Question 1100355
.
 Sam has $3.15 in quarters nickels and dimes. He has half as many quarters as he has dimes and nickels together.
He also has two more nickels than dimes. How many of each coin are there 
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<pre>
Let d be the number of dimes.  

Then the number of nickels "n" is n = (d+2),   according to the condition.

Also, the number of quarters is {{{(d + n)/2}}} = {{{(d+(d+2))/2}}} = {{{(2d+2)/2}}} = d+1,   according to the condition.


Thus dimes contribute  10*d cents to the total.

Nickels contribute      5*(d+2) cents to the total.    (<<<---=== because the number of nickels is (d+2), as it was said above)

Quarters contribute     25*(d+1) cents to the total.   (<<<---=== because the number of quarters is (d+1) as it was shown above)


Then your money equation is

Dimes + Nickels + Quarters = 315   cents,   or

10d   + 5*(d+2) + 25(d+1)  = 315.


Simplify and solve for d:

10d + 5d + 10 + 25d + 25    = 315

40d = 315 - 10 - 25 = 280  ====>  d = {{{280/40}}} = 7.


<U>Answer</U>.  There were 7 dimes,  d+2 = 7+2 = 9 nickels and  d+1 = 7+1 = 8 quarters. 


<U>Check</U>.   7*10 + 5*9 + 8*25 = 315.    ! Correct !
</pre>


<U>The lesson to learn from this solution</U>

<pre>
    This problem is FOR ONE unknown, NOT FOR THREE.

    Therefore your task is carefully choose the leading/the major unknown;

        then express other unknowns via this leading one;

        then to make your equation;

        then carefully solve it.
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It also can be solved mentally, without using any equations.



<pre>
For a moment, let's place aside two nickels and one quarter.


Then we will have a collection, which is worth  315-2*5-1*25 = 280 cents and has equal number 
of dimes and nickels, while the number of dimes and nickels together is twice the number of quarters.


From it, we logically conclude that in this modified collection, the number of quarters is the same 
as the number of dimes and is the same as the number of nickels.


It means that we can group the coins in the sets in a way that each set contains 
one quarter, one dime and one nickel.


It is easy to compute mentally that each such set is worth 25 + 10 + 5 = 40 cents.

Hence, the number of the sets is 280/40 = 7.


Thus, the modified collection has 7 quarters, 7 dimes and 7 nickels


Now, returning to the original collection, we find that it has 7+1 = 8 quarters, 7 dimes and 7+2 = 9 nickels.
</pre>

In this way, the problem is solved mentally.



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Here, in connection to this problem, some tutors propose to use three equations
in three unknown. Others propose to use online Internet solvers to ease suffering.


Some of tutors even propose to use existing AI.


While and although it is possible, I still think that it is wrong way teaching.


To me, it is straight way to raise many-many idiots everywhere and nobody else except of idiots.