Fractions

Write the following division as fractions:

(i) 3 ÷ 7

(ii) 11 ÷ 78

(iii) 113 ÷ 128

Answer

(i) 3 ÷ 7

A division a ÷ b can be written as the fraction $\dfrac{a}{b}$.

So, 3 ÷ 7 = $\dfrac{3}{7}$.

Hence, 3 ÷ 7 = $\dfrac{3}{7}$.

(ii) 11 ÷ 78

A division a ÷ b can be written as the fraction $\dfrac{a}{b}$.

So, 11 ÷ 78 = $\dfrac{11}{78}$.

Hence, 11 ÷ 78 = $\dfrac{11}{78}$.

(iii) 113 ÷ 128

A division a ÷ b can be written as the fraction $\dfrac{a}{b}$.

So, 113 ÷ 128 = $\dfrac{113}{128}$.

Hence, 113 ÷ 128 = $\dfrac{113}{128}$.

Write the following fractions in words:

(i) $\dfrac{2}{7}$

(ii) $\dfrac{3}{10}$

(iii) $\dfrac{15}{28}$

Answer

(i) $\dfrac{2}{7}$

Hence, $\dfrac{2}{7}$ in words is two-seventh.

(ii) $\dfrac{3}{10}$

Hence, $\dfrac{3}{10}$ in words is three-tenth.

(iii) $\dfrac{15}{28}$

Hence, $\dfrac{15}{28}$ in words is fifteen-twenty eighth.

Write the following fractions in number form:

(i) one-sixth

(ii) three-eleventh

(iii) seven-fortieth

(iv) thirteen-one hundred twenty fifth

Answer

(i) one-sixth

Hence, one-sixth in number form is $\dfrac{1}{6}$.

(ii) three-eleventh

Hence, three-eleventh in number form is $\dfrac{3}{11}$.

(iii) seven-fortieth

Hence, seven-fortieth in number form is $\dfrac{7}{40}$.

(iv) thirteen-one hundred twenty fifth

Hence, thirteen-one hundred twenty fifth in number form is $\dfrac{13}{125}$.

What fraction of each of the following figures is shaded part?

Answer

(i) Total parts = 7

Shaded parts = 4

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{4}{7}$.

Hence, fraction = $\dfrac{4}{7}$.

(ii) Total parts = 8

Shaded parts = 3

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{3}{8}$.

Hence, fraction = $\dfrac{3}{8}$.

(iii) Total parts = 4

Shaded parts = $\dfrac{1}{2}$

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{\dfrac{1}{2}}{4} = \dfrac{1}{2 \times 4} = \dfrac{1}{8}$.

Hence, fraction = $\dfrac{1}{8}$.

(iv) Total parts = 4

Shaded parts = 1

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{1}{4}$.

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

(v) Total parts = 6

Shaded parts = 1

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{1}{6}$.

Hence, fraction = $\dfrac{1}{6}$.

(vi) Total parts = 10

Shaded parts = 3

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

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

(vii) Total parts = 7

Shaded parts = 3

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{3}{7}$.

Hence, fraction = $\dfrac{3}{7}$.

(viii) Total parts = 4

Shaded parts = 2

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{2}{4}$.

Hence, fraction = $\dfrac{2}{4}$.

(ix) Total parts = 9

Shaded parts = 4

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{4}{9}$.

Hence, fraction = $\dfrac{4}{9}$.

Shade the parts of the following figures according to given fractions:

Answer

(i) $\dfrac{3}{4}$

The figure (square) is divided into 4 equal parts. Shade 3 parts out of 4.

(ii) $\dfrac{1}{6}$

The figure (circle) is divided into 6 equal parts. Shade 1 part out of 6.

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

The figure (triangle) is divided into 4 equal parts. Shade 1 part out of 4.

(iv) $\dfrac{4}{9}$

The figure (square) is divided into 9 equal parts. Shade 4 parts out of 9.

(v) $\dfrac{1}{3}$

The figure (hexagon) is divided into 6 equal parts (i.e., 2 parts make $\dfrac{1}{3}$). Shade 2 parts out of 6.

(vi) $\dfrac{5}{8}$

The figure (8 circles) shows 8 equal parts. Shade 5 circles out of 8.

Write the fraction in which

(i) numerator = 5 and denominator = 13

(ii) denominator = 23 and numerator = 17

Answer

(i) numerator = 5 and denominator = 13

Fraction = $\dfrac{\text{numerator}}{\text{denominator}} = \dfrac{5}{13}$.

Hence, the required fraction = $\dfrac{5}{13}$.

(ii) denominator = 23 and numerator = 17

Fraction = $\dfrac{\text{numerator}}{\text{denominator}} = \dfrac{17}{23}$.

Hence, the required fraction = $\dfrac{17}{23}$.

Shabana has to stitch 35 dresses. So far, she has stitched 21 dresses. What fraction of dresses has she stitched?

Answer

Total number of dresses to be stitched = 35

Number of dresses stitched so far = 21

Fraction of dresses stitched = $\dfrac{\text{Dresses stitched}}{\text{Total dresses}}$

= $\dfrac{21}{35}$

Hence, the required fraction = $\dfrac{21}{35}$.

What fraction of a day is 8 hours?

Answer

We know, 1 day = 24 hours.

Fraction = $\dfrac{8 \text{ hours}}{24 \text{ hours}}$

= $\dfrac{8}{24}$

Hence, the required fraction = $\dfrac{8}{24}$.

What fraction of an hour is 45 minutes?

