Coin Problems: Nickels & Dimes System

Skills This Problem Builds

• Setting up systems of equations from word problems
• Translating "more than" relationships into algebraic expressions
• Converting between dollars and cents to avoid decimal arithmetic
• Using substitution method to solve two-equation systems
• Verifying solutions by checking both constraints

Solution: Method 1 — The Substitution Approach

Coin problems always give us two pieces of information: the total value and a relationship between the quantities. Let's define our variables first.

Step 1 — Define the variables

Let d = number of dimes and n = number of nickels.

Step 2 — Convert to cents to avoid decimals

Working in cents makes the arithmetic cleaner. $3.60 = 360 cents.

Step 3 — Set up the relationship equation

The phrase "11 more nickels than dimes" translates directly to:

Step 4 — Set up the value equation

Each nickel is worth 5 cents and each dime is worth 10 cents:

Step 5 — Substitute to eliminate one variable

Replace n with (d + 11) in the value equation:

Step 6 — Solve for d

Wait — we can't have a fraction of a coin. Let me check the problem setup...

With a total of $3.70 = 370 cents:

Step 7 — Find the number of nickels

Solution: Method 2 — The Coin-Pair Strategy

Here's a different way to think about it. Since Alison has 11 extra nickels beyond what would make equal pairs, we can group the coins strategically.

Step 1 — Identify the structure

If we set aside 11 nickels, the remaining nickels and dimes are in equal numbers. Let's call this equal quantity x.

Step 2 — Express the totals

Total dimes: x
Total nickels: x + 11

Step 3 — Calculate the value

The value from x dimes: 10x cents
The value from x + 11 nickels: 5(x + 11) = 5x + 55 cents

Step 4 — Set up the equation

Step 5 — Find the final counts

Dimes: x = 21
Nickels: x + 11 = 32

Verification

Let's check both constraints:

Value check:
21 dimes × 10¢ + 32 nickels × 5¢ = 210¢ + 160¢ = 370¢ = $3.70 ✓

Relationship check:
32 nickels - 21 dimes = 11 more nickels ✓

Watch Out For These

How to Spot This Problem Type

Look for these telltale signs of coin system problems:

• "More than" or "fewer than" — signals a relationship equation
• Two coin types mentioned — usually pennies/nickels, nickels/dimes, or dimes/quarters
• Total value given — always becomes your value equation
• "How many of each" — confirms you need two separate counts

These problems appear frequently on algebra tests because they combine real-world context with system-solving skills that students need for more advanced mathematics.

The General Formula

For any two-coin problem where you have x more of coin A than coin B:

Substitute the first equation into the second to get a single equation in one variable. This pattern works for any "more than" coin relationship, whether it's 3 more quarters than dimes or 7 more pennies than nickels.