Question

Tim has a total of 19 nickels and dimes worth $1.65 in his coin bag. how many of each type of coin does Tim have?

Answer 1

To solve the problem, we can set up a system of equations based on the information provided.

Let:
- \( n \) be the number of nickels
- \( d \) be the number of dimes

From the problem, we have the following two equations:

1. The total number of coins equation:
\[
n + d = 19
\]

2. The total value equation. Nickels are worth $0.05 and dimes are worth $0.10, which gives us:
\[
0.05n + 0.10d = 1.65
\]

To eliminate the decimals in the value equation, we can multiply the entire equation by 100:
\[
5n + 10d = 165
\]

Now we have the following system of equations:
1. \( n + d = 19 \)
2. \( 5n + 10d = 165 \)

We can simplify the second equation by dividing everything by 5:
\[
n + 2d = 33
\]

Now, we can write the system as:
1. \( n + d = 19 \)
2. \( n + 2d = 33 \)

Next, we can subtract the first equation from the second:
\[
(n + 2d) - (n + d) = 33 - 19
\]
This simplifies to:
\[
d = 14
\]

Now that we have the value of \( d \), we can substitute it back into the first equation to find \( n \):
\[
n + 14 = 19
\]
Subtracting 14 from both sides gives us:
\[
n = 5
\]

Thus, Tim has:
- **5 nickels**
- **14 dimes**

To verify:
- The total number of coins is \( 5 + 14 = 19 \), which is correct.
- The total value is \( 5 \times 0.05 + 14 \times 0.10 = 0.25 + 1.40 = 1.65 \), which is also correct.

Therefore, the solution is:
- Nickels: 5
- Dimes: 14