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A pile of coins, consisting of nickels, dimes, and quarters, is worth $4.55. There are 4 more dimes than nickels and 3 quarters more than dimes. How many of each are there?

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n=number of nickles
q=number of quarters
d=number of dimes


4 more dmes than nickles, 3 more quarters than dimes
d+4=n
q=3+d

total value is 455 cents
10d+5n+25q=455
divide both sides by 5 to simplify
2d+n+5q=91

now we convert all to one coin
q=3+d and n+4=d
subsitute 3+d for q and n+4 for d

2(n+4)+n+5(3+d)=91
2(n+4)+n+5(3+n+4)=91
2(n+4)+n+5(n+7)=91
2n+8+n+5n+35=91
8n+43=91
minus 43 both sides
8n=48
divide both sides by 8
n=6

sub back
n+4=d
6+4=d
10=d

q=3+d
q=3+10
q=13


6 nickles
10 dimes
13 quarters

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