SOLUTION: What are the algebraic formulas for this (in a collection of nickels, dimes, and quarters, there are twice as many dimes as nickels and 3 fewer quarters than dimes. if the total va

Algebra ->  Finance -> SOLUTION: What are the algebraic formulas for this (in a collection of nickels, dimes, and quarters, there are twice as many dimes as nickels and 3 fewer quarters than dimes. if the total va      Log On


   



Question 297266: What are the algebraic formulas for this (in a collection of nickels, dimes, and quarters, there are twice as many dimes as nickels and 3 fewer quarters than dimes. if the total value is $4.50 how many of each coin are there)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let n = number of nickels
Let d = number of dimes
Let q = number of quarters
given:
d+=+2n
q+=+d+-+3
5n+%2B+10d+%2B+25q+=+450 (in cents)
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This is 3 equations and 3 unknowns, so it's solvable
By substitutions:
5n+%2B+10d+%2B+25%2A%28d+-+3%29+=+450
And, substituting again:
5%2A%28d%2F2%29+%2B+10d+%2B+25%2A%28d+-+3%29+=+450
5d%2F2+%2B+10d+%2B+25d+-+75+=+450
5d+%2B+20d+%2B+50d+-+150+=+900
75d+=+1050
d+=+14
and, since
q+=+d+-+3
q+=+11
and
n+=+d%2F2
n+=+7
There are 14 dimes, 11 quarters, and 7 nickels
check:
5n+%2B+10d+%2B+25q+=+450
5%2A7+%2B+10%2A14+%2B+25%2A11+=+450
35+%2B+140+%2B+275+=+450
450+=+450
OK