Percentile problem

[andy123]

Angela's grade was in the 90th percentile out of 80 grades in her class. In another class 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


Tried searching ...has not yet been solved..

I solved it doiing:

number of students less then angela = 71 ..in class 1

no of students less then angela = 80-19 = 61 -60 = 60 below her in class 2

thus : 71 + 60 = 131

( x/100)*160 = 131

=> x = (131 * 100 ) / 160 = 81 .87 .. = 81

but the answer is not 81 .

[Talkativetree]

I've seen this problem on here, the answer I think was 85. What people did was add 8+19=27; 160-27=133; 133/160=x/100; x~83.1%, which would round up to 85%...haven't seen a better answer than basically that

Here's a pretty good post that goes through percentiles
https://www.beatthegmat.com/percentile-t1696.html

[Leon1984]

I'm also new to this, but I'll try to think out loud,
although I may be wrong.

Basically, we don't know the size of the 2nd class so we can't really determine. It is possible that the class is huge and she is in the 99th percentile, or the opposite.

However, if the classes are equal, we can just work with percentiles. We know she is in the 90th in the 1st class, and little bit over 75th percentile in the 2nd.
(61 out of 80 is little more that 3/4, or little over 75th percentile). We can average 90+75=165/2=82.5
But since we know it's not 75 but more than that, we actually get over 83, and therefore it's rounded to 85th percentile.

Hope it helped