SOLUTION: Help! I'm stuck on this problem that looks pretty simple but I can't can't figure out the answer:
There were 13 more nickels than dimes and the total value of the coins was $1.8
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: Help! I'm stuck on this problem that looks pretty simple but I can't can't figure out the answer:
There were 13 more nickels than dimes and the total value of the coins was $1.8
Log On
Question 50144: Help! I'm stuck on this problem that looks pretty simple but I can't can't figure out the answer:
There were 13 more nickels than dimes and the total value of the coins was $1.85. Find the number of nickels and the number of dimes.
I know this problem has to do with solving a system with two equations but I can't find the correct two equations.
Thanks a bunch
-Molly Answer by lkwtcnj(8) (Show Source):
You can put this solution on YOUR website! Since there's more nickels, let the number of dimes be n and the number of nickels is then n + 13. Then the equation becomes:
(0.05 * (n + 13)) + (0.10 * n) = 1.85
Just reduce the equation to get n:
0.05n + 0.65 + 0.10n = 1.85
0.15n = 1.2
n = 8
Then there are 8 dimes and 21 nickels. Hope that helps!