1. Joe has 37 coins (all nickels, dimes and quarters) worth $5.50. She has 4 more quarters than nickels. How many dimes does Joe have?
To solve this problem, let's set up some equations. Let's represent the number of nickels, dimes, and quarters as "n", "d", and "q" respectively.
We know that Joe has a total of 37 coins, so we can write the first equation as:
n + d + q = 37
We also know that the total value of all the coins is $5.50. We need to convert this to cents since all our coin values are in cents:
5n + 10d + 25q = 550
Additionally, we're given that Joe has 4 more quarters than nickels, which can be expressed as:
q = n + 4
Now we have a system of three equations. We can solve this system using substitution or elimination to find the values of n, d, and q. Then we can determine how many dimes Joe has.
Let's use substitution to solve the system:
1. Substitute q = n + 4 into the other equations:
n + d + (n + 4) = 37
5n + 10d + 25(n + 4) = 550
2. Simplify the first equation:
2n + d + 4 = 37
5n + 10d + 25n + 100 = 550
3. Rearrange the first equation to solve for d:
d = 33 - 2n
4. Substitute d = 33 - 2n into the second equation:
5n + 10(33 - 2n) + 25n + 100 = 550
5. Simplify and solve for n:
5n + 330 - 20n + 25n + 100 = 550
10n + 430 = 550
10n = 550 - 430
10n = 120
n = 120 / 10
n = 12
Now that we have the value of n, we can substitute it back into the first equation to find d:
12 + d + (12 + 4) = 37
d + 28 = 37
d = 37 - 28
d = 9
Therefore, Joe has 9 dimes.
number of nickels ---- x
number of quarters -- x+4
number of dimes = 37-(x+x+4) = 33 - 2x
5x + 25(x+4) + 10(33-2x) = 550
solve for x
let me know what you get
I got:
Nickels - 12
Quarters - 16
Dimes - 9
I was not sure how to set up the problem. I'm good with two variables but not three.
thx