We have three pumps filling a tank: The first one fills the whole tank in a particular time, the second one is twice as fast as the first one, and the third one is three time as fast.
All the pumps together can fill the tank in two minutes.
How long would each pump take on its own?
On
Denote the volume of the Tank by $V$. The time that each takes to fill it by $t$.
Now the first one fills a volume $V$ in time $t$ hence it's rate is $V/t$. Similarly the others rates are : $2V/t$ and $3V/t$.
Now to find $t$ you use your second condition : "All the pumps together can fill the tank in two minutes"
This means : $(V/t+2V/t+3V/t)\cdot 2=V$.
I hope this will make you start.
On
Let $g$ be the capacity of the tank in gallons, and let $t$ be the number of minutes it takes the first pump to fill the tank. Then the first pump pumps $\frac{g}t$ gallons a minute. The second pump is twice as fast, so it pumps $\frac{2g}t$ gallons a minute. Similarly, the third pump pumps $\frac{3g}t$ gallons a minute. Thus, working together they pump
$$\frac{g}t+\frac{2g}t+\frac{3g}t=\frac{6g}t$$
gallons a minute. It takes them $2$ minutes to pump the $g$ gallons required to fill the tank, so working together they pump $\frac{g}2$ gallons a minute. Thus,
$$\frac{6g}t=\frac{g}2\;,$$
and from here it’s straightforward algebra to find $t$ and then the times required by the second and third pumps.
The general principle in these problems is to work in terms of rates; here that means gallons per minute. Note too that I really didn’t need to introduce gallons: I could have used a tankfull as my unit of quantity. That’s as if all of my $g$’s were equal to $1$. Since the $g$’s disappear in the end anyway, I decided to stick with more familiar units.
Let $\,T_i\;$ be the proportion of the tank filled in one minute by pump $\;i\;,\;i=1,2,3\;$ , then
$$T_2=2T_1\;,\;\;T_3=3T_1\;\;\text{and}\;\; 2(T_1+T_2+T_3)=12T_1=1\implies$$
$$T_1=\frac1{12}\;,\;\;T_2=\frac16\;,\;\;T_3=\frac14\;,\;\;\text{so}\ldots\ldots$$