Chad has $1.80 in coins consisting of nickels and dimes only. If the total value
of his nickels is twice that of his dimes, how many coins does Chad have?
Let’s assume that Chad has x nickels and y dimes.
The value of the nickels is 5x cents, and the value of the dimes is 10y cents.
We can create a system of equations based on the given information:
Equation 1: x + y = the total number of coins
Equation 2: 5x + 10y = 180 cents (since Chad has $1.80, which is equal to 180 cents)
Since the equation 1 is linear and easy to solve for x, let’s solve it first.
From Equation 1, we can solve for x:
x = the total number of coins - y
Substituting x in Equation 2 with the above equation:
5(total number of coins - y) + 10y = 180
Simplifying the equation:
5(total number of coins) - 5y + 10y = 180
5(total number of coins) + 5y = 180
5(total number of coins) = 180 - 5y
Now we can solve for the total number of coins in terms of y:
(total number of coins) = (180 - 5y)/5
(total number of coins) = 36 - y
Since the total number of coins must be a positive integer, (180 - 5y) must be divisible by 5. We can find all the possible values of y that satisfy this condition.
180 - 5y = 5 * 36 => 180 - 5y = 180 => -5y = 0
Since there is no solution for this equation, we can conclude that there is no possible value for y.
Therefore, Chad does not have any coins.