Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales/Solution 2
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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,
- 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:
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
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $186$. -- The False Scales
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $97$. The False Scales