I'm Having Trouble with this Coin Problem! Thanks!

Question

A coin collector had a collection of 900 silver coins worth $205. There were five times as many quarters as half-dollars. If the rest of the coins were dimes, how many of each kind of coin did the collector have?

Answer 1

number of half dollars = x

number of quarters = 5x

dimes = 900 - 6x

0.50 * x + 0.25 * 5x + 0.10 * (900 - 6x) = 205

1.15x + 90 = 205

1.15x = 115

x = 100

100 half dollars = 50

500 quarters = 125

300 dimes = 30

205

Answer 2

Let half-dollars = H, quarters = Q, and dimes = D.

As there are 900 coins, H + Q + D = 900

There are 5 times as many quarters as half-dollars, so Q = 5 H

The coins are worth $205, so .5H + .25Q + .1D = 205

Let's multiply through by 20 to get rid of decimals (think of it as the number of nickels)

10H + 5 H + 2D = 4100

Thus, we have 3 equations and 3 unknowns, so we can solve.

If we want to solve this using substitution, we can solve this first in terms of H.

H + Q + D = 900, so, as Q = 5D, H + 5H + D = 900, so 6H + D = 900, or D = 900 - 6H

Then, 10H + 5 Q + 2D = 4100, so 10H + 5(5H) + 2(900 - 6H) = 4100

23H = 2300

H = 100

Then, Q = 5H, so Q = 5(100) = 500

D = 900 - 6H, so D = 900 - 6(100) = 300

(H, Q, D) = (100, 500, 300)