Question 1170563
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Certainly, if a formal algebraic solution is required, taking the time to analyze the problem to set up the problem using a single variable is much better than using several variables.<br>
But if a formal algebraic solution is not required, an informal solution using logical analysis and simple arithmetic can get you to the answer faster; and it can give you some very good brain exercise (i.e., valuable problem solving experience).<br>
(1) Because the number of quarters is twice the number of dimes, we can group the quarters and dimes into groups each containing two quarters and one dime.
(2) The value of each such group is 60 cents; that means the total value of the quarters and dimes is a multiple of 60 cents.
(3) The largest multiple of 60 cents that is less than the total of $4.10 (410 cents) is 6 times 60 cents, or 360 cents, or $3.60.  That leaves 50 cents to be made using the nickels.<br>
Check that answer.  The 6 groups of quarters and dimes contain a total of 6*3=18 coins; the 50 cents in nickels contains 50/5 = 10 coins.  18+10 = 28; the number of coins is right, so the solution is correct.<br>
ANSWER: 6 dimes and 12 quarters; and 10 nickels.<br>