We can also solve this problem algebraically using the combined rate equation:
$$\frac{1}{A}+\frac{1}{B}=\frac{1}{A\ and\ B}$$
Where A and B are the time it would take each person (or machine, pump, printer, etc. depending on the question) to finish a job alone, and A and B is the time it would take both to finish the same job together.
From our equation, we know:
$$\frac{1}{A\ and\ B}=20$$ $$\frac{1}{\ B\ and\ C}=30$$ $$\frac{1}{\ A\ and\ C}=40$$ This means $$\frac{1}{\ A\ and\ B}<\frac{1}{B\ and\ C}<\frac{1}{A\ and\ C}$$
Using the combined rate equation, we can express each term as the sum of the reciprocals of the time it would take each person to complete the task individually:
$$\frac{1}{\ A}+\frac{1}{B}<\frac{1}{B}+\frac{1}{C}<\frac{1}{A}+\frac{1}{C}$$ or separated out $$\frac{1}{\ A}+\frac{1}{B}<\frac{1}{B}+\frac{1}{C}\ \ and\ \ \frac{1}{B}+\frac{1}{C}<\frac{1}{A}+\frac{1}{C}$$ Then subtracting like terms from both sides $$\frac{1}{\ A}<\frac{1}{C}\ \ and\ \ \frac{1}{B}<\frac{1}{A}$$ We can then cross-multiply without changing the direction of the sign, since we know all of the variables are positive (since a task can't be completed in negative time) to give

C < A and A < B

or combined

B > A > C

AAPL wrote:A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order.

$$A.\ A>B>C$$
$$B.\ A>C>B$$
$$C.\ B>A>C$$
$$D.\ B>C>A$$
$$E.\ C>A>B$$

Since B is in the two least number of days: 20 and 30, it must mean B is the most efficient. Similarly, since C is in the two most number of days: 30 and 40, it must mean C is the least efficient. That leaves A's efficiency between those of B and C. Thus, their efficiencies in descending order, is: B > A > C.

Answer: C