SOLUTION: Margaux has coins in quarter, dimes, and nickels worth $58.5 in all. If the worth of the nickels is the same as that of the quarters, then her money became $77. And if the number o

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Question 1200117: Margaux has coins in quarter, dimes, and nickels worth $58.5 in all. If the worth of the nickels is the same as that of the quarters, then her money became $77. And if the number of nickels is the same as that of the dimes, then the worth of her money either becomes $52.5 or $50.5. How many coins does she have for each denomination?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


The problem involves quarters, dimes, and nickels, so I assume the problem comes from the US.

But the language is not standard English....

We are first told that the total value of the coins is $58.50. That is clear.

But then I have no idea what this means: "If the worth of the nickels is the same as that of the quarters, then her money became $77." Apparently some coins were added to make a new total. Conceptually that is fine; but I can't tell from the wording what coins were added.

And then there is a similar statement saying that (apparently) some coins were taken away, making the new total "either $52.50 or $50.50". I can't make any sense out of how changing the coins can result in two possible different new totals.

Re-post the problem, using standard English....

Answer by ikleyn(52886) About Me  (Show Source):
You can put this solution on YOUR website!
.
Margaux has coins in highlight%28cross%28quarter%29%29 quarters, dimes, and nickels worth $58.5 in all.
If the worth of the nickels is the same as that of the quarters,
then her money became $77. And if the number of nickels is the same
as that of the dimes, then the worth of her money either becomes $52.5 or $50.5.
How many coins does she have for each denomination?
~~~~~~~~~~~~~~~~~~~


        It is really non-standard formulation,  which is difficult to understand,
        but I will try to do my best. From my analysis,  you will see the truth. 


Let the number of nickels  be n and let N be worth of nickels.

Let the number of dimes    be d and let D be worth of dimes.

Let the number of quarters be q and let Q be worth of quarters.


First equation is obvious

    N + D + Q = 5850  cents.        (1)


To get second equation, we should replace in equation (1) N by Q
and replace 5850 cents there by 7700 cents.  So, the second equation is

    Q + D + Q = 7700  cents.        (2)   <<<---=== it is the literal translation
                                                    of the second statement.


We can rewrite equation (1) in the form

    5n + 10d + 25q = 5850,  or, reducing by factor 5,   n + 2d + 5q = 1170.     (1')


Similarly, we can rewrite equation (2) in the form

    25q + 10d + 25q = 7700,  or, reducing by factor 5,  5q + 2d +  5q = 1540.   

or, combining like terms,                                    2d + 10q = 1540,

or, reducing by factor 2

     d + 5q = 770.                 (2')


Third equation is

    5d + 10d + 25q = 5250 cents    (3a)
or                                          <<<---=== this "or" is from the problem's third statement
    5d + 10d + 25q = 5050 cents.   (3b)


Reducing (3a) and (3b) by 5, we get equations (3a) and (3b) simplified

    d + 2d + 5q = 1050             (3a')
or
    d + 2d + 5q = 1010.            (3b')


Combining like terms in equations (3a') and (3b'), we get

        3d + 5q = 1050             (3a")
or
        3d + 5q = 1010.            (3b")


This service word "OR" between equations (3a") and (3b") is misterious and suspicious, 
but it can be interpreted as if the system of equations (1'), (2') and (3") has two possible solutions.

I will keep it in mind and will see later what does it mean and how to work with it.


So, my system of equations for now is combination of equations (1'), (2') and (3")

    n + 2d +  5q = 1170,            (4)    (as (1'))

         d +  5q =  770,            (5)    (as (2'))

        3d +  5q = 1050.            (6)    (as (3a"))


To solve it, subtract equation (5) from (6) and get

        2d       = 1050 - 770 = 280  --->  d = 280/2 = 140.


Then from equation (5)

              5q =  770 - d = 770 - 140 = 630  --->  q = 630/5 = 126.


Then from equation (4) 

    n = 1170 - 2d - 5q = 1170 - 2*140 - 5*126 = 260.


ANSWER.  260 nickels;  140 dimes and 126 quarters.

Solved.

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

The highlight%28highlight%28diagnosis%29%29 to the problems formulation is THIS:

        this way formulating with the service word "OR"
        in the last problem' statement is incorrect/absurdist.

Probably, it was an attempt to pack two different problems in one package; but in true Math,
nobody and never does it. It is the first time in my life I see such absurdist formulation.


///////////////////


For the clarity: the problem is solved in this formulation 
    Margaux has coins in quarters, dimes, and nickels worth $58.5 in all. 
    Had the worth of the nickels be the same as that of the quarters, 
    then her money would be $77. And had the number of nickels would be 
    the same as that of the dimes, then the worth of her money would be $52.5. 
    How many coins does she have for each denomination?


The original formulation in the post was wrong, absurdist, incorrect and nonsensical.

Therefore, I changed and edited it to make sense (from nonsense).