Question

Two taps together can fill a tank completely in $3\frac{1}{13}$ minutes. The smaller tap takes 3 minutes more than the bigger tap to fill the tank. How much time does each tap take to fill the tank completely?

Answer 1

Let larger tap A takes $x$ minutes to fill the tank completely.

Thus, in 1 minute, tap A will fill $\frac{1}{x}$ portion of the tank.

By given condition, smaller tap B takes $x+3$ minutes to fill tank completely.

Thus, in 1 minute, tap B will fill $\frac{1}{x+3}$ portion of tank

Time taken by both taps to fill tank completely is $3\frac{1}{13}=\frac{40}{13}$ minutes

Thus, in 1 minute, both taps will fill $=\frac{13}{40}$ portion of the tank.

Thus, we can write,

$\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}$

$\therefore \frac{x+x+3}{x(x+3)}=\frac{13}{40}$

$\therefore \frac{2x+3}{x(x+3)}=\frac{13}{40}$

$\therefore \frac{2x+3}{x^2+3x}=\frac{13}{40}$

$\therefore 40(2x+3)=13(x^2+3x)$

$\therefore 80x+120=13x^2+39x$

$\therefore 13x^2-41x-120=0$

$\therefore x=\frac{-(-41)\pm \sqrt{{(-41)}^2-4\times 13\times -120}}{2\times 13}$

$\therefore x=\frac{41\pm \sqrt{1681+6240}}{26}$

$\therefore x=\frac{41\pm \sqrt{7921}}{26}$

$\therefore x=\frac{41\pm 89}{26}$

We will only consider positive root as time is never negative.

$\therefore x=\frac{41+89}{26}$

$\therefore x=\frac{130}{26}$

$\therefore x=5minutes$

Thus, larger tap will take $5minutes$ to fill the tank completely.

Smaller tap will take $5+3=8minutes$ to fill the tank complete