Answer

We know, 1 hour = 60 minutes.

Fraction = $\dfrac{45 \text{ minutes}}{60 \text{ minutes}}$

= $\dfrac{45}{60}$

Hence, the required fraction = $\dfrac{45}{60}$.

How many natural numbers are there from 87 to 97? What fraction of them are prime numbers?

Answer

Natural numbers from 87 to 97 are :

87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97 and their number = 11.

Out of these, the prime numbers are 89, 97.

Fraction of prime numbers = $\dfrac{\text{Number of primes}}{\text{Total numbers}} = \dfrac{2}{11}$.

Hence, total natural numbers = 11 and fraction of prime numbers = $\dfrac{2}{11}$.

Show the fractions $\dfrac{2}{5}, \dfrac{3}{5}, \dfrac{4}{5}$ and $\dfrac{5}{5}$ on a number line.

Answer

Draw a straight line and mark a point O on it representing 0. Mark a point A on the right of O at unit length so that OA represents 1.

Now divide the line segment OA into 5 equal parts. Each part represents $\dfrac{1}{5}$.

Starting from O, mark the points at $\dfrac{1}{5}, \dfrac{2}{5}, \dfrac{3}{5}, \dfrac{4}{5}$ and $\dfrac{5}{5}$ (= 1).

Hence, the fractions $\dfrac{2}{5}, \dfrac{3}{5}, \dfrac{4}{5}$ and $\dfrac{5}{5}$ are represented on the number line as shown above.

Show $\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}$ and $\dfrac{7}{8}$ on a number line.

Answer

Draw a straight line and mark a point O on it representing 0. Mark a point A on the right of O at unit length so that OA represents 1.

Now divide the line segment OA into 8 equal parts. Each part represents $\dfrac{1}{8}$.

Starting from O, mark the points at $\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}$ and $\dfrac{7}{8}$.

Hence, the fractions $\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}$ and $\dfrac{7}{8}$ are represented on the number line as shown above.

Show $\dfrac{0}{10}, \dfrac{1}{10}, \dfrac{3}{10}, \dfrac{5}{10}, \dfrac{7}{10}$ and $\dfrac{10}{10}$ on a number line.

Answer

Draw a straight line and mark a point O on it representing 0. Mark a point A on the right of O at unit length so that OA represents 1.

Now divide the line segment OA into 10 equal parts. Each part represents $\dfrac{1}{10}$.

Starting from O, mark the points at $\dfrac{0}{10}$ (= 0), $\dfrac{1}{10}, \dfrac{3}{10}, \dfrac{5}{10}, \dfrac{7}{10}$ and $\dfrac{10}{10}$ (= 1).

Hence, the fractions $\dfrac{0}{10}, \dfrac{1}{10}, \dfrac{3}{10}, \dfrac{5}{10}, \dfrac{7}{10}$ and $\dfrac{10}{10}$ are represented on the number line as shown above.

State which of the following fractions are proper, improper or mixed:

(i) $\dfrac{15}{26}$

(ii) $\dfrac{17}{12}$

(iii) $5\dfrac{2}{3}$

(iv) $\dfrac{6}{8}$

(v) $11\dfrac{5}{7}$

(vi) $\dfrac{117}{8}$

(vii) $\dfrac{222}{333}$

(viii) $\dfrac{531}{247}$

Answer

(i) $\dfrac{15}{26}$

Here, numerator (15) < denominator (26).

Hence, $\dfrac{15}{26}$ is a proper fraction.

(ii) $\dfrac{17}{12}$

Here, numerator (17) > denominator (12).

Hence, $\dfrac{17}{12}$ is an improper fraction.

(iii) $5\dfrac{2}{3}$

It consists of a natural number 5 and a proper fraction $\dfrac{2}{3}$.

Hence, $5\dfrac{2}{3}$ is a mixed fraction.

(iv) $\dfrac{6}{8}$

Here, numerator (6) < denominator (8).

Hence, $\dfrac{6}{8}$ is a proper fraction.

(v) $11\dfrac{5}{7}$

It consists of a natural number 11 and a proper fraction $\dfrac{5}{7}$.

Hence, $11\dfrac{5}{7}$ is a mixed fraction.

(vi) $\dfrac{117}{8}$

Here, numerator (117) > denominator (8).

Hence, $\dfrac{117}{8}$ is an improper fraction.

(vii) $\dfrac{222}{333}$

Here, numerator (222) < denominator (333).

Hence, $\dfrac{222}{333}$ is a proper fraction.

(viii) $\dfrac{531}{247}$

Here, numerator (531) > denominator (247).

Hence, $\dfrac{531}{247}$ is an improper fraction.

Convert the following improper fractions into mixed numbers:

(i) $\dfrac{17}{3}$

(ii) $\dfrac{119}{15}$

(iii) $\dfrac{961}{13}$

(iv) $\dfrac{117}{32}$

Answer

(i) $\dfrac{17}{3}$

Dividing 17 by 3, we get quotient = 5 and remainder = 2.

$\Rightarrow \dfrac{17}{3} = 5\dfrac{2}{3}.$

Hence, $\dfrac{17}{3} = 5\dfrac{2}{3}$.

(ii) $\dfrac{119}{15}$

Dividing 119 by 15, we get quotient = 7 and remainder = 14.

$\Rightarrow \dfrac{119}{15} = 7\dfrac{14}{15}.$

Hence, $\dfrac{119}{15} = 7\dfrac{14}{15}$.

