Cup Sales Problem Solver

Visualize the relationship between "Sets" (Buy 5 Get 1 Free) and "Loose" cups to find the original number of cups in the crate.

Adjust Sales

Sets Sold (Buy 5, Get 1 Free) 10 sets

10 sets × 6 cups = 60 cups total (Only pay for 50)

Loose Cups Sold 30 cups

30 cups × 1 = 30 cups total (Pay for all)

Results

Total Revenue

$0.00

Target: $1174.80

Distribution Check

Sets

Current Sets: 0 cups (0%) Target: 66.7%

Total Cups in Crate

0.0

Visual Breakdown

Sets Pile (2/3)

10 Sets × $44.50

Loose Pile (1/3)

30 Cups × $8.90

The Math Behind It

We can treat this as a "Unit" problem. Let's find the smallest group of cups that satisfies the 2/3 vs 1/3 ratio.

Constraint: For every 2 cups in the "Sets" pile, there is 1 cup in the "Loose" pile.
A single "Set" contains 6 cups. To balance the ratio (2:1), we need 3 loose cups for every 1 set (because 6 cups : 3 cups = 2 : 1).
The "Magic Unit": 1 Set (6 cups) + 3 Loose Cups = 9 Cups Total.
Revenue per Unit: (1 Set × $44.50) + (3 Loose × $8.90) = $71.20.
Total Units: $1174.80 (Total Revenue) ÷ $71.20 (Unit Revenue) = 16.5 Units.
Total Cups: 16.5 Units × 9 Cups/Unit = 148.5 Cups.

Note: The answer results in a decimal (148.5 cups), which physically implies a "half cup" or "half set" was sold. In real life, this might mean a data entry error in the question, but the algebraic method remains valid.