Filling a swimming pool with two hoses Hose A can fill a pool in 4 days. You add a second hose B, now both A and B can fill it in 3 days. How many days will it take to fill the pool with just hose B?
I would like to see a formula.
 A: Hose A filled 1/4 pools per day. So in 3 days it filled 3/4. That means in 3 days hose b filled the other 1/4 so it will take hose B 12 days.
A: We assume that the hoses are independent and that their fill rates are invariant.
The rate hose$_A$ fills up the pool is $1 / 4$  [pool day$^{-1}$] and the rate of hose$_A$ and hose$_B$ combined is $1/3$ [pool day$^{-1}$], we then substitue the known rate of hose$_A$ to obtain
$1/4$ + hose$_B$.rate = 1/3;
i.e. hose$_B$.rate = 1/3 - 1/4 = 4/12 - 3/12 = 1/12 [pool days$^{-1}$],
and we are done.
A: Let $V$ be the volume of the pool. Then the flow rate of the hoses are $r_A = \frac{V}{4}, r_{A}+r_B = \frac{V}{3}$.
Hence the rate of $B$ alone is $r_B = (r_{A}+r_{B})-r_A$. Hence the time taken to fill the pool with hose B is $\frac{V}{r_B} = \frac{V}{(r_{A}+r_{B})-r_A}$. Substitute the values for $r_A, r_B$ to get the answer.
A: Define:
$ F_A $: flow rate of the hose A
$ F_B $: flow rate of the hose B
$ T_A $: total time in which hose A would fill the empty pool
$ T_B $: total time in which hose B would fill the empty pool
$ V $: volume of the pool
$$T_A = \frac{V}{F_A} = 4 \, days  \\
Given: \; \frac{V}{F_A + F_B} = 3 \, days \\
\implies \frac{1}{4}\frac{V}{F_A} = \frac{1}{3}\frac{V}{F_A + F_B} \\
4F_A = 3(F_A + F_B) \\
F_B = \frac{1}{3}F_A \\
T_B = \frac{V}{F_B} = \frac{V}{\frac{1}{3}F_A} = 3\frac{V}{F_A} = 3T_A \\
T_B = 12 \, days
$$
