20}.\\[1em] \Rightarrow \dfrac{1}{2} - \dfrac{2}{5} + \dfrac{1}{7} = \dfrac{35}{70} - \dfrac{28}{70} + \dfrac{10}{70} = \dfrac{17}{70}.\\[1em] \Rightarrow \dfrac{1}{6} \times \dfrac{-1}{20} \div \dfrac{17}{70} = \dfrac{-1}{120} \times \dfrac{70}{17} \\[1em] = \dfrac{-70}{2040} \\[1em] = \dfrac{-7}{204}.$

Hence, $\left(\dfrac{1}{2} - \dfrac{1}{3}\right)\left(\dfrac{3}{4} - \dfrac{4}{5}\right) \div \left(\dfrac{1}{2} - \dfrac{2}{5} + \dfrac{1}{7}\right) = \dfrac{-7}{204}$ .

Question 5(ix)

Simplify:

$\dfrac{5}{6} - \dfrac{3}{5}\left(\dfrac{1}{3} + \dfrac{2}{11}\right)$

Answer

$\dfrac{5}{6} - \dfrac{3}{5}\left(\dfrac{1}{3} + \dfrac{2}{11}\right)$

Using BODMAS, we simplify the bracket first,

$\Rightarrow \dfrac{1}{3} + \dfrac{2}{11} = \dfrac{11}{33} + \dfrac{6}{33} = \dfrac{17}{33}.\\[1em] \Rightarrow \dfrac{5}{6} - \dfrac{3}{5} \times \dfrac{17}{33} = \dfrac{5}{6} - \dfrac{51}{165} \\[1em] = \dfrac{5}{6} - \dfrac{17}{55} \\[1em] = \dfrac{275}{330} - \dfrac{102}{330} \\[1em] = \dfrac{173}{330}.$

Hence, $\dfrac{5}{6} - \dfrac{3}{5}\left(\dfrac{1}{3} + \dfrac{2}{11}\right) = \dfrac{173}{330}$ .

Exercise 6(E)

Question 1

From a rope $10\dfrac{1}{2}$ m long, $4\dfrac{5}{8}$ m is cut off. Find the length of the remaining rope.

Answer

Total length of the rope = $10\dfrac{1}{2}$ m.

Length cut off = $4\dfrac{5}{8}$ m.

Length of remaining rope = Total length − Length cut off

$= 10\dfrac{1}{2} - 4\dfrac{5}{8} \\[1em] = \dfrac{21}{2} - \dfrac{37}{8} \\[1em] \text{L.C.M. of 2 and 8 = 8}\\[1em] = \dfrac{84}{8} - \dfrac{37}{8} \\[1em] = \dfrac{47}{8} \\[1em] = 5\dfrac{7}{8} \text{ m}.$

Hence, the length of the remaining rope = $5\dfrac{7}{8}$ m.

Question 2

A piece of cloth is 5 m long. After washing, it shrinks by $\dfrac{1}{25}$ of its length. What is the length of the cloth after washing?

Answer

Length of the cloth = 5 m.

Shrinkage = $\dfrac{1}{25}$ of its length

$= \dfrac{1}{25} \times 5 \\[1em] = \dfrac{5}{25} \\[1em] = \dfrac{1}{5} \text{ m}.$

Length of the cloth after washing = 5 − $\dfrac{1}{5}$

$= \dfrac{25}{5} - \dfrac{1}{5} \\[1em] = \dfrac{24}{5} \\[1em] = 4\dfrac{4}{5} \text{ m}.$

Hence, the length of the cloth after washing = $4\dfrac{4}{5}$ m.

Question 3

I bought wheat worth ₹ $12\dfrac{1}{2}$ , rice worth ₹ $25\dfrac{3}{4}$ and vegetables worth ₹ $10\dfrac{1}{4}$ . I gave a hundred-rupee note to the shopkeeper; how much money did he return to me?

Answer

Price of wheat = ₹ $12\dfrac{1}{2}$

Price of rice = ₹ $25\dfrac{3}{4}$

Price of vegetables = ₹ $10\dfrac{1}{4}$

Money given to Shopkeeper = ₹ 100

Total money spent = $12\dfrac{1}{2} + 25\dfrac{3}{4} + 10\dfrac{1}{4}$

$= \dfrac{25}{2} + \dfrac{103}{4} + \dfrac{41}{4} \\[1em] = \dfrac{50}{4} + \dfrac{103}{4} + \dfrac{41}{4} \\[1em] = \dfrac{194}{4} \\[1em] = \dfrac{97}{2} \\[1em] = 48\dfrac{1}{2}.$

So, total money spent = ₹ $48\dfrac{1}{2}$ .

