Question

Two water taps together can fill a tank in $8\frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Hint:

Always cross-verify that the positive roots you obtain lead to positive, physically possible times for BOTH taps.
In this case, rejecting the smaller fractional root was necessary because a negative time is physically impossible!

Answer 1

Step 1: Let the larger tap alone take $x$ hours, so the smaller tap takes $x+4$ hours (working from the faster tap instead of the slower one).
Their combined rate must equal the joint rate of $\frac{9}{80}$ tank per hour (since $8\frac89=\frac{80}{9}$ hours together).
Step 2: Set up the rate equation.
\[ \frac1x + \frac{1}{x+4} = \frac{9}{80} \]
Step 3: Clear denominators and simplify to a quadratic.
\[ 80(2x+4) = 9x(x+4) \implies 160x+320 = 9x^2+36x \]
\[ 9x^2 -124x -320 = 0 \]
Solving by splitting the middle term: $9x^2-144x+20x-320=0 \implies 9x(x-16)+20(x-16)=0 \implies (x-16)(9x+20)=0$, giving $x=16$ (rejecting the negative root $x=-\frac{20}{9}$).
Step 4: State both individual times and conclude.
The larger tap takes $16$ hours, and the smaller tap takes $16+4=20$ hours, matching the original working.
\[ \boxed{\text{Larger tap: 16 hours, Smaller tap: 20 hours}} \]