Money Question!

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

Hi, fun question!

Number of quarters = q

number of half dollars = H

dumber of dimes = d

Let us first write out what we know:

The number quarters is 5 times the number of half dollars

q=5H

There is a total of 900 coins:

q+H+d=900

The total value is $205:

q(0.25)+H(0.5)+d(0.1)=205

Let's take our second equation and substitute in for q:

5H+H+d=900

Thus,

6H+d=900

Let's solve this equation for d so that we can put everything in terms of H:

d=900-6H

Now lets solve q(0.25)+H(0.5)+d(0.1)=205 by putting everything in terms of H:

5H(0.25)+H(0.5)+(900-6H)(0.1)=205

Simplify this and solve for H:

H=100

Now you can use this to solve for q and d :)

Answer 2

Let's set up the system of linear equations.

Let d = number of dimes, q = number of quarters, and h = number of half-dollars

10d + 25q + 50h = 20500

d + q + h = 900

Set up the possible combinations of dimes, quarters, and half-dollars so they will add up to $205 (20,500 cents).

25(500) + 50(100) + 10(300) = 12500 + 5000 + 3000 = 20500

The collector has 500 quarters, 100 half-dollars, and 300 dimes that will add up to $205.