Fractions

What fractions do the shaded parts in each of the following figures represent?

Answer

(i) Total parts = 8

Shaded parts = 5

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}}$

= $\dfrac{5}{8}$.

Hence, fraction = $\dfrac{5}{8}$.

(ii) Total parts = 12

Shaded parts = 7

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}}$

= $\dfrac{7}{12}$.

Hence, fraction = $\dfrac{7}{12}$.

(iii) Total parts = 7

Shaded parts = 2

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}}$

= $\dfrac{2}{7}$

Hence, fraction = $\dfrac{2}{7}$.

Write a fraction for each of the following :

(i) 4 parts out of 9 equal parts

(ii) 5 parts out of 11 equal parts

(iii) two-fifths

(iv) four-sevenths

(v) seven-tenths

(vi) six-eighths

Answer

(i) Total parts = 9

Given parts = 4

Fraction = $\dfrac{\text{Given parts}}{\text{Total parts}}$

= $\dfrac{4}{9}$

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

(ii) Total parts = 11

Given parts = 5

Fraction = $\dfrac{\text{Given parts}}{\text{Total parts}}$

= $\dfrac{5}{11}$

Hence, fraction = $\dfrac{5}{11}$.

(iii) Total parts = 5

Given parts = 2

Fraction = $\dfrac{\text{Given parts}}{\text{Total parts}}$

= $\dfrac{2}{5}$

Hence, fraction = $\dfrac{2}{5}$.

(iv) Total parts = 7

Given parts = 4

Fraction = $\dfrac{\text{Given parts}}{\text{Total parts}}$

= $\dfrac{4}{7}$

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

(v) Total parts = 10

Given parts = 7

Fraction = $\dfrac{\text{Given parts}}{\text{Total parts}}$

= $\dfrac{7}{10}$

Hence, fraction = $\dfrac{7}{10}$.

(vi) Total parts = 8

Given parts = 6

Fraction = $\dfrac{\text{Given parts}}{\text{Total parts}}$

= $\dfrac{6}{8}$

Hence, fraction = $\dfrac{6}{8}$.

Point out the numerator and denominator in each of the following fractions :

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

(ii) $\dfrac{3}{7}$

(iii) $\dfrac{9}{14}$

(iv) $\dfrac{1}{20}$

(v) $\dfrac{12}{25}$

Answer

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

Hence, numerator = 5 and denominator = 6.

(ii) $\dfrac{3}{7}$

Hence, numerator = 3 and denominator = 7.

(iii) $\dfrac{9}{14}$

Hence, numerator = 9 and denominator = 14.

(iv) $\dfrac{1}{20}$

Hence, numerator = 1 and denominator = 20.

(v) $\dfrac{12}{25}$

Hence, numerator = 12 and denominator = 25.

Write five fractions equivalent to each of the following :

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

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

(iii) $\dfrac{5}{6}$

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

(v) $\dfrac{3}{14}$

Answer

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\dfrac{2 \times 2}{3 \times 2} = \dfrac{4}{6}$

Multiply by 3: $\dfrac{2 \times 3}{3 \times 3} = \dfrac{6}{9}$

Multiply by 4: $\dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12}$

Multiply by 5: $\dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

Multiply by 6: $\dfrac{2 \times 6}{3 \times 6} = \dfrac{12}{18}$

Hence, the equivalent fractions are : $\dfrac{4}{6}, \dfrac{6}{9}, \dfrac{8}{12}, \dfrac{10}{15}, \dfrac{12}{18}$.

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\dfrac{3 \times 2}{4 \times 2} = \dfrac{6}{8}$

Multiply by 3: $\dfrac{3 \times 3}{4 \times 3} = \dfrac{9}{12}$

Multiply by 4: $\dfrac{3 \times 4}{4 \times 4} = \dfrac{12}{16}$

Multiply by 5: $\dfrac{3 \times 5}{4 \times 5} = \dfrac{15}{20}$

Multiply by 6: $\dfrac{3 \times 6}{4 \times 6} = \dfrac{18}{24}$

Hence, the equivalent fractions are: $\dfrac{6}{8}, \dfrac{9}{12}, \dfrac{12}{16}, \dfrac{15}{20}, \dfrac{18}{24}$.

(iii) $\dfrac{5}{6}$

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\dfrac{5 \times 2}{6 \times 2} = \dfrac{10}{12}$

Multiply by 3: $\dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18}$

Multiply by 4: $\dfrac{5 \times 4}{6 \times 4} = \dfrac{20}{24}$

Multiply by 5: $\dfrac{5 \times 5}{6 \times 5} = \dfrac{25}{30}$

Multiply by 6: $\dfrac{5 \times 6}{6 \times 6} = \dfrac{30}{36}$

Hence, the equivalent fractions are: $\dfrac{10}{12}, \dfrac{15}{18}, \dfrac{20}{24}, \dfrac{25}{30}, \dfrac{30}{36}$.

