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Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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28 Jul 2010, 08:07

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Amy's grade was 90th percentile of the 80 grades for her class. Of the 100 grades from another class, 19 were higher than Amy's, and the rest were lower. If no other grade was the same as Amy's grade, then Amy's grade was what percentile of the grades of the two classes of two classes combined?

Amy's grade was 90th percentile of the 80 grades for her class. Of the 100 grades from another class, 19 were higher than Amy's, and the rest were lower. If no other grade was the same as Amy's grade, then Amy's grade was what percentile of the grades of the two classes of two classes combined?

A) 72nd B) 80th C) 81st D) 85th E) 92nd

If someone's grade is in \(x_{th}\) percentile of the \(n\) grades, this means that \(x%\) of people out of \(n\) has the grades less than this person.

Being in 90th percentile out of 80 grades means Amy outscored \(80*0.9=72\) classmates.

In another class she would outscored \(100-19=81\) students (note: Amy herself is not in this class).

So, in combined classes she outscored \(72+81=153\). As there are total of \(80+100=180\) students, so Amy is in \(\frac{153}{180}=0.85=85%\), or in 85th percentile.

Re: Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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12 Oct 2013, 04:29

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I solved this question using this logic:

If there are 19 grades above Amy in the second class, and no grades are repeated, that means that Amy has the 20th highest grade which means there are 80 grades beneath her, which means that she is in the 80th percentile.

Solving using weighted average, using the number of students as weights = [90th (80)+ 80th (100)] / 180 =85%
_________________

I was trying to do the math quickly so I did this:

I added the percents 90%+81% = 171% then I divided it by 2 to get a estimated average of 85...

Can i do this?

No, you cannot. What would be your answer if one of the options were 86%? Also, if the numbers in the question were different, your approximation could give wrong answer.
_________________

Re: Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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09 Nov 2015, 10:51

Is this ok? 10% = 8 of 80 = 10 of 100 = 19 of 100 so we get 29th of 200... divide by 2 = aprox. 14.5 which leads to 100-14.5 = aprox. 85 the closest answer.

Re: Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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26 Nov 2016, 17:17

Percentile questions do appear from time to time so I wanted to check whether my solution is correct. I did it the other way round and the numbers are quite comfortable.

1) 90th percentile = 10% better than her 10% * 80 = 8 better than her

2) 19 better than her in the other class

3) total 80 + 100

4) 8 + 19 / 180 = 3/20 (divided by 9) = 15% that are better than her - that is equivalent to the 85th percentile

Amy's grade was 90th percentile of the 80 grades for her class. Of the 100 grades from another class, 19 were higher than Amy's, and the rest were lower. If no other grade was the same as Amy's grade, then Amy's grade was what percentile of the grades of the two classes of two classes combined?

A) 72nd B) 80th C) 81st D) 85th E) 92nd

Since Amy’s grade is in the 90th percentile of the 80 grades in her class, 10% of the number of total grades in her class are not lower than Amy’s grade; i.e., 8 grades. Note that these 8 grades include Amy’s own grade. When combined with the number of grades from the other class, there are 8 + 19 = 27 grades that are not lower than Amy’s grade, including Amy’s own, in a total of 80 + 100 = 180 grades. Since 27 / 180 = 0.15, this amounts to 15% of all the grades. Therefore, Amy’s grade is the 100 - 15 = 85th percentile of the combined grades of the two classes.

Answer: D
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