Money returned by the shopkeeper = 100 − $48\dfrac{1}{2}$

$= \dfrac{200}{2} - \dfrac{97}{2} \\[1em] = \dfrac{103}{2} \\[1em] = 51\dfrac{1}{2}.$

Hence, the shopkeeper returned ₹ $51\dfrac{1}{2}$ .

Question 4

Out of 500 oranges in a box, $\dfrac{3}{25}$ are rotten and $\dfrac{1}{5}$ are kept for some guests. How many oranges are left in the box?

Answer

Total oranges = 500.

Number of rotten oranges = $\dfrac{3}{25} \text{ of } 500 = \dfrac{3}{25} \times 500$ = 60.

Number of oranges kept for guests = $\dfrac{1}{5} \text{ of } 500 = \dfrac{1}{5} \times 500$ = 100.

Oranges used = 60 + 100 = 160.

Oranges left = Total oranges − Oranges used

Oranges left = 500 − 160 = 340.

Hence, the number of oranges left in the box = 340.

Question 5

An ornament piece is made of gold and copper. Its total weight is 96 g. If $\dfrac{1}{12}$ of the ornament is copper, find the weight of gold in it.

Answer

Total weight of the ornament = 96 g.

Weight of copper = $\dfrac{1}{12} \text{ of } 96 = \dfrac{1}{12} \times 96$ = 8 g.

Weight of gold = Total weight − Weight of copper

= 96 − 8

= 88 g.

Hence, the weight of gold = 88 g.

Question 6

A girl did half of some work on Monday and one-third of it on Tuesday. How much will she have to do on Wednesday in order to complete the work?

Answer

Let the total work = 1.

Work done on Monday = $\dfrac{1}{2}$ .

Work done on Tuesday = $\dfrac{1}{3}$ .

Total work done in two days = $\dfrac{1}{2} + \dfrac{1}{3}$

$= \dfrac{3}{6} + \dfrac{2}{6} \\[1em] = \dfrac{5}{6}.$

Work left for Wednesday = Total work - Work done in two days

Work left for Wednesday = 1 − $\dfrac{5}{6}$

$= \dfrac{6}{6} - \dfrac{5}{6} \\[1em] = \dfrac{1}{6}.$

Hence, she will have to do $\dfrac{1}{6}$ of the work on Wednesday.

Question 7

A man spends $\dfrac{3}{8}$ of his money and still has ₹ 720 left with him. How much money did he have at first?

Answer

Part of money spent = $\dfrac{3}{8}$ .

Part of money left = 1 − $\dfrac{3}{8} = \dfrac{8}{8} - \dfrac{3}{8} = \dfrac{5}{8}$ .

Given, $\dfrac{5}{8}$ of the money = ₹ 720.

$\Rightarrow \text{Total money} = 720 \div \dfrac{5}{8} \\[1em] = 720 \times \dfrac{8}{5} \\[1em] = 144 \times 8 \\[1em] = 1152.$

Hence, the man had ₹ 1,152 at first.

Question 8

In a school, $\dfrac{4}{5}$ of the students are boys, and the number of girls is 100. Find the number of boys.

Answer

Part of students who are boys = $\dfrac{4}{5}$ .

Part of students who are girls = 1 − $\dfrac{4}{5} = \dfrac{1}{5}$ .

Given, $\dfrac{1}{5}$ of the students = 100 girls.

$\Rightarrow \text{Total students} = 100 \div \dfrac{1}{5} \\[1em] = 100 \times 5 \\[1em] = 500.$

Number of boys = $\dfrac{4}{5} \text{ of } 500 = \dfrac{4}{5} \times 500$ = 400.

Hence, the number of boys = 400.

Question 9

After finishing $\dfrac{3}{4}$ of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?

Answer

Given, $\dfrac{3}{4}$ of the journey = 12 km.

$\Rightarrow \text{Total journey} = 12 \div \dfrac{3}{4} \\[1em] = 12 \times \dfrac{4}{3} \\[1em] = 16 \text{ km}.$

Distance left = Total journey − Distance covered

= 16 − 12

= 4 km.

Hence, the distance still left to be covered = 4 km.

Question 10

When Ajit travelled 15 km, he found that one-fourth of his journey was still left. What was the full length of the journey?

Answer

Part of journey left = $\dfrac{1}{4}$ .

Part of journey travelled = 1 − $\dfrac{1}{4} = \dfrac{3}{4}$ .

Given, $\dfrac{3}{4}$ of the journey = 15 km.