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\dfrac{7 \times 2}{9 \times 2} = \dfrac{14}{18}$

Multiply by 3: $\dfrac{7 \times 3}{9 \times 3} = \dfrac{21}{27}$

Multiply by 4: $\dfrac{7 \times 4}{9 \times 4} = \dfrac{28}{36}$

Multiply by 5: $\dfrac{7 \times 5}{9 \times 5} = \dfrac{35}{45}$

Multiply by 6: $\dfrac{7 \times 6}{9 \times 6} = \dfrac{42}{54}$

Hence, the equivalent fractions are: $\dfrac{14}{18}, \dfrac{21}{27}, \dfrac{28}{36}, \dfrac{35}{45}, \dfrac{42}{54}$.

(v) $\dfrac{3}{14}$

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\dfrac{3 \times 2}{14 \times 2} = \dfrac{6}{28}$

Multiply by 3: $\dfrac{3 \times 3}{14 \times 3} = \dfrac{9}{42}$

Multiply by 4: $\dfrac{3 \times 4}{14 \times 4} = \dfrac{12}{56}$

Multiply by 5: $\dfrac{3 \times 5}{14 \times 5} = \dfrac{15}{70}$

Multiply by 6: $\dfrac{3 \times 6}{14 \times 6} = \dfrac{18}{84}$

Hence, the equivalent fractions are: $\dfrac{6}{28}, \dfrac{9}{42}, \dfrac{12}{56}, \dfrac{15}{70}, \dfrac{18}{84}$.

Which of the following are the pairs of equivalent fractions?

(i) $\dfrac{4}{5}$ and $\dfrac{24}{30}$

(ii) $\dfrac{6}{8}$ and $\dfrac{18}{24}$

(iii) $\dfrac{9}{16}$ and $\dfrac{3}{4}$

(iv) $\dfrac{8}{12}$ and $\dfrac{14}{21}$

(v) $\dfrac{7}{11}$ and $\dfrac{21}{32}$

(vi) $\dfrac{5}{12}$ and $\dfrac{35}{84}$

Answer

(i) $\dfrac{4}{5}$ and $\dfrac{24}{30}$

By cross multiplication,

So, 4 x 30 = 120 and 5 x 24 = 120

Here, the cross products are equal.

Hence, $\dfrac{4}{5} \text{ and } \dfrac{24}{30}$ are equivalent fractions.

(ii) $\dfrac{6}{8}$ and $\dfrac{18}{24}$

By cross multiplication,

So, 6 x 24 = 144 and 8 x 18 = 144

Here, the cross products are equal.

Hence, $\dfrac{6}{8}\text{ and } \dfrac{18}{24}$ are equivalent fractions.

(iii) $\dfrac{9}{16}$ and $\dfrac{3}{4}$

By cross multiplication,

So, 9 x 4 = 36 and 16 x 3 = 48

Here, the cross products are not equal.

Hence, $\dfrac{9}{16}\text{ and } \dfrac{3}{4}$ are not equivalent fractions.

(iv) $\dfrac{8}{12}$ and $\dfrac{14}{21}$

By cross multiplication,

So, 8 x 21 = 168 and 12 x 14 = 168

Here, the cross products are equal.

Hence, $\dfrac{8}{12}\text{ and } \dfrac{14}{21}$ are equivalent fractions.

(v) $\dfrac{7}{11}$ and $\dfrac{21}{32}$

By cross multiplication,

So, 7 x 32 = 224 and 11 x 21 = 231

Here, the cross products are not equal.

Hence, $\dfrac{7}{11}\text{ and } \dfrac{21}{32}$ are not equivalent fractions.

(vi) $\dfrac{5}{12}$ and $\dfrac{35}{84}$

By cross multiplication,

So, 5 x 84 = 420 and 12 x 35 = 420

Here, the cross products are equal.

Hence, $\dfrac{5}{12}\text{ and } \dfrac{35}{84}$ are equivalent fractions.

Write an equivalent fraction of :

(i) $\dfrac{4}{5}$ with numerator 32

(ii) $\dfrac{7}{9}$ with numerator 42

(iii) $\dfrac{3}{7}$ with denominator 63

(iv) $\dfrac{10}{13}$ with denominator 78

Answer

(i) $\dfrac{4}{5}$ with numerator 32

To find an equivalent fraction whose numerator is 32, we multiply the fraction $\dfrac{4}{5}$ with 8,

$\dfrac{4}{5} = \dfrac{4 \times 8}{5 \times 8} = \dfrac{32}{40}$.

Hence, an equivalent fraction of $\dfrac{4}{5} = \dfrac{32}{40}$.

(ii) $\dfrac{7}{9}$ with numerator 42

To find an equivalent fraction whose numerator is 42, we multiply the fraction $\dfrac{7}{9}$ with 6,

$\dfrac{7}{9} = \dfrac{7 \times 6}{9 \times 6} = \dfrac{42}{54}$.

Hence, an equivalent fraction of $\dfrac{7}{9} = \dfrac{42}{54}$.

(iii) $\dfrac{3}{7}$ with denominator 63

To find an equivalent fraction whose denominator is 63, we multiply the fraction $\dfrac{3}{7}$ with 9,

$\dfrac{3}{7} = \dfrac{3 \times 9}{7 \times 9} = \dfrac{27}{63}$.

Hence, an equivalent fraction of $\dfrac{3}{7} = \dfrac{27}{63}$.

(iv) $\dfrac{10}{13}$ with denominator 78

To find an equivalent fraction whose denominator is 78, we multiply the fraction $\dfrac{10}{13}$ with 6,

$\dfrac{10}{13} = \dfrac{10 \times 6}{13 \times 6} = \dfrac{60}{78}$.

