## at the end of her shift a cashier has a total of \$6.30 in dimes and quarters. There are 7 more dimes than quarters. how many of each of thes

Question

at the end of her shift a cashier has a total of \$6.30 in dimes and quarters. There are 7 more dimes than quarters. how many of each of these coins does she have?

0

The cashier has 23 dimes and 16 quarters.

Step-by-step explanation:

Let the amount of dimes = d.

And the amount of quarters = q.

We can write 2 equations with the given information.

d = q + 7 — the amount of dimes is 7 more than the amount of quarters

0.1d + 0.25q = 6.3 — a dime is 0.1 dollars and a quarter is 0.25 dollars

0.1(q + 7) + 0.25q = 6.3 — we can substitute q + 7 for d because they have the same value and we need to have only 1 variable to solve

0.1q + 0.7 + 0.25q = 6.3 — distribute multipliers outside the parenthesis

0.35q + 0.7 = 6.3 — combine like terms

0.35q = 5.6 — subtract 0.7 from both sides

q = 16 — divide both sides by 0.35

The cashier has 16 quarters. We can use this to solve for dimes.

d = q + 7

d = 16 + 7

d = 23

The cashier has 23 dimes and 16 quarters.

0.1d + 0.25q = 6.3

0.1(23) + 0.25(16) = 6.3

2.3 + 4 = 6.3

6.3 = 6.3