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: Understanding the Question:
We are given that two water taps together fill a tank in $8\frac{8}{9}$ hours.
The larger tap takes 4 hours less than the smaller tap to fill the tank individually.
We need to calculate the individual times taken by each tap to fill the tank completely.

Step 2: Key Formula or Approach:
Let the smaller tap take $y$ hours to fill the tank.
Then, the larger tap takes $y - 4$ hours.
The sum of their individual hourly rates equals their joint hourly rate:
\[ \frac{1}{y} + \frac{1}{y - 4} = \frac{1}{\text{Total Joint Time}} \]
We will simplify and solve the resulting quadratic equation.

Step 3: Detailed Explanation:

• Convert joint time into a standard fraction:
\[ \text{Total Joint Time} = 8\frac{8}{9} = \frac{80}{9}\text{ hours} \]
The joint hourly rate is $\frac{9}{80}$.

• Formulate the rate equation:
\[ \frac{1}{y} + \frac{1}{y - 4} = \frac{9}{80} \]

• Simplify the equation:
\[ \frac{y - 4 + y}{y(y - 4)} = \frac{9}{80} \]
\[ \frac{2y - 4}{y^2 - 4y} = \frac{9}{80} \]
\[ 80(2y - 4) = 9(y^2 - 4y) \]
\[ 160y - 320 = 9y^2 - 36y \]
Rearrange all terms to one side to form a quadratic equation:
\[ 9y^2 - 196y + 320 = 0 \]

• Solve the quadratic equation by splitting the middle term:
\[ 9y^2 - 180y - 16y + 320 = 0 \]
\[ 9y(y - 20) - 16(y - 20) = 0 \]
\[ (9y - 16)(y - 20) = 0 \]
This gives:
\[ y = 20 \quad \text{or} \quad y = \frac{16}{9} \]
If $y = \frac{16}{9} \approx 1.78\text{ hours}$, then the larger tap takes $1.78 - 4 = -2.22\text{ hours}$, which is impossible.
Therefore, we reject $y = \frac{16}{9}$ and accept $y = 20\text{ hours}$.

• Find the time for the larger tap:
\[ \text{Time for larger tap} = y - 4 = 20 - 4 = 16\text{ hours} \]


Step 4: Final Answer:
The smaller tap takes $20\text{ hours}$ and the larger tap takes $16\text{ hours}$ to fill the tank separately.