Questions involving change with algebra 1

Question

A storekeeper goes to the bank to get $10 worth of change . She requests twice as many quarters as half dollars, twice as many dimes as quarters, three times as many nickles as dimes, and no pennies or dollars. How many of each coin did the storekeeper get?
For this problem  can you give an explination on how you would do it  with all the twice as many  i get confused.

Answer 1

Let the first letter of each denomination stand for that coin. Then we have Q = 2xH, D = 2xQ, N + 3xD. Then we can get all denominations in terms of Q, letting Q be the number of quarters, and the number of other coins after conversion.

N = 3xD, and D= 2xQ, so  N = 3 x 2Q = 6Q. D = 2Q, Q = Q. H = (1/2)Q. Now, we know total of all coins is $10, so we add all coins thus:  $0.5 H + $0.25 Q + $0.1 D + $0.05 N = $10.00. Substituting our converted coins in terms of Q gives:

0.5 x (1/2)Q + 0.25 Q + 0.1 x 2Q + 0.05 x 6Q = 10.00.  0.25 Q + 0.25 Q + 0.2 Q + 0.3 Q = 10.00. Then 1Q = 10.00, so we have 10 quarters. We have half as many half dollars, so 5 of them. We have twice as many dimes as quarters, so that number is 20. We have 3 times as many nickels as dimes, so that is 60.  5 half dollars is $2.50, 10 quarters is $2.50.

20 dimes is $2.00. 60 nickels is $3.00. These all add up to $10.00.