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?
dimes--- x
quarters -- x+2
nickels --- 3(x+2) = 3x + 6
10x + 25(x+2) + 5(3x+6) = 180
10x + 25x + 50 + 15x + 30 = 180
50x =100
x = 2
so 2 dimes
and 4 quarters
and 12 nickels
check:
10(2) + 4(25) + 5(12) = 180
all is good
Thanks so much Reiny.
To solve this problem, let's break it down step by step:
Let's assume the number of dimes the newspaper carrier has is 'x'.
Since the newspaper carrier has two more quarters than dimes, the number of quarters he has is 'x + 2'.
Since the newspaper carrier has three times as many nickels as quarters, the number of nickels he has is '3(x + 2)' or '3x + 6'.
Now, let's calculate the values of the coins in terms of cents:
The value of a dime is 10 cents, so the value of 'x' dimes is 10x cents.
The value of a quarter is 25 cents, so the value of 'x + 2' quarters is 25(x + 2) cents.
The value of a nickel is 5 cents, so the value of '3x + 6' nickels is 5(3x + 6) cents.
We know that the total change the newspaper carrier has is $1.80 or 180 cents.
So, we can create the equation:
10x + 25(x + 2) + 5(3x + 6) = 180
Now, let's solve the equation:
10x + 25x + 50 + 15x + 30 = 180
50x + 80 = 180
50x = 100
x = 2
Now that we have found 'x' (the number of dimes), we can substitute it back into our equations to find the number of quarters and nickels:
Number of dimes = x = 2
Number of quarters = x + 2 = 2 + 2 = 4
Number of nickels = 3(x + 2) = 3(2 + 2) = 3(4) = 12
Therefore, the newspaper carrier has 2 dimes, 4 quarters, and 12 nickels.