A newspaper carrier has $1.80 in change. He has two more quarters than dimes but three times as many nickels as quarters. How many coins of each type does he have?
My! You have a major identity problem, John/Brian/Sam/Nancy/Jessica/Mark/Kaela!
Please use only one name for your posts.
To find out how many coins of each type the newspaper carrier has, we can set up a system of equations based on the given information.
Let's represent the number of quarters as "q", the number of dimes as "d", and the number of nickels as "n".
From the given information, the first equation states that the newspaper carrier has two more quarters than dimes:
q = d + 2
The second equation states that there are three times as many nickels as quarters:
n = 3q
We also know the values of the coins:
quarters (q) = $0.25
dimes (d) = $0.10
nickels (n) = $0.05
Now, let's convert the equation into cents (since we're working with coins):
quarters (q) = 25 cents
dimes (d) = 10 cents
nickels (n) = 5 cents
From the given information, the newspaper carrier has $1.80 in change, which is equal to 180 cents.
Now we can solve the system of equations:
1. q = d + 2
2. n = 3q
Substitute equation 2 into equation 1:
3q = d + 2
Now, substitute the known values:
3(25) = 10 + 2
75 = 12 + 2
75 = 14
This is not a valid equation, so let's adjust it:
3q = d + 2
3(25) = 10 + 2
75 = 10 + 2
75 = 12
This equation is also not valid, so let's adjust it again:
Multiply equation 1 by 3:
3q = 3(d + 2)
Now substitute equation 2 into the adjusted equation 1:
3(25) = 3(10 + 2)
75 = 3(10 + 2)
75 = 3(12)
75 = 36
Again, this equation is not valid. We made a mistake in setting up the equations. Let's reassess the information given.
The correct equation to represent the given information is:
q = d + 2
Since we know that there are three times as many nickels as quarters, we can set up another equation:
n = 3q
Now we can solve the system of equations:
1. q = d + 2
2. n = 3q
We also know the total value of the coins is $1.80 or 180 cents, so we can write another equation:
25q + 10d + 5n = 180
Now let's substitute equation 1 and equation 2 into the third equation:
25q + 10d + 5n = 180
25(d + 2) + 10d + 5(3q) = 180
25d + 50 + 10d + 15q = 180
35d + 15q + 50 = 180
35d + 15q = 130
Dividing this equation by 5, we get:
7d + 3q = 26
Now we can solve this equation simultaneously with equation 1:
q = d + 2
Substituting this back into the previous equation:
7d + 3(d + 2) = 26
7d + 3d + 6 = 26
10d + 6 = 26
10d = 20
d = 2
Now, substitute the value of d into equation 1:
q = 2 + 2
q = 4
Finally, we can use equation 2 to find n:
n = 3q
n = 3(4)
n = 12
Therefore, the newspaper carrier has 4 quarters, 2 dimes, and 12 nickels.