Convert the following mixed fractions into improper fractions:

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

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

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

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

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 $7\dfrac{2}{3} = \dfrac{23}{3}$.