 # 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 $$\8$$.