SOLUTION: There are total of 41 coins in the piggy bank that consists of nickels, dimes, and quarters. The total value is $4.95. The number of dimes is seven less than the number of quarters

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Question 994316: There are total of 41 coins in the piggy bank that consists of nickels, dimes, and quarters. The total value is $4.95. The number of dimes is seven less than the number of quarters. Find the number of nickels dimes and quarters in the piggy bank
Found 3 solutions by CubeyThePenguin, ikleyn, josgarithmetic:
Answer by CubeyThePenguin(3113)   (Show Source): You can put this solution on YOUR website!
n = number of nickels
d = number of dimes
q = number of quarters

n + d + q = 41
5n + 10d + 25q = 495
d = q - 7

Substitute the third equation into the first and second equations.

n + 2q = 48
5n + 35q = 565 ---> n + 7q = 113

n + 2q = 48
-n - 7q = -113
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-5q = -65
q = 13 ---> d = 6 ---> n = 22

There are 22 nickels, 6 dimes, and 13 quarters.
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
There are total of 41 coins in the piggy bank that consists of nickels, dimes, and quarters.
The total value is $4.95. The number of dimes is seven less than the number of quarters.
Find the number of nickels dimes and quarters in the piggy bank
~~~~~~~~~~~~~


            It is a typical problem to be solved using only one single unknown and only one single equation. 


    
Let x be the number of quarters.

Then the number of dimes is (x-7), according to the condition.

The nickels are the rest coins, and their number is  (41-x-(x-7)) = (48-2x).


Having it, you write the total money equation


   25x + 10(x-7) + 5*(48-2x) = 495   cents.


Simplify and solve


    25x + 10x - 70 + 5*48 - 10x = 495

          25x                   = 495 + 70 - 5*48 = 325

            x                                     =  = 13.


ANSWER.  13 quarters, 6 dimes and the rest 41 - 13 - 6 = 22  coins are nickels.


CHECK.  22*5 + 6*10 + 13*25 = 495 cents, in total.    ! Correct !

Solved.


--------------------


It is how this problem is intended and is expected to be solved.


It is intended for young students, who didn't learn the conception of system of equations yet,
and the major goal of such problems is to teach such young students TO THINK.



Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!




E1 and E2:









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