If tommy has 175 dollars in 1, 5, 10 and 25 cent coins, how many coins of each would he have if he has 3576 coins total.
+561 Let w = number of 1 cent coins, x = number of 5 cent coins, y = number of 10 cent coins, and z = number of 25 cent coins.
We know two things.
\(0.01w+0.05x+0.1y+0.25z=175\)
\(w+x+y+z=3576\)
Let's eliminate w because we can.
\(0.01(3576-x-y-z)+0.05x+0.1y+0.25z=175\)
\(0.04x+0.09y+0.24z=139.24\)
There will be more than one solution for this problem. Let's settle for finding just one of them by saying z=0 and y=0.
\(0.04x=139.24\)
\(x=3481\)
\(w+3481=3576\)
\(w=95\)
One solution to this problem would be 95 one cent coins, 3481 five cent coins 0 ten cent coins, 0 twenty-five cent coins.
+37146 Don't really think that answered the question, Will ...it says there ARE 10 and 25 cent coins. We need more info for a discreet answer to this question....there are 4 variables and only two equations relating them, so there are possibly multiple solutions.
There are MANY solutions to this problem. We can, however, proceed as follows:
Set quarters to =$50 x 4=200 quarters
Set dimes to =$50 x 10=500 dimes
Amount remaining in nickels and pennies
=$75 for a total of 2,876 nickels and pennies
n + p =2876
.05n +.01p=7500, solve for n, p
p=1,720 x .01 =$17.20
n=1,156 x .05 =$57.80
d=500 x .10 =$50.00
q=200 x .25 =$50.00
Totals: 3,576 coins for a total of $175.00
+118677 Here are a whole lot more answers :)
http://www.wolframalpha.com/input/?i=Diophantine+equation+4b%2B9c%2B24d%3D14924
