question2answer

Answered 2021-08-04
Author has **28686** answers

Let q be the number of quarters and d be the number of dimes. Hecall that 1 quarter = 0.25 and 1 dime = 0.10

Fred has 5.05 consisting of quarters and dimes:

\(0.25q + 0.10d = 5.05\)

There are 5 more dimes than quarters:

\(d=q+5\)

Solve by substitution. Substitute (2) to (1) and solve for q: \(0.25q + 0.10(q + 5) = 5.05\)

\(0.25q + 0.10q + 0.50 = 5.05\)

\(0.35q + 0.50 = 5.05\)

\(0.35q = 4.55\)

\(q=13\)

Solve for d using (2):

\(d=13+5\)

\(d=18\)

Hence, Fred has 13 quarters and 18 dimes

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