# 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

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

$10d + 5n = 1065$

$-5d - 5n = -740$

$5d + 0n = 325$ 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:

$d + n = 148$

$65 + n = 148$

$n = 148 - 65$

$n = 83$

**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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