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.
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 \(14\,000 \times \$2.50= \$35\,000\).