Solving Coin Problems: A Practical Example with Dimes and Nickels

In this article, we will explore a common type of algebraic problem involving coins with a specific focus on solving a problem related to dimes and nickels. Such problems are frequent in basic algebra textbooks and are useful for reinforcing concepts in algebra and problem-solving skills.

Problem Presentation

The problem presented is as follows: Octavia’s coin holder contains 6.75 in dollars worth of dimes and nickels. There are 12 fewer nickels than dimes. How many of each coin are in the holder?

Step-by-Step Solution

Let's denote:
- N as the number of nickels;
- D as the number of dimes.

We know the following:
1. The total value of the coins is 675 cents (since 6.75 dollars 675 cents).

We can write the following equations based on the problem statement:

1. Value equation: 5N 10D 675 2. Quantity equation: N D - 12

First, we substitute 5N 10D 675 and N D - 12 into each other:

Substituting N D - 12 into the value equation:

5(D - 12) 10D 675

This simplifies to:

5D - 60 10D 675

Combining like terms:

15D - 60 675

Add 60 to both sides:

15D 735

Dividing by 15:

D 49

Now, using the second equation, substitute D 49 back into N D - 12:

N 49 - 12 37

Verification

To verify, we check the total value of the coins:

Total value 5N 10D 5(37) 10(49) 185 490 675 cents

Thus, the solution is confirmed to be correct.

Alternative Solution

Another approach is to solve the problem without using formulas. Add 12 nickels to the problem, then you get:

Total value (6.75 12 * 0.05) 7.35

Now, if the total value is 7.35 dollars which equals 735 cents, and we know that 735/15 (each pair of nickel and dime equals 15 cents) is 49 pairs of nickels and dimes.

Therefore, the original number of dimes is 49, and the number of nickels is 49 - 12 37.

The solution can be summarized as: Octavia has 49 dimes and 37 nickels.

Conclusion

In conclusion, we have solved the coin problem successfully using both traditional algebraic methods and a more intuitive approach. Both methods provide the same solution, and the practical steps can be used for various similar problems in algebra.