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