# Archimedes bath problem

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• Nov 22nd 2012, 03:17 PM
Natasha1
Archimedes bath problem
Could anyone help me with this problem please?

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

The rates of flow of B is twice that of A, the rates of flow of C is three times that of A.

Yesterday, Archimedes filled his bath as follows:

Firstly 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 (it's a very big bath).

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

How long will this take him?
• Nov 22nd 2012, 05:15 PM
skeeter
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
• Nov 22nd 2012, 06:32 PM
Soroban
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 $\displaystyle a$ = number of hours for A to fill the bath alone.
In one hour, A can fill $\displaystyle \tfrac{1}{a}$ of the bath.

Then B can fill the bath alone in $\displaystyle \tfrac{a}{2}$ hours.
In one hour, B can fill $\displaystyle \frac{1}{\frac{a}{2}} = \tfrac{2}{a}$ of the bath.

And C can fill the bath alone in $\displaystyle \tfrac{a}{3}$ hours.
In one hour, C can fill $\displaystyle \frac{1}{\frac{a}{3}} = \tfrac{3}{a}$ of the bath.

In $\displaystyle 2$ hours, the three taps can fill: $\displaystyle 2(\tfrac{1}{a}) + 2(\tfrac{2}{a}) + 2(\tfrac{3}{a}) \,=\,\tfrac{12}{a}$ of the bath.

In $\displaystyle 2\tfrac{3}{4} = \tfrac{11}{4}$ hours, A and C can fill: $\displaystyle \tfrac{11}{4}(\tfrac{1}{a}) + \tfrac{11}{4}(\tfrac{3}{a}) \,=\,\tfrac{11}{a}$ of the bath.

These two quantities can fill the entire bath: .$\displaystyle \tfrac{12}{a} + \tfrac{11}{a} \:=\:1$

. . Hence: .$\displaystyle a \,=\,23$

A takes $\displaystyle 23$ hours to fill the bath.
. . In one hour, A fills $\displaystyle \tfrac{1}{23}$ of the bath.
. . In $\displaystyle x$ hours, A fills $\displaystyle \tfrac{x}{23}$ of the bath.

B takes $\displaystyle \tfrac{23}{2}$ hours to fill the bath.
. . In one hour, B fills $\displaystyle \tfrac{2}{23}$ of the bath.
. . In $\displaystyle x$ hours, B fills $\displaystyle \tfrac{2x}{23}$ of the bath.

C takes $\displaystyle \tfrac{23}{3}$ hours to fill the bath.
. . In one hour, C fills $\displaystyle \tfrac{3}{23}$ of the bath.
. . In $\displaystyle x$ hours, C fills $\displaystyle \tfrac{3x}{23}$ of the bath.

In $\displaystyle x$ hours, the three taps will fill the entire bath.

. . $\displaystyle \frac{x}{23} + \frac{2x}{23} + \frac{3x}{23} \:=\:1\quad\Rightarrow\quad \frac{6x}{23} \:=\:1$

. . $\displaystyle 6x \:=\:23 \quad\Rightarrow\quad x \:=\:\frac{23}{6}$

Therefore, the three taps will take $\displaystyle 3\tfrac{5}{6}$ hours.