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,

$\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}\Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}\Rightarrow \frac{27}{27x-x^2}=\frac{3}{20}\Rightarrow 27x-x^2=180\Rightarrow x^2-27x+180=0\Rightarrow x^2-15x-12x+180=0x(x-15)-12(x-15)=0(x-12)(x-15)=0x=12or15Whenx=12,27-x=27-12=15Whenx=15,27-x=27-15=12$

Hence, the required parts are 12 and 15.