SOLUTION: Jane has a collection of 175 coins in her piggybank valued at $44.25 if she has 30 less dimes than Nicolas and 10 morehalf dollars than nickels , how many coins of each type ate in

Algebra ->  Test -> SOLUTION: Jane has a collection of 175 coins in her piggybank valued at $44.25 if she has 30 less dimes than Nicolas and 10 morehalf dollars than nickels , how many coins of each type ate in      Log On


   

Question 501491: Jane has a collection of 175 coins in her piggybank valued at $44.25 if she has 30 less dimes than Nicolas and 10 morehalf dollars than nickels , how many coins of each type ate in jane's piggybank?
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
With money problems you have to keep track of the number of coins and their values.
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For example,
d = number of dimes
10d = value of the dimes in cents
n = number of nickels
5d = value of the nickels in cents
h = number of half dollars
50h = value of the half dollars in cents
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n + d + h = 175 coins
5n + 10d + 50h = 4425 cents
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d = n - 30
h = n + 10
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substitute
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n + n-30 + n+10 = 175
3n -20 = 175
3n = 195
n = 65
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d = n-30
d = 65 -30
d = 35
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h = n +10
h = 65 +10
h = 75
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Always check your work.
In this case, what is the value of the coins?
5(65) = 325 cents
10(35) = 350 cents
50(75) = 3750 cents
325+350+3750= 4425 cents
Correct.
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Answer: Jane has 65 nickels, 35 dimes, and 75 half dollars.
.
Done.