Dimes and Nickels: Solving a Coin Problem

Question

PROBLEM

Thirty coins, all dimes and nickels, are worth $2.60. How many nickels are there?

What This Problem Teaches

Setting up systems of equations from real-world constraints
Converting between dollars and cents to avoid decimal arithmetic
Recognizing that coin problems always yield whole number solutions
Using substitution method to solve 2×2 systems
Verification strategies for multi-constraint problems

Let's Draw It

Before we dive into the algebra, let's visualize what we're working with:

Thirty coins, all dimes and nickels, are worth $2.60. How many nickels are there?

This diagram shows our setup: we have some number of nickels (worth 5¢ each) plus some number of dimes (worth 10¢ each), and we know the total count is 30 and the total value is 260 cents.

Solution: Method 1 — System of Equations Approach

The standard approach treats this as a system with two unknowns and two constraints.

Step 1 — Define the variables

Let n = number of nickels and d = number of dimes. We're asked to find n.

Step 2 — Write the coin count equation

The total number of coins is 30:

n + d = 30

Step 3 — Write the value equation

Convert $2.60 to 260 cents to avoid decimals. Each nickel contributes 5 cents, each dime contributes 10 cents:

5n + 10d = 260

Step 4 — Solve using substitution

From the first equation: d = 30 - n. Substitute this into the value equation:

5n + 10(30 - n) = 260
5n + 300 - 10n = 260
-5n + 300 = 260
-5n = -40
n = 8

Step 5 — Find the number of dimes

Using n = 8 in the first equation: d = 30 - 8 = 22

Solution: Method 2 — "All Dimes" Assumption

This approach uses a clever shortcut: assume all coins are one type, then adjust.

Step 1 — Imagine all coins were dimes

If all 30 coins were dimes, their total value would be:

30 × 10¢ = 300¢

Step 2 — Compare to actual value

The actual value is 260¢, so our assumption gives us:

300¢ - 260¢ = 40¢ too much

Step 3 — Calculate the adjustment needed

Each time we replace a dime (10¢) with a nickel (5¢), we reduce the total value by 5¢. To reduce by 40¢:

40¢ ÷ 5¢ per replacement = 8 replacements

So we need 8 nickels and 22 dimes.

There are 8 nickels in the collection.

Verification

Let's check our answer against both constraints:

Coin Type	Count	Value Each	Total Value
Nickels	8	5¢	40¢
Dimes	22	10¢	220¢
Total	30	—	260¢ = $2.60

✓ Count check: 8 + 22 = 30 coins
✓ Value check: 40¢ + 220¢ = 260¢ = $2.60

Watch Out For These

✗ Mixing dollars and cents: Using 5n + 10d = 2.60 leads to messy decimal arithmetic. Always convert to cents first: 5n + 10d = 260.

✗ Setting up the value equation incorrectly: Writing "n + d = 2.60" confuses the count constraint with the value constraint. The value equation must account for each coin's worth.

✗ Forgetting to answer the right question: The problem asks for nickels, not dimes. Always re-read what the question is asking for before you write your final answer.

✗ Getting a fractional answer and continuing: If your algebra gives you 7.3 nickels, stop and check your work. You can't have fractional coins in real life — there's an error somewhere.

The Pattern Behind This

All coin problems with two types follow this structure:

If x = number of first coin type, y = number of second coin type:

Count equation: x + y = total coins
Value equation: (value₁)x + (value₂)y = total value in cents

Key insight: The "all one type" method works because it exploits the difference in coin values. If all coins had the same value, we wouldn't have enough information to solve the problem. The fact that nickels and dimes are worth different amounts is what makes the system solvable.

How to Spot This Problem Type

Coin problems have distinctive markers that signal the approach:

"All [coin type] and [coin type]" — tells you there are exactly two types
A total count — gives you the first equation
A total value — gives you the second equation
"How many..." — asks for one of the unknowns

Variations you might see: "pennies and quarters," "stamps worth 37¢ and 60¢," "tickets costing $8 and $12." The structure is always the same: two types, total count, total value, find one quantity.