Hence, an equivalent fraction of $\dfrac{10}{13} = \dfrac{60}{78}$.

Write an equivalent fraction of :

(i) $\dfrac{32}{48}$ with numerator 2

(ii) $\dfrac{21}{28}$ with numerator 3

(iii) $\dfrac{24}{56}$ with denominator 7

(iv) $\dfrac{121}{132}$ with denominator 12

Answer

(i) $\dfrac{32}{48}$ with numerator 2

The H.C.F. of 32 and 48 is 16. Thus,

To find an equivalent fraction whose numerator is 2, we divide the numerator and denominator of the given fraction by 16,

$\dfrac{32}{48} = \dfrac{32 ÷ 16}{48 ÷ 16} = \dfrac{2}{3}$.

Hence, an equivalent fraction of $\dfrac{32}{48} = \dfrac{2}{3}$.

(ii) $\dfrac{21}{28}$ with numerator 3

The H.C.F. of 21 and 28 is 7. Thus,

To find an equivalent fraction whose numerator is 3, we divide the numerator and denominator of the given fraction by 7,

$\dfrac{21}{28} = \dfrac{21 ÷ 7}{28 ÷ 7} = \dfrac{3}{4}$.

Hence, an equivalent fraction of $\dfrac{21}{28} = \dfrac{3}{4}$.

(iii) $\dfrac{24}{56}$ with denominator 7

The H.C.F. of 24 and 56 is 8. Thus,

To find an equivalent fraction whose denominator is 7, we divide the numerator and denominator of the given fraction by 8,

$\dfrac{24}{56} = \dfrac{24 ÷ 8}{56 ÷ 8} = \dfrac{3}{7}$.

Hence, an equivalent fraction of $\dfrac{24}{56} = \dfrac{3}{7}$.

(iv) $\dfrac{121}{132}$ with denominator 12

The H.C.F. of 121 and 132 is 11. Thus,

To find an equivalent fraction whose denominator is 12, we divide the numerator and denominator of the given fraction by 11,

$\dfrac{121}{132} = \dfrac{121 ÷ 11}{132 ÷ 11} = \dfrac{11}{12}$.

Hence, an equivalent fraction of $\dfrac{121}{132} = \dfrac{11}{12}$.

Write an equivalent fraction of :

(i) $\dfrac{45}{54}$ with numerator 20.

(ii) $\dfrac{75}{90}$ with denominator 42.

Answer

(i) $\dfrac{45}{54}$ with numerator 20.

The H.C.F. of 45 and 54 is 9. Thus,

To find an equivalent fraction whose numerator is 20,

We divide the numerator and denominator of the given fraction by 9,

$\dfrac{45}{54} = \dfrac{45 ÷ 9}{54 ÷ 9} = \dfrac{5}{6}$.

Now, to get a fraction with numerator 20, we multiply both the numerator and denominator by 4,

$\dfrac{5}{6} = \dfrac{5 \times 4}{6 \times 4} = \dfrac{20}{24}$.

Hence, the equivalent fraction = $\dfrac{20}{24}$.

(ii) $\dfrac{75}{90}$ with denominator 42.

The H.C.F. of 75 and 90 is 15. Thus,

To find an equivalent fraction whose denominator is 42,

We divide the numerator and denominator of the given fraction by 15,

$\dfrac{75}{90} = \dfrac{75 ÷ 15}{90 ÷ 15} = \dfrac{5}{6}$.

Now, to get a fraction with denominator 42, we multiply both the numerator and denominator by 7,

$\dfrac{5}{6} = \dfrac{5 \times 7}{6 \times 7} = \dfrac{35}{42}$.

Hence, the equivalent fraction = $\dfrac{35}{42}$.

Write the missing numerals in the place-holders :

(i) $\dfrac{4}{9}$ = $\dfrac{\boxed{\phantom{63}}}{63}$

(ii) $\dfrac{3}{4}$ = $\dfrac{18}{\boxed{\phantom{63}}}$

(iii) $\dfrac{42}{70}$ = $\dfrac{3}{\boxed{\phantom{63}}}$

(iv) $\dfrac{52}{78}$ = $\dfrac{\boxed{\phantom{63}}}{6}$

Answer

(i) $\dfrac{4}{9}$ = $\dfrac{\boxed{\phantom{63}}}{63}$

$\dfrac{4}{9} = \dfrac{4 \times 7}{9 \times 7} = \dfrac{28}{63}$

Hence, $\dfrac{4}{9}$ = $\dfrac{\boxed{28}}{63}$.

(ii) $\dfrac{3}{4}$ = $\dfrac{18}{\boxed{\phantom{63}}}$

$\dfrac{3}{4} = \dfrac{3 \times 6}{4 \times 6} = \dfrac{18}{24}$

Hence, $\dfrac{3}{4}$ = $\dfrac{18}{\boxed{24}}$.

(iii) $\dfrac{42}{70}$ = $\dfrac{3}{\boxed{\phantom{63}}}$

Simplifying the fraction,

$\dfrac{42}{70} = \dfrac{42 ÷ 14}{70 ÷ 14} = \dfrac{3}{5}$

Hence, $\dfrac{42}{70}$ = $\dfrac{3}{\boxed{5}}$.

