SOLUTION: Part of the proceeds from a garage sale was $410 worth of $10 and $20 bills. If there were 2 more $10 bills than $20 bills, find the number of each denomination.

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Question 1123771: Part of the proceeds from a garage sale was $410 worth of $10 and $20 bills. If there were 2 more $10 bills than $20 bills, find the number of each denomination.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
Let x be the number of the $20 dollars bills.

Then the number of the $10 dollars bills is (x+2), according to the condition.


The money equation is


    20x + 10*(x+2) = 410


Simplify and solve for x.  


    20x + 10x + 20 = 410

    30x = 410-20 = 390  ====>  x =  = 13.


Answer.  13 of the $20-dollars bills and 13+2 = 15  of the $10-dollars bills.


Check.   20*13 + 10*15 = 410 dollars.   ! Correct !


Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Here is a solution using logical reasoning that uses virtually the same calculations as the formal algebraic solution shown by the other tutor.

(1) First count the two "extra" $10 bills. That leaves equal numbers of $10 and $20 bills, with a total value of $390.

(2) Think of the remaining bills in groups of one $10 bill and one $20 bill. The value of one $10 bill and one $20 bill is $30.

(3) The number of groups, each worth $30, required to make the remaining $390 of the total is 390/30 = 13.

So the number of $20 bills is 13; and the number of $10 bills is 13+2 = 15.
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