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Practice Question 1.21 PreAlgebra ratio and percents

 
 
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Practice Question 1.21 PreAlgebra ratio and percents
by Anata Bezman - Sunday, March 15, 2020, 7:37 PM
 

It seems to me that there is missing information in the question or I don't quite understand the solution. Could you please clarify? 

 
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Re: Practice Question 1.21 PreAlgebra ratio and percents
by Areteem Professor - Monday, March 16, 2020, 11:22 AM
 

The first goal of the solution is to find the total number of attendees to the Math Olympiad classes. Noting that $\dfrac{8}{17}$ of them are in 6th grade and that $\dfrac{9}{23}$ are in 7th grade, $\dfrac{8}{17} + \dfrac{9}{23} = \dfrac{337}{391}$ of the attendees are in the 6th or 7th grade. As this fraction represents an integer number of students, and is in lowest terms, the total number of students who attend the Math Olympiad classes has to be a multiple of the denominator of the fraction. Thus, the possible number of students who attend the Math Olympiad classes are $391 \times 1 = 391$, $391 \times 2 = 782$, etc. Seeing that there are only $780$ students in the school, it is not possible for $782$ (or more) of them to be enrolled in the Math Olympiad classes. Thus, there are exactly $391$ students attending the Math Olympiad classes.