|
Question 1085718: Tap A can fill a tank in 15 minutes.
Tap B can fill a tank in 20 minutes.
How long will it take to fill the tank when both taps are turned on?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Table
| Time | Rate |
|---|
| Tap A | 15 | 1/15 | | Tap B | 20 | 1/20 | | Total | | 7/60 |
The table represents the rates and times for both taps.
The time values are in minutes. The rates are in jobs per minute
A rate like "1/20" means tap B can get 1/20th of a job done per minute. Each minute, tap B fills up 1/20 = 0.05 = 5% of the sink.
The combined rate of 7/60 is the result of adding the fractions 1/15 and 1/20 like so
(1/15) + (1/20) = (4/4)*(1/15) + (3/3)*(1/20)
(1/15) + (1/20) = 4/60 + 3/60
(1/15) + (1/20) = (4+3)/60
(1/15) + (1/20) = 7/60
This means that if the taps are opened together, then the sink is filled at a rate of 7/60 jobs per minute. Put another way, the two taps work together to fill the sink 7/60 of the way for each minute.
Let t be the time it takes for both sinks to get the job done together. In this case,
(combined rate)*(time) = 1
where "1" represents the fact that 1 job is completed, ie the sink is full 100%
The combined rate found earlier is 7/60. The time is unknown. Let's solve for t
(combined rate)*(time) = 1
(7/60)*(t) = 1
60*(7/60)*(t) = 60*1
7*t = 60
7*t/7 = 60/7
t = 60/7 <--- Exact answer as a fraction
t = 8.571429 <--- Approximate answer in decimal form
It takes roughly 8.571429 minutes for the two taps to completely fill the sink. This is assuming that one tap doesn't hinder the other in any way.
|
|
|
| |
