Question 1188035
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Cindy, Derrick and Edmund had $410 altogether. Cindy gave 1/3 of her money to
Edmund, Derrick then gave 2/5 of his money to Cindy. As a result, Derrick had $20
less than Cindy and $30 less than Edmund. How much money did Cindy have at first?
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            Let solve the problem in a backward manner.



<pre>
Let x be what Derrick had finally.

Then Cindy had (x+20) dollars at the end;  Edmund had (x+30) dollars at the end.


The total is $410;  it did not change after these interior changes.  So


    x + (x+20) + (x+30) = 410

           3x + 50 = 410

           3x = 410 - 50 = 360

            x = 360/3 = 120  dollars.


Thus at final, Derrick had  120 dollars;  Cindy had 120+20 = 140 dollars  and  Edmund had 120+30 = 150 dollars.



Now we will make one step back, from the end to the beginning.



    As the problem says, Derrick gave 2/5 of his money to Cindy.

    Hence, Derrick left with only 3/5 of his money, and this 3/5 of his original money is 120 dollars.

    It means that originally Derrick had  {{{(5/3)*120}}} = 5*40 = 200 dollars.


       HENCE, Derrick gave  2/5  of $200,  or 80 dollars to Cindy.


   
    The problem says: "Cindy gave 1/3 of her money to Edmund".

    So, Cindy left with 2/3 of her original money, PLUS 80 dollars she got from Derrick.


    So, we write this equation  for the Cindy original amount "C"

        {{{(2/3)*C + 80}}} = 140  dollars.


    We solve this equation easily and get  {{{(2/3)*C}}} = 140-80 = 60;  hence  C = {{{(3/2)*60}}} = 90.


<U>ANSWER</U>.  Cindy had 90 dollars, at first.
</pre>

Solved.


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Solving backward allows to untangle the problem and obtain the solution in a simple manner.