SOLUTION: Sue counted her money and found that her 25 coins which were nickels, dimes, and quarters were worth $3.20. The number of dimes exceeded the number of nickels by 4. How many coin
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Question 23736: Sue counted her money and found that her 25 coins which were nickels, dimes, and quarters were worth $3.20. The number of dimes exceeded the number of nickels by 4. How many coins of each kind did she have?
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let x = no of nickels
x+4 = no of dimes
The total of the nickels and dimes is 2x + 4, and the total number of coins is 25. Therefore the number of quarters is actually 25 - (total of nickels and dimes).
25-(2x+4) = no of quarters
25 - 2x - 4 = no quarters
21 - 2x = no quarters
Values of the coins (in cents) is the value of each coin (in cents) times the number of coins:
5*x = value of nickels
10*(x+4)= value of dimes
25*(21-2x) = value of quarters
Equation is the total value of the coins in CENTS = 320:
5x + 10(x+4) + 25(21-2x) = 320
5x + 10x + 40 + 525 - 50x = 320
-35x + 565 = 320
Subtract 565 from each side:
-35x + 565 - 565 = 320-565
-35x = -245
Divide both sides by -35 and hope it comes out even!!
Nickels
Dimes
Quarters
Check the Values:
x = 7 Nickels = $0.35
x+4 = 11 Dimes= $1.10
21-2x = 7 Quarters = 1.75
TOTAL Value of coins = $3.20
It checks!!
R^2 at SCC
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