While grocery shopping, Terry has a way to approximate the total cost of his purchases. He simply approximates that each item will cost \(\$3.00\).
One day, Terry purchased \(20\) items. He purchased items that each had an actual cost of either \(\$1.00\), \(\$3.00\), or \(\$7.50\). Exactly seven of the purchased items had an actual cost of \(\$3.00\). If the total actual cost of the \(20\) items was the same as the total approximated cost, how many items had an actual cost of \(\$7.50\)?
The total approximated cost for the \(20\) items is \(20 \times \$3= \$60\). Since the total actual cost is the same as the total approximated cost, the total actual cost for the \(20\) items is \(\$60\). Since \(7\) of the items cost \(\$3.00\), it cost Terry \(7\times\$3= \$21\) to buy these items. Therefore, the remaining \(20-7 = 13\) items cost \(\$60 - \$21= \$39\).
From this point, we will continue with two different solutions.
Solution 1
In this solution, we will use systematic trial-and-error to solve the problem.
Let \(s\) represent the number of items Terry bought with an actual cost of \(\$7.50\) and \(d\) represent the number of items that Terry bought with an actual cost \(\$1.00\). Then the total cost of the \(\$7.50\) items would be \(7.5s\). Also, the total cost of the \(\$1.00\) items would be \(1d = d\). Since Terry’s total remaining cost was \(\$39\), then \(7.5s+d=39\). We also know that \(s+d=13\).
At this point we can systematically pick values for \(s\) and \(d\) that add to \(13\) and substitute into the equation \(7.5s+d=39\) to find the combination that works. (We can observe that \(s<6\) since \(7.5\times 6 = 45 > 39\). If this were the case, then \(d\) would have to be a negative number.)
Let’s start with \(s=3\). Then \(d=13-3=10\). The cost of these items would be \(7.5\times 3+ 10 =22.50+10= \$32.50\), which is less than \(\$39\).
So let’s try \(s=4\). Then \(d=13-4=9\). The cost of these items would be \(7.5\times 4+9=30 + 9 = \$39\), which is the amount we want.
Therefore, Terry purchased \(4\) items that cost \(\$7.50\).
Solution 2
In this solution, we will use algebra to solve the problem.
Let \(s\) represent the number of items that cost \(\$7.50\). Therefore, \((13-s)\) represents the number of items that cost \(\$1.00\). Also, the total cost of the \(\$7.50\) items would be \(7.5s\), the total cost of the \(\$1.00\) items would be \(1\times (13-s) = 13- s\), and the total of these two is \(7.5s + 13 - s = 6.5s+13\).
Since Terry’s total remaining cost was \(\$39.00\), we must have \[\begin{aligned} 6.5s + 13&=39\\ 6.5s + 13 - 13 &= 39 - 13 \\ 6.5s &= 26\\ \frac{6.5s}{6.5}& = \frac{26}{6.5}\\ s & = 4 \end{aligned}\]
Therefore, Terry purchased \(4\) items that cost \(\$7.50\).