Expert: Sherry Wallin - 1/5/2010

Question
there were 50 nickels, quarters, and dimes in the pot. their total value was $6.40 there were four more quarters than nickels, how many coins of each kind were there?

Answer
Each nickel has a value of $.05 and each quarter has a value of $.25 and each dime has a value of $.10
There are three equations here so you have a system of three equations and three unknowns:
Let n= # of nickels, d = # of dimes and q = # of quarters. You have 50 coins so n + d + q  = 50 (call this equation 1)
and .05n + .10d + .25q = 6.40 . I prefer to get rid of decimals so I multiply this last equation by 100 to get 5n + 10d + 25q = 640, (call this equation 2). Since there are 4 more quarters than nickels this means that n + 4 = q. I can use n + 4 for my q in both my equations. n + d + 4 + n  = 50 which simpifies to 2n + d + 4 = 50 -> 2n + d = 46 (this is my new equation 1) and 5n + 10d + 25(n + 4) = 640 which simplifies to 30n + 10d + 100 = 640 -> 30n + 10d = 540 (this is my new equation 2).

Let's examine what we now have:

2n + d = 46
30n + 10d = 540 which is now 2 equations with just 2 unknowns

multiply the first equation by -10 to get rid of the dimes
-20n -10d = -460
30n + 10d = 540  add these

10n = 80 so n = 8, now substitute n=8 into either of the equations above to find the number of dimes
2(8) + d = 46 so d = 30

Don't forget that n + 4 = q so 8 + 4 = q so q = 12

Now what you have found is that you have 8 nickels, 30 dimes and 12 quarters. Check this to make sure you actually have 50 coins and that the sum total of money is $6.40

8+30+12 = 50 coins
$.40 + $3.00 + $3.00 is $6.40

Math prof