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

Here, we need to divide 27 into two parts. Let the parts be 𝑥 and 27 − 𝑥 Now, the sum of their reciprocals is $\frac{3}{20}$.

$\begin{array}{r}\Rightarrow \frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}\\ \Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}\\ \Rightarrow \frac{27}{27x-x^2}=\frac{3}{20}\\ \Rightarrow x^2-27x+180=0\end{array}$

Factorising by splitting the middle term, we get

$x^2-15x-12x+180=0$

Grouping common terms,

𝑥(𝑥 − 15) − 12(𝑥 − 15) = 0 

⇒ (𝑥 − 15)(𝑥 − 12) = 0 

⇒ 𝑥 − 15 = 0 or, (𝑥 − 12) = 0 

⇒ 𝑥 = 15 or 12

Therefore, the two parts are 15 and 12.