Coin Mixture: Find Number of Dimes and Quarters

Coin Problems 8th-9th Grade
PROBLEM
Joaquín has 31 coins in his pocket, all of which are dimes and quarters. If the total value of his change is 535 cents, how many dimes and how many quarters does he have?

What This Problem Teaches

  • Setting up systems of linear equations from word problems with two constraints
  • Translating coin values into algebraic expressions (10d + 25q)
  • Using substitution method to solve 2×2 systems efficiently
  • Recognizing that coin problems always have unique integer solutions when well-constructed
  • Checking answers by verifying both the count and value constraints simultaneously

Let's Visualize the Setup

Joaqu n has 31 coins in his pocket, all of which are dimes and quarters. If the total value of his change is 535...

Solution: Method 1 — The Substitution Approach

The most direct method for coin problems is substitution: solve the simpler equation for one variable, then substitute into the value equation.

Step 1 — Define variables

Let d = number of dimes and q = number of quarters.

Step 2 — Write the system of equations

We have two constraints to work with:

Count constraint: d + q = 31
Value constraint: 10d + 25q = 535

Step 3 — Solve the count equation for one variable

From the first equation: d + q = 31, we get d = 31 - q

Step 4 — Substitute into the value equation

Replace d with (31 - q) in the value equation:

10(31 - q) + 25q = 535
310 - 10q + 25q = 535
310 + 15q = 535
15q = 225
q = 15

Step 5 — Find the number of dimes

Now that we know q = 15, we substitute back:

d = 31 - q = 31 - 15 = 16

Solution: Method 2 — The Elimination Method

Alternatively, we can solve this system using elimination by multiplying one equation to eliminate a variable.

Step 1 — Set up the system

d + q = 31 ... (1)
10d + 25q = 535 ... (2)

Step 2 — Eliminate the dimes

Multiply equation (1) by -10 to eliminate d:

-10d - 10q = -310 ... (1) × (-10)
10d + 25q = 535 ... (2)

Step 3 — Add the equations

(-10d - 10q) + (10d + 25q) = -310 + 535
15q = 225
q = 15

Step 4 — Back-substitute

Using q = 15 in equation (1): d + 15 = 31, so d = 16

Joaquín has 16 dimes and 15 quarters.

Verification

Let's check both constraints with our answer:

Count check

16 dimes + 15 quarters = 31 coins ✓

Value check

16 × 10¢ + 15 × 25¢ = 160¢ + 375¢ = 535¢ ✓

Both constraints are satisfied, confirming our solution is correct.

Watch Out For These

✗ Mixing up the coin values
Writing 25d + 10q = 535 instead of 10d + 25q = 535. Always remember: dimes = 10¢, quarters = 25¢.
✗ Converting to dollars incorrectly
Converting 535 cents to $53.50 instead of $5.35. When dealing with mixed units, stick to cents throughout the problem.
✗ Setting up only one equation
Using just d + q = 31 and guessing from there. You need both the count constraint AND the value constraint to get a unique solution.

How to Spot This Problem Type

Coin problems have distinctive features that make them recognizable:

  • "X coins, all of which are..." — signals a count constraint
  • "total value is Y cents/dollars" — signals a value constraint
  • Two coin types mentioned — creates exactly 2 unknowns
  • Specific coin values — pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢)
Clinical context: The same mathematical structure appears in pharmacy problems: "X tablets of two different strengths with total active ingredient Y mg."

The Pattern Behind This

All two-coin problems follow the same template:

General form:
x + y = total_count
(value₁)x + (value₂)y = total_value

The solution always exists and is unique when:

  • The coin values are different (ensuring the lines aren't parallel)
  • The total value is achievable with the given count
  • Both answers come out as non-negative integers
Why the answer is always clean: Textbook coin problems are carefully constructed so the math works out to whole numbers of coins. In real life, you might have to round or determine that no exact solution exists.

What If?