(iii) $\dfrac{961}{13}$

Dividing 961 by 13, we get quotient = 73 and remainder = 12.

$\Rightarrow \dfrac{961}{13} = 73\dfrac{12}{13}.$

Hence, $\dfrac{961}{13} = 73\dfrac{12}{13}$.

(iv) $\dfrac{117}{32}$

Dividing 117 by 32, we get quotient = 3 and remainder = 21.

$\Rightarrow \dfrac{117}{32} = 3\dfrac{21}{32}.$

Hence, $\dfrac{117}{32} = 3\dfrac{21}{32}$.

Convert the following mixed fractions into improper fractions:

(i) $7\dfrac{2}{11}$

(ii) $3\dfrac{5}{48}$

(iii) $13\dfrac{7}{64}$

(iv) $7\dfrac{2}{3}$

Answer

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

(i) $7\dfrac{2}{11}$

$\Rightarrow 7\dfrac{2}{11} = \dfrac{7 \times 11 + 2}{11} \\[1em] = \dfrac{77 + 2}{11} \\[1em] = \dfrac{79}{11}.$

Hence, the required fraction is $7\dfrac{2}{11} = \dfrac{79}{11}$.

(ii) $3\dfrac{5}{48}$

$\Rightarrow 3\dfrac{5}{48} = \dfrac{3 \times 48 + 5}{48} \\[1em] = \dfrac{144 + 5}{48} \\[1em] = \dfrac{149}{48}.$

Hence, the required fraction is $3\dfrac{5}{48} = \dfrac{149}{48}$.

(iii) $13\dfrac{7}{64}$

$\Rightarrow 13\dfrac{7}{64} = \dfrac{13 \times 64 + 7}{64} \\[1em] = \dfrac{832 + 7}{64} \\[1em] = \dfrac{839}{64}.$

Hence, the required fraction is $13\dfrac{7}{64} = \dfrac{839}{64}$.

(iv) $7\dfrac{2}{3}$

$\Rightarrow 7\dfrac{2}{3} = \dfrac{7 \times 3 + 2}{3} \\[1em] = \dfrac{21 + 2}{3} \\[1em] = \dfrac{23}{3}.$

Hence, the required fraction is $7\dfrac{2}{3} = \dfrac{23}{3}$.

Write the fractions representing the shaded regions and pair up the equivalent fractions from each row:

Answer

The fractions representing the shaded regions are:

(i) Total parts = 2

Shaded parts = 1

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{1}{2}$.

Hence, fraction = $\dfrac{1}{2}$

(ii) Total parts = 6

Shaded parts = 4

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{4}{6}$.

Hence, fraction = $\dfrac{4}{6}$

(iii) Total parts = 9

Shaded parts = 3

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{3}{9}$.

Hence, fraction = $\dfrac{3}{9}$

(iv) Total parts = 8

Shaded parts = 2

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

Hence, fraction = $\dfrac{2}{8}$

(v) Total parts = 4

Shaded parts = 3

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{3}{4}$.

Hence, fraction = $\dfrac{3}{4}$

(a) Total parts = 8

Shaded parts = 2

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

Hence, fraction = $\dfrac{2}{8}$

(b) Total parts = 12

Shaded parts = 8

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{8}{12}$.

Hence, fraction = $\dfrac{8}{12}$

(c) Total parts = 16

Shaded parts = 12

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{12}{16} = \dfrac{3}{4}$.

Hence, fraction = $\dfrac{3}{4}$

(d) Total parts = 8

Shaded parts = 4

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{4}{8}$.

Hence, fraction = $\dfrac{4}{8}$

(e) Total parts = 9

Shaded parts = 3

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{3}{9}$.

Hence, fraction = $\dfrac{3}{9}$

Hence, the equivalent pairs are:

(i) ↔ (d), (ii) ↔ (b), (iii) ↔ (e), (iv) ↔ (a), (v) ↔ (c).

(i) Find the equivalent fraction of $\dfrac{15}{35}$ with denominator 7

(ii) Find the equivalent fraction of $\dfrac{2}{9}$ with denominator 63

Answer

(i) Equivalent fraction of $\dfrac{15}{35}$ with denominator 7

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

So, divide both numerator and denominator by 5.

$\Rightarrow \dfrac{15}{35} = \dfrac{15 \div 5}{35 \div 5} \\[1em] = \dfrac{3}{7}.$

Hence, the required equivalent fraction = $\dfrac{3}{7}$.

(ii) Equivalent fraction of $\dfrac{2}{9}$ with denominator 63

To get 63 from 9, we have to multiply 9 by 7.

So, multiply both numerator and denominator by 7.

$\Rightarrow \dfrac{2}{9} = \dfrac{2 \times 7}{9 \times 7} \\[1em] = \dfrac{14}{63}.$

Hence, the required equivalent fraction = $\dfrac{14}{63}$.

Find the equivalent fraction of $\dfrac{3}{5}$ having

(i) denominator 30

(ii) numerator 27

Answer

(i) Equivalent fraction of $\dfrac{3}{5}$ with denominator 30

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

So, multiply both numerator and denominator by 6.

$\Rightarrow \dfrac{3}{5} = \dfrac{3 \times 6}{5 \times 6} \\[1em] = \dfrac{18}{30}.$

Hence, the required equivalent fraction = $\dfrac{18}{30}$.

(ii) Equivalent fraction of $\dfrac{3}{5}$ with numerator 27

To get 27 from 3, we have to multiply 3 by 9.

So, multiply both numerator and denominator by 9.

