Question

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

Answer 1

Let the two parts be x and  ${\displaystyle (27-x)}$ 

According to the given condition, 

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

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

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

⇒${\displaystyle 27x-x^2=180}$ 

⇒${\displaystyle x^2-27x+180=0}$ 

⇒${\displaystyle x^2-15x-12x+180=0}$ 

⇒${\displaystyle x(x-15)-12(x-15)=0}$ 

⇒${\displaystyle (x-12)(x-15)=0}$ 

⇒${\displaystyle x-12=0 \text{or}x-15=0}$ 

⇒${\displaystyle x=12 \text{or} x-15}$ 

When x=12, 

${\displaystyle 27-x=27-12=15}$ 

When${\displaystyle x=15}$ 

${\displaystyle 27-x=27-15=15}$ 

Hence, the required parts are 12 and 15.