SOLUTION: Hello, i am having problems setting up this problem that consists of coins. 1. The amount of money in a bag is $7.70. There are 62 coins consisting of nickels, dimes, and quarte

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Question 920612: Hello, i am having problems setting up this problem that consists of coins.
1. The amount of money in a bag is $7.70. There are 62 coins consisting of nickels, dimes, and quarters. If there are three times as many dimes as nickels, how many coins of each are there?



Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
1. The amount of money in a bag is $7.70. There are 62 coins consisting of nickels, dimes, and quarters. If there are three times as many dimes as nickels, how many coins of each are there?
Value Equation:: 5n + 10d + 25q = 770 cents
Quantity Eq::::: n + d + q = 62
Quandity Eq::::: d = 3n
-----------------------------
Modify::
n + 2d + 5q = 154
n + d + q = 62
d = 3n
----------------
Substitute::
n + 6n + 5q = 154
n + 3n + q = 62
--------
Simplified Form::
7n + 5q = 154
4n + q = 62
-----
Modify for elimination::
7n + 5q = 154
20n + 5q = 310
-------------------
Subtract and solve for "n":
13n = 156
n = 12 (# of nickels)
d = 3n = 36 (# of dimes)
n + d + q = 62
12 + 36 + q = 62
q = 14 (3 of quarters)
==============
Cheers,
Stan H.
--------------

Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

Hello, i am having problems setting up this problem that consists of coins.
1. The amount of money in a bag is $7.70. There are 62 coins consisting of nickels, dimes, and quarters. If there are three times as many dimes as nickels, how many coins of each are there?


Let number of nickels, and quarters, be N, and Q, respectively

Since there are 3 times as many dimes as nickels, then: D = 3N 

N + 3N + Q = 62 ------- Equation expressing the number of coins

    4N + Q = 62 ------- eq (i)



.05N + .1(3N) + .25Q = 7.7 ----- Equation expressing the value  of all the coins

.05N + .3N + .25Q = 7.7

      .35N + .25Q = 7.7 -------- eq (ii)



You now have 2 equations (i & ii) that you can solve. My choice would be either:

1) to solve equation (i) for Q, and substitute that value for Q in eq (ii), or

2) to apply the elimination method of solving simultaneous equations

Either way, the value derived for N: the number of nickels, MUST BE an INTEGER,

and so MUST the value for Q: the number of quarters.



Then do a check!! 

===================

If you need a complete and detailed solution, let me know!!



Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.

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