$\Rightarrow \dfrac{3}{5} = \dfrac{3 \times 9}{5 \times 9} \\[1em] = \dfrac{27}{45}.$

Hence, the required equivalent fraction = $\dfrac{27}{45}$.

Replace '$\boxed{\phantom{653}}$' in each of the following by the correct number:

(i) $\dfrac{2}{3} = \dfrac{\boxed{\phantom{653}}}{15}$

(ii) $\dfrac{7}{18} = \dfrac{42}{\boxed{\phantom{653}}}$

(iii) $\dfrac{4}{\boxed{\phantom{653}}} = \dfrac{12}{15}$

(iv) $\dfrac{\boxed{\phantom{653}}}{11} = \dfrac{70}{154}$

Answer

(i) $\dfrac{2}{3} = \dfrac{\boxed{\phantom{653}}}{15}$

To get 15 from 3, we have to multiply 3 by 5.

So, multiply both numerator and denominator by 5.

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

Hence, replace $\boxed{\phantom{653}}$ by 10.

(ii) $\dfrac{7}{18} = \dfrac{42}{\boxed{\phantom{653}}}$

To get 42 from 7, we have to multiply 7 by 6.

So, multiply both numerator and denominator by 6.

$\Rightarrow \dfrac{7}{18} = \dfrac{7 \times 6}{18 \times 6} \\[1em] = \dfrac{42}{108}.$

Hence, replace $\boxed{\phantom{653}}$ by 108.

(iii) $\dfrac{4}{\boxed{\phantom{653}}} = \dfrac{12}{15}$

To get 4 from 12, we have to divide 12 by 3.

So, divide both numerator and denominator by 3.

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

Hence, replace $\boxed{\phantom{653}}$ by 5.

(iv) $\dfrac{\boxed{\phantom{653}}}{11} = \dfrac{70}{154}$

To get 11 from 154, we have to divide 154 by 14.

So, divide both numerator and denominator by 14.

$\Rightarrow \dfrac{70}{154} = \dfrac{70 \div 14}{154 \div 14} \\[1em] = \dfrac{5}{11}.$

Hence, replace $\boxed{\phantom{653}}$ by 5.

Check whether the given pairs of fractions are equivalent:

(i) $\dfrac{3}{10}, \dfrac{12}{40}$

(ii) $\dfrac{5}{8}, \dfrac{30}{48}$

(iii) $\dfrac{4}{6}, \dfrac{30}{20}$

(iv) $\dfrac{7}{13}, \dfrac{5}{11}$

Answer

(i) $\dfrac{3}{10}, \dfrac{12}{40}$

By cross multiplication,

3 × 40 = 120 and 10 × 12 = 120

Here, the cross products are equal.

Hence, $\dfrac{3}{10}$ and $\dfrac{12}{40}$ are equivalent fractions.

(ii) $\dfrac{5}{8}, \dfrac{30}{48}$

By cross multiplication,

5 × 48 = 240 and 8 × 30 = 240

Here, the cross products are equal.

Hence, $\dfrac{5}{8}$ and $\dfrac{30}{48}$ are equivalent fractions.

(iii) $\dfrac{4}{6}, \dfrac{30}{20}$

By cross multiplication,

4 × 20 = 80 and 6 × 30 = 180

Here, the cross products are not equal.

Hence, $\dfrac{4}{6}$ and $\dfrac{30}{20}$ are not equivalent fractions.

(iv) $\dfrac{7}{13}, \dfrac{5}{11}$

By cross multiplication,

7 × 11 = 77 and 13 × 5 = 65

Here, the cross products are not equal.

Hence, $\dfrac{7}{13}$ and $\dfrac{5}{11}$ are not equivalent fractions.

Reduce the following fractions to simplest form:

(i) $\dfrac{12}{27}$

(ii) $\dfrac{150}{350}$

(iii) $\dfrac{18}{81}$

(iv) $\dfrac{276}{115}$

Answer

(i) $\dfrac{12}{27}$

By prime factorisation,

$\Rightarrow \dfrac{12}{27} = \dfrac{2 \times 2 \times 3}{3 \times 3 \times 3} \\[1em] = \dfrac{2 \times 2}{3 \times 3} \\[1em] = \dfrac{4}{9}.$

Hence, $\dfrac{12}{27}$ in simplest form is $\dfrac{4}{9}$.

(ii) $\dfrac{150}{350}$

By prime factorisation,

$\Rightarrow \dfrac{150}{350} = \dfrac{2 \times 3 \times 5 \times 5}{2 \times 5 \times 5 \times 7} \\[1em] = \dfrac{3}{7}.$

Hence, $\dfrac{150}{350}$ in simplest form is $\dfrac{3}{7}$.

(iii) $\dfrac{18}{81}$

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 simplest form is $\dfrac{2}{9}$.

(iv) $\dfrac{276}{115}$

By prime factorisation,

$\Rightarrow \dfrac{276}{115} = \dfrac{2 \times 2 \times 3 \times 23}{5 \times 23} \\[1em] = \dfrac{2 \times 2 \times 3}{5} \\[1em] = \dfrac{12}{5} \\[1em] = 2\dfrac{2}{5}.$

Hence, $\dfrac{276}{115}$ in simplest form is $\dfrac{12}{5}$ or $2\dfrac{2}{5}$.

