Henry Ernest Dudeney/Modern Puzzles/84 - The Three Brothers/Solution
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Modern Puzzles by Henry Ernest Dudeney: $84$
- The Three Brothers
- The discussion arose before one of the tribunals as to which of a tradesman's three sons could best be spared for service in the Army.
- "All I know as to their capabilities," said the father, "is this:
- Arthur and Benjamin can do a certain quantity of work in eight days,
- which Arthur and Charles will do in nine days,
- and which Benjamin and Charles will take ten days over."
- Of course, it was at once seen that as longer time was taken over the job whenever Charles was one of the pair,
- he must be the slowest worker.
- This was all they wanted to know, but it is an interesting puzzle to ascertain just how long each son would be required to do that job alone.
- Can you discover?
Solution
Arthur, Benjamin and Charles would take respectively $14 \tfrac {34} {49}$ days, $17 \tfrac {23} {41}$ days, and $23 \tfrac 7 {31}$ days.
Proof
Let $a, b, c$ be the rate of working in jobs per day of (respectively) Arthur, Benjamin and Charles.
Let $t_a, t_b, t_c$ be the number of days it would take (respectively) Arthur, Benjamin and Charles to do the job alone.
We have:
\(\ds a + b\) | \(=\) | \(\ds \dfrac 1 8\) | Arthur and Benjamin can do a certain quantity of work in eight days, | |||||||||||
\(\ds a + c\) | \(=\) | \(\ds \dfrac 1 9\) | which Arthur and Charles will do in nine days, | |||||||||||
\(\ds b + c\) | \(=\) | \(\ds \dfrac 1 {10}\) | and which Benjamin and Charles will take ten days over. |
and so:
\(\ds \paren {a + c} - \paren {b + c}\) | \(=\) | \(\ds \dfrac 1 9 - \dfrac 1 {10}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a - b\) | \(=\) | \(\ds \dfrac 1 {90}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a + b} - \paren {a - b}\) | \(=\) | \(\ds \dfrac 1 8 - \dfrac 1 {90}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 b\) | \(=\) | \(\ds \dfrac {45 - 4} {360}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b\) | \(=\) | \(\ds \dfrac {41} {720}\) | |||||||||||
\(\ds t_b\) | \(=\) | \(\ds \dfrac 1 b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {720} {41}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 17 \tfrac {23} {41}\) |
\(\ds \paren {a + b} - \paren {b + c}\) | \(=\) | \(\ds \dfrac 1 8 - \dfrac 1 {10}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a - c\) | \(=\) | \(\ds \dfrac 1 {40}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a + c} - \paren {a - c}\) | \(=\) | \(\ds \dfrac 1 9 - \dfrac 1 {40}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 c\) | \(=\) | \(\ds \dfrac {40 - 9} {360}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(=\) | \(\ds \dfrac {31} {720}\) | |||||||||||
\(\ds t_c\) | \(=\) | \(\ds \dfrac 1 c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {720} {31}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 23 \tfrac 7 {31}\) |
\(\ds \leadsto \ \ \) | \(\ds \paren {a + c} + \paren {a - c}\) | \(=\) | \(\ds \dfrac 1 9 + \dfrac 1 {40}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 a\) | \(=\) | \(\ds \dfrac {40 + 9} {360}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds \dfrac {49} {720}\) | |||||||||||
\(\ds t_a\) | \(=\) | \(\ds \dfrac 1 a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {720} {49}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 14 \tfrac {34} {49}\) |
Hence and so.
$\blacksquare$
Also see
- Henry Ernest Dudeney: Puzzles and Curious Problems: $138$ - The Three Workmen, exactly the same problem but with different names.
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $84$. -- The Three Brothers
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $167$. The Three Brothers