(iv) $\dfrac{52}{78}$ = $\dfrac{\boxed{\phantom{63}}}{6}$

Simplifying the fraction,

$\dfrac{52}{78} = \dfrac{52 ÷ 13}{78 ÷ 13} = \dfrac{4}{6}$

Hence, $\dfrac{52}{78}$ = $\dfrac{\boxed{4}}{6}$.

What fraction of an hour is 25 minutes?

Answer

Given minutes = 25 minutes

Total minutes = 60 minutes

Fraction = $\dfrac{\text{Given minutes}}{\text{Total minutes}} = \dfrac{25}{60}$.

Hence, required fraction = $\dfrac{25}{60}$.

What fraction of a day is 9 hours?

Answer

Given hours = 9 hours

Total hours = 24 hours

Fraction = $\dfrac{\text{Given hours}}{\text{Total hours}} = \dfrac{9}{24}$.

Hence, required fraction = $\dfrac{9}{24}$.

Which of the following fractions are in simplest form?

(i) $\dfrac{21}{40}$

(ii) $\dfrac{35}{49}$

(iii) $\dfrac{42}{54}$

(iv) $\dfrac{64}{81}$

(v) $\dfrac{56}{65}$

(vi) $\dfrac{23}{92}$

(vii) $\dfrac{102}{119}$

(viii) $\dfrac{91}{114}$

Answer

A fraction is in its simplest form when the highest common factor (HCF) of its numerator and denominator is 1.

(i) $\dfrac{21}{40}$

First, we find the H.C.F. of 21 and 40.

$\begin{array}{l} 21\overline{\smash{\big)}\quad 40\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-21} \\ \phantom{{x^2 =)}} 19 \overline{\smash{\big)}\quad 21\smash{\big(}} 1 \\ \phantom{ac = sc)}\phantom{)}\underline{-19} \\ \phantom{{x^2 )} + 2x =}2\overline{\smash{\big)}\quad 19 \smash{\big(}} 9 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-18} \\ \phantom{{x^2 )} + 2x = + dca}1\overline{\smash{\big)}\quad 2 \smash{\big(}} 2 \\ \phantom{ac = sc + sa + sa)}\phantom{)}\underline{-2} \\ \phantom{{x^2 + 3x - 54)} + 2x)}\times \\ \end{array}$

So, HCF of 21 and 40 = 1.

Hence, $\dfrac{21}{40}$ is in simplest form.

(ii) $\dfrac{35}{49}$

$\begin{array}{l} 35\overline{\smash{\big)}\quad 49\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-35} \\ \phantom{{x^2 =)}} 14 \overline{\smash{\big)}\quad 35\smash{\big(}} 2 \\ \phantom{ac = sc)}\phantom{)}\underline{-28} \\ \phantom{{x^2 )} + 2x =}7\overline{\smash{\big)}\quad 14 \smash{\big(}} 2 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-14} \\ \phantom{{x^2 + 3x - 54)} + }\times \\ \end{array}$

So, HCF of 35 and 49 = 7.

Hence, $\dfrac{35}{49}$ is not in simplest form.

(iii) $\dfrac{42}{54}$

$\begin{array}{l} 42\overline{\smash{\big)}\quad 54\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-42} \\ \phantom{{x^2 =)}} 12 \overline{\smash{\big)}\quad 42\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-36} \\ \phantom{{x^2 )} + 2x =}6\overline{\smash{\big)}\quad 12 \smash{\big(}} 2 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-12} \\ \phantom{{x^2 + 3x - 54)} + }\times \\ \end{array}$

So, HCF of 42 and 54 = 6.

Hence, $\dfrac{42}{54}$ is not in simplest form.

(iv) $\dfrac{64}{81}$

$\begin{array}{l} 64\overline{\smash{\big)}\quad 81\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-64} \\ \phantom{{x^2 =)}} 17 \overline{\smash{\big)}\quad 64\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-51} \\ \phantom{{x^2 )} + 2x =}13\overline{\smash{\big)}\quad 17 \smash{\big(}} 1 \\ \phantom{ac = sc + sa)))}\phantom{)}\underline{-13} \\ \phantom{{x^2 )} + 2x = acdvu}4\overline{\smash{\big)}\quad 13 \smash{\big(}} 3 \\ \phantom{ac = sc + sa+ df)}\phantom{)}\underline{-12} \\ \phantom{{x^2 )} + 2x = acdv + df}1\overline{\smash{\big)}\quad 4 \smash{\big(}} 4 \\ \phantom{ac = sc + sa+ df + dc)}\phantom{)}\underline{-4} \\ \phantom{{x^2 + 3x - 54)} + advnscj }\times \\ \end{array}$

So, HCF of 64 and 81 = 1.

Hence, $\dfrac{64}{81}$ is in simplest form.

