Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales/Solution 2

From ProofWiki
Jump to navigation Jump to search

Modern Puzzles by Henry Ernest Dudeney: $186$

The False Scales
A pudding, when placed into one of the pans of a balance, appeared to weigh $4$ ounces more than $\tfrac 9 {11}$ of its true weight,
but when placed into the other pan it appeared to weigh $3$ pounds more than in the first pan.
What was its true weight?


Solution

Suppose that the false balance has unequal arms.

That is, the arms are of different length.

Then the true weight of the pudding is $143$ ounces.


Proof

Let $W$ ounces be the true weight of the pudding.

We are given that:

in one pan the apparent weight was $\dfrac {9 W} {11} + 4$
in the other pan the apparent weight was $\dfrac {9 W} {11} + 52$.


Hence we have:

\(\ds W^2\) \(=\) \(\ds \paren {\dfrac {9 W} {11} + 4} \paren {\dfrac {9 W} {11} + 52}\) True Weight from False Balance with Unequal Arms
\(\ds \leadsto \ \ \) \(\ds 121 W^2\) \(=\) \(\ds \paren {9 W + 44} \paren {9 W + 572}\) simplifying
\(\ds \) \(=\) \(\ds 81 W^2 + 396 W + 5148 W + 25 \, 168\) simplifying
\(\ds \leadsto \ \ \) \(\ds 40 W^2 - 5544 W - 25 \, 168\) \(=\) \(\ds 0\) rearranging into the form of a quadratic equation
\(\ds \leadsto \ \ \) \(\ds 5 W^2 - 693 W - 3146\) \(=\) \(\ds 0\) dividing both sides by $8$
\(\ds \leadsto \ \ \) \(\ds W\) \(=\) \(\ds \dfrac {693 \pm \sqrt {693^2 + 4 \times 5 \times 3146} } {2 \times 5}\) Quadratic Formula
\(\ds \) \(=\) \(\ds \dfrac {693 \pm \sqrt {543 \, 169} } {10}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {693 \pm 737} {10}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {1430} {10}\) taking the positive square root to make the answer positive

Hence the result.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Henry_Ernest_Dudeney/Modern_Puzzles/186_-_The_False_Scales/Solution_2&oldid=595158"