question_answer

Two taps running together can fill a tank in \[3\frac{1}{13}\] hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?

Answer:

Let tank fill by one tap = x hrs.

Other tap \[=(x+3)\text{ }hrs.\]

Together they fill by \[3\frac{1}{13}=\frac{40}{13}\,hrs.\]

Now,

\[\Rightarrow \]               \[\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}\]

\[\Rightarrow \]               \[\frac{x+3+x}{(x)(x+3)}=\frac{13}{40}\]

\[\Rightarrow \]               \[\frac{2x+3}{{{x}^{2}}+3x}=\frac{13}{40}\]

\[\Rightarrow \]               \[13{{x}^{2}}+39x=80x+120\]

\[\Rightarrow \]               \[13{{x}^{2}}-41x-120=0\]

\[\Rightarrow \]               \[13{{x}^{2}}-65x+24x-120=0\]

\[\Rightarrow \]               \[13x(x-5)+24(x-5)=0\]

\[\Rightarrow \]               \[(x-5)(13x+24)=0\]

Either        \[x-5=0\] or \[13x+24=0\]

\[x=5,x=-24/13\] (Rejected)

One tap fill the tank in 5 hrs.

So other tap fill the tank in \[5+3=8\,hrs\].