Henry Ernest Dudeney/Puzzles and Curious Problems/32 - Apple Transactions/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $32$

Apple Transactions
A man was asked what price per $100$ he paid for some apples, and his reply was as follows:
"If they had been $4 \oldpence$ more per $100$ I should have got $5$ less for $10 \shillings$"
Can you say what was the price per $100$?


Solution

$8 \shillings$


Proof

Let $p$ be the price per apple in pence.

Let $n$ be the number of apples that were bought for $10 \shillings$, that is, $120 \oldpence$

Then we have:

\(\text {(1)}: \quad\) \(\ds n\) \(=\) \(\ds \dfrac {120} p\) the number of apples that can be bought for $120 \oldpence$
\(\text {(2)}: \quad\) \(\ds \paren {p + \dfrac 4 {100} } \paren {n - 5}\) \(=\) \(\ds 120\) "If they had been $4 \oldpence$ more per $100$ I should have got $5$ less for $10 \shillings$"
\(\ds \leadsto \ \ \) \(\ds \paren {p + \dfrac 4 {100} } \paren {\dfrac {120} p - 5}\) \(=\) \(\ds 120\) substituting for $n$ in $(2)$
\(\ds \leadsto \ \ \) \(\ds \paren {\dfrac {100 p + 4} {100} } \paren {\dfrac {120 - 5 p} p}\) \(=\) \(\ds 120\) common denominators
\(\ds \leadsto \ \ \) \(\ds \paren {100 p + 4} \paren {120 - 5 p}\) \(=\) \(\ds 120 \times 100 p\) multiplying through by $100 p$
\(\ds \leadsto \ \ \) \(\ds 100 \times 120 p + 4 \times 120 - 500 p^2 - 20 p\) \(=\) \(\ds 120 \times 100 p\) multiplying out
\(\ds \leadsto \ \ \) \(\ds 25 p^2 + p - 24\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds \dfrac {-1 \pm \sqrt {1^2 + 4 \times 25 \times 24} } {2 \times 25}\) Quadratic Formula
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds \dfrac {-1 \pm 49} {50}\) simplifying
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds \dfrac {96} {100} \text { or } -1\) simplifying

Only the positive root is appropriate here.

Hence:

$p = \dfrac {96} {100}$

That is, the price for $100$ apples is $96 \oldpence$, which works out as $\dfrac {96} {12} = 8 \shillings$

$\blacksquare$


Sources

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