Convert the following fractions into equivalent like fractions:

(i) $\dfrac{7}{8}, \dfrac{5}{14}$

(ii) $\dfrac{5}{6}, \dfrac{7}{16}$

(iii) $\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}$

Answer

(i) $\dfrac{7}{8}, \dfrac{5}{14}$

L.C.M. of 8 and 14 is :

LCM of 8 and 14 = 2 x 2 x 2 x 7 = 56

$\Rightarrow \dfrac{7}{8} = \dfrac{7 \times 7}{8 \times 7} = \dfrac{49}{56} \\[1em] \Rightarrow \dfrac{5}{14} = \dfrac{5 \times 4}{14 \times 4} = \dfrac{20}{56}$

Hence, the equivalent like fractions are $\dfrac{49}{56}$ and $\dfrac{20}{56}$.

(ii) $\dfrac{5}{6}, \dfrac{7}{16}$

L.C.M. of 6 and 16 is :

LCM of 6 and 16 = 2 x 2 x 2 x 2 x 3 = 48.

$\Rightarrow \dfrac{5}{6} = \dfrac{5 \times 8}{6 \times 8} = \dfrac{40}{48} \\[1em] \Rightarrow \dfrac{7}{16} = \dfrac{7 \times 3}{16 \times 3} = \dfrac{21}{48}$

Hence, the equivalent like fractions are $\dfrac{40}{48}$ and $\dfrac{21}{48}$.

(iii) $\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}$

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

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

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

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

Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the black bold lines in the respective boxes or in your notebook.

Answer

Each unit is divided into 5 equal parts, so each part represents $\dfrac{1}{5}$.

The bold line shown in the figure starts from 0 and goes up to a length of 2 parts, so it represents $\dfrac{2}{5}$.

The remaining boxes are filled as follows (counting parts from 0 to the end of bold line):

The fraction in the first box (after $\dfrac{1}{5}$) = $\dfrac{2}{5}$.

The fraction in the second box (after $\dfrac{3}{5}$) = $\dfrac{4}{5}$.

The fraction in the third box (after 1) = $\dfrac{6}{5}$ or $1\dfrac{1}{5}$.

Hence, the missing fractions are $\dfrac{2}{5}, \dfrac{4}{5}$ and $\dfrac{6}{5}$ (i.e. $1\dfrac{1}{5}$).

Find the missing numbers:

(i) 3 glasses of juice shared equally among 4 friends is the same as _____ glasses of juice shared equally among 8 friends.

So, $\dfrac{3}{4} = \dfrac{\boxed{\phantom{653}}}{8}$

(ii) 4 kg of sugar divided equally in 3 bags is the same as 12 kg of sugar divided equally in _____ bags.

So, $\dfrac{4}{3} = \dfrac{12}{\boxed{\phantom{653}}}$

(iii) 7 gulabjamuns divided among 5 children is the same as _____ gulabjamuns divided among _____ children.

So, $\dfrac{7}{5} = \dfrac{\boxed{\phantom{653}}}{\boxed{\phantom{653}}}$

Answer

(i) $\dfrac{3}{4} = \dfrac{\boxed{\phantom{653}}}{8}$

To get 8 from 4, we have to multiply 4 by 2.

So, multiply numerator and denominator by 2.

$\Rightarrow \dfrac{3}{4} = \dfrac{3 \times 2}{4 \times 2} = \dfrac{6}{8}.$

Hence, the missing number is 6 i.e. 6 glasses of juice shared equally among 8 friends.

(ii) $\dfrac{4}{3} = \dfrac{12}{\boxed{\phantom{653}}}$

To get 12 from 4, we have to multiply 4 by 3.

So, multiply numerator and denominator by 3.

$\Rightarrow \dfrac{4}{3} = \dfrac{4 \times 3}{3 \times 3} = \dfrac{12}{9}.$

Hence, the missing number is 9 i.e. 12 kg of sugar divided equally in 9 bags.

(iii) $\dfrac{7}{5} = \dfrac{\boxed{\phantom{653}}}{\boxed{\phantom{653}}}$

We can multiply both numerator and denominator by 2.

$\Rightarrow \dfrac{7}{5} = \dfrac{7 \times 2}{5 \times 2} = \dfrac{14}{10}.$

Hence, the missing numbers are 14 and 10 i.e. 14 gulabjamuns divided among 10 children.

Raghu got a job at the age of 24 years and got retired from the job at the age of 60 years. What fraction of his age till retirement was he in job?

Answer

Age at which Raghu got the job = 24 years.

Age of retirement = 60 years.

Number of years he was in job = 60 - 24 = 36 years.

Total age till retirement = 60 years.

Fraction of his age in job = $\dfrac{\text{Years in job}}{\text{Total age}}$

$= \dfrac{36}{60} \\[1em] = \dfrac{3}{5}.$

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

Compare the given fractions and replace '$\boxed{\phantom{653}}$' by an appropriate sign '< or >':

(i) $\dfrac{3}{6} \boxed{\phantom{653}} \dfrac{5}{6}$

(ii) $\dfrac{2}{7} \boxed{\phantom{653}} \dfrac{2}{5}$

(iii) $\dfrac{3}{5} \boxed{\phantom{653}} \dfrac{4}{5}$

(iv) $\dfrac{4}{7} \boxed{\phantom{653}} \dfrac{4}{9}$

Answer

(i) $\dfrac{3}{6}$ and $\dfrac{5}{6}$

These are like fractions (same denominator 6).

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

Since 3 < 5, therefore, $\dfrac{3}{6} \lt \dfrac{5}{6}$.

Hence, $\dfrac{3}{6} \lt \dfrac{5}{6}$.