(v) $\dfrac{56}{65}$

$\begin{array}{l} 56\overline{\smash{\big)}\quad 65\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-56} \\ \phantom{{x^2 =)}} 9 \overline{\smash{\big)}\quad 56\smash{\big(}} 6 \\ \phantom{ac = sc)}\phantom{)}\underline{-54} \\ \phantom{{x^2 )} + 2x =}2\overline{\smash{\big)}\quad 9 \smash{\big(}} 4 \\ \phantom{ac = sc + sa))}\phantom{)}\underline{-8} \\ \phantom{{x^2 )} + 2x = acdv}1\overline{\smash{\big)}\quad 2 \smash{\big(}} 2 \\ \phantom{ac = sc + sa+ df)}\phantom{)}\underline{-2} \\ \phantom{{x^2 + 3x - 54)} + aax}\times \\ \end{array}$

So, HCF of 56 and 65 = 1.

Hence, $\dfrac{56}{65}$ is in simplest form.

(vi) $\dfrac{23}{92}$

$\begin{array}{l} 23\overline{\smash{\big)}\quad 92\smash{\big(}} 4 \\ \phantom{ac)}\phantom{)}\underline{-92} \\ \phantom{{x^2 + 3}}\times \\ \end{array}$

So, HCF of 23 and 92 = 23.

Hence, $\dfrac{23}{92}$ is not in simplest form.

(vii) $\dfrac{102}{119}$

$\begin{array}{l} 102\overline{\smash{\big)}\quad 119\smash{\big(}} 1 \\ \phantom{ac =)}\phantom{}\underline{-102} \\ \phantom{{x^2 = aa)}} 17 \overline{\smash{\big)}\quad 102\smash{\big(}} 6 \\ \phantom{ac = sc =s)}\phantom{)}\underline{-102} \\ \phantom{{x^2 + 3x - 54)} x}\times \\ \end{array}$

So, HCF of 102 and 119 = 17.

Hence, $\dfrac{102}{119}$ is not in simplest form.

(viii) $\dfrac{91}{114}$

$\begin{array}{l} 91\overline{\smash{\big)}\quad 114\smash{\big(}} 1 \\ \phantom{ac +)}\phantom{)}\underline{-91} \\ \phantom{{x^2 =)))}} 23 \overline{\smash{\big)}\quad 91\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-69} \\ \phantom{{x^2 )} + 2x =}22\overline{\smash{\big)}\quad 23 \smash{\big(}} 1 \\ \phantom{ac = sc + sa =)}\phantom{)}\underline{-22} \\ \phantom{{x^2 )} + 2x = ac + ad}1\overline{\smash{\big)}\quad 22 \smash{\big(}} 22 \\ \phantom{ac = sc + sa+ df -)}\phantom{)}\underline{-22} \\ \phantom{{x^2 + 3x - 54)} + aax+ s} \times \\ \end{array}$

So, HCF of 91 and 114 = 1.

Hence, $\dfrac{91}{114}$ is in simplest form.

Reduce each of the following fraction to its lowest terms :

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

(ii) $\dfrac{45}{54}$

(iii) $\dfrac{38}{95}$

(iv) $\dfrac{58}{87}$

(v) $\dfrac{85}{153}$

(vi) $\dfrac{105}{168}$

(vii) $\dfrac{117}{143}$

(viii) $\dfrac{135}{150}$

Answer

(i) Prime factorizing,

$\dfrac{27}{36} = \dfrac{3 \times 3 \times 3}{2 \times 2 \times 3 \times 3} = \dfrac{3}{4}$.

Hence, $\dfrac{27}{36}$ in lowest terms is $\dfrac{3}{4}$.

(ii) Prime factorizing,

$\dfrac{45}{54} = \dfrac{3 \times 3 \times 5}{3 \times 3 \times 3 \times 2} = \dfrac{5}{6}$.

Hence, $\dfrac{45}{54}$ in lowest terms is $\dfrac{5}{6}$.

(iii) Prime factorizing,

$\dfrac{38}{95} = \dfrac{2 \times 19}{5 \times 19} = \dfrac{2}{5}$.

Hence, $\dfrac{38}{95}$ in lowest terms is $\dfrac{2}{5}$.

(iv) Prime factorizing,

$\dfrac{58}{87} = \dfrac{2 \times 29}{3 \times 29} = \dfrac{2}{3}$.

Hence, $\dfrac{58}{87}$ in lowest terms is $\dfrac{2}{3}$.

(v) Prime factorizing,

$\dfrac{85}{153} = \dfrac{5 \times 17}{3 \times 3 \times 17} = \dfrac{5}{9}$.

Hence, $\dfrac{85}{153}$ in lowest terms is $\dfrac{5}{9}$.

(vi) Prime factorizing,

$\dfrac{105}{168} = \dfrac{3 \times 5 \times 7}{2 \times 2 \times 2\times 3 \times 7} = \dfrac{5}{8}$.

Hence, $\dfrac{105}{168}$ in lowest terms is $\dfrac{5}{8}$.

(vii) Prime factorizing,

$\dfrac{117}{143} = \dfrac{3 \times 3 \times 13}{11 \times 13} = \dfrac{9}{11}$.

Hence, $\dfrac{117}{143}$ in lowest terms is $\dfrac{9}{11}$.

(viii) Prime factorizing,

$\dfrac{135}{150} = \dfrac{3 \times 3 \times 3 \times 5}{2 \times 3 \times 5 \times 5} = \dfrac{9}{10}$.

Hence, $\dfrac{135}{150}$ in lowest terms is $\dfrac{9}{10}$.

