SOLUTION: Tracey is counting all the change she has been saving in her car. She only collects silver coins and finds that she has eight less dimes than nickels, and has four less than twice

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Question 1170767: Tracey is counting all the change she has been saving in her car. She only collects silver coins and finds that she has eight less dimes than nickels, and has four less than twice as many quarters as nickels. If she has $9.25 in her car all together, how many of each coin does she have?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52866) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x = # of nickels.

then the number of dimes = x-8 and

     the number of quarters is 2x-4.


You then write the total money equation

    
    5x + 10*(x-8) + 25*(2x-4) = 925  cents.


Simplify and solve


    5x + 10x - 80 + 50x - 100 = 925

    5x + 10x      +50x        = 925 + 80 + 100

         65x                  = 1105

           x                  = 1105/65 = 17.


ANSWER.  17 nickels, 17-8 = 9 dimes and 2*17-4 = 30 quarters.


CHECK.   17*5 + 9*10 + 30*25 = 925 cents.   ! Precisely correct !

Solved.



Answer by greenestamps(13206) About Me  (Show Source):
You can put this solution on YOUR website!


Compare the good formal algebraic solution from tutor @ikleyn to the following informal solution.

(1) Add 8 more dimes so that the number of dimes is the same as the number of nickels; the total value is now $9.25+$0.80 = $10.05.
(2) Add 4 more quarters so that the number of quarters is twice the number of nickels; the total value is now $10.05+$1 = $11.05.

(3) Now the numbers of nickels and dimes are the same, and the number of quarters is twice that number. Group the coins into groups of 2 quarters, 1 dime, and 1 nickel. Each of those groups has a value of $0.65; the number of groups to make the total $11.05 is 11.05/0.65 = 17.

So at this point we have 17 nickels, 17 dimes, and 2*17 = 34 quarters.

Now take away the 8 dimes and the 4 quarters we added to the original collection of coins to get the final answer of 17 nickels, 9 dimes, and 30 quarters.

The informal solution uses virtually the same computations as the formal algebraic solution. So the comparison of the two solution methods does two things:

(1) It shows you that the formal algebra is not magic -- it actually does exactly what common sense and logical reasoning says you should do; and
(2) it shows you that, if a formal algebraic solution to a problem like this is not needed, common sense and logical reasoning can get you to the answer.