(ii) $\dfrac{2}{7}$ and $\dfrac{2}{5}$

These are unlike fractions with same numerator 2.

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

Since 7 > 5, therefore, $\dfrac{2}{7} \lt \dfrac{2}{5}$.

Hence, $\dfrac{2}{7} \lt \dfrac{2}{5}$.

(iii) $\dfrac{3}{5}$ and $\dfrac{4}{5}$

These are like fractions (same denominator 5).

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

Since 3 < 4, therefore, $\dfrac{3}{5} \lt \dfrac{4}{5}$.

Hence, $\dfrac{3}{5} \lt \dfrac{4}{5}$.

(iv) $\dfrac{4}{7}$ and $\dfrac{4}{9}$

These are unlike fractions with same numerator 4.

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

Since 7 < 9, therefore, $\dfrac{4}{7} \gt \dfrac{4}{9}$.

Hence, $\dfrac{4}{7} \gt \dfrac{4}{9}$.

Replace '$\boxed{\phantom{653}}$' by an appropriate sign '< , = or >' between the given fractions:

(i) $\dfrac{1}{2} \boxed{\phantom{653}} \dfrac{1}{5}$

(ii) $\dfrac{2}{4} \boxed{\phantom{653}} \dfrac{3}{6}$

(iii) $\dfrac{7}{9} \boxed{\phantom{653}} \dfrac{3}{9}$

(iv) $\dfrac{3}{4} \boxed{\phantom{653}} \dfrac{2}{8}$

Answer

(i) $\dfrac{1}{2}$ and $\dfrac{1}{5}$

These are unlike fractions with same numerator 1.

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

Since 2 < 5, therefore, $\dfrac{1}{2} \gt \dfrac{1}{5}$.

Hence, $\dfrac{1}{2} \gt \dfrac{1}{5}$.

(ii) $\dfrac{2}{4}$ and $\dfrac{3}{6}$

By cross multiplication,

2 × 6 = 12 and 4 × 3 = 12

The cross products are equal.

Hence, $\dfrac{2}{4} = \dfrac{3}{6}$.

(iii) $\dfrac{7}{9}$ and $\dfrac{3}{9}$

These are like fractions (same denominator 9).

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

Since 7 > 3, therefore, $\dfrac{7}{9} \gt \dfrac{3}{9}$.

Hence, $\dfrac{7}{9} \gt \dfrac{3}{9}$.

(iv) $\dfrac{3}{4}$ and $\dfrac{2}{8}$

By cross multiplication,

3 × 8 = 24 and 4 × 2 = 8

Since 24 > 8, therefore, $\dfrac{3}{4} \gt \dfrac{2}{8}$.

Hence, $\dfrac{3}{4} \gt \dfrac{2}{8}$.

Compare the following pairs of fractions:

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

(ii) $\dfrac{9}{16}$ and $\dfrac{5}{9}$

Answer

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

LCM of 9 and 5 = 45.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{5}{9} = \dfrac{5 \times 5}{9 \times 5} = \dfrac{25}{45} \\[1em] \Rightarrow \dfrac{4}{5} = \dfrac{4 \times 9}{5 \times 9} = \dfrac{36}{45}$

As 25 < 36, $\dfrac{25}{45} \lt \dfrac{36}{45} \Rightarrow \dfrac{5}{9} \lt \dfrac{4}{5}$.

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

(ii) $\dfrac{9}{16}$ and $\dfrac{5}{9}$

LCM of 16 and 9 = 144.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{9}{16} = \dfrac{9 \times 9}{16 \times 9} = \dfrac{81}{144} \\[1em] \Rightarrow \dfrac{5}{9} = \dfrac{5 \times 16}{9 \times 16} = \dfrac{80}{144}$

As 81 > 80, $\dfrac{81}{144} \gt \dfrac{80}{144} \Rightarrow \dfrac{9}{16} \gt \dfrac{5}{9}$.

Hence, $\dfrac{9}{16} \gt \dfrac{5}{9}$.

Fill in the boxes by the symbol < or > to make the given statements true:

(i) $\dfrac{5}{11} \boxed{\phantom{653}} \dfrac{3}{7}$

(ii) $\dfrac{8}{15}$ $\boxed{\phantom{653}}$ $\dfrac{3}{5}$

(iii) $\dfrac{11}{14}$ $\boxed{\phantom{653}}$ $\dfrac{29}{35}$

(iv) $\dfrac{13}{27}$ $\boxed{\phantom{653}}$ $\dfrac{15}{48}$

Answer

(i) $\dfrac{5}{11}$ and $\dfrac{3}{7}$

By cross multiplication,

5 × 7 = 35 and 11 × 3 = 33

Since 35 > 33, therefore, $\dfrac{5}{11} \gt \dfrac{3}{7}$.

Hence, $\dfrac{5}{11} \gt \dfrac{3}{7}$.

(ii) $\dfrac{8}{15}$ and $\dfrac{3}{5}$

By cross multiplication,

8 × 5 = 40 and 15 × 3 = 45

Since 40 < 45, therefore, $\dfrac{8}{15} \lt \dfrac{3}{5}$.

Hence, $\dfrac{8}{15} \lt \dfrac{3}{5}$.

(iii) $\dfrac{11}{14}$ and $\dfrac{29}{35}$

LCM of 14 and 35 = 70.

