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In a certain class, 1/5 of the boys are shorter than

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In a certain class, 1/5 of the boys are shorter than [#permalink] New post 21 Aug 2017, 12:09
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In a certain class, \(\frac{1}{5}\) of the boys are shorter than the shortest girl in the class, and \(\frac{1}{3}\) of the girls are taller than the tallest boy in the class. If there are 16 stu­dents in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

A. 25%

B. 50%

C. 62.5%

D. 66.7%

E. 75%
[Reveal] Spoiler: OA

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Re: In a certain class, 1/5 of the boys are shorter than [#permalink] New post 19 Sep 2017, 08:18
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I've done it in many ways but I am continuing to get choice E. How is it possible?
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Re: In a certain class, 1/5 of the boys are shorter than [#permalink] New post 19 Sep 2017, 09:44
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Given the information about the fraction of boys who are shorter than the shortest girl and the fraction of girls who are taller than the tallest boy, the number of boys in the class must be a multiple of 5 while the number of girls must be a multiple of 3 (since the number of boys and girls must take on integer values). The only combination of a multiple of 5 and multiple of 3 that sum to 16 is 2 x 5 = 10 boys and 2 x 3 = 6 girls. Of the 10 boys, 2 are shorter than the shortest girl. Of the 6 girls, 2 are taller than the tallest boy. Therefore, the order of heights must be as follows:

B,B,G _,_,_,_,_,_, B,G,G

This implies that there must be exactly 10 students that are taller than the shortest girl and shorter than the tallest boy. This equals \(\frac{10}{16} = \frac{5}{8} = 62.5 %\)


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Re: In a certain class, 1/5 of the boys are shorter than   [#permalink] 19 Sep 2017, 09:44
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