Three pipes A, B and C are installed to fill a tank. Pipes A and B opened together can fill the tank in the same time in which C can alone fill the tank. If pipe B can fill the tank 15 minutes faster than pipe A and 5 minutes slower than pipe C, then the time required by pipe A to fill the tank alone is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30 minutes
Let time taken by pipe $A=x$ minutes.
Then time taken by $B=x-15$ minutes.
Since $B$ is $5$ minutes slower than $C$, time taken by $C=x-20$ minutes.
Rates:
$A=\frac{1}{x},\;B=\frac{1}{x-15},\;C=\frac{1}{x-20}$
Given $A$ and $B$ together fill the tank in the same time as $C$ alone:
$\frac{1}{x}+\frac{1}{x-15}=\frac{1}{x-20}$
Solve:
$\frac{2x-15}{x(x-15)}=\frac{1}{x-20}$
$(2x-15)(x-20)=x(x-15)$
$2x^{2}-55x+300=x^{2}-15x$
$x^{2}-40x+300=0$
$(x-30)(x-10)=0$
$x=30$ or $x=10$
Only $x=30$ is valid.
final answer: $30$ minutes