Lesson Coin Problem

Source code of 'Coin Problem'

This Lesson (Coin Problem) was created by by oberobic(2304)

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About oberobic: MBA/Ph.D. University Administrator

Problem: Joan has $4.50 in nickels, dimes, and quarters. There are twice as many dimes as nickels, and two more quarters than nickels. How many of each type of coin does she have?
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Solution: To solve coin problems, you have to keep track of both the number of coins and the value of the coins. In this case, we are told she has only nickels, dimes, and quarters. So, we can define our variables as:
n = number of nickels
5n = value of nickels in cents
d = number of dimes
10d = value of dimes in cents
q = number of quarters
25q = value of quarters in cents
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The total value is $4.50, but we'll work only in cents.
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5n + 10d + 25q = 450 cents
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We also are told certain relationships among the different types of coins:
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There are twice as many nickels as there are dimes.
d = 2n
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There are two more quarters than nickels.
q = n+2
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Substitute.
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5n + 10(2n) + 25(n+2) = 450
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5n + 20n +25n +50 = 450
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50n = 400
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n = 8
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Substitute the number of nickels to find the number of dimes and quarters.
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d = 2(8) = 16
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q = n+2 = 10
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Always check the solution to ensure it is the correct answer.
In this case, check the values against the required total.
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5n = 5*8 = 40 cents
10d = 10*16 = 160 cents
25q = 25*10 = 250 cents
40 cents + 160 cents + 250 cents = 450 cents
450 cents = $4.50
Correct.
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Answer: Joan has 8 nickels, 16 dimes, and 10 quarters.