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Amy's grade was 90th percentile of the 80 grades for her

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

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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?

A) 72nd
B) 80th
C) 81st
D) 85th
E) 92nd
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Re: Help [#permalink]

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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?

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.

Answer: D.
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Re: Help [#permalink]

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New post 28 Jul 2010, 10:21
thanks for the correction and also for explanation
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Re: Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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

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New post 13 Oct 2013, 23:05
D: 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? :shock:
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Re: Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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New post 14 Oct 2013, 00:16
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cowashington wrote:
D: 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? :shock:


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

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I took the following approach:

For 80 grades -> 90 percentile
For 100 grades -> 81 percentile

For 180 -> Her percentile had to be higher than 81 and lower than 90 -> The only answer satisfying this was 85% D

Please correct me if I'm wrong
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Re: Amy's grade was 90th percentile of the 80 grades for her [#permalink]

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

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New post 09 Nov 2015, 11:51
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.

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

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

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New post 26 Nov 2016, 18:17
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
Re: Amy's grade was 90th percentile of the 80 grades for her   [#permalink] 26 Nov 2016, 18:17
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