Math question Algebra need help

Question

Michael has a jar full of dimes and quarters. There are a total of 120 coins with a total value of $19.20. The system of equations below represents this situation.

How many MORE dimes than quarters are in Michael's jar?

A.12

B.15

C.24

D.48

Answer 1

Let's say we have d dimes and q quarters.

We know that we have in total 120 coins, so d + q = 120

We also know that the total amount of money is $19.20, so 10d + 25q = 1920 (to avoid dealing with decimals, I converted everything in cents).

We need to solve the following system of equation:

d + q = 120

10d + 25q = 1920

We isolate the d in the first equation and substitute it in the second one:

d = 120 - q

10(120 - q) + 25q = 1920

1200 - 10q + 25q = 1920

15q = 720

q = 48

We have 48 quarters and d=120-48= 72 dimes.

The difference between the dimes and the quarters is:

72 - 48 = 24