Pete needs help

Hello, Pete!

This is not a typical "work" problem . . . it's trickier!
. . I had to baby-talk my way through it . . .

An oil refinery has a series of empty tanks, all of the same size.
Pump A, working by itself, can fill one of these tanks in 18 hours.
If pumps A and B work together, they can fill an empty tank in 8 hours.
How long would it take pump B, working by itself, to fill an empty tank?

Click to expand...

A fills a tank in 18 hours.
. . In 1 hour, A fills \(\displaystyle \frac{1}{18}\) of a tank.
. . In 8 hours, A fills \(\displaystyle \frac{8}{18}\,=\,\frac{4}{9}\) of a tank.

B fills tank in \(\displaystyle x\) hours.
. . In 1 hour, B fills \(\displaystyle \frac{1}{x}\) of a tank.
. . In 8 hours, B fills \(\displaystyle \frac{8}{x}\) of a tank.

In 8 hours, they both fill one (1) tank.
. . We have: .\(\displaystyle \L\frac{4}{9}\,+\,\frac{8}{x}\:=\:1\) . . . There!

Multiply by \(\displaystyle 9x:\;\;4x\,+\,72\:=\:9x\;\;\Rightarrow\;\;5x\,=\,72\;\;\Rightarrow\;\;x\,=\,\frac{72}{5}\)

Therefore, pump B will take \(\displaystyle 14.4\) hours to fill a tank alone.

I think this way is also ok:
one hour:
1/18 + 1/B = 1/8
1/B = 1/8 - 1/18
1/B = 5/72
B = 72/5 = 14.4

soroban said:

In 8 hours, they both fill one (1) tank. We have:
. . . . .\(\displaystyle \L\frac{4}{9}\,+\,\frac{8}{x}\:=\:1\)

Click to expand...

Technical note: The "1" doesn't stand for "one tank (or whatever) being filled", so much as "one job being done". If the "job" had been filling a tank farm, you still would have used "1" (in this method) to represent 100% of the job being completed.



Eliz.