Question

Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}$.

Answer 1

Let the two parts be x and (27 − x).

According to the given condition,
$$\begin{gathered} \frac{1}{x}+\frac{1}{27-x}=\frac{3}{20} \\ \Rightarrow \frac{27-x+x}{x\left(27-x\right)}=\frac{3}{20} \\ \Rightarrow \frac{27}{27x-x^2}=\frac{3}{20} \\ \Rightarrow 27x-x^2=180 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2-27x+180=0 \\ \Rightarrow x^2-15x-12x+180=0 \\ \Rightarrow x\left(x-15\right)-12\left(x-15\right)=0 \\ \Rightarrow \left(x-12\right)\left(x-15\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-12=0\mathrm{or}x-15=0 \\ \Rightarrow x=12\mathrm{or}x=15 \end{gathered}$$

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.