$\Rightarrow \text{Total journey} = 15 \div \dfrac{3}{4} \\[1em] = 15 \times \dfrac{4}{3} \\[1em] = 20 \text{ km}.$

Hence, the full length of the journey = 20 km.

Question 11

In a particular month, a man earns ₹ 7,200. Out of this income, he spends $\dfrac{3}{10}$ on food, $\dfrac{1}{4}$ on house rent, $\dfrac{1}{10}$ on insurance and $\dfrac{2}{25}$ on holidays. How much did he save in that month?

Answer

Total income = ₹ 7,200.

Fraction of income spent = $\dfrac{3}{10} + \dfrac{1}{4} + \dfrac{1}{10} + \dfrac{2}{25}$

LCM of 10, 4, 10 and 25 = 100.

$= \dfrac{30}{100} + \dfrac{25}{100} + \dfrac{10}{100} + \dfrac{8}{100} \\[1em] = \dfrac{73}{100}.$

Fraction of income saved = 1 − $\dfrac{73}{100} = \dfrac{27}{100}$ .

Money saved = $\dfrac{27}{100}$ of 7,200

$= \dfrac{27}{100} \times 7,200 \\[1em] = 27 \times 72 \\[1em] = 1,944.$

Hence, the man saved ₹ 1,944 in that month.

Multiple Choice Questions

Question 1

In the given number line, identify the values of A and B.

Mark all the given integers on a number line: Mathematics Solutions ICSE Class 7.

$A = \dfrac{2}{3}, B = \dfrac{4}{5}$
$A = \dfrac{2}{3}, B = 1\dfrac{3}{4}$
$A = 3 \text{ and } B = 7$
$A = 2 \text{ and } B = 6$

Answer

On the number line, point A lies between 0 and 1. Therefore, its value is greater than 0 and less than 1.

The interval between 0 and 1 is divided into 3 equal parts, and point A lies on the second division mark.

$\therefore A = \dfrac{2}{3}$

Point B lies between 1 and 2. Therefore, its value is greater than 1 and less than 2.

The interval between 1 and 2 is divided into 4 equal parts, and point B lies on the third division mark after 1.

$\therefore B = 1 + \dfrac{3}{4} = 1\dfrac{3}{4}$

Hence, option 2 is the correct option.

Question 2

In $\dfrac{3}{5}, \dfrac{2}{5}, \dfrac{3}{4}, \dfrac{2}{7} \text{ and } \dfrac{4}{5}$ , like fractions are:

$\dfrac{2}{5}, \dfrac{2}{7}$
$\dfrac{3}{5}, \dfrac{3}{4}$
$\dfrac{3}{5}, \dfrac{2}{5}$
$\dfrac{3}{5}, \dfrac{2}{5}, \dfrac{4}{5}$

Answer

Like fractions have the same denominator. The fractions with denominator 5 are $\dfrac{3}{5}, \dfrac{2}{5} \text{ and } \dfrac{4}{5}$ .

Hence, option 4 is the correct option.

Question 3

Out of $\dfrac{15}{7}, \dfrac{15}{8}, \dfrac{15}{11} \text{ and } \dfrac{15}{13}$ , the largest fraction is:

$\dfrac{15}{7}$
$\dfrac{15}{8}$
$\dfrac{15}{11}$
$\dfrac{15}{13}$

Answer

These fractions have the same numerator 15. Among fractions with the same numerator, the one with the smallest denominator is the largest. The smallest denominator is 7.

Hence, option 1 is the correct option.

Question 4

Out of $\dfrac{7}{15}, \dfrac{8}{15}, \dfrac{11}{15} \text{ and } \dfrac{13}{15}$ , the least fraction is:

$\dfrac{7}{15}$
$\dfrac{8}{15}$
$\dfrac{11}{15}$
$\dfrac{13}{15}$

Answer

These are like fractions (same denominator 15). Among like fractions, the one with the smallest numerator is the least. The smallest numerator is 7.

Hence, option 1 is the correct option.

Question 5

$\dfrac{7}{16} + \dfrac{13}{16} - \dfrac{5}{16}$ is equal to:

$\dfrac{15}{16}$
$\dfrac{25}{16}$
0
1

Answer

$\Rightarrow \dfrac{7}{16} + \dfrac{13}{16} - \dfrac{5}{16} = \dfrac{7 + 13 - 5}{16} \\[1em] = \dfrac{15}{16}.$

Hence, option 1 is the correct option.

