Question 176671This question is from textbook
: Sam found a number of nickels, dimes and quarters in his room. He found 3 more dimes than nickels but twice as many quarters as dimes. The total value of the coins was $5.05 how many coins of each type did Sam find?
This question is from textbook
Answer by gonzo(654)   (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.
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d = n+3
q = 2*d = 2*(n+3) = 2*n+6
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since:
$5.05 * 100 = 505 cents
and:
5*n = number of nickels * 5 cents per nickel
10*d = number of dimes * 10 cents per dime
25*q = number of quarters * 25 cents per quarter
then:
5*n + 10*d + 25*q = 505 cents
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since d = n+3,
and q = 2d = 2*n+6,
then:
5*n + 10*(n+3) + 25*(2*n+6) = 505
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remove parentheses and this equation becomes:
5*n + 10*n + 30 + 50*n + 150 = 505
combine like terms to get:
65*n + 180 = 505
subtract 180 from both sides to get:
65*n = 325
divide both sides by 65 to get:
n = 5
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number of nickels = 5
number of dimes is number of nickels + 3 = 8
number of quarters is 2 times number of dimes = 16
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5 nickels times 5 cents each is 25 cents
8 dimes times 10 cents each is 80 cents
16 quarters times 25 cents each is 400 cents
400 cents + 80 cents + 25 cents = 505 cents.
divide 505 cents by 100 to get:
$5.05
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answer is:
sam found 5 nickels, 8 dimes, and 16 quarters.
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