What If?

1

Different Total Value

Twenty-four coins, all dimes and nickels, are worth $2.00. How many nickels are there?

Step 1 — Set up the system

Let n = nickels, d = dimes. Then: n + d = 24 and 5n + 10d = 200 cents.

Step 2 — Use substitution

From the first equation: d = 24 - n. Substituting: 5n + 10(24 - n) = 200

Step 3 — Solve for n

5n + 240 - 10n = 200, so -5n = -40, giving n = 8

Step 4 — Verify

8 nickels and 16 dimes. Check: 8 + 16 = 24 coins, and 40¢ + 160¢ = 200¢ = $2.00 ✓

2

Three Coin Types

A collection has 30 coins: pennies, nickels, and dimes. There are twice as many pennies as nickels. The total value is $1.57. How many of each coin are there?

Step 1 — Define variables

Let n = nickels, p = pennies, d = dimes. Given: p = 2n (twice as many pennies as nickels).

Step 2 — Set up equations

Count: p + n + d = 30. Value: p + 5n + 10d = 157 cents. Substituting p = 2n: 2n + n + d = 30 and 2n + 5n + 10d = 157

Step 3 — Simplify

3n + d = 30 so d = 30 - 3n. Also: 7n + 10d = 157

Step 4 — Solve

7n + 10(30 - 3n) = 157 gives 7n + 300 - 30n = 157, so -23n = -143 and n = 6.22... Wait — this should be a whole number!

Step 5 — Check the problem

This problem has no whole-number solution — the given constraints are inconsistent with having whole coins. In a real problem, we'd need to verify the given values.

3

Reverse the Unknown

A collection of dimes and nickels is worth $2.60. There are 8 nickels. How many coins are there in total?

Step 1 — Find the value of nickels

8 nickels are worth 8 × 5¢ = 40¢

Step 2 — Find the value of dimes

Total value is 260¢, so dimes are worth 260¢ - 40¢ = 220¢

Step 3 — Count the dimes

Number of dimes = 220¢ ÷ 10¢ = 22 dimes

Step 4 — Find total coins

Total coins = 8 + 22 = 30 coins

Step 5 — Verify

Check: 8 nickels (40¢) + 22 dimes (220¢) = 30 coins worth 260¢ = $2.60 ✓

4

The Impossible Case

Thirty coins, all dimes and nickels, are worth $2.55. How many nickels are there?

Step 1 — Set up the system

n + d = 30 and 5n + 10d = 255 cents

Step 2 — Solve by substitution

d = 30 - n, so 5n + 10(30 - n) = 255 gives 5n + 300 - 10n = 255

Step 3 — Complete the algebra

-5n = -45, so n = 9 and d = 21

Step 4 — Check the answer

9 nickels (45¢) + 21 dimes (210¢) = 30 coins worth 255¢ = $2.55 ✓

Step 5 — The lesson

Answer: 9 nickels. This problem works fine! The "impossible" was a red herring — some combinations of count and value do work out to whole numbers.

Frequently Asked Questions

How do you set up equations for coin word problems?+

Write two equations: one for the total count of coins and one for the total value. In this problem, if n is nickels and d is dimes, then n + d = 30 (count equation) and 5n + 10d = 260 cents (value equation). Always convert dollars to cents to avoid decimals.

What's the "all one type" method for solving coin problems?+

Assume all coins are one type, calculate the hypothetical value, then compare to the actual value. Here, if all 30 coins were dimes, they'd be worth 300 cents. The actual value is 260 cents, so we're 40 cents too high. Since each nickel is 5 cents less than a dime, we need 8 nickels to reduce the value by 40 cents.

Why do coin problems always have whole number answers?+

You can't have fractional coins in real life - you either have 5 nickels or 6 nickels, never 5.7 nickels. If your setup leads to a fractional answer, check your arithmetic or verify that the given values are consistent with having only whole coins.

NJ

Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-06-11