1
Different Total Value
Joaquín still has 31 coins (dimes and quarters), but now the total value is exactly $4.90. How many of each coin does he have?
Step 1 — Set up equations

Count: d + q = 31
Value: 10d + 25q = 490 cents

Step 2 — Substitute

From d = 31 - q: 10(31 - q) + 25q = 490

Step 3 — Solve for q

310 - 10q + 25q = 490
15q = 180
q = 12

Step 4 — Find d

d = 31 - 12 = 19

Verification

Count: 19 + 12 = 31 ✓
Value: 19(10) + 12(25) = 190 + 300 = 490¢ ✓

Answer: 19 dimes, 12 quarters

2
Three Coin Types
Joaquín has 31 coins: nickels (5¢), dimes, and quarters. The total value is 535 cents. He has twice as many dimes as nickels. How many of each?
Step 1 — Define variables

Let n = nickels, d = dimes, q = quarters
Given: d = 2n

Step 2 — Set up equations

Count: n + d + q = 31
Value: 5n + 10d + 25q = 535

Step 3 — Substitute d = 2n

n + 2n + q = 31 → 3n + q = 31
5n + 10(2n) + 25q = 535 → 25n + 25q = 535

Step 4 — Solve the system

From q = 31 - 3n: 25n + 25(31 - 3n) = 535
25n + 775 - 75n = 535
-50n = -240
n = 4.8

Wait — check this!

Since we need whole coins, let's try n = 5:
d = 10, q = 16
Check: 5(5) + 10(10) + 25(16) = 25 + 100 + 400 = 525¢ (close, but not 535)

This particular problem has no integer solution with the given constraints.

3
Reverse the Unknown
Joaquín has only dimes and quarters. The total value is 535 cents. The number of dimes is one more than the number of quarters. How many coins does he have in total?
Step 1 — Set up with given relationship

Given: d = q + 1
Value: 10d + 25q = 535

Step 2 — Substitute

10(q + 1) + 25q = 535
10q + 10 + 25q = 535

Step 3 — Solve for q

35q = 525
q = 15

Step 4 — Find d and total

d = q + 1 = 16
Total coins = 16 + 15 = 31

Verification

Dime check: 16 = 15 + 1 ✓
Value: 16(10) + 15(25) = 535¢ ✓

Answer: 31 coins total

4
Optimization Challenge
Joaquín wants exactly 535 cents using only dimes and quarters. What's the minimum number of coins he could use? What's the maximum?
Step 1 — For minimum coins, maximize quarters

535 ÷ 25 = 21 remainder 10
So: 21 quarters + 1 dime = 525 + 10 = 535¢

Step 2 — Check this solution

21 quarters + 1 dime = 22 coins total
Value: 21(25) + 1(10) = 525 + 10 = 535¢ ✓

Step 3 — For maximum coins, maximize dimes

If all dimes: 535 ÷ 10 = 53.5 (not possible)
Try 53 dimes: 53(10) = 530¢, need 5¢ more (impossible with quarters)

Step 4 — Work backwards from quarters

1 quarter + ? dimes: 25 + 10d = 535 → d = 51
Check: 1 quarter + 51 dimes = 52 coins

Verification

Min: 21(25) + 1(10) = 535¢ ✓
Max: 1(25) + 51(10) = 535¢ ✓

Minimum: 22 coins (21Q + 1D)
Maximum: 52 coins (1Q + 51D)

Frequently Asked Questions

How do you set up a system of equations for coin problems? +
Write one equation for the total count of coins and another for the total value in cents. Here, d + q = 31 (total coins) and 10d + 25q = 535 (total value), where d is dimes and q is quarters. Then solve using substitution or elimination.
What's the fastest way to solve coin problems with two types of coins? +
Use substitution: solve the count equation for one variable, then substitute into the value equation. In this problem, d = 31 - q, so 10(31 - q) + 25q = 535, which simplifies to 15q = 225, giving q = 15 quarters and d = 16 dimes.
Why do coin problems always have exactly one solution? +
Because you have two independent constraints (total count and total value) with two unknowns. The different coin values ensure the lines intersect at exactly one point with integer coordinates, provided the problem is well-constructed.
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.

2026-05-21

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