Point out the proper and improper fractions from the following :

(i) $\dfrac{7}{9}$

(ii) $\dfrac{16}{11}$

(iii) $\dfrac{18}{25}$

(iv) $\dfrac{10}{10}$

(v) $\dfrac{37}{23}$

(vi) $\dfrac{21}{8}$

(vii) $\dfrac{56}{57}$

(viii) $\dfrac{137}{105}$

(ix) $\dfrac{2}{1}$

(x) $\dfrac{100}{101}$

Answer

(i) $\dfrac{7}{9}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 7 < 9

Hence, $\dfrac{7}{9}$ is a proper fraction.

(ii) $\dfrac{16}{11}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 16 > 11

Hence, $\dfrac{16}{11}$ is an improper fraction.

(iii) $\dfrac{18}{25}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 18 < 25

Hence, $\dfrac{18}{25}$ is a proper fraction.

(iv) $\dfrac{10}{10}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 10 = 10

Hence, $\dfrac{10}{10}$ is an improper fraction.

(v) $\dfrac{37}{23}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 37 > 23

Hence, $\dfrac{37}{23}$ is an improper fraction.

(vi) $\dfrac{21}{8}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 21 > 8

Hence, $\dfrac{21}{8}$ is an improper fraction.

(vii) $\dfrac{56}{57}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 56 < 57

Hence, $\dfrac{56}{57}$ is a proper fraction.

(viii) $\dfrac{137}{105}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 137 > 105

Hence, $\dfrac{137}{105}$ is an improper fraction.

(ix) $\dfrac{2}{1}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 2 > 1

Hence, $\dfrac{2}{1}$ is an improper fraction.

(x) $\dfrac{100}{101}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 100 < 101

Hence, $\dfrac{100}{101}$ is a proper fraction.

Convert each of the following mixed fractions into an improper fraction :

(i) $8\dfrac{3}{4}$

(ii) $5\dfrac{11}{13}$

(iii) $10\dfrac{7}{9}$

(iv) $33\dfrac{1}{3}$

(v) $9\dfrac{7}{16}$

Answer

(i) $8\dfrac{3}{4}$

$8\dfrac{3}{4} = \dfrac{8 \times 4 + 3}{4} = \dfrac{35}{4}$

Hence, the improper fraction = $\dfrac{35}{4}$.

(ii) $5\dfrac{11}{13}$

$5\dfrac{11}{13} = \dfrac{5 \times 13 + 11}{13} = \dfrac{76}{13}$

Hence, the improper fraction = $\dfrac{76}{13}$.

(iii) $10\dfrac{7}{9}$

$10\dfrac{7}{9} = \dfrac{10 \times 9 + 7}{9} = \dfrac{97}{9}$

Hence, the improper fraction = $\dfrac{97}{9}$.

(iv) $33\dfrac{1}{3}$

$33\dfrac{1}{3} = \dfrac{33 \times 3 + 1}{3} = \dfrac{100}{3}$

Hence, the improper fraction = $\dfrac{100}{3}$.

(v) $9\dfrac{7}{16}$

$9\dfrac{7}{16} = \dfrac{9 \times 16 + 7}{16} = \dfrac{151}{16}$

Hence, the improper fraction = $\dfrac{151}{16}$.

Convert each of the following improper fractions into a mixed fraction :

(i) $\dfrac{31}{5}$

(ii) $\dfrac{80}{7}$

(iii) $\dfrac{107}{3}$

(iv) $\dfrac{115}{13}$

(v) $\dfrac{200}{9}$

Answer

(i) $\dfrac{31}{5}$

On dividing 31 by 5, we get quotient = 6 and remainder = 1.

$\therefore \dfrac{31}{5} = 6 + \dfrac{1}{5} = 6\dfrac{1}{5}$.

Hence, the mixed fraction is $6\dfrac{1}{5}$.

(ii) $\dfrac{80}{7}$

On dividing 80 by 7, we get quotient = 11 and remainder = 3.

$\therefore \dfrac{80}{7} = 11 + \dfrac{3}{7} = 11\dfrac{3}{7}$.

Hence, the mixed fraction is $11\dfrac{3}{7}$.

(iii) $\dfrac{107}{3}$

On dividing 107 by 3, we get quotient = 35 and remainder = 2.

$\therefore \dfrac{107}{3} = 35 + \dfrac{2}{3} = 35\dfrac{2}{3}$.

Hence, the mixed fraction is $35\dfrac{2}{3}$.

(iv) $\dfrac{115}{13}$

On dividing 115 by 13, we get quotient = 8 and remainder = 11.

$\therefore \dfrac{115}{13} = 8 + \dfrac{11}{13} = 8\dfrac{11}{13}$.

Hence, the mixed fraction is $8\dfrac{11}{13}$.

(v) $\dfrac{200}{9}$

On dividing 200 by 9, we get quotient = 22 and remainder = 2.

$\therefore \dfrac{200}{9} = 22 + \dfrac{2}{9} = 22\dfrac{2}{9}.$

Hence, the mixed fraction is $22\dfrac{2}{9}$.

