Re: Archimedes bath problem

let A = flow rate for tap A in units per hour

(2 hrs)(A + 2A + 3A) + (2.75 hrs)(A + 3A) = 1 tub filled

23A = 1

A = (1 tub filled)/(23 hrs)

using all 3 ...

(A + 2A + 3A)t = 1 tub filled

solve for t

Re: Archimedes bath problem

Hello, Natasha1!

Quote:

Archimedes has a bath which is filled using three taps labelled A, B and C.

The rate of flow of B is twice that of A; the rate of flow of C is three times that of A.

Yesterday, Archimedes filled his bath as follows:

First, he used all three taps for 2 hours. Then he turned off B and finished filling the bath

in an additional 2 hours and 45 mins using taps A and C only.

Today Archimedes is going to fill his bath (currently empty) using all three taps.

How long will this take him?

Let = number of hours for A to fill the bath alone.

In one hour, A can fill of the bath.

Then B can fill the bath alone in hours.

In one hour, B can fill of the bath.

And C can fill the bath alone in hours.

In one hour, C can fill of the bath.

In hours, the three taps can fill: of the bath.

In hours, A and C can fill: of the bath.

These two quantities can fill the entire bath: .

. . Hence: .

A takes hours to fill the bath.

. . In one hour, A fills of the bath.

. . In hours, A fills of the bath.

B takes hours to fill the bath.

. . In one hour, B fills of the bath.

. . In hours, B fills of the bath.

C takes hours to fill the bath.

. . In one hour, C fills of the bath.

. . In hours, C fills of the bath.

In hours, the three taps will fill the entire bath.

. .

. .

Therefore, the three taps will take hours.