Question 1124531: Marilyn has $2.25 in change. The number of dimes ($0.10) is double that of the quarters ($0.25), And the number of nickels ($0.05) is triple the number of dimes. How many nickels, dimes, and quarters does she have?
Answer by greenestamps(13200)   (Show Source):
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The problem can be solved informally using logical reasoning:
(1) Take one quarter -- value $0.25.
(2) Since the number of dimes is twice the number of quarters, now take two dimes -- value $0.20. The total value of the coins is now $0.25+$0.20 = $0.45.
(3) Since the number of nickels is three times the number of dimes, now take 6 nickels -- value $0.30. The total value of the coins is now $0.45+$0.30 = $0.75.
The total Marilyn has is $2.25, which is 3 times $0.75. So what Marilyn has is three groups of coins, each containing 1 quarter, 2 dimes, and 6 nickels.
So what Marilyn has is 3 quarters, 6 dimes, and 18 nickels.
Solving the problem with formal algebra takes virtually the same path:
let x = number of quarters
then 2x = number of dimes
then 6x = number of nickels
x(.25)+2x(.10)+6x(.05) = 2.25
.25x + .20x + .30x = 2.25
.75x = 2.25
x = 2.25/.75 = 3
quarters: x = 3
dimes: 2x = 6
nickels: 6x = 18
You should, of course, understand the formal algebraic solution; for more complicated problems, the formal algebra will often be easier to use than informal logical reasoning.
But thinking through problems like this using logical analysis is good brain exercise.
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