Coin Problems: Nickels & Dimes System
Alison has $3.60 worth of nickels and dimes in her coin purse. She has 11 more nickels than dimes. How many of each coin does she have?
A note before we start: this problem, exactly as written, has no valid answer. The algebra produces a fractional number of dimes (about 20.33), and no purse holds a third of a coin. The most likely explanation is a one-digit typo: the total was probably meant to be $3.70. This page does three things: it solves the problem as written so you can see exactly where it breaks, it solves the corrected version, and it uses the failure to teach you how to test a coin problem for solvability before spending time on it.
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Setup:n = d + 11 and 5n + 10d = total (in cents).
As written ($3.60):15d + 55 = 360 → d = 61/3. Not a whole number, so there is no valid solution.
With $3.70:15d + 55 = 370 → d = 21 dimes, n = 32 nickels.
Quick check:32 × 5 + 21 × 10 = 160 + 210 = 370 ✓ and 32 − 21 = 11 ✓.
If you got something like "20 dimes and 31 nickels," read the Common Pitfalls section. That answer is a rounding trap.
Skills This Problem Builds
- Translating "more than" correctly. "11 more nickels than dimes" is
n = d + 11, notn = 11d. - Working in cents. Converting every value to whole cents keeps the arithmetic clean and avoids decimal slips.
- Substitution in a two-variable system. One equation describes how the counts relate, the other describes how the values add up.
- Treating context as a constraint. A coin count must be a non-negative whole number. An algebraically valid answer that violates this is a failed problem, not a rounding exercise.
- Diagnosing a broken problem. You will learn a quick divisibility test that predicts whether a nickel-dime problem can work before you solve it.
Solution: Method 1 — The Substitution Approach
This is the standard algebra path for a two-unknown coin problem: name both counts, write one equation per clue, and collapse the system into a single equation.
Step 1 — Name the unknowns
The question asks for two quantities, so we use two variables:
n = number of nickels
Step 2 — Translate the "more than" clue
"She has 11 more nickels than dimes" means the nickel count is the dime count plus 11. A quick sanity test: if she had 20 dimes, she would have 31 nickels, which is 11 more. So:
Step 3 — Translate the money clue
A nickel is worth 5 cents and a dime is worth 10 cents. Each coin type contributes (value per coin) × (number of coins). $3.60 is 360 cents:
Step 4 — Substitute and solve
Replace n in the value equation with d + 11:
5d + 55 + 10d = 360
15d + 55 = 360
15d = 305
d = 305 ÷ 15 = 61/3 ≈ 20.33
Step 5 — Check the answer against the story
The algebra was done correctly, but d = 61/3 cannot be the number of dimes in a purse. There is no valid solution to the problem as stated. This is not an arithmetic error. The numbers $3.60 and "11 more" are simply incompatible for nickels and dimes. We will prove that in Method 2.
Step 6 — Solve the corrected version ($3.70)
Changing the total to 370 cents changes only one line of the solution:
15d + 55 = 370
15d = 315
d = 21
n = d + 11 = 32
Now the dime count is a whole number, and Alison has 21 dimes and 32 nickels.
Solution: Method 2 — Remove the Extras and Count Pairs
This method uses no variables. It also explains why $3.60 fails, which Method 1 only reveals by accident.
Step 1 — Set aside the extra nickels
Alison has 11 more nickels than dimes. Take those 11 "extra" nickels out of the purse. They are worth 11 × 5 = 55 cents. What remains is an equal number of nickels and dimes.
Step 2 — Find the value of what remains
As written: 360 − 55 = 305 cents remain.
Step 3 — Group the remaining coins into nickel-and-dime pairs
Since the counts are now equal, every nickel can be paired with a dime. Each pair is worth 5 + 10 = 15 cents. The number of pairs is:
There is a remainder, which means the leftover money cannot be split into whole pairs. Twenty pairs would be worth 300 cents, and one more pair would be 315 cents. The target of 305 falls between them. So the problem as written is impossible.