$\Rightarrow \dfrac{11}{14} = \dfrac{11 \times 5}{14 \times 5} = \dfrac{55}{70} \\[1em] \Rightarrow \dfrac{29}{35} = \dfrac{29 \times 2}{35 \times 2} = \dfrac{58}{70}$

As 55 < 58,

$\dfrac{55}{70} \lt \dfrac{58}{70} \Rightarrow \dfrac{11}{14} \lt \dfrac{29}{35}$.

Hence, $\dfrac{11}{14} \lt \dfrac{29}{35}$.

(iv) $\dfrac{13}{27}$ and $\dfrac{15}{48}$

By cross multiplication,

13 × 48 = 624 and 27 × 15 = 405

Since 624 > 405, therefore, $\dfrac{13}{27} \gt \dfrac{15}{48}$.

Hence, $\dfrac{13}{27} \gt \dfrac{15}{48}$.

Arrange the given fractions in descending order:

(i) $\dfrac{5}{17}, \dfrac{4}{9}, \dfrac{7}{12}$

(ii) $\dfrac{7}{12}, \dfrac{11}{36}, \dfrac{37}{72}$

Answer

(i) $\dfrac{5}{17}, \dfrac{4}{9}, \dfrac{7}{12}$

LCM of 17, 9 and 12 = 612.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{5}{17} = \dfrac{5 \times 36}{17 \times 36} = \dfrac{180}{612} \\[1em] \Rightarrow \dfrac{4}{9} = \dfrac{4 \times 68}{9 \times 68} = \dfrac{272}{612} \\[1em] \Rightarrow \dfrac{7}{12} = \dfrac{7 \times 51}{12 \times 51} = \dfrac{357}{612}$

As 357 > 272 > 180,

$\dfrac{357}{612} \gt \dfrac{272}{612} \gt \dfrac{180}{612} \Rightarrow \dfrac{7}{12} \gt \dfrac{4}{9} \gt \dfrac{5}{17}$.

Hence, the given fractions in descending order are $\dfrac{7}{12}, \dfrac{4}{9}, \dfrac{5}{17}$.

(ii) $\dfrac{7}{12}, \dfrac{11}{36}, \dfrac{37}{72}$

LCM of 12, 36 and 72 = 72.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{7}{12} = \dfrac{7 \times 6}{12 \times 6} = \dfrac{42}{72} \\[1em] \Rightarrow \dfrac{11}{36} = \dfrac{11 \times 2}{36 \times 2} = \dfrac{22}{72} \\[1em] \Rightarrow \dfrac{37}{72} = \dfrac{37}{72}$

As 42 > 37 > 22,

$\dfrac{42}{72} \gt \dfrac{37}{72} \gt \dfrac{22}{72} \Rightarrow \dfrac{7}{12} \gt \dfrac{37}{72} \gt \dfrac{11}{36}$.

Hence, the given fractions in descending order are $\dfrac{7}{12}, \dfrac{37}{72}, \dfrac{11}{36}$.

Arrange the given fractions in ascending order:

(i) $\dfrac{7}{8}, \dfrac{15}{16}, \dfrac{5}{6}$

(ii) $\dfrac{3}{4}, \dfrac{15}{22}, \dfrac{26}{33}$

(iii) $\dfrac{5}{12}, \dfrac{1}{4}, \dfrac{7}{8}, \dfrac{5}{6}$

Answer

(i) $\dfrac{7}{8}, \dfrac{15}{16}, \dfrac{5}{6}$

LCM of 8, 16 and 6 = 48.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{7}{8} = \dfrac{7 \times 6}{8 \times 6} = \dfrac{42}{48} \\[1em] \Rightarrow \dfrac{15}{16} = \dfrac{15 \times 3}{16 \times 3} = \dfrac{45}{48} \\[1em] \Rightarrow \dfrac{5}{6} = \dfrac{5 \times 8}{6 \times 8} = \dfrac{40}{48}$

As 40 < 42 < 45,

$\dfrac{40}{48} \lt \dfrac{42}{48} \lt \dfrac{45}{48} \Rightarrow \dfrac{5}{6} \lt \dfrac{7}{8} \lt \dfrac{15}{16}$.

Hence, the given fractions in ascending order are $\dfrac{5}{6}, \dfrac{7}{8}, \dfrac{15}{16}$.

(ii) $\dfrac{3}{4}, \dfrac{15}{22}, \dfrac{26}{33}$

LCM of 4, 22 and 33 = 132.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{3}{4} = \dfrac{3 \times 33}{4 \times 33} = \dfrac{99}{132} \\[1em] \Rightarrow \dfrac{15}{22} = \dfrac{15 \times 6}{22 \times 6} = \dfrac{90}{132} \\[1em] \Rightarrow \dfrac{26}{33} = \dfrac{26 \times 4}{33 \times 4} = \dfrac{104}{132}$

As 90 < 99 < 104,

$\dfrac{90}{132} \lt \dfrac{99}{132} \lt \dfrac{104}{132} \Rightarrow \dfrac{15}{22} \lt \dfrac{3}{4} \lt \dfrac{26}{33}$.

Hence, the given fractions in ascending order are $\dfrac{15}{22}, \dfrac{3}{4}, \dfrac{26}{33}$.

