Question

The first tap can fill the pool in 10 hours. When the second tap was opened, the empty pool was filled in 6 hours. How long will it take the second tap to fill the pool alone?

Answer 1

Answer:

Time taken the second tap to fill the pool alone = 15 hours.

Step-by-step explanation:

Let the volume of pool be V.

The first tap can fill the pool in 10 hours.

$\texttt{Rate of tap 1,}r_1=\frac{V}{10}$

When the second tap was opened, the empty pool was filled in 6 hours. Here both taps are open

That is

          $V=(r_1+r_2)\times 6\\\\V=(\frac{V}{10}+r_2)\times 6\\\\r_2=\frac{4V}{60}=\frac{V}{15}$    

We need to find how long will it take the second tap to fill the pool alone.

         $t_2=\frac{V}{r_2}=\frac{V}{\frac{V}{15}}=15hours$

Time taken the second tap to fill the pool alone = 15 hours.    

Answer 2

Answer: $15\ hours$

Step-by-step explanation:

We can use the following formula to solve this exercise:

$\frac{T}{A}+\frac{T}{B}=1$

In this case "T" is the time in hours it takes to both taps fill the pool working together, "A" is the time in hours for the first tap filling the pool alone an "B" is the time in hours for the second tap filling the pool alone.

We can identify that:

$T=6\\\\A=10$

Then, we must subsitute the known values into the formula:

$\frac{6}{10}+\frac{6}{B}=1$

And finally we must solve for "B".

Then we get:

$\frac{6}{10}+\frac{6}{B}=1\\\\\frac{6}{B}=1-\frac{6}{10}\\\\\frac{6}{B}=0.4\\\\6=0.4B\\\\\frac{6}{0.4}=B\\\\B=15$