System of Equations: Sorting 5¢ and 10¢ Bottles

What You Will Learn

This is a two-unknown counting problem. The skill is not the arithmetic, which is gentle, but translating one story into two different equations that each carry a different piece of information.

• Translating "how many" and "how much" separately. One equation tracks quantity (items); the other tracks value (money). Keeping those two ideas apart is the heart of every coin, ticket, and mixture problem.
• Choosing a single unit. Why $30.55 should become 3055¢ before it goes anywhere near an equation.
• Substitution as a tool. Eliminating one variable by expressing it through the other.
• Reasoning without variables. The "pretend they're all 5¢" argument and the weighted-average argument reach the same answer by pure logic, which is a powerful way to check algebra.
• Judging reasonableness. Bounding an answer between the "all cheap" and "all expensive" extremes before computing anything.

Let's Organize What We Know

Before writing any equation, sort the information into a table. Each row is a type of container; each column is a different "measurement." The unknowns get variables, and the table practically writes the equations for you.

Read the table by columns. The "How many" column must add to 405, and the "Total value" column must add to 3055. Those two column sums are our two equations.

Solution: Method 1 — The Two-Unknown System with Substitution

Step 1 — Name the unknowns

Let x be the number of 5-cent bottles and cans, and let y be the number of 10-cent bottles and cans. Writing this down in words first prevents the classic mix-up of which letter means what.

Step 2 — Write the count equation

Every container is one type or the other, so the two counts must add to the total number collected:

Step 3 — Write the value equation in cents

A 5-cent item contributes 5 cents, so x of them contribute 5x cents. Likewise y ten-cent items contribute 10y cents. The total of $30.55 is the same as 3055 cents (since $1 = 100¢, multiply 30.55 by 100).

Step 4 — Substitute

The count equation is easy to rearrange: x = 405 − y. Put that expression in place of x in the value equation. Notice that the parentheses matter, because the 5 multiplies the entire expression.

Step 5 — Back-solve for the other unknown

Now that y is known, return to the simple equation:

There are 199 five-cent items and 206 ten-cent items.

Solution: Method 2 — The "Pretend They're All 5¢" Deficit Method

You can solve this with no variables at all. This reasoning was the standard tool long before algebra notation existed, and it builds real intuition for what the system is doing.

Step 1 — Start with a deliberately wrong assumption

Imagine every one of the 405 containers is worth only 5¢. The pile would be worth:

Step 2 — Measure the shortfall

The real pile is worth 3055¢, so our pretend pile is short by:

Step 3 — Fix the shortfall one container at a time

Upgrading a single container from 5¢ to 10¢ adds exactly 5¢ to the total. To make up 1030¢ we need to upgrade:

Those 206 are the ten-cent items. The rest, 405 − 206 = 199, remain five-cent items. Same answer, no unknowns required.

Solution: Method 3 — Weighted Average (the Balance-Point View)

Here is a third lens, and it is useful whenever a problem gives you a total count and a total value. Ask: what is the average value of one container?

Step 1 — Compute the average value

The average falls between 5¢ and 10¢, which it must. It sits only slightly above the midpoint of 7.5¢, so there should be slightly more 10¢ items than 5¢ items.

Step 2 — Measure distances from the average, working in whole cents

The ratio of 5¢ items to 10¢ items equals the ratio of the distances on the opposite sides of the average. A cheaper-than-average item is balanced by an expensive-than-average one, like a seesaw. Using exact values (average = 3055/405):

Step 3 — Scale the ratio up to 405

The ratio 199 : 206 already totals 199 + 206 = 405, so the scale factor is 1. The counts are exactly 199 and 206.

Verification

We check the answer against both original statements, not against the rearranged versions we created along the way.

Here is the same answer as a picture. The bar is one row of 405 containers, cut into two parts:

Sanity Check Before You Move On

A good habit is to bracket the answer before solving. The pile's value is trapped between two extremes:

$30.55 sits well inside the $20.25 to $40.50 range, so a solution exists. It is just a hair above the "half and half" value of $30.375, so we expect the two counts to be close, with the 10¢ group a little larger. We found 199 and 206. That is exactly the pattern predicted.

Also note that every total made from 5¢ and 10¢ items is a multiple of 5 cents, and 3055 is. A total like $30.52 could never come from this pile, which is a nice quick test for whether a problem like this is even possible.

What Trips People Up

How to Spot This Problem Type

Look for these structural clues. Wording changes, but the skeleton does not:

• Two kinds of item with different per-item values (cents, dollars, points, prices, legs).
• A total count ("405 bottles and cans in all") anda total value ("worth $30.55").
• Questions that ask "how many of each."
• A number of equally disguised cousins: adult and child tickets, 2-point and 3-point baskets, nickels and dimes, chickens and rabbits.

The Pattern Behind This Problem

Any problem of this type has the same four ingredients: a total count N, a lower per-item value a, a higher per-item value b, and a total value V. Method 2 generalizes to a direct formula:

For our numbers: (3055 − 5·405) ÷ (10 − 5) = 1030 ÷ 5 = 206.

The same idea also drives the weighted-average view in Method 3, where V/N is the mean value per item. The more the mean leans toward b, the more items must be of type b. Once you see this, mixture problems, investment splits, and even "what grade do I need on the final?" questions all look like one family.

Beyond the Textbook

• Cash-drawer audits. A cashier who knows the number of bills and their total can check how many are $5s versus $10s without sorting them.
• Recycling and deposit programs. Collection centers reconcile item counts with payouts, just like this problem, only with more container categories.
• Blending in chemistry and food science. Mixing a 5% solution with a 10% solution to hit a target concentration uses exactly the weighted-average reasoning of Method 3.

A Brief History: Chickens and Rabbits

The "deficit" idea in Method 2 appears in the Sunzi Suanjing, a Chinese mathematical text from roughly the 4th–5th century. It asks: a cage holds some chickens and rabbits with 35 heads and 94 feet in total, so how many of each? Solvers assume every animal is a chicken (two feet each), then upgrade animals to rabbits at two extra feet per swap. Our bottles-and-cans problem has precisely the same logic. Such problems have survived for over 1,500 years because they are the cleanest training ground for turning a story into a pair of constraints.

Try These Variations

Each card below changes one thing about the original, and each has a full solution waiting behind the button. Try it on paper first. The sequence gets harder as it goes.