Question 1108410
<br>
Solution by logical analysis....<br>
If all 24 items were boxes of pens, the total cost would have been $240.
The actual cost, $200, is $40 less than that.
Each time one box of pens is replaced by a box of rubber bands, the cost goes down by $5.
To bring the cost down by the required $40, the number of boxes of rubber bands must be 40/5 = 8.<br>
Answer: 8 boxes of rubber bands and 16 boxes of pens.<br>
Standard algebraic solution....<br>
let x = number of boxes of pens
let y = number of boxes of rubber bands
then...
{{{x+y = 24}}}  the total number of boxes was 24
{{{10x+5y = 200}}}  the total cost was $200<br>
multiply the first equation by 5 and subtract from the second equation to eliminate y:
{{{10x+5y = 200}}}
{{{5x+5y = 120}}}
{{{5x = 80}}}
{{{x = 16}}}<br>
As before, of course, the answer is 16 boxes of pens and 8 of rubber bands.<br>
You can also solve this kind of "mixture" problem (you are mixing boxes of pens at one price and boxes of rubber bands at another price) using the method of alligation.  Here is how it works....<br>
{{{matrix(3,3,240,"",80,"",200,"",120,"",40)}}}<br>
The numbers in the first column are the total cost if all 24 boxes were either pens or rubber bands; the number in the middle column is the actual cost of the 24 boxes.  The numbers in the third column are the differences, calculated diagonally, between the numbers in the first and second columns.  When the calculations are performed this way, the numbers in the third column show the ratio in which the two kinds of boxes must be "mixed".<br>
That ratio is 80:40, or 2:1; that means you need twice as many boxes of pens as rubber bands.  And with 24 boxes total, that means 16 boxes of pens and 8 boxes of rubber bands.