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.
R=1/T S=1/U
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
1/t=1/t1+1/t2+1/t3+...+1/tn.
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.
1/T+1/(T+15)=1/15
multiply both sides by the denominators to remove the fractions.
15(T+15)+15T=T(T+15)
15T+225+15T=T^2+15T
225+15T=T^2
T^2-15T-225=0
Solve this and the results is as Nathan said.
Craig