Convert each of the following sets of unlike fractions into that of like fractions :

(i) $\dfrac{4}{5}, \dfrac{7}{10}, \dfrac{11}{15}, \dfrac{13}{20}$

(ii) $\dfrac{2}{3}, \dfrac{1}{4}, \dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12}$

(iii) $\dfrac{1}{3}, \dfrac{3}{4}, \dfrac{5}{12}, \dfrac{9}{16}, \dfrac{17}{24}$

(iv) $\dfrac{2}{3}, \dfrac{1}{6}, \dfrac{5}{9}, \dfrac{7}{12}, \dfrac{13}{18}$

(v) $\dfrac{1}{2}, \dfrac{4}{7}, \dfrac{9}{14}, \dfrac{11}{21}, \dfrac{37}{42}$

(vi) $\dfrac{2}{7}, \dfrac{5}{8}, \dfrac{11}{14}, \dfrac{9}{16}, \dfrac{3}{4}$

Answer

(i) $\dfrac{4}{5}, \dfrac{7}{10}, \dfrac{11}{15}, \dfrac{13}{20}$

Denominators: 5, 10, 15, 20

LCM = 60.

Equivalent fractions:

$\Rightarrow \dfrac{4 \times 12}{5 \times 12}, \dfrac{7 \times 6}{10 \times 6}, \dfrac{11 \times 4}{15 \times 4}, \dfrac{13 \times 3}{20 \times 3}\\[1em] \Rightarrow \dfrac{48}{60}, \dfrac{42}{60}, \dfrac{44}{60}, \dfrac{39}{60}$

Hence, the like fractions are $\dfrac{48}{60}, \dfrac{42}{60}, \dfrac{44}{60}, \dfrac{39}{60}$.

(ii) $\dfrac{2}{3}, \dfrac{1}{4}, \dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12}$

Denominators: 3, 4, 6, 8 and 12

LCM = 24.

Equivalent fractions:

$\Rightarrow \dfrac{2 \times 8}{3 \times 8}, \dfrac{1 \times 6}{4 \times 6}, \dfrac{5 \times 4}{6 \times 4}, \dfrac{7 \times 3}{8 \times 3}, \dfrac{11 \times 2}{12 \times 2}\\[1em] \Rightarrow \dfrac{16}{24}, \dfrac{6}{24}, \dfrac{20}{24}, \dfrac{21}{24}, \dfrac{22}{24}$

Hence, the like fractions are $\dfrac{16}{24}, \dfrac{6}{24}, \dfrac{20}{24}, \dfrac{21}{24}, \dfrac{22}{24}$.

(iii) $\dfrac{1}{3}, \dfrac{3}{4}, \dfrac{5}{12}, \dfrac{9}{16}, \dfrac{17}{24}$

Denominators: 3, 4, 12, 16, 24

LCM = 48.

Equivalent fractions:

$\Rightarrow \dfrac{1 \times 16}{3 \times 16}, \dfrac{3 \times 12}{4 \times 12}, \dfrac{5 \times 4}{12 \times 4}, \dfrac{9 \times 3}{16 \times 3}, \dfrac{17 \times 2}{24 \times 2}\\[1em] \Rightarrow \dfrac{16}{48}, \dfrac{36}{48}, \dfrac{20}{48}, \dfrac{27}{48}, \dfrac{34}{48}$

Hence, the like fractions are $\dfrac{16}{48}, \dfrac{36}{48}, \dfrac{20}{48}, \dfrac{27}{48}, \dfrac{34}{48}$.

(iv) $\dfrac{2}{3}, \dfrac{1}{6}, \dfrac{5}{9}, \dfrac{7}{12}, \dfrac{13}{18}$

Denominators: 3, 6, 9, 12, 18

LCM = 36.

Equivalent fractions:

$\Rightarrow \dfrac{2 \times 12}{3 \times 12}, \dfrac{1 \times 6}{6 \times 6}, \dfrac{5 \times 4}{9 \times 4}, \dfrac{7 \times 3}{12 \times 3}, \dfrac{13 \times 2}{18 \times 2}\\[1em] \Rightarrow \dfrac{24}{36}, \dfrac{6}{36}, \dfrac{20}{36}, \dfrac{21}{36}, \dfrac{26}{36}$

Hence, the like fractions are $\dfrac{24}{36}, \dfrac{6}{36}, \dfrac{20}{36}, \dfrac{21}{36}, \dfrac{26}{36}$.

(v) $\dfrac{1}{2}, \dfrac{4}{7}, \dfrac{9}{14}, \dfrac{11}{21}, \dfrac{37}{42}$

Denominators: 2, 7, 14, 21, 42

LCM = 42.

Equivalent fractions:

$\Rightarrow \dfrac{1 \times 21}{2 \times 21}, \dfrac{4 \times 6}{7 \times 6}, \dfrac{9 \times 3}{14 \times 3}, \dfrac{11 \times 2}{21 \times 2}, \dfrac{37 \times 1}{42 \times 1}\\[1em] \Rightarrow \dfrac{21}{42}, \dfrac{24}{42}, \dfrac{27}{42}, \dfrac{22}{42}, \dfrac{37}{42}$

Hence, the like fractions are $\dfrac{21}{42}, \dfrac{24}{42}, \dfrac{27}{42}, \dfrac{22}{42}, \dfrac{37}{42}$.

(vi) $\dfrac{2}{7}, \dfrac{5}{8}, \dfrac{11}{14}, \dfrac{9}{16}, \dfrac{3}{4}$

Denominators: 7, 8, 14, 16, 4

LCM = 112.

