Example

Solving a Coin Mixture Problem: Pennies and Nickels

Apply the seven-step problem-solving strategy and the total value of coins model to a coin mixture problem in which the relationship between the two coin counts involves both multiplication and addition.

Problem: Danny has $2.14 worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have?

  1. Read the problem and identify the coin types involved: pennies (worth $0.01 each) and nickels (worth $0.05 each). The total value of all coins is $2.14.
  2. Identify what to find: the number of pennies and the number of nickels.
  3. Name the unknowns using a single variable. Since the number of nickels is defined in terms of the number of pennies, start with pennies. Let pp = the number of pennies. The phrase "two more than ten times" combines multiplication by 1010 with adding 22, so the number of nickels is 10p+210p + 2. Organize everything in a table:
TypeNumberValue ($)Total Value ($)
Penniespp0.010.01p0.01p
Nickels10p+210p + 20.050.05(10p+2)0.05(10p + 2)
2.14
  1. Translate into an equation by adding the total values and setting the sum equal to the overall total:

0.01p+0.05(10p+2)=2.140.01p + 0.05(10p + 2) = 2.14

  1. Solve the equation:
  • Distribute 0.050.05: 0.01p+0.50p+0.10=2.140.01p + 0.50p + 0.10 = 2.14
  • Combine like terms: 0.51p+0.10=2.140.51p + 0.10 = 2.14
  • Subtract 0.100.10 from both sides: 0.51p=2.040.51p = 2.04
  • Divide both sides by 0.510.51: p=4p = 4

Find the number of nickels: 10(4)+2=4210(4) + 2 = 42.

  1. Check: 4(0.01)+42(0.05)=?2.144(0.01) + 42(0.05) \stackrel{?}{=} 2.140.04+2.10=2.140.04 + 2.10 = 2.14 \checkmark
  2. Answer: Danny has four pennies and 42 nickels.

This example extends the coin mixture technique to a case where the relationship between the two coin counts involves both multiplication and addition — "two more than ten times" translates to 10p+210p + 2 — rather than a simple additive relationship like "nine more than." The expression 10p+210p + 2 introduces a larger coefficient (1010) inside the parentheses, so distributing 0.050.05 across 10p+210p + 2 produces 0.50p0.50p rather than a small decimal term. After distribution, combining 0.01p+0.50p0.01p + 0.50p yields the non-obvious coefficient 0.510.51, making the final division step (2.04÷0.512.04 \div 0.51) more challenging than in simpler coin problems.

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Updated 2026-04-21

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