Greek Anthology Book XIV: 7. - Problem/Variant
Jump to navigation
Jump to search
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
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Light Reflected off a Mirror: $20$