Question 1114184
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A box contains $6.15 in nickels , dimes, and quarters. There are 36 coins in all, and the sum of the numbers of nickels and dimes 
is 2 less than the number of quarters. How many coins of each kind are there? 
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(1)  nickels + dimes + quarters = 36      ( <<<---=== given )

     N       + D     + Q        = 36      ( the same )


(2)  N       + D   = Q - 2                ( <<<---=== given )


========>  Q + (Q-2) = 36

          2Q - 2 = 36  ====>  2Q = 36 + 2 = 38  ====>  Q = 19


So, we found number of quarters.  It is 19.

Next, you can reduce the problem from 3 unknowns to only two of them:


    we have 36-19 = 17 nickels and dimes, that are worth  6.15 - 19*0.25 = 1.40 dollars.


 N +   D =  17      (coins)    (1)
5N + 10D = 140      (cents)    (2)


Simplify

 N +   D =  17                 (1')
 N +  2D =  28                 (2')


Subtract eq(1') from eq(2').


      D = 28-17 = 11.


<U>Answer</U>.  11 dimes,  19 quarters  and  17-11 = 6 nickels.


<U>Check</U>.   5*6 + 11*10 + 19*25 = 615 cents.   ! Correct !
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