# Problem of the Week Problem E and Solution Coffee Sales

## Problem

For the months of April, May and June, Coffee Only sold coffee for $$\2.50$$ per cup.

In May, they sold $$y\%$$ more cups of coffee than in April, where $$y\geq 0$$. In June, they sold $$y\%$$ fewer cups of coffee than in May.

Their records for sales in May were misplaced. They sold $$\31\,250$$ worth of coffee in April. In June they sold $$\30\,800$$ worth of coffee.

Determine the total value of the coffee they sold in May.

## Solution

Since the company had sales of $$\31\,250$$ in April, the amount of coffee sold in April was $$\frac{31\,250}{2.50} = 12\,500$$ cups.

Since the company had sales of $$\30\,800$$ in June, the amount of coffee sold in June was $$\frac{30\,800}{2.50} = 12\,320$$ cups.

In May, Coffee Only sold $$y\%$$ more cups of coffee than in April. In other words, they sold $12\,500 + 12\,500\left(\frac{y}{100}\right) = 12\,500\left(1+\frac{y}{100}\right)$ cups of coffee in May.

In June, they sold $$y\%$$ fewer cups of coffee than in May. In other words, they sold $\left[12\,500\left(1+\frac{y}{100}\right)\right] - \left [12\,500\left(1+\frac{y}{100}\right)\right]\left(\frac{y}{100}\right) = \left[12\,500\left(1+\frac{y}{100}\right)\right]\left(1-\frac{y}{100}\right)$ cups of coffee in June.

We also know that Coffee Only sold $$12\,320$$ cups of coffee in June. Therefore, \begin{align} 12\,500\left(1+\frac{y}{100}\right)\left(1-\frac{y}{100}\right) &=12\,320\\ \left(1+\frac{y}{100}\right)\left(1-\frac{y}{100}\right) &= \frac{616}{625}\\ 1-\frac{y^2}{10\,000} &= \frac{616}{625}\\ \frac{y^2}{10\,000} &= \frac{9}{625}\\ y^2 &= 144 \end{align} Since $$y\geq 0$$, we have $$y=12$$. Therefore, in May Coffee Only sold $12\,500\left(1+\frac{y}{100}\right) = 12\,500\left(1+\frac{12}{100}\right) = 14\,000$ cups of coffee. The total value of the coffee sold in May was $$14\,000 \times \2.50= \35\,000$$.