Step 4 — Repeat with $3.70
370 − 55 = 315 cents remain, and 315 ÷ 15 = 21 pairs. That gives 21 dimes and 21 nickels, plus the 11 extra nickels we set aside: 21 + 11 = 32 nickels.
This confirms the Method 1 result without any equations, and it exposes the structure of the problem: the leftover value must be a multiple of 15 cents.
The Answer
As written ($3.60): no valid solution. The problem would require 61/3 ≈ 20.33 dimes.
With the likely intended total of $3.70:
Dimes: 21 | Nickels: 32
(21 dimes = $2.10, 32 nickels = $1.60, total = $3.70.)
Another single-number repair also works: keeping $3.60 and changing "11 more" to "12 more" gives 20 dimes and 32 nickels.
Verification
Corrected problem ($3.70, 11 more nickels):
Count clue: 32 − 21 = 11 ✓
Value clue: 32 × 5 = 160 cents, 21 × 10 = 210 cents, and 160 + 210 = 370 cents = $3.70 ✓
Alternate repair ($3.60, 12 more nickels):
n = 32, d = 20, 32 − 20 = 12 ✓ and 160 + 200 = 360 cents = $3.60 ✓
Original problem, independent check that no answer exists: if Alison had d dimes, she would have d + 11 nickels, worth 15d + 55 cents. For whole d this equals 55, 70, 85, 100, … , 355, 370, … and never 360. Every possible total ends in 0 or 5, and the totals climb in steps of 15, so 360 is skipped.
Reality Check: Why $3.60 Gets Skipped
The totals that are possible when there are 11 more nickels than dimes form a sequence with a gap of 15 cents between neighbors. Here are the values near our target:
| Dimes | Nickels (d + 11) | Total value |
|---|---|---|
| 19 | 30 | $3.40 |
| 20 | 31 | $3.55 |
| 21 | 32 | $3.70 |
| 22 | 33 | $3.85 |
The totals jump from $3.55 to $3.70, which skips $3.60 entirely. Adding one dime and one nickel always adds 15 cents, so you can never land on a value that is not 15 cents away from another valid total. That is exactly why d = 20.33 sits between the 20-dime row and the 21-dime row.
Three Mistakes That Are Easy to Make
✗ Rounding 20.33 to 20 and finishing.
With 20 dimes and 31 nickels you get 200 + 155 = 355 cents, which is $3.55, not $3.60. Rounding "fixes" the fraction but breaks the money clue. In a counting problem, a fractional answer is a diagnosis, not a decimal to tidy up.
✗ Writing n = 11d or d = n + 11.
"11 more than" is addition on the larger quantity's side. The group with more (nickels) equals the smaller group plus 11. Multiplying by 11 would mean "eleven times as many," which is a different statement. A quick test catches it: plug in a small number and ask whether the relationship reads back correctly in English.
✗ Mixing dollars and cents: 5n + 10d = 3.60
Nickels are worth $0.05, not 5, if the total is in dollars. Either write 0.05n + 0.10d = 3.60 or convert everything to cents: 5n + 10d = 360. Mixing the two silently multiplies your coin values by 100 and gives nonsense.
The Pattern Behind This: A Divisibility Test You Can Run in Ten Seconds
Every "k more nickels than dimes" problem has the same skeleton. If the total is T cents, then:
So a valid whole-number answer exists exactly when T − 5k is a non-negative multiple of 15. Plug in the original numbers: 360 − 55 = 305, and 305 is not divisible by 15. Plug in $3.70: 370 − 55 = 315 = 15 × 21.
The structure is a general principle. A system with whole-number unknowns is a Diophantine problem, and the context (counting coins) restricts which equation results are acceptable. This kind of check shows up across coin, ticket, age, and mixture problems. Whenever the unknowns are counts, solve the equation and then ask whether the answer is a whole number. Before you start, you can often predict whether it will be.
If You See These Words...
- "worth," "value," "in coins," "total of $…" signal a value equation: (value per item) × (count), summed.
- "k more … than …," "k fewer … than …" signal a relationship between counts, giving you a substitution equation.
- "how many of each" signals two unknowns and therefore two equations.
The same structure appears with adult and child tickets, two kinds of stamps, or two sizes of bills. The coins are just a story wrapped around "count equation + value equation."
