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Problem of the Week
Problem C and Solution
Meal Deal

Problem

Jessica and Callista go the local burger joint. They both want to buy the meal deal. Jessica has \(\frac{3}{4}\) of the money needed to buy the meal deal and Callista has half of the money needed to buy the meal deal. If the meal deal was \(\$3\) cheaper, then together they would have exactly enough money to buy two of the meal deals.

What is the original price of the meal deal?

Solution

Solution 1:
Suppose that the cost of the meal deal, in dollars, is \(C\). Then Jessica has \(\frac{3}{4} C\) and Callista has \(\frac{1}{2} C\). Combining their money, together Jessica and Callista have \[\frac{3}{4} C + \frac{1}{2} C = \frac{3}{4} C + \frac{2}{4} C = \frac{5}{4} C\]

If the meal deal was \(\$3\) cheaper, then the cost to buy one meal deal would be \(C - 3\). If the cost of one meal deal was \(C - 3\), then the cost to buy two meal deals at this price would be \(2(C - 3) = (C-3) + (C-3) = 2C - 6\).

Combined, Jessica and Callista would have enough money to buy exactly two meal deals at this reduced price. Thus, \(2C - 6 = \frac{5}{4} C\).

Solving for \(C\), \[\begin{aligned} 2C - 6 &= \frac{5}{4} C\\ 2C - \frac{5}{4} C & = 6\\ \frac{8}{4} C - \frac{5}{4} C & = 6\\ \frac{3}{4} C &= 6\\ 3C & = 24\\ C & = 8 \end{aligned}\]

Therefore, the original price of the meal deal is \(\$8\).

Solution 2:

Since the new price of the meal deal is \(\$3\) cheaper than the original price, then the original price must be more than \(\$3\). We will use systematic trial and error to figure out the original price.

Suppose the original price of the meal deal was \(\$6\). Then the reduced price would be \(\$3\). Also, Jessica has \(\frac{3}{4}\times \$6 = \$4.50\) and Callista has \(\frac{1}{2}\times \$6 = \$3\), and in total they have \(\$4.50 + \$3 = \$7.50\). With \(\$7.50\), they could buy exactly \(7.50 \div 3 = 2.5\) meal deals at a price of \(\$3\) each.

Suppose the original price of the meal deal was \(\$12\). Then the reduced price would be \(\$9\). Also, Jessica has \(\frac{3}{4}\times \$12 = \$9\) and Callista has \(\frac{1}{2}\times \$12 = \$6\), and in total they have \(\$9 + \$6 = \$15\). With \(\$15\), they could buy \(15\div 9 \approx 1.67\) meal deals at a price of \(\$9\) each.

We can see that the original price of the meal deal lies somewhere between \(\$6\) and \(\$12\).

Let’s suppose the original price of the meal deal was \(\$8\). Then the reduced price would be \(\$5\). Also, Jessica has \(\frac{3}{4}\times \$8 = \$6\) and Callista has \(\frac{1}{2}\times \$8 = \$4\), and in total they have \(\$6 + \$4 = \$10\). With \(\$10\), they could buy exactly \(10\div 5 = 2\) meal deals at a price of \(\$5\) each.

Thus, we can see that the original price of the meal deal is \(\$8\).