SOLUTION: John has a collection of nickels,dimes, and quarters. Altogether, there are 241 coins worth $26.20. There are 4 more nickels than dimes. How many of each type of coin are in the co

Algebra ->  Equations -> SOLUTION: John has a collection of nickels,dimes, and quarters. Altogether, there are 241 coins worth $26.20. There are 4 more nickels than dimes. How many of each type of coin are in the co      Log On


   

Question 742235: John has a collection of nickels,dimes, and quarters. Altogether, there are 241 coins worth $26.20. There are 4 more nickels than dimes. How many of each type of coin are in the collection.
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
# of Coins: N-nickels, D-dimes, Q-quarters
there are 241 coins - means N%2BD%2BQ=241
worth $26.20 - means 0.05N%2B0.10D%2B0.25Q=%2426.20, but if we multiply 20 we get
N%2B2D%2B5Q=524
There are 4 more nickels than dimes means D%2B4=N
So we have 3 equations and 3 unknowns:
N%2B2D%2B5Q=524
N%2BD%2BQ=241
D%2B4=N
So we begin...
5 times the second equation minus the first equation will eliminate Q:
5%28N%2BD%2BQ%29-%28N%2B2D%2B5Q%29=5%28241%29-524
5N%2B5D%2B5Q-N-2D-5Q=681
4N%2B3D=681
Since N=D+4 we can write 4N%2B3D=681 as
4%28D%2B4%29%2B3D=681
7D%2B16=681
7D=665
D=95 (95 dimes)
This means that there D+4=95+4=99 nickels
Finally we can see that there must be 241-99-95=47 quarters
(so there are 99 nickels, 95 dimes and 47 quarters)