Question

Find the fraction equivalent of $\frac{35}{42}$, having:

(i) numerator 15
(ii) denominator 18
(iii) denominator 30
(iv) numerator 30

Answer 1

$$\begin{gathered} \mathrm{Firstly},\mathrm{we}\mathrm{will}\mathrm{reduce}\frac{35}{42}\mathrm{into}\mathrm{the}\mathrm{lowest}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{divide}\mathrm{both}\mathrm{the}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}\mathrm{the}\mathrm{HCFs}\mathrm{of}35\&42. \\ \Rightarrow \frac{35÷7}{42÷7}=\frac{5}{6} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{i}\right)\frac{5}{6}=\frac{15}{\mathrm{Denominator}} \\ \mathrm{We}\mathrm{will}\mathrm{consider}\mathrm{the}\mathrm{numerators}. \\ \mathrm{As}5\times 3=15,\mathrm{we}\mathrm{will}\mathrm{multiply}\mathrm{both}\mathrm{the}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}3. \\ \Rightarrow \frac{5\times 3}{6\times 3}=\frac{15}{18} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{5}{6}=\frac{\mathrm{Numerator}}{18} \\ \mathrm{We}\mathrm{will}\mathrm{consider}\mathrm{the}\mathrm{denominators}. \\ \mathrm{As}6\times 3=18,\mathrm{we}\mathrm{will}\mathrm{multiply}\mathrm{both}\mathrm{the}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}3. \\ \Rightarrow \frac{5\times 3}{6\times 3}=\frac{15}{18} \end{gathered}$$


$$\begin{gathered} \left(iii\right)\frac{5}{6}=\frac{\mathrm{numerator}}{30} \\ Wewillconsider\mathrm{the}\mathrm{denominators}, \\ \mathrm{Since},6\times 5=30,\mathrm{therefore}\mathrm{multiplying}\mathrm{both}\mathrm{the}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}5 \\ \Rightarrow \frac{5\times 5}{6\times 5}=\frac{25}{30} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iv}\right)\frac{5}{6}=\frac{30}{\mathrm{Denominator}} \\ \mathrm{We}\mathrm{will}\mathrm{consider}\mathrm{the}\mathrm{numerators}. \\ \mathrm{As}5\times 6=30,\mathrm{we}\mathrm{will}\mathrm{multiply}\mathrm{both}\mathrm{the}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}6. \\ \Rightarrow \frac{5\times 6}{6\times 6}=\frac{30}{36} \end{gathered}$$