On 10/2/00 5:21:24 PM, Stormkill wrote: >Two taps turned on together >can fill a tank in 15 minutes. >By themselves, one takes 15 >minutes longer than the other >to fill the tank. find the >time taken to fill the tank by >each tap alone.
Let A be the fraction of the tank filled by the first tap. Let B be the fraction filled by the second tap. Thus, A+B represents the entire take, so it is 1. A+B=1.
Let t represent the time taken to fill the tank together. In this case, t=15 since we want t to be the time taken to fill the tank together. Let R and S be the rates at which both takes are filled.
Rt=A St=B A+B=1
Those are the equations I have so far. R is the rate at which it is filled. ie, rate = (fraction of tank) / (time to fill it). since the rate is calculated by the time taken to fill the tank, rate = 1 / time. Let T and U be the times required for each tap to fill it alone.
Rt+St=1 combine the first set of equations
t/T+t/U=1 or 1/T+1/U=1/t
The formula is well known and is easily generalized. if n taps working together can perform the task in t1, t2, t3, ..., tn time, the time taken if they work together is
This formula applies in many occasions. (eg, people mowing lawns)
Back to your problem.
Because one tap is 15 minutes faster, let U=T+15. Of course, t=15.
multiply both sides by the denominators to remove the fractions.