Greek Anthology Book XIV: 7. - Problem/Variant

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Problem

$A$ and $B$ together can do a piece of work in $6$ days.
$B$ and $C$ together can do it in $20$ days.
$C$ and $A$ together can do it in $7 \frac 1 2$ days.

How many days will each require to do the job separately?


Solution

It takes:

$8$ days for $A$ to do the job
$24$ days for $B$ to do the job
$120$ days for $C$ to do the job.


Proof

Let $a$, $b$ and $c$ denote the number of days it takes $A$, $B$ and $C$ to complete the job respectively.

Then each completes $\dfrac 1 a$, $\dfrac 1 b$ and $\dfrac 1 c$ of the job each day, respectively.

Thus:

\(\text {(1)}: \quad\) \(\ds \dfrac 1 a + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 6\) as $a$ and $b$ together take $6$ days to do the job
\(\text {(2)}: \quad\) \(\ds \dfrac 1 b + \dfrac 1 c\) \(=\) \(\ds \dfrac 1 {20}\) as $b$ and $c$ together take $20$ days to do the job
\(\text {(3)}: \quad\) \(\ds \dfrac 1 a + \dfrac 1 c\) \(=\) \(\ds \dfrac 2 {15}\) as $a$ and $c$ together take $7 \frac 1 2 = \frac {15} 2$ days to do the job


Thus we have:

\(\ds \dfrac 1 c\) \(=\) \(\ds \dfrac 2 {15} - \dfrac 1 a\) from $(3)$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 b + \paren {\dfrac 2 {15} - \dfrac 1 a}\) \(=\) \(\ds \dfrac 1 {20}\) substituting in $(2)$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 a - \dfrac 1 b\) \(=\) \(\ds \dfrac 2 {15} - \dfrac 1 {20}\) rearranging
\(\ds \) \(=\) \(\ds \dfrac {8 - 3} {60}\) common denominator
\(\text {(4)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac 1 a - \dfrac 1 b\) \(=\) \(\ds \dfrac 1 {12}\) after arithmetic
\(\ds \leadsto \ \ \) \(\ds \dfrac 2 a\) \(=\) \(\ds \dfrac 1 6 + \dfrac 1 {12}\) $(4)$ + $(1)$
\(\ds \) \(=\) \(\ds \dfrac {2 + 1} {12}\) common denominator
\(\ds \) \(=\) \(\ds \dfrac 1 4\) arithmetic
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 a\) \(=\) \(\ds \dfrac 1 8\) arithmetic
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 8 + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 6\) substituting for $\dfrac 1 a$ in $(1)$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 b\) \(=\) \(\ds \dfrac 1 6 - \dfrac 1 8\)
\(\ds \) \(=\) \(\ds \dfrac {4 - 3} {24}\) common denominator
\(\ds \) \(=\) \(\ds \dfrac 1 {24}\) arithmetic
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 c\) \(=\) \(\ds \dfrac 1 {20}\) substituting for $\dfrac 1 b$ in $(2)$
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 c\) \(=\) \(\ds \dfrac 1 {20} - \dfrac 1 {24}\)
\(\ds \) \(=\) \(\ds \dfrac {6 - 5} {120}\) common denominator
\(\ds \) \(=\) \(\ds \dfrac 1 {120}\) arithmetic

$\blacksquare$


Sources

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