SOLUTION: I need some help with this problem: Justin has $6.50 in quarters, dimes, and nickels. If he has a total of 42 coins and he has twice as many dimes as nickels, how many of each coin
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: I need some help with this problem: Justin has $6.50 in quarters, dimes, and nickels. If he has a total of 42 coins and he has twice as many dimes as nickels, how many of each coin
Log On
Question 838753: I need some help with this problem: Justin has $6.50 in quarters, dimes, and nickels. If he has a total of 42 coins and he has twice as many dimes as nickels, how many of each coin does he have?
I think that I have the right equations, but everytime I try to solve, the answers either don't make sense or are wrong.
The equations I know are:
n+d+q=42 and 5n+10d+25q=650
The equation that I'm not sure about is the part in the problem where is says he has twice as many dimes as nickles. I've tried 2d-n=0 and then I solved for n which gave me: n=2d, but like I said, none of the answers I get when I try to solve make sense or are right.
Please help! Answer by ewatrrr(24785) (Show Source):
Hi
n+d+q=42
he has twice as many dimes as nickels
nickels: x
dimes: 2x
quarters (42-2x - x) 0r (42-3x)
5n+10d+25q=650 |Yes! CENTS make sense. Good work.
5x + 10(2x) + 25(42-3x) = 650¢ *** |Solving for x
5x + 20x + 1050 - 75x = 650
-50x = -400
x = 8, number oF Nickels. 16 Dimes and 18 Quarters (42-24)
CHECKING our answer***
40 + 160 + 25*18 = 40 + 160 + 450 = 650¢ 0r $6.50
Wish You the Best in your Studies.
