# Problem of the Week Problem C and Solution Gone Shopping

## Problem

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 $$\7.50$$?

## Solution

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 $$\7.50$$.