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.
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?
1 answer