Henry Ernest Dudeney/Puzzles and Curious Problems/295 - Blending the Teas/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $295$

Blending the Teas
A grocer buys two kinds of tea --
one at $2 \shillings 8 \oldpence$ per pound,
and the other, a better quality, at $3 \shillings 4 \oldpence$ per pound.
He mixes together some of each, which he proposes to sell at $3 \shillings 7 \oldpence$ a pound,
and so make a profit of $25$ per cent on the cost.
How many pounds of each kind must he use to make a mixture of $100$ pounds weight?


Solution

$70$ pounds of cheaper tea and $30$ pounds of the more expensive one.


Proof

To simplify the analysis, we express all prices in pence:

$2 \shillings 8 \oldpence$ per pound is $2 \times 12 + 8 = 32 \oldpence$ per pound
$3 \shillings 4 \oldpence$ per pound is $3 \times 12 + 4 = 40 \oldpence$ per pound
$3 \shillings 7 \oldpence$ per pound is $3 \times 12 + 7 = 43 \oldpence$ per pound
$25 \%$ profit on $32 \oldpence$ per pound is $8 \oldpence$, hence the cheaper brand will be selling at $40 \oldpence$ per pound
$25 \%$ profit on $40 \oldpence$ per pound is $10 \oldpence$, hence the dearer brand will be selling at $50 \oldpence$ per pound.


Let $a$ be the weight of cheaper tea going into making the blend.

Let $b$ be the weight of dearer tea going into making the blend.


We have:

\(\text {(1)}: \quad\) \(\ds a + b\) \(=\) \(\ds 100\) to make a mixture of $100$ pounds weight
\(\text {(2)}: \quad\) \(\ds 40 a + 50 b\) \(=\) \(\ds 4300\)
\(\ds \leadsto \ \ \) \(\ds 40 \paren {100 - b} + 50 b\) \(=\) \(\ds 4300\)
\(\ds \leadsto \ \ \) \(\ds 10 b\) \(=\) \(\ds 4300 - 4000\)
\(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds 30\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds 70\)

$\blacksquare$


Sources

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