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# Angela’s grade was in the 90th percentile out of 80 grades

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Angela’s grade was in the 90th percentile out of 80 grades [#permalink]  09 Dec 2010, 07:22
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Angela’s grade was in the 90th percentile out of 80 grades in her class. In another class of 100 students there were 19 grades higher than Angela’s. If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

A. 72
B. 80
C. 81
D. 85
E. 92
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Jun 2013, 03:04, edited 2 times in total.
Edited the stem.
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Re: Some PS problems Part 2 [#permalink]  09 Dec 2010, 07:33
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gdk800 wrote:
Angela’s grade was in the 90th percentile out of 80 grades in her class. In another class of 100 students there were 19 grades higher than Angela’s. If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

a. 72
b. 80
c. 81
d. 85
e. 92

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 Angela outscored $$80*0.9=72$$ classmates.

In another class she would outscored $$100-19=81$$ students (note: Angela 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 Angela is in $$\frac{153}{180}=0.85=85%$$, or in 85th percentile.

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Re: Some PS problems Part 2 [#permalink]  09 Dec 2010, 12:56
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gdk800 wrote:
Angela’s grade was in the 90th percentile out of 80 grades in her class. In another class of 100 students there were 19 grades higher than Angela’s. If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

a. 72
b. 80
c. 81
d. 85
e. 92

She is in 90th percentile in one class and 81st in the other. We need to find her 'average percentile'. The average will lie between 81 and 90 so it has to be 85!

If we had closer answer options e.g. 84, 86 etc, we would need to calculate the weighted average.
Using weights as 80 and 100 respectively, you can find their average using weighted average formula which is as given below:
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$

$$C_{avg} = \frac{90*80 + 81*100}{180}$$ = 85

In case you are not comfortable with weighted averages, check this discussion:
http://gmatclub.com/forum/tough-ds-105651.html#p828579
and get back if there is a doubt.
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Re: Some PS problems Part 2 [#permalink]  09 Dec 2010, 18:55
Thanks Karishma, I put it to good use on this question. To all, the discussion in the link Karishma provided is very helpful.
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Re: Some PS problems Part 2 [#permalink]  10 Dec 2010, 10:57
Thanx a ton to Bunuel & Karishma for such detailed explanations.

Special thanx to Bunuel for editing the question part i missed n also to Karishma for the helpful link on weighted averages.
Senior Manager
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Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
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Re: Some PS problems Part 2 [#permalink]  10 Dec 2010, 16:45
VeritasPrepKarishma wrote:
gdk800 wrote:
Angela’s grade was in the 90th percentile out of 80 grades in her class. In another class of 100 students there were 19 grades higher than Angela’s. If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

a. 72
b. 80
c. 81
d. 85
e. 92

She is in 90th percentile in one class and 81st in the other. We need to find her 'average percentile'. The average will lie between 81 and 90 so it has to be 85!

If we had closer answer options e.g. 84, 86 etc, we would need to calculate the weighted average.
Using weights as 80 and 100 respectively, you can find their average using weighted average formula which is as given below:
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$

$$C_{avg} = \frac{90*80 + 81*100}{180}$$ = 85

In case you are not comfortable with weighted averages, check this discussion:
http://gmatclub.com/forum/tough-ds-105651.html#p828579
and get back if there is a doubt.

This formula helped tremendously.

Thanks
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Re: Angela’s grade was in the 90th percentile out of 80 grades [#permalink]  10 Jun 2013, 04:09
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: Angela’s grade was in the 90th percentile out of 80 grades [#permalink]  10 Jun 2013, 06:26
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gdk800 wrote:
Angela’s grade was in the 90th percentile out of 80 grades in her class. In another class of 100 students there were 19 grades higher than Angela’s. If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

A. 72
B. 80
C. 81
D. 85
E. 92

Angela is 90th percentile in her class and would be 81st percentile in the other class.
The ratio of her class size to the ratio of the other class size is 80:100 or 4:5. Because of that, we know that her congregate percentile is weighted more heavily towards the 81st percentile because there are more students in the class, at the same ratio as the class size.
Draw a number line between 81 and 90 and apply the reverse of the class size ratio to account for the weight of each individual percentile. Thus, the congregate percentile is 4 away from 81 and 5 away from 90
81----85-----90

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Re: Angela’s grade was in the 90th percentile out of 80 grades [#permalink]  13 Jul 2014, 10:34
90th Percentile means she has scored more then 90% of the people.
==> (90/100)*80 = 72 ( she scored more then 72 people)

In the other class of 100, we are given that 19 people scored more then Angela
==> 100 - 19 = 81 ( she scored more then 81 people)

Since we are told to find her percentile in the combined class, we need to find how many people did she score over in a combined class of 100 + 80 = 180.

(81+72)/180 = 85 %

One could also use the logic that if in the first class her percentile was 90 and in the second her percentile was 81, the average would be greater then 81 and less than 90. Hence the Answer has to be D
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Re: Angela’s grade was in the 90th percentile out of 80 grades [#permalink]  10 Aug 2015, 04:55
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Re: Angela’s grade was in the 90th percentile out of 80 grades   [#permalink] 10 Aug 2015, 04:55
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# Angela’s grade was in the 90th percentile out of 80 grades

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