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The problem is as follows:

The cook said "I paid $12$ usd for the apples I bought" and they gave me two apples because I claimed that the apples were too small, that made me pay one dollar less per dozen. How many apples did the cook bring?

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{16 apples}\\ 2.&\textrm{18 apples}\\ 3.&\textrm{12 apples}\\ 4.&\textrm{15 apples}\\ \end{array}$

This problem is the typical word problem regarding a linear equation but I'm confused because I don't know the right intepretation.

The way how I put the equations is as follows.

The labels are $x$ the number of apples and $y$ the price of each apple.

Thus:

$12+2y=xy$

Because it mentions that a dozen price was one dollar less.

$y=12(y-1)$

The replacing this in the earlier equation yields.

$x=13$

This means that she ended up with $13$ plus the two additional apples that she had thus is $15$.

To me the most confusing part was to assume:

$y=12(y-1)$

I don't know if this is correct, thus I'd like this to be discussed in the answer please. The reason for which I am confused is that,

I think that the new prize of the apples (which ended up being used is one dollar less).

I don't know if this is the right answer, or if my math was correctly used. Hence I would like someone could answer in a more logical way because I'm confused.

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3 Answers 3

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Here is how I would solve it. Say the cook came back home with $x$ apples. But they bought only $(x-2)$ apples and paid $12$ dollars for it. The other two apples came free.

So original price per apple $\displaystyle = \frac{12}{x-2}$

But since they finally got $x$ apples, now the effective price per apple is $ = \displaystyle \frac{12}{x}$

For each apple, they paid $\displaystyle \frac{1}{12}$ dollar less ($1$ dollar less per dozen).

So, $\displaystyle \frac{12}{x-2} - \frac{12}{x} = \frac{1}{12}$

$\implies x^2-2x = 288$

Or, $(x-1)^2 = 289 = 17^2 \implies x = 18$.

So the cook brought $18$ apples.

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  • $\begingroup$ I think your approach its easier to follow than the other two as you get the answer straight. But you have to use fractions for that. $\endgroup$ Mar 4, 2021 at 5:28
  • $\begingroup$ Chris, as you mentioned you were confused about it, I felt the best way was to show that it was easier to think everything in terms of unit apple. Original price of one apple, effective price of one apple and how much less was paid on each apple. $\endgroup$
    – Math Lover
    Mar 4, 2021 at 5:52
  • $\begingroup$ You finally get a quadratic equation to solve which seems straightforward. $\endgroup$
    – Math Lover
    Mar 4, 2021 at 5:56
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Let $p$ be the price per dozen. After the reduction we have that

the number of apples is $$\left(\frac{12}{p-1}\right)\cdot 12=\frac{144}{p-1}$$ which is the number of apples they had got without the reduction plus $2$ apples. $$\left(\frac{12}{p}\right)\cdot 12+2=\frac{144}{p}+2$$ We have the equation $$\frac{144}{p-1}=\frac{144}{p}+2\to (p=-8);\;p=9$$ So they payed the apples $\$8$ per dozen, as they spent $\$12$ they got $144/8=18$ apples.

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Let $c$=original cost per dozen in dollars

The cook paid $c-1$ so the number of apples (in dozens) that were bought was $\frac{12}{c-1}$

the number of dozens that would have been bought at the original price was $\frac{12}{c}$

the number from the two ways of buying should be the same

$\frac{12}{c-1} = \frac{12}{c} +\frac{2}{12}$

solving this gives $c=9$ and so 12/8 or 1.5 dozen apples, which is 18.

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  • $\begingroup$ I believe that a way to understand your answer is that in the equation you're adding dozens, hence you must use $\frac{2}{12}$ not just $2$ because that would not be adding dozens isn't it?. $\endgroup$ Mar 4, 2021 at 5:27
  • $\begingroup$ Yes, that's right, the 2 extra apples are $\frac{2}{12}$ of a dozen. $\endgroup$ Mar 4, 2021 at 19:49

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