Divide 27 into two parts such that the sum of their reciprocals is

Let the two parts be x and (27 − x).

According to the given condition,

$\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}$

$\Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}$

$\Rightarrow \frac{27}{27 x-x^{2}}=\frac{3}{20}$

$\Rightarrow 27 x-x^{2}=180$

$\Rightarrow x^{2}-27 x+180=0$

$\Rightarrow x^{2}-15 x-12 x+180=0$

$\Rightarrow x(x-15)-12(x-15)=0$

$\Rightarrow(x-12)(x-15)=0$

$\Rightarrow x-12=0$ or $x-15=0$

$\Rightarrow x=12$ or $x=15$

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.