Lornz can finish checking a set of papers in 2/3 hours as Butch can and Butch can do it in 3/4 as many hours as Fe. working together, they can do it in 36/13 hours. In how many hours can each of them alone, finish checking the set?
let $\displaystyle t$ be the number of hours Fe takes to check the papers. then Butch takes $\displaystyle \frac 34t$, and Lornz takes $\displaystyle \frac 23 \cdot \frac 34t = \frac 12t$
So, Lornz' rate is $\displaystyle \frac {1 \text{ Job}}{\frac 12t \text{ Hours}}$
Butch's rate is $\displaystyle \frac {1 \text{ Job}}{\frac 34t \text{ Hours}}$
and Fe's rate is $\displaystyle \frac {1 \text{ Job}}{t \text{ Hours}}$
together, they do $\displaystyle \frac {1 \text{ Job}}{\frac {36}{13} \text{ Hours}}$
so add up their rates:
$\displaystyle \frac 1{\frac 12t} + \frac 1{\frac 34t} + \frac 1t = \frac 1{\frac {36}{13}}$
now solve for $\displaystyle t$ and you can find the required times
Basically,
Problems involving rates, the key to solving these typical, or 'prosaic' type of questions is to find the individual rates. The sum of the individual rates should equal the total rate.
Jhevon brilliantly shows this reasoning though his mathematical steps.
Apply this principle to the commonplace questions, and you will master them. Good luck.