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Intern  Joined: 20 Oct 2010
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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

Originally posted by gdk800 on 09 Dec 2010, 08:22.
Last edited by Bunuel on 10 Jun 2013, 04:04, edited 2 times in total.
Edited the stem.
Math Expert V
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Re: Some PS problems Part 2  [#permalink]

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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]

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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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Posts: 103
Re: Some PS problems Part 2  [#permalink]

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Thanks Karishma, I put it to good use on this question. To all, the discussion in the link Karishma provided is very helpful.
Intern  Joined: 20 Oct 2010
Posts: 10
Re: Some PS problems Part 2  [#permalink]

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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.
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Re: Some PS problems Part 2  [#permalink]

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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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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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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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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Nice Question =>
Here is what i did in this one =>
Number of students in class 1 with score greater than angela => 10/100 *80=> 8
Number if students in class with score greater than angela => 19

Total students with score >angela => 19+8=27
Combined class strength = 100+80=180
27 is 15 percent of 180

Hence 85 percent scored less than angela.
In other words => Angela's Rank would be 85 percentile

Hence D

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Bunuel 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

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.

Bunuel
I think it would have been simpler to calculate: if we add up people ahead of her 8+19 = 27, therefore 27/180*100 = 15% ahead of her, therefore 85% is percentile of both classes.
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Bunuel 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

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.

Bunuel why did you exclude 19 students who scored higher than angela? the questions is If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

we are not asked what percentage of students she outscored Senior Manager  G
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dave13 wrote:
Bunuel 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

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.

Bunuel why did you exclude 19 students who scored higher than angela? the questions is If nobody had Angela’s grade, then Angela was what percentile of the two classes combined?

we are not asked what percentage of students she outscored dave13

The 19 students in other class are those who have scored higher than Angela, so if we are calculating percentile of angela's we need to know how many students did angela outscored in the other class, which is 100 - 19 = 81 students. In other class angela outscored 81 students.

To calculate percentile of two classes combined, we need to calculate how many students have been outscored by angela out of total number of students.
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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

$$?\,\, = \,\,\frac{{\# \,\,{\text{below}}\,\,{\text{Angela}}}}{{\# \,\,{\text{total}}}}\, \cdot \,\,100\,\,\,\left( \% \right)$$

Angela was in the "10th best grades" (but not better) in her class (80 students), hence (she is "number 8" and) there are 7 grades above her.

From the question stem, there are 7+19 = 26 grades above her in 180 grades, therefore 180-26-1 = 153 below her.

$$? = \frac{{153}}{{180}}\, \cdot \,\,100 = \,\underleftrightarrow {\frac{{\left( {90 + 63} \right)}}{9} \cdot \frac{{10}}{2}} = \,\,17 \cdot 5 = 85\,\,\,\left( \% \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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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

Alternate approach (similar to Karishma´s nice explanations above!):

$$?\,\,\,\,:\,\,\,{\text{Angela}}\,\,{\text{percentile}}$$

The approximations presented are validated a posteriori, because there are no alternative choices "near" the correct one!

$$\left\{ \begin{gathered} \,80\,\,{\text{people}}\,\,\,:\,\,\,90\% \,\,{\text{below}} \hfill \\ \,10\boxed1\,\,\,{\text{people}}\,\,\,:\,\,\,\, \cong 80\% \,\,{\text{below}}\,\,\, \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \begin{gathered} \,\,4\,\,{\text{people}}\,\,\,:\,\,\,90\% \,\,{\text{below}} \hfill \\ \,\,5\,\,\,{\text{people}}\,\,\,:\,\,\,80\% \,\,{\text{below}}\,\, \hfill \\ \end{gathered} \right.$$

$$\boxed1\,\,\,:\,\,\,{\text{if}}\,\,{\text{Angela}}\,\,{\text{there!}}$$

$$?\,\,\, \cong \,\,\,\frac{{4 \cdot 90\% + 5 \cdot 80\% }}{9}\,\,\, \cong \,\,\,\frac{{4 \cdot 90\% + 5 \cdot 81\% }}{9}\,\,\, = \,\,\,4 \cdot 10\% + 5 \cdot 9\% \,\,\, = \,\,\,85\%$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Hi All,

We know that Angela is the 90th percentile in the first class. With 100 students in the second class, and 19 of them 'ahead' of Angela, this means that she's the 100 - 19 = 81st percentile in that class. Thus, when you combine those two classes, Angela will be somewhere in the 81st to 90th percentile (but not 81st nor 90th). There's only one answer that fits that range...

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In the first class, Angela scored 90th Percentile of 80 that means you calculate 90% of which is 72

Second class of 100, only 19 grades were higher than her, this means she scored 81 out of 100.

Combined percentile = 72 + 81/180 *100% = 153/180 *100% = 85%

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Bunuel
Hi I have a small doubt with the use of 81st percentile, Shouldn't his be 80th percentile:
Lets say there are 5 students in class and Angela scored at 3rd place which means there are two people who scored above Angela:
Diagram:
S1 S2 A S4 S5

Now here when we subtract 2( the no of people who scored above angela) we get 5-2 = 3, however this represents the position of Angela and thus the Number of people who scored below should be 2 and thus the angela's score should be : 100*2 / 5 percentile= 40th percentile

Kindly help me out with my reasoning Re: Angela’s grade was in the 90th percentile out of 80 grades   [#permalink] 02 Aug 2019, 21:46
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