# Algebra

A boy was sent with \$2.10 to buy oranges.
He found the price 3 cents higher per dozen than he had expected to pay, and so he bought 6 fewer oranges than he had intended to buy and received 1 cent in change. How much did he pay for a dozen oranges?

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1. anticipated cost/doz --- x
anticipated number of dozen bought = 210/x

actual cost/doz = x+3
number of dozen he bought = 210/(x+3)

210/x - 210/(x+3) = 1/2
multiply each term by 2x(x+3)
420(x+3) - 420x = x^2 + 3x

x^2 + 3x - 1260 = 0
x = appr 34.03 or some negative value

210/34 = 6.1...
so he probably intended to buy 6 dozen
that would have been 210/6 or 35 cents/doz

But at 3 cents/doz more would have made it 38 cents/doz
so he could only buy 5.5 dozen
5.5(38) = \$ 2.09 , leaving him with 1 penny change.

As you can see, the numbers did not come out nice from the equation, but after "fudging" them and using only "half dozen" multiples, it worked out

Must be using a textbook from the 1920's
Today you couldn't even get 1 orange for 35 cents

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2. If he intended to buy x dozen, the price was evidently 210/x cents/dozen. Instead,

(x-1/2)(210/x + 3) = 209
x = 6

So, the price per dozen was expected to be \$.35

Instead, he wound up paying \$.38/doz, meaning he bought 5 1/2 dozen, which cost \$2.09

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