Cfrgtkky

I'm stuck with a question from school, not even the teacher knew this one:

Question: Ben goes to the market and pays $12 for apples (amount unknown), but they were so small that the cashier gave Ben two apples for free. In this way, the price for a dozen apples dropped by exactly $1. How many apples did Ben get for $12?

Let's define

$p$ - regular prize of the apples per apple

$p^*$ - reduced prize of the apples per apple

$n$ - number of apples originally bought

Then we have the following equations:

$ncdot p=12$ - he payed 12 dollars

$(n+2)cdot p^*=12$ - he payed 12 dollars for two more apples with the fictional reduced prize

$pcdot 12 = p^*cdot12+1$ - the price for a dozen apples is reduced by 1

Solving these equations leads to a quadatric equation which has two possible solutions $n=16$ or $n = -18$.

Please feel free to comment if you like to see the way to actually solve the equations not just the set up.

Sketch

• Let $x$ be the number of apples and $p$ the price per apple. In this case, $12p$ is the price per dozen.




• Ben received $x+2$ apples at a price of $xp=12$. Therefore, the price per apple is $frac{xp}{x+2}$. The cost for a dozen is now $frac{12xp}{x+2}$.




• The $$1$ price drop means that
  $$
  12p=frac{xp}{x+2}+1
  $$

Now, combine this equation with $xp=12$ and solve. Can you take it from here?

Let the original price of each apple be x.

Let $n$ be the number of apples Ben got before getting the $2$ apples for free.

$begin{equation}label{eq:a}tag{1}therefore nx=$12end{equation}$

Now Ben got $n+2$ apples.

$therefore$ The new price per apple will be $dfrac{12}{n+2}=x'(say)$

given that $12x-12x'=1$ (difference given is $$1$)

Now from equation $(1)$ $x=dfrac{12}{n}$

solving we get $n=16$