You find 66 coins consisting only of nickels, dimes, and quarters, with a face value of $11.55. However, the coins all date from 1919 and are worth considerably more than their face value. A coin dealer offers you $10 for each nickel, $20 for each dime, and $15 for each quarter, for a total of $945. How many nickels did you find? Group of answer choices

You find 19 nickels.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

x is the number of nickels.

y is the number of dimes.

z is the number of quarters.

66 coins

This means that x + y + z = 66.

With a face value of $11.55.

Nickel is worth $0.05, dime worth $0.1 and quarter worth $0.25. So

0.05x + 0.1y + 0.25z = 11.55.

A coin dealer offers you $10 for each nickel, $20 for each dime, and $15 for each quarter, for a total of $945.

This means that 10x + 20y + 15z = 945.

How many nickels did you find?

We have to find x.

From the first equation:

z = 66 - x - y

Replacing in the second:

0.05x + 0.1y + 0.25z = 11.55.

0.05x + 0.1y + 0.25 (66 - x - y) = 11.55

0.05x + 0.1y + 16.5 - 0.25x - 0.25y = 11.55

0.2x + 0.15y = 4.95

So

0.15y = 4.95 - 0.2x

y = 33 - 1.33x

On the final equation:

10x + 20y + 15z = 945

10x + 20 (33 - 1.33x) + 15 (66 - x - (33 - 1.33x)) = 945

10x + 660 - 26.6x + 15 (66 - x - 33 + 1.33x) = 945

-16x + 660 + 15 (33 + 0.33x) = 945

-16x + 660 + 495 + 5x = 945

11x = 210

x = 210/11

x = 19

You find 19 nickels.