Question 6

$31\dfrac{2}{3} - 16\dfrac{2}{3} - 15\dfrac{2}{3}$ is equal to:

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

Answer

$\Rightarrow 31\dfrac{2}{3} - 16\dfrac{2}{3} - 15\dfrac{2}{3} \\[1em] \text{Changing mixed fraction into improper fraction}\\[1em] = \dfrac{95}{3} - \dfrac{50}{3} - \dfrac{47}{3} \\[1em] = \dfrac{95 - 50 - 47}{3} \\[1em] = \dfrac{-2}{3}.$

Hence, option 4 is the correct option.

Question 7

12 ÷ 12 = 1, 0 ÷ 12 = 0; then 12 ÷ 0 is:

1
0
not defined

Answer

Division by zero is not defined.

Hence, option 3 is the correct option.

Question 8

$\dfrac{1}{4} \times \dfrac{1}{4} \div \dfrac{1}{4} \text{ of } \dfrac{1}{4}$ is equal to:

1
16
$\dfrac{1}{16}$
$\dfrac{1}{4}$

Answer

Solving,

$\Rightarrow \dfrac{1}{4} \times \dfrac{1}{4} \div \dfrac{1}{4} \text{ of } \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{4} \times \dfrac{1}{4} \div \dfrac{1}{16} \\[1em] \Rightarrow \dfrac{1}{4} \times \dfrac{1}{4} \times 16 \\[1em] \Rightarrow \dfrac{16}{16} \\[1em] \Rightarrow 1.$

Hence, option 1 is the correct option.

Question 9

$\dfrac{1}{4} \text{ of } \dfrac{1}{4} \div \dfrac{1}{4} \times \dfrac{1}{4}$ is equal to:

1
16
$\dfrac{1}{16}$
$\dfrac{1}{4}$

Answer

Solving 'of' first,

$\Rightarrow \dfrac{1}{4} \text{ of } \dfrac{1}{4} = \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{16}.\\[1em] \Rightarrow \dfrac{1}{16} \div \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{16} \times \dfrac{4}{1} \times \dfrac{1}{4} \\[1em] = \dfrac{1}{16}.$

Hence, option 3 is the correct option.

Question 10

If $\dfrac{x}{3} = 15$ , then $\dfrac{x}{5}$ is:

45
9
75
none of these

Answer

$\Rightarrow \dfrac{x}{3} = 15 \\[1em] \Rightarrow x = 15 \times 3 = 45.\\[1em] \Rightarrow \dfrac{x}{5} = \dfrac{45}{5} = 9.$

Hence, option 2 is the correct option.

Question 11

Assertion (A): 7 ÷ 7 x 7 - 7 = 0

Reason (R): $7 \times \dfrac{1}{7} \times 7 - 7 = 7 - 7 = 0$

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

According to BODMAS,

Checking A: 7 ÷ 7 x 7 - 7 = 1 x 7 - 7 = 7 - 7 = 0. So A is true.

Checking R: $7 \times \dfrac{1}{7} \times 7 - 7 = 7 - 7 = 0$ . So R is true and correctly explains A.

Hence, option 3 is the correct option.

Question 12

Assertion (A): $\dfrac{1}{2}$ of a fraction is 15, then $\dfrac{1}{5}$ of the same fraction is 6.

Reason (R): The fraction is 30.

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

If $\dfrac{1}{2}$ of a fraction is 15, then the fraction = 15 × 2 = 30, so R is true.

Now, $\dfrac{1}{5} \text{ of } 30 = \dfrac{1}{5} \times 30$ = 6, so A is true.

Both A and R are true, and R correctly explains A.

Hence, option 3 is the correct option.

Question 13

Statement 1: If $x = 1\dfrac{1}{2} \div 3\dfrac{1}{2} \times 2\dfrac{1}{3}$ , then the value of $x$ is 1.

Statement 2: $1\dfrac{1}{2} \div 3\dfrac{1}{2} \times 2\dfrac{1}{3} = \dfrac{3}{2} \times \dfrac{2}{7} \times \dfrac{7}{3} = 1$

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

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

So, the value of $x$ is 1, and both statements are true with statement 2 showing the correct working.

Hence, option 1 is the correct option.

Chapter 6

Fractions

Class - 6 Concise Mathematics Selina

Exercise 6(A)

Question 1

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 3

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 4(iv)

Question 5(i)

Question 5(ii)

Question 5(iii)

Question 5(iv)

Question 6(i)

Question 6(ii)

Question 6(iii)

Exercise 6(B)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 5(i)

Question 5(ii)

Question 5(iii)

Question 6(i)

Question 6(ii)

Question 6(iii)

Question 7(i)

Question 7(ii)

Question 7(iii)

Question 7(iv)

Exercise 6(C)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

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)

Exercise 6(D)

Question 1(i)

Question 1(ii)

Question 1(iii)

Question 1(iv)

Question 1(v)

Question 1(vi)