Question 1100038
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<pre>
n + d + q = 500,   (1)

2n = 3d            (2)      ("For every 2 nickels, there are 3 dimes.")

2d = 5q            (3)      ("For every 2 dimes, there are 5 quarters.")


From (2),  n = {{{(3/2)*d}}}.

From (3),  q = {{{(5/2)*d}}}.


Substitute  (4)  and  (5) into (1).  You will get

{{{3/2)*d}}} + d + {{{(5/2)*d}}} = 500.

Multiply by 2 both sides:


3d + 2d + 5d = 1000  ===>  10d = 1000  ====>  d = {{{1000/10}}} = 100.


<U>Answer</U>.  100 dimes,  150 nickels  and  250  quarters.
</pre>

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O-o-o-p-s !


The other tutor CORRECTLY DETERMINED that the above solution is wrong.


I agree, it is WRONG !


So, below I place the corrected solution.  Thanks,  @richwmiller !
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<pre>
n + d + q = 500,   (1)

{{{n/2}}} = {{{d/3}}}            (2)      ("For every 2 nickels, there are 3 dimes.")

{{{d/2}}} = {{{q/5}}}            (3)      ("For every 2 dimes, there are 5 quarters.")


From (2),  n = {{{(2/3)*d}}}.

From (3),  q = {{{(5/2)*d}}}.


Substitute  (4)  and  (5) into (1).  You will get

{{{(2/3)*d}}} + d + {{{(5/2)*d}}} = 500.

Multiply by 6 both sides:


4d + 6d + 15d = 3000  ===>  25d = 3000  ====>  d = {{{3000/25}}} = 120.


<U>Answer</U>.  120 dimes,  {{{(2/3)*120}}} = 80 nickels  and  {{{(5/2)*120}}} = 300 quarters.
</pre>