SOLUTION: A coin collector had a collection of silver coins worth $205. There were 5 times as many quarters as half-dollars. There were 200 fewer dimes than quarters. How many of each kin

Question 174777: A coin collector had a collection of silver coins worth $205. There were 5 times as many quarters as half-dollars. There were 200 fewer dimes than quarters. How many of each kind of coin did the collector have?
Found 2 solutions by stanbon, checkley75:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A coin collector had a collection of silver coins worth $205. There were 5 times as many quarters as half-dollars. There were 200 fewer dimes than quarters. How many of each kind of coin did the collector have?
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dimes,quarters,half-dollars
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Value equation:
10d + 25q + 50h = 20500 cents
Quantity Equation:
q = 5h
d = q-200
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Rearrange:
h = (1/5)q
d = q-200
----
Substitute into the Value equation to solve for q:
10(q-200) + 25q + 50(1/5)q = 20500
45q = 20700
q = 460 (# of quarters)
h = (1/5)460 = 92 (# of half-dollars)
d = 460-200 = 260 (# of dimes)
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Cheers,
Stan H.

Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
Q=5H OR H=Q/5
D=Q-200 OR Q=D+200
.50H+.25Q+.10D=205
.50(Q/5)+.25Q+.10(Q-200)=205
.10Q+.25Q+.10Q-20=205
.45Q=205+2-
.45Q=225
Q=225/.45
Q=500 ANS. FOR THE NUMBER OF QUARTERS.
500=5H
H=500/5
H=100 ANS FOR THE NUMBER OF HALVES.
D=500-200
D=300 ANS. FOR THE NUMBEROF DIMES.
PROOF:
.50*100+.25*500+.10*300=205
50+125+30=205
205=205
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