A newspaper carrier has $4.00 in change. He has two more quarters than dimes but five times as many nickels as quarters. How many of coins of each type does he have.
This is what I got so far and then I get lost.
dimes = x
quarters = x+2
nickels = 5(x+2)
10x+25(x+2)+5*4(x+2)= 400
55x + 90 -90 = 400 - 90
55x= 310
55x/55 = 310/55
x=
I am doing some wrong somewhere and I do not know where.
10x+25(x+2)+5*5(x+2)= 400
10x + 25x +50 + 25x + 50 = 400
55x = 300
x = 300/55 = 5.45
I found one mistake in you equation above (bold), but that still does come out right. Do you have any typos?
Your approach to setting up the equations is correct, but there seems to be a small mistake in your equation. Let's go through it step by step to figure out where the error is:
1. Let's assume that the number of dimes is represented by x.
2. According to the problem, the number of quarters is two more than the number of dimes, so it should be x+2.
3. The number of nickels is given as five times the number of quarters, so it should be 5(x+2).
Now, let's form the equation based on the values of the coins:
10x + 25(x+2) + 5(5(x+2)) = 400
Let's simplify this equation:
10x + 25x + 50 + 25x + 50 = 400
60x + 100 = 400
60x = 300
x = 300/60
x = 5
So, the number of dimes is 5, the number of quarters is x+2 = 5+2 = 7, and the number of nickels is 5(5+2) = 5(7) = 35.
Therefore, the newspaper carrier has 5 dimes, 7 quarters, and 35 nickels.