A parking meter contains $8.25 in dimes and quarters. after emptying the meter, the worker counts 51 coins. how many quarters and dimes were in the meter

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

1 dime = $ 0.10

1 quarter = $ 0.25

The worker counts 51 coins

Therefore,

d + q = 51

d = 51 - q eqn 1

A parking meter contains $8.25 in dimes and quarters

Therefore,

0.10d + 0.25q = 8.25 eqn 2

Substitute eqn 1 in eqn 2

0.10(51 - q) + 0.25q = 8.25

5.1 - 0.10q + 0.25q = 8.25

0.15q = 3.15

Divide both sides by 0.15

Substitute q = 21 in eqn 1

d = 51 - 21

Thus 30 dimes and 21 quarters are present

step-by-step explanation:

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