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