Equivalent fractions:

$\Rightarrow \dfrac{2 \times 16}{7 \times 16}, \dfrac{5 \times 14}{8 \times 14}, \dfrac{11 \times 8}{14 \times 8}, \dfrac{9 \times 7}{16 \times 7}, \dfrac{3 \times 28}{4 \times 28}\\[1em] \Rightarrow \dfrac{32}{112}, \dfrac{70}{112}, \dfrac{88}{112}, \dfrac{63}{112}, \dfrac{84}{112}$

Hence, the like fractions are $\dfrac{32}{112}, \dfrac{70}{112}, \dfrac{88}{112}, \dfrac{63}{112}, \dfrac{84}{112}$.

Fill in the placeholders with > or < :

(i) $\dfrac{7}{9} {\boxed{\phantom{63}}} \dfrac{5}{9}$

(ii) $\dfrac{9}{13} {\boxed{\phantom{63}}} \dfrac{11}{13}$

(iii) $\dfrac{3}{5} {\boxed{\phantom{63}}} \dfrac{3}{4}$

(iv) $\dfrac{7}{9} {\boxed{\phantom{63}}} \dfrac{7}{11}$

(v) $\dfrac{5}{8} {\boxed{\phantom{63}}} \dfrac{5}{6}$

(vi) $\dfrac{7}{9} {\boxed{\phantom{63}}} \dfrac{6}{11}$

(vii) $\dfrac{6}{5} {\boxed{\phantom{63}}} \dfrac{5}{4}$

(viii) $\dfrac{7}{11} {\boxed{\phantom{63}}} \dfrac{8}{13}$

(ix) $\dfrac{10}{13} {\boxed{\phantom{63}}} \dfrac{13}{16}$

(x) $\dfrac{2}{9} {\boxed{\phantom{63}}} \dfrac{3}{14}$

(xi) $\dfrac{7}{12} {\boxed{\phantom{63}}} \dfrac{5}{9}$

(xii) $\dfrac{15}{19} {\boxed{\phantom{63}}} \dfrac{3}{4}$

Answer

(i) $\dfrac{7}{9} {\boxed{\phantom{63}}} \dfrac{5}{9}$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

Hence, $\dfrac{7}{9} {\boxed{\gt}} \dfrac{5}{9}$.

(ii) $\dfrac{9}{13} {\boxed{\phantom{63}}} \dfrac{11}{13}$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

Hence, $\dfrac{9}{13} {\boxed \lt} \dfrac{11}{13}$.

(iii) $\dfrac{3}{5} {\boxed{\phantom{63}}} \dfrac{3}{4}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

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

(iv) $\dfrac{7}{9}$ ${\boxed{\phantom{63}}}$ $\dfrac{7}{11}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

Hence, $\dfrac{7}{9}$ ${\boxed{\gt}}$ $\dfrac{7}{11}$

(v) $\dfrac{5}{8} {\boxed{\phantom{63}}} \dfrac{5}{6}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

Hence, $\dfrac{5}{8} {\boxed{\lt}} \dfrac{5}{6}$.

(vi) $\dfrac{7}{9} {\boxed{\phantom{63}}} \dfrac{6}{11}$

Cross multiply; 7 x 11 and 6 x 9

⇒ 77 and 54

Now, comparing both numbers;

⇒ 77 > 54

Hence, $\dfrac{7}{9} {\boxed{\gt}} \dfrac{6}{11}$.

(vii) $\dfrac{6}{5} {\boxed{\phantom{63}}} \dfrac{5}{4}$

Cross multiply; 6 x 4 and 5 x 5

⇒ 24 and 25

Now, comparing both numbers;

⇒ 24 < 25

Hence, $\dfrac{6}{5} {\boxed{\lt}} \dfrac{5}{4}$.

(viii) $\dfrac{7}{11} {\boxed{\phantom{63}}} \dfrac{8}{13}$

Cross multiply; 7 x 13 and 8 x 11

⇒ 91 and 88

Now, comparing both numbers;

⇒ 91 > 88

Hence, $\dfrac{7}{11} {\boxed{\gt}} \dfrac{8}{13}$.

(ix) $\dfrac{10}{13} {\boxed{\phantom{63}}} \dfrac{13}{16}$

Cross multiply; 10 x 16 and 13 x 13

⇒ 160 and 169

Now, comparing both numbers;

⇒ 160 < 169

Hence, $\dfrac{10}{13} {\boxed{\lt}} \dfrac{13}{16}$.

(x) $\dfrac{2}{9} {\boxed{\phantom{63}}} \dfrac{3}{14}$

Cross multiply; 2 x 14 and 3 x 9

⇒ 28 and 27

Now, comparing both numbers;

⇒ 28 > 27

Hence, $\dfrac{2}{9} {\boxed{\gt}} \dfrac{3}{14}$.

(xi) $\dfrac{7}{12} {\boxed{\phantom{63}}} \dfrac{5}{9}$

Cross multiply; 7 x 9 and 12 x 5

⇒ 63 and 60

Now, comparing both numbers;

⇒ 63 > 60

Hence, $\dfrac{7}{12} {\boxed{\gt}} \dfrac{5}{9}$.