Bunuel wrote:
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.
Bunuel, I have a doubt.
I failed to understand this, "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"
19 scores are higher; thus, 100 – 19 = 81 are lower since it is given that the rest were lower. So, where is the positioning of Amy's score? She's neither in 19 or in 81 given that no other grade was the same as Amy's grade.