Question 201020
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Right off the bat you can tell that 116 is not correct.  116 is not evenly divisible by either 3 or 6, so you would have a fractional part of a student bringing their lunch and another fractional part of a student going home.  Basically, someone's head and shoulders went home and the rest of their body stayed at school and ate the lunch they brought.


Let *[tex \Large x] represent the number of students in the two classes.  One third of them bring their lunch, *[tex \Large \frac{x}{3}], one fourth of them buy, *[tex \Large \frac{x}{4}], one-sixth go home, *[tex \Large \frac{x}{6}], and 18 don't eat lunch.  And the sum of all these pieces is *[tex \Large x], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{x}{3} + \frac{x}{4} + \frac{x}{6} + 18]


The lowest common denominator of 3, 4, and 6 is 12, so multiply both sides of the equation by 12:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x = 4x + 3x + 2x + (12\cdot18)] 


Collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x = 9x + (12\cdot18)]


Add *[tex \Large -9x] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x = 12\cdot18]


Multiply both sides by *[tex \Large \frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 4\cdot18 = 72]


Check the answer:  One-third of 72 is 24, one-fourth of 72 is 18, one-sixth of 72 is 12.  24 plus 18 plus 12 plus 18 is 72.  Answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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