SOLUTION: 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

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Question 1114184: 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?
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

-
The last two equations allow you to quickly find q.




--------19 quarters

Revise to simpler system:


----------11 dimes

---------6 nickels
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~ 

(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.


Answer.  11 dimes,  19 quarters  and  17-11 = 6 nickels.


Check.   5*6 + 11*10 + 19*25 = 615 cents.   ! Correct !


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
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? 
Let the number of nickels, dimes, and quarters be N, D, and Q, respectively

Then we get: .05N + .1D + .25Q = 6.15 ------ eq (i)

Also, N + D + Q = 36 ------- eq (ii)

In addition, N + D = Q - 2______N + D - Q = - 2 ------ eq (iii)

2Q = 38 ------ Subtracting eq (iii) from eq (ii)





.05N + .1D + .25(19) = 6.15 ------ Substituting 19 for Q in eq (i)

.05N + .1D + 4.75 = 6.15

.05N + .1D = 1.4 ------- eq (iv)



N + D = 19 - 2 ------- Substituting 19 for Q in eq (iii)

N + D = 17 ------ eq (v)

- .1N - .1D = - 1.7 ------- Multiplying eq (v) by - .1 ------ eq (vi)

- .05N = - .3 ------ Adding eqs (vi) & (iv)



6 + D = 17 ------ Substituting 6 for N in eq (v)

 

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