Henry Ernest Dudeney/Puzzles and Curious Problems/138 - The Three Workmen/Solution

From ProofWiki
Jump to navigation Jump to search

Puzzles and Curious Problems by Henry Ernest Dudeney: $138$

The Three Workmen
"Me and Bill," said Casey, "can do the job for you in ten days,
but give me Alec instead of Bill, and we can get it done in nine days."
"I can do better than that," said Alec. "Let me take Bill as a partner, and we will do the job for you in eight days."
Then how long would each man take over the job alone?


Solution

Alfred, Bill and Casey 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) Alec, Bill and Casey.

Let $t_a, t_b, t_c$ be the number of days it would take (respectively) Alec, Bill and Casey to do the job alone.


We have:

\(\ds b + c\) \(=\) \(\ds \dfrac 1 {10}\) "Me and Bill," said Casey, "can do the job for you in ten days,
\(\ds a + c\) \(=\) \(\ds \dfrac 1 9\) but give me Alec instead of Bill, and we can get it done in nine days.
\(\ds a + b\) \(=\) \(\ds \dfrac 1 8\) "I can do better than that," said Alec. "Let me take Bill as a partner, and we will do the job for you in eight days."


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


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Henry_Ernest_Dudeney/Puzzles_and_Curious_Problems/138_-_The_Three_Workmen/Solution&oldid=561748"