Question

You have $4.10 in nickels, dimes, and quarters. If you have twice as many dimes as quarters, and five more nickels than dimes, how many coins of each type do you have?

Answer 1

For this case, the first thing to do is define variables.

We have then:

• x: number of nickels
• y: number of dimes
• z: number of quarters

We write now the system of equations:

$0.05x + 0.1y + 0.25z = 4.10\\y = 2z\\x = y + 5$

Then rewriting the equations we have:

$0.05 (2z + 5) + 0.1 (2z) + 0.25z = 4.1\\0.05 (2z + 5) + 0.1 (2z) + 0.25z = 4.1\\0.1z + 0.25 + 0.2z + 0.25z = 4.1\\0.55z = 4.1-0.25$

$z = \frac {3.85} {0.55}\\z = 7$

Then, substituting values we have:

$y = 2 (7)\\y = 14$

Finally:

$x = 14 + 5\\x = 19$

Answer:

19 nickels

14 dimes

7 quarters