Let q stand for the number of quarters we have. We have twice as many nickels as quarters, so let nickels be 2q. The problem doesn't tell us how many dimes we have, but we know the remainder has to be dimes. So we start with the total number of coins and subtract the total of the rest of the coins leaving us with 56-(2q+q) to equal the dimes. Then I set up a table as follows to keep track of everything:
#of coins cents per coin total cents
nickels 2q 5 5(2q)
dimes 56-(2q+q) 10 10(56-(2q+q))
quarters q 25 25q
total 56 620
Now I can simply add the total cents of all coins and set it equal to 620 ($6.20 stated as cents)
5(2q)+10(56-(2q+q))+25q=620
Simplify and Solve:
10q+10(56-3q)+25q=620
10q+560-30q+25q=620
35q-30q+560=620
5q=60
q=12
I can determine the number of dimes and nickels by plugging the number back into # of coins and I get:
12 quarters, 20 dimes, 24 nickels
ALWAYS CHECK YOUR WORK!!!!!
I can check my work by adding the # of each coin multiplied times the cents per coin. The answer should be 620 (given in problem)
CHECK:
12*25+20*10+24*5=
300+200+120=620
