Henry Ernest Dudeney/Puzzles and Curious Problems/70 - The Donkey Cart/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $70$

The Donkey Cart
Atkins, Brown and Cranby had to go an journey of $40$ miles.
Atkins could walk $1$ mile an hour,
Brown could walk $2$ miles an hour,
and Cranby could go in his donkey-cart at $8$ miles an hour.
Cranby drove Atkins a certain distance, and, dropping him to walk the remainder,
drove back to meet Brown on the way and carried him to their destination,
where they all arrived at the same time.
How long did the journey take?
Of course, each went at a uniform rate throughout.


Solution

The whole journey takes $10 \tfrac 5 {41}$ hours.


Proof

Let Atkins, Brown and Cranby be denoted by $A$, $B$ and $C$ respectively.

We have that:

$C$ sets off with $A$
drops him off $d_1$ miles from the start
then returns to pick up $B$ at a point $d_2$ miles from the start.

Let $t_1$ be the time $C$ and $A$ reach $d_1$.

Let $t_2$ be the time $C$ reaches $d_2$ to pick up $B$.

Let $t$ be the time they all arrive at their destination.


We have:

\(\text {(1)}: \quad\) \(\ds t_1\) \(=\) \(\ds \dfrac {d_1} 8\) $C$ sets off with $A$
\(\text {(2)}: \quad\) \(\ds t_2 - t_1\) \(=\) \(\ds \dfrac {d_1 - d_2} 8\) $C$ returns to pick up $B$
\(\text {(3)}: \quad\) \(\ds t - t_2\) \(=\) \(\ds \dfrac {40 - d_2} 8\) $C$ travels to destination with $B$
\(\text {(4)}: \quad\) \(\ds t_2\) \(=\) \(\ds \dfrac {d_2} 2\) $B$ sets off walking to $d_2$
\(\text {(5)}: \quad\) \(\ds t - t_1\) \(=\) \(\ds \dfrac {40 - d_1} 1\) $A$ walks from $d_1$ to the final destination
\(\ds \leadsto \ \ \) \(\ds \dfrac {d_2} 2 - \dfrac {d_1} 8\) \(=\) \(\ds \dfrac {d_1 - d_2} 8\) substituting for $t_1$ and $t_2$ from $(1)$ and $(4)$ into $(2)$
\(\text {(6)}: \quad\) \(\ds \leadsto \ \ \) \(\ds d_2\) \(=\) \(\ds \dfrac {2 d_1} 5\) simplifying
\(\text {(7)}: \quad\) \(\ds \leadsto \ \ \) \(\ds t - \dfrac {d_1} 8\) \(=\) \(\ds 40 - d_1\) substituting for $t_1$ from $(1)$ into $(5)$
\(\text {(8)}: \quad\) \(\ds \leadsto \ \ \) \(\ds t - \dfrac {d_2} 2\) \(=\) \(\ds \dfrac {40 - d_2} 8\) substituting for $t_2$ from $(4)$ into $(3)$
\(\text {(9)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 40 - d_1 + \dfrac {d_1} 8\) \(=\) \(\ds \dfrac {40 - d_2} 8 + \dfrac {d_2} 2\) eliminating $t$ from between $(7)$ and $(8)$
\(\ds \leadsto \ \ \) \(\ds 8 \paren {40 - d_1} + d_1\) \(=\) \(\ds \paren {40 - d_2} + 4 d_2\) multiplying by $8$ to clear fractions
\(\text {(10)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 280\) \(=\) \(\ds 7 d_1 + 3 d_2\) simplifying
\(\ds \) \(=\) \(\ds \paren {7 + \dfrac 6 5} d_1 = \dfrac {41} 5 d_1\) substituting for $d_2$ from $(6)$ into $(10)$
\(\ds \leadsto \ \ \) \(\ds d_1\) \(=\) \(\ds \dfrac {5 \times 280} {41} = \dfrac {1400} {41} = 34 \tfrac 6 {41}\) simplifying


Then we have:

\(\ds d_2\) \(=\) \(\ds \dfrac 2 5 \times \dfrac {1400} {41} = \dfrac {560} {41} = 13 \tfrac {27} {41}\) from $(6)$: $d_2 = \dfrac {2 d_1} 5$
\(\ds t\) \(=\) \(\ds 40 - \dfrac {1400} {41} + \dfrac 1 8 \times \dfrac {1400} {41}\) from $(5)$ and $(1)$: $t = 40 - d_1 + \dfrac {d_1} 8$
\(\ds \) \(=\) \(\ds \dfrac {1640 - 1400 + 175} {41}\)
\(\ds \) \(=\) \(\ds \dfrac {415} {41} = 10 \tfrac 5 {41}\)

$\blacksquare$


Sources

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