Try These Variations
The four problems below get steadily harder. Each one has been built so that it has a clean whole-number solution. The last asks you to reason about which totals are possible.
Marcus has $2.60 in nickels and dimes. He has 7 more nickels than dimes. How many of each coin does he have?
Let d = dimes and n = nickels.
Count clue: n = d + 7. Value clue in cents: 5n + 10d = 260.
5(d + 7) + 10d = 260 → 15d + 35 = 260 → 15d = 225.
d = 15, so n = 15 + 7 = 22. Marcus has 15 dimes and 22 nickels.
22 × 5 = 110 and 15 × 10 = 150, and 110 + 150 = 260 cents = $2.60 ✓. Also 22 − 15 = 7 ✓.
Priya has 45 coins, all nickels and dimes, worth $3.15 in total. How many nickels and how many dimes does she have?
Let n = nickels and d = dimes.
Total coins: n + d = 45. Value in cents: 5n + 10d = 315.
From the first equation, n = 45 − d.
5(45 − d) + 10d = 315 → 225 − 5d + 10d = 315 → 5d = 90 → d = 18. Then n = 45 − 18 = 27. Priya has 27 nickels and 18 dimes.
27 + 18 = 45 coins ✓. 27 × 5 = 135 and 18 × 10 = 180, and 135 + 180 = 315 cents = $3.15 ✓.
Jo has $4.30 in quarters, dimes, and nickels. She has 5 more nickels than dimes and 3 fewer quarters than dimes. How many of each coin does she have?
Let d = dimes. Then nickels n = d + 5 and quarters q = d − 3. Everything is now in terms of d.
10d + 5(d + 5) + 25(d − 3) = 430.
10d + 5d + 25 + 25d − 75 = 430 → 40d − 50 = 430.
40d = 480 → d = 12. Then n = 17 and q = 9. Jo has 12 dimes, 17 nickels, and 9 quarters.
Dimes: 12 × 10 = 120. Nickels: 17 × 5 = 85. Quarters: 9 × 25 = 225. Sum: 120 + 85 + 225 = 430 cents = $4.30 ✓. Relationships: 17 − 12 = 5 ✓ and 12 − 9 = 3 ✓.
Maya has nickels and dimes, with 5 more nickels than dimes. She says the total is exactly $2.40. Show that this is impossible, then list every total between $2.00 and $3.00 that could be correct.
With d dimes and d + 5 nickels, the total is 10d + 5(d + 5) = 15d + 25 cents.
Set 15d + 25 = 240. Then 15d = 215 and d = 215/15 = 43/3 ≈ 14.33. This is not a whole number, so $2.40 is impossible. (Equivalently, 215 is not a multiple of 15.)
We need 200 ≤ 15d + 25 ≤ 300, so 175 ≤ 15d ≤ 275, which gives 11.67 ≤ d ≤ 18.33. The whole numbers in that range are d = 12, 13, …, 18.
d = 12: 205 d = 13: 220 d = 14: 235 d = 15: 250 d = 16: 265 d = 17: 280 d = 18: 295 (all in cents).
The possible totals are $2.05, $2.20, $2.35, $2.50, $2.65, $2.80, and $2.95, seven in all, spaced 15 cents apart. $2.40 is not among them.
Check the endpoints: d = 11 gives 190 cents (below $2.00) and d = 19 gives 310 cents (above $3.00), so no valid totals were missed. Spot check d = 15: 15 dimes and 20 nickels give 150 + 100 = 250 cents ✓.
Frequently Asked Questions
d be the number of dimes, then write the nickel count in terms of d (for "k more nickels than dimes," n = d + k). Next, write a value equation in cents: 5n + 10d = total. Substitute the first equation into the second and solve for d. In this example, with n = d + 11 and a total of 370 cents, 5(d + 11) + 10d = 370 gives 15d = 315, so d = 21 dimes and n = 32 nickels.d = 61/3 ≈ 20.33 dimes, which is impossible. Changing the total to $3.70 gives exactly 21 dimes and 32 nickels.2026-08-03