Question

One tank can be filled up by two taps in 6 hours. The Smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required (in hours) by the smaller tap to fill the tank separately.

• A. 15
• B. 12
• C. 5
• D. 14

Answer 1

The correct option is A 15

Let time taken to fill tank by small tap be $x$ hours
Time taken to fill tank by big tap be $y$ hours
Then, $x-y=5$ or $x=5+y$
Smaller tap can fill in one hour $=\frac{1}{x}$
Bigger tap can fill in one hour $=\frac{1}{y}$
Then, $\frac{1}{x}+\frac{1}{y}=\frac{1}{6}$
$\frac{x+y}{xy}=\frac{1}{6}$
$xy=6(x+y)$
$y(5+y)=6(5+y+y)$
$5y+y^2=30+12y$
$y^2-7y-30=0$
$y=\frac{7\pm \sqrt{49+120}}{2}$
$y=\frac{7\pm 13}{2}$
$y=10,-3$
$x=5+10=5+10=15$ hours
Hence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours