Amy's grade was 90th percentile

[gphil]

Could you please help to solve the problem?
Thanks!

Practice Test 2 from MBA software

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 - correct
E) 92nd

[GregS&P]

This is a weighted average problem:

90 (80/180) + 81 (100/180) =85

We know that her grade falls in the 81st percentile within the other class, because within that class of 100, 19 were ahead of her.

[RonPurewal]

There's also the brute force method, which doesn't take that much time here.

Amy's grade is 90th percentile, meaning that she outperformed 90% of her classmates. That means that she outperformed 72 of her classmates.
She also outperformed 81 of the students in the other class (note: not 80, because Amy herself is not in the other class).

This means that she outperformed 72 + 81 = 153 of the combined 180 students in the two classes.
Since 153/180 = 0.85, she outperformed 85% of the students in the combined pool, putting her in the overall 85th percentile.

[Amit]

Both above answer are great, so I'll throw in my two cents also.

Amy's grade was 90th percentile of the 80 grades for her class.
>> 90 % in the first class
Of the 100 grades from another class, 19 were higher than Amy's, and the rest were lower.
>> As mentioned above, this puts her at 81%

She outperformed 90% of her class and 81% of the other class. If you think about what the overall ranking should be (considering weighted averaged) you can conclude that the overall %ile ranking should be somewhere BETWEEN 90 and 81. That only leaves D - no need to solve it! :)

A) 72nd
B) 80th
C) 81st
D) 85th - correct
E) 92nd

[RonPurewal]

She outperformed 90% of her class and 81% of the other class. If you think about what the overall ranking should be (considering weighted averaged) you can conclude that the overall %ile ranking should be somewhere BETWEEN 90 and 81. That only leaves D - no need to solve it! :)

nicely done.

this is a takeaway that's much more commonly applicable to data sufficiency, but is valuable for both problem types: the weighted average of 2 sub-averages must lie between the sub-averages.
this is common sense in most real-life situations - for instance, a committee whose women average 45 years old and whose men average 51 years old can't have an average age of, say, 42 - but most students wouldn't think to dredge it up on an exam problem.

[imanemekouar]

can you please help me.
How did you know it's a weighted average problem.

[RonPurewal]

can you please help me.
How did you know it's a weighted average problem.

it's basically equivalent to having 80 grades with an overall average of 90, and 100 more grades with an overall average of 81, and then throwing those 180 grades into a common pool (and finding the resultant weighted average).

making this analogy requires a lot of experience with these types of problems, though. i wouldn't have mentioned it -- the only reason it's under discussion is because it was mentioned by a poster.

if you don't see this interpretation, then you should just go ahead and grind out the brute-force solution (see my post above).