<PRE><B>Question 364357</B>
warren has 40 coins (all nickels, dimes, and quarters) worth 4.05. He has 7 more nicles than dimes. how many quarters does he have?

Let Q = number of quarters
    N = number of nickels
    D = number of dimes
    N = D + 7

Equation 1 is

       Q + N + D = 40
Substite       N = D + 7
   Q + (D+7) + D = 40
        Q + 2D   = 40-7
        Q + 2D   = 33  
            2D   = 33 - Q
             D   = 16.5 - .5Q
  

Equation 2 is

     .25Q + .05N + .10D = 4.05
Substite  N = D + 7
 .25Q + .05(D+7) + .10D = 4.05
       .25Q +.05D + .35 + .10D = 4.05
       .25Q + .15D = 4.05 - .35
       .25Q + .15D = 3.7

Since D  = 16.5 - .5Q in equation 1
      .25Q + .15D = 3.7  (in equation 2) becomes

 .25Q + .15(16.5-.5Q) = 3.7
 .25Q + 2.475 - .075Q = 3.7
               .175Q  = 3.7 - 2.475
                   Q  = 1.225/.175
                   Q  = 7

      D = 16.5 - .5(7) = 13
      N = D + 7 = 13 + 7 = 20

To check:

.25(7) + .05(20) + .10(13) = 4.05

Warren has 7 quarters             


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