Equation: Apple problem 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?
 A: 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:
$n\cdot p=12$ - he payed 12 dollars
$(n+2)\cdot p^*=12$ - he payed 12 dollars for two more apples with the fictional reduced prize
$p\cdot 12 = p^*\cdot12+1$ - 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.
A: 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?
A: 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.
$\begin{equation}\label{eq:a}\tag{1}\therefore nx=\$12\end{equation}$ 
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 $n=16$ 
