A student buys a sandwich for $4.20 and pays the exact amount using 50 cent coins, 20 cent coins and 10 cent coins.

Question

**second half of question**

If the number of 20 cent coins is the same as half the 10 cent coins plus four times the number of 50 cent coins and the student pays the cashier with 22 coins in total, how many coins of each type did the student use?

Answer 1

Hi Sarah M.,

Let: x = # of 50 cent coins, with .50x equal to the an amount for 50 cent coins.

y = # of 20 cent coins, with .20y equal to the an amount for 20 cent coins.

z = # of 10 cent coins, with .10z equal to the an amount for 10 cent coins..

For the first part of the question we can write an equation, .50x + .20y + .10z = 4.20.

And, the # of 20 cent coins (y) is the same as (=) half the # of 10 cent coins (.5z) plus (+) four times the # of 50 cent coins (4x). Or, y = .5z + 4x which we can rearrange: -4x + y - .5z = 0.

And, the total # of coins (x + y + z) is equal to 22 coins, x + y + z = 22.

With these 3 equations you can solve the system. I hope you can take it from here.

I solved the system using a matrix and the answers are x = 2, y = 12, and z = 8.

(.50*8 + .20*12 + .10*8 = 4.20)

Let me know if you need help solving the system.

I hope this helps, Joe.