Systems of Equations: Lemonade Stand Cup Problem

What You Will Learn

• Two quantities, two equations. The problem gives you a total count and a total amount. Each needs its own equation, and keeping these two ideas separate is the central skill.
• Turning "per-item size × number of items" into algebra. The term 12s means "12 ounces per cup, times s cups," which is total ounces from small cups.
• Substitution as a tool. When one equation says "these two add to a total," you can rewrite one variable in terms of the other and reduce the problem to one unknown.
• Reasoning without algebra. The "assume everything is small" trick and the weighted-average view reach the same answer using only arithmetic, and they explain why the algebra works.
• Checking against context. Counts of cups must be whole, non-negative numbers. A result like 12.5 cups tells you an equation is wrong.

Picture This

Before writing any equations, it helps to organize what the problem actually gives you. Two cup sizes, two totals:

Read the "Total" row across: the "Number sold" column must add to 60, and the "Ounces from this size" column must add to 920. Those are the two equations, already waiting for us.

Solution: Method 1 — Substitution, One Variable at a Time

This is the standard classroom approach. We write the two facts from the table as equations, use the simpler one to eliminate a variable, and solve what's left.

Step 1 — Define the variables

Let s = the number of small (12 oz) cups sold and L = the number of large (16 oz) cups sold. Choosing letters that remind you of the quantity ("s" for small, "L" for large) helps avoid mix-ups later.

Step 2 — Write the cup-count equation

Salma sold 60 cups in total, so the small cups plus the large cups must equal 60:

Step 3 — Write the total-ounces equation

Each small cup contributes 12 ounces, so s small cups contribute 12s ounces. Each large cup contributes 16 ounces, so L large cups contribute 16L ounces. Together they total 920 ounces:

Step 4 — Solve the count equation for one variable

The first equation is the easiest to rearrange, since neither variable has a coefficient to divide out. Solving for L:

Step 5 — Substitute into the ounces equation

Replace L in 12s + 16L = 920 with (60 − s). The parentheses matter, because the 16 must multiply the entire expression:

Step 6 — Find the other variable

Go back to the simple equation: L = 60 − s = 60 − 10 = 50.

So Salma sold 10 small cups and 50 large cups.

Solution: Method 2 — The "Pretend They're All Small" Approach

You can solve this problem with arithmetic alone, no variables, by running an imaginary experiment. This is the same idea as the algebra above, stripped to its core.

Step 1 — Imagine all 60 cups were small

If every cup held 12 ounces, the total would be 60 × 12 = 720 ounces.

Step 2 — Measure the gap

The real total is 920 ounces, which is 920 − 720 = 200 ounces more than our imaginary total. Those 200 "extra" ounces have to come from somewhere.

Step 3 — Count what one swap adds

Replacing one small cup with one large cup adds 16 − 12 = 4 ounces. Every large cup in reality accounts for exactly one such swap.

Step 4 — Divide to find the number of swaps

200 ÷ 4 = 50 swaps, meaning 50 large cups. The remaining 60 − 50 = 10 cups are small.

The table below makes the idea tangible. Start at the bottom (all small, 720 oz) and move up one row at a time. Each block of 10 swaps adds 40 ounces:

Notice that the total climbs in a perfectly steady pattern. That steadiness is exactly what a linear system describes: every change of one cup between sizes shifts the total by a fixed 4 ounces.

Solution: Method 3 — The Weighted-Average (Balance Point) Approach

Here's another way to look at the data: what was the average cup size? The total ounces divided by the number of cups is

That average must lie between 12 and 16. It acts as a balance point on the number line, and it sits much closer to 16 than to 12, which tells us large cups dominate.

The average sits 3⅓ oz from the small cup size but only ⅔ oz from the large cup size.

Step 1 — Measure the distances to the average

From 12 to 15⅓ is 10/3. From 15⅓ to 16 is 2/3.

Step 2 — Flip the ratio

On a balance beam, the side with more weight sits closer to the pivot. So the counts are in the reverse ratio of the distances:

Step 3 — Split the 60 cups

A 1 : 5 ratio means 6 equal parts, each worth 60 ÷ 6 = 10 cups. That gives 1 part small = 10 small and 5 parts large = 50 large.

This is the same technique used to blend solutions of different strengths in chemistry (the "alligation" method), and it works whenever you know an overall average and two component values.

Verification

Substitute s = 10 and L = 50 into both original equations. Checking only one of them can hide an error.

Both statements from the problem hold true, and both cup counts are whole, positive numbers, as they must be for physical cups. As an independent check, the three methods above (substitution, all-small baseline, and weighted average) arrived at the same pair of numbers by completely different routes.

Sanity Check

Could the answer have been, say, 30 small and 30 large? Look back at the table: an even split gives only 840 ounces. The actual total of 920 is only 40 ounces short of the maximum possible 960 (all large), so the mix must be heavily large-cup.

The average cup of 15⅓ ounces tells the same story: it's nearly 16. Any answer with more small cups than large cups should immediately look suspicious. This quick estimate catches the most common mistake in this problem, which is swapping the two sizes.

Common Pitfalls

How to Spot This Problem Type

This is a two-types, two-totals problem, sometimes called a "mixture-style" or "total-count and total-value" problem. Look for these signals:

• Two categories of item, each with a fixed "per item" amount (ounces per cup, dollars per ticket, cents per coin).
• A total number of items, and a total amount (volume, money, weight, points).
• A question asking for "how many of each."

The same structure hides in other costumes:

The Pattern Behind This

For any problem with two item types of sizes a and b, a total count N, and a total amount T:

The formula for y is Method 2 in symbols: a·N is the "all type-a" baseline, T − a·N is the excess, and b − a is the gain per swap. With our numbers: (920 − 12·60) ÷ (16 − 12) = 200 ÷ 4 = 50.

The broader idea is that a pair of linear equations describes two constraints that must hold at once. Here, every point on the line s + L = 60 uses 60 cups, and every point on 12s + 16L = 920 pours 920 ounces. Only one point, (10, 50), sits on both lines.

Beyond the Textbook

• Blending and pharmacy: Mixing a 12% solution with a 16% solution to hit a target concentration is Method 3 exactly. Pharmacists and chemists call it alligation.
• Inventory and logistics: A warehouse knows the total number of boxes on a truck and the total weight, and must work out how many are heavy versus light.
• Event pricing: Ticket sales totals (adult vs. student) are routinely reconstructed from a headcount and revenue, the same two-equation structure.

Push Further

Try each problem on your own first, then click to compare your work. They get harder as you go.