(iii) $\dfrac{5}{12}, \dfrac{1}{4}, \dfrac{7}{8}, \dfrac{5}{6}$

LCM of 12, 4, 8 and 6 = 24.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{5}{12} = \dfrac{5 \times 2}{12 \times 2} = \dfrac{10}{24} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{1 \times 6}{4 \times 6} = \dfrac{6}{24} \\[1em] \Rightarrow \dfrac{7}{8} = \dfrac{7 \times 3}{8 \times 3} = \dfrac{21}{24} \\[1em] \Rightarrow \dfrac{5}{6} = \dfrac{5 \times 4}{6 \times 4} = \dfrac{20}{24}$

As 6 < 10 < 20 < 21,

$\dfrac{6}{24} \lt \dfrac{10}{24} \lt \dfrac{20}{24} \lt \dfrac{21}{24} \Rightarrow \dfrac{1}{4} \lt \dfrac{5}{12} \lt \dfrac{5}{6} \lt \dfrac{7}{8}$.

Hence, the given fractions in ascending order are $\dfrac{1}{4}, \dfrac{5}{12}, \dfrac{5}{6}, \dfrac{7}{8}$.

Calculate the following:

(i) $\dfrac{8}{15} + \dfrac{3}{15}$

(ii) $\dfrac{12}{15} - \dfrac{7}{15}$

(iii) $1 - \dfrac{2}{3}$

(iv) $\dfrac{7}{13} + \dfrac{2}{13} - \dfrac{5}{13}$

(v) $2\dfrac{4}{5} + 3\dfrac{3}{5}$

(vi) $3\dfrac{2}{7} - 1\dfrac{4}{7}$

Answer

(i) $\dfrac{8}{15} + \dfrac{3}{15}$

$\Rightarrow \dfrac{8 + 3}{15}\\[1em] \Rightarrow \dfrac{11}{15}$

Hence, $\dfrac{8}{15} + \dfrac{3}{15} = \dfrac{11}{15}$.

(ii) $\dfrac{12}{15} - \dfrac{7}{15}$

$\Rightarrow \dfrac{12 - 7}{15}\\[1em] \Rightarrow \dfrac{5}{15}\\[1em] \Rightarrow \dfrac{1}{3}$

Hence, $\dfrac{12}{15} - \dfrac{7}{15} = \dfrac{1}{3}$.

(iii) $1 - \dfrac{2}{3}$

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

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

(iv) $\dfrac{7}{13} + \dfrac{2}{13} - \dfrac{5}{13}$

$\Rightarrow \dfrac{7 + 2 - 5}{13}\\[1em] \Rightarrow \dfrac{4}{13}$

Hence, $\dfrac{7}{13} + \dfrac{2}{13} - \dfrac{5}{13} = \dfrac{4}{13}$.

(v) $2\dfrac{4}{5} + 3\dfrac{3}{5}$

$\Rightarrow \dfrac{14}{5} + \dfrac{18}{5}\\[1em] \Rightarrow \dfrac{14 + 18}{5}\\[1em] \Rightarrow \dfrac{32}{5}\\[1em] \Rightarrow 6\dfrac{2}{5}$

Hence, $2\dfrac{4}{5} + 3\dfrac{3}{5} = 6\dfrac{2}{5}$.

(vi) $3\dfrac{2}{7} - 1\dfrac{4}{7}$

$\Rightarrow \dfrac{23}{7} - \dfrac{11}{7}\\[1em] \Rightarrow \dfrac{23 - 11}{7}\\[1em] \Rightarrow \dfrac{12}{7}\\[1em] \Rightarrow 1\dfrac{5}{7}$

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

Calculate the following:

(i) $\dfrac{2}{3} + \dfrac{3}{4}$

(ii) $\dfrac{5}{7} - \dfrac{4}{9}$

(iii) $\dfrac{1}{2} + \dfrac{3}{5}$

(iv) $1\dfrac{4}{9} + 3\dfrac{5}{12}$

(v) $2\dfrac{1}{4} - 1\dfrac{7}{10}$

(vi) $3\dfrac{5}{6} - 2\dfrac{7}{15}$

Answer

(i) $\dfrac{2}{3} + \dfrac{3}{4}$

LCM of 3 and 4 = 12.

$\Rightarrow \dfrac{2 \times 4}{3 \times 4} + \dfrac{3 \times 3}{4 \times 3}\\[1em] \Rightarrow \dfrac{8}{12} + \dfrac{9}{12}\\[1em] \Rightarrow \dfrac{8 + 9}{12}\\[1em] \Rightarrow \dfrac{17}{12}\\[1em] \Rightarrow 1\dfrac{5}{12}$

Hence, $\dfrac{2}{3} + \dfrac{3}{4} = 1\dfrac{5}{12}$.

(ii) $\dfrac{5}{7} - \dfrac{4}{9}$

LCM of 7 and 9 = 63.

$\Rightarrow \dfrac{5 \times 9}{7 \times 9} - \dfrac{4 \times 7}{9 \times 7}\\[1em] \Rightarrow \dfrac{45}{63} - \dfrac{28}{63}\\[1em] \Rightarrow \dfrac{45 - 28}{63}\\[1em] \Rightarrow \dfrac{17}{63}$

Hence, $\dfrac{5}{7} - \dfrac{4}{9} = \dfrac{17}{63}$.

(iii) $\dfrac{1}{2} + \dfrac{3}{5}$

LCM of 2 and 5 = 10.

$\Rightarrow \dfrac{1 \times 5}{2 \times 5} + \dfrac{3 \times 2}{5 \times 2}\\[1em] \Rightarrow \dfrac{5}{10} + \dfrac{6}{10}\\[1em] \Rightarrow \dfrac{5 + 6}{10}\\[1em] \Rightarrow \dfrac{11}{10}\\[1em] \Rightarrow 1\dfrac{1}{10}$

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