Henry Ernest Dudeney/Modern Puzzles/187 - Weighing the Goods/Solution

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Modern Puzzles by Henry Ernest Dudeney: $187$

Weighing the Goods
A tradesman whose morals had become corrupted during the war by a course of profiteering went to the length of introducing a pair of false scales.
It will be seen from the diagram that one arm is longer than the other,
though they are purposely drawn so as to give no clue as to the answer.
Dudeney-Modern-Puzzles-187.png
As a consequence, it happened that in one of the cases exhibited eight of the little packets
(it does not matter what they contain)
exactly balanced three of the canisters,
while in the other case one packet appeared to be of the same weight as $6$ canisters.
Now, as the true weight of one canister was known to be exactly one ounce, what was the true weight of the eight packets?


Solution

$8$ packets weigh $12$ ounces.


Proof

Let the arms of the balance be $x$ and $y$.

Let $P$ and $C$ ounces be the weights of package and canister respectively.


From the laws of physics and the rubric of the question:

\(\ds 3 C x\) \(=\) \(\ds 8 P y\) eight of the little packets ... exactly balanced three of the canisters,
\(\ds P x\) \(=\) \(\ds 6 C y\) one packet appeared to be of the same weight as $6$ canisters.
\(\ds \leadsto \ \ \) \(\ds P\) \(=\) \(\ds 6 C \dfrac y x\)
\(\ds 3 C\) \(=\) \(\ds 8 P \dfrac y x\)
\(\ds \leadsto \ \ \) \(\ds 3 C\) \(=\) \(\ds 8 \paren {6 C \dfrac y x} \dfrac y x\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds 16 \dfrac {y^2} {x^2}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac y x\) \(=\) \(\ds \dfrac 1 4\)
\(\ds \leadsto \ \ \) \(\ds P\) \(=\) \(\ds \dfrac {6 C} 4\)
\(\ds \) \(=\) \(\ds \dfrac {3 C} 2\)


We are given that $C = 1$.

So $P = \dfrac 3 2$ and so $8 P = 12$.

$\blacksquare$


Sources

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