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# Dime and Nickel

Dime and Nickel Problem

A parking meter coin slot receives dimes and nickels. When the meter box is emptied, there were 148 coins and a total of $10.65. How many dimes and how many nickels were there? Solution Before we can derive a valid algebraic equations, we have to first define what values are represented with a dime and a nickel. In US currency, the following coins have the following monetary values: 1 US dollar = 100 cents 1 nickel = 5 cents 1 dime = 10 cents Therefore, based on this definition and the problem statement, we can derive the following representation: Let d = dime n = nickel Hence,$d + n = 148$Eq. 1 Total number of coins in the slot$0.1d + 0.05n = 10.65$Eq. 2 Total amount of all coins It will be easier to solve this system of equations in two unknowns by first eliminating the decimals in Eq. 2. Then using addition/elimination method we get$(0.1d + 0.05n = 10.65)(100)10d + 5n = 1065(d + n = 148)(-5)$Multiply Eq. 1 with -5 to eliminate n during addition$10d + 5n = 1065-5d - 5n = -7405d + 0n = 325$Divide both sides by 5$d = \frac{325}{5}d = 65$Substituting this value of d in Eq. 1, we will get the value of n as follows:$d + n = 14865 + n = 148n = 148 - 65n = 83$Therefore, there were 65 dimes and 83 nickels in the meter box. Check There were 148 coins: 65 (dimes) + 83 (nickels) = 148 The total money is$10.65: 65(0.1) + 83(0.05) = 10.65
6.5 + 4.15 = 10.65
10.65 = 10.65

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