Critical Reasoning

Cheever College offers several online courses via remote computer connection, in addition to traditional classroom-based courses. A study of student performance at Cheever found that, overall, the average student grade for online courses matched that for classroom-based courses. In this calculation of the average grade, course withdrawals were weighted as equivalent to a course failure, and the rate of withdrawal was much lower for students enrolled in classroom-based courses than for students enrolled in online courses.

If the statements above are true, which of the following must also be true of Cheever College?

(A) Among students who did not withdraw, students enrolled in online courses got higher grades, on average, than students enrolled in classroom-based courses.
(B) The number of students enrolled per course at the start of the school term is much higher, on average, for the online courses than for the classroom-based courses.
(C) There are no students who take both an online and a classroom-based course in the same school term.
(D) Among Cheever College students with the best grades, a significant majority take online, rather than classroom-based, courses.
(E) Courses offered online tend to deal with subject matter that is less challenging than that of classroom-based courses.
[spoiler]
OA-A
someone pls explain A and D
I thought A is just opposite
i.e. if no. of students in online is less than classroom
then for average to be same average grades for online should be lower than classroom one.[/spoiler]

If we consider numbers A is the clear winner.

10 students take online course and 10 again for classroom course.

Course withdrawal is low for classroom courses, so consider 2 students from Online course withdrew or considered as Failure.

Lets say all 10 students score 75 in Classroom course, avg = 75 * 10/10

Then to match the average with Class room course the students should score higher to match the same average as in to get the 750/10. Two withdrawals will result in 0 and the rest 8 should score more than 75 to match the average.

Students involved in online course score the best grades whereas it is stated the opposite in D.

Hope this helped.

sam2304 wrote:If we consider numbers A is the clear winner.

10 students take online course and 10 again for classroom course.

Course withdrawal is low for classroom courses, so consider 2 students from Online course withdrew or considered as Failure.

Lets say all 10 students score 75 in Classroom course, avg = 75 * 10/10

Then to match the average with Class room course the students should score higher to match the same average as in to get the 750/10. Two withdrawals will result in 0 and the rest 8 should score more than 75 to match the average.

Students involved in online course score the best grades whereas it is stated the opposite in D.

Hope this helped.

i proceeded as

A/X=B/Y
A/X is the average grade of online ones
and B/Y is the average grade of classroom guys

NOW if X decreases then A has to decrease to make the ratio equal which means opposite of option A.

WHERE AM I GOING WRONG??

killer1387 wrote: i proceeded as

A/X=B/Y
A/X is the average grade of online ones
and B/Y is the average grade of classroom guys

NOW if X decreases then A has to decrease to make the ratio equal which means opposite of option A.

WHERE AM I GOING WRONG??

-> Here

NOW if X decreases then A has to decrease to make the ratio equal which means opposite of option A.

It is said withdrawals are considered as Failures so we don't decrease X i.e the # ppl taking online courses remains same only their marks are reduced to zero or considered as failure. So to match B/Y ratio A increases for the rest of the people.

that's something real clever...

Hello Experts,

Please help explaining this problem.