Data Sufficiency

Hi j_shreyans,

This DS question can be solved with a combination of TESTing VALUES and math concepts.

We're told that each class (Class A and Class B) has 5 students in it. We're asked if the MEDIAN score for Class A is greater than the MEDIAN score for Class B? This is a YES/NO question.

Fact 1: Range of scores for Class A = 40 to 80, Range of scores for Class B = 50 to 90

This means that Class A has a 40, an 80 and 3 other scores:

40 _ _ _ 80

And Class B has a 50, a 90 and 3 other scores:

50 _ _ _ 90

The missing scores in each class can be ANY number in the range, so we could have....

Class A: 40, 40, 40, 40, 80 --> Median = 40
Class B: 50, 50, 50, 50, 90 --> Median = 50
The answer to the question is NO

Class A: 40, 80, 80, 80, 80 --> Median = 80
Class B: 50, 50, 50, 50, 90 --> Median = 50
The answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: If the students are paired in study teams such that each student from Dr. Adams' class has a partner from Dr. Brown's class, there is a way to pair the 10 students such that the higher scorer in each pair is one of Dr. Brown's students.

This tells us that every score in Class A will be less than AT LEAST one score in Class B.

For example:
Class A: 40, 40, 40, 40, 80
Class B: 50, 50, 50, 50, 90

Since every score in Class A is less than at least one score in Class B, the MEDIAN of Class B MUST be bigger than the median of Class A. Even if the numbers were "out of order", the median of A would still be paired with a larger number in B.

For example:
Class A: 40, 41, 42, 43, 80 --> Median = 42
Class B: 50, 80, 70, 60, 90 --> Median = 70
Median of B is ALWAYS greater than Median of A, so the answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

Rich does a great job of laying out values to test the statements. But we could also just think CONCEPTUALLY about the different metrics given.

As a general rule, knowing one statistical metric about a set will not tell you about any others, unless you have more information given:

- MEDIAN will not tell you about the MEAN... unless it's a consecutive set
- RANGE will not tell you about MEAN
- STANDARD DEVIATION will not tell you what the MEDIAN is
- MEAN will not tell you what the RANGE is
etc.

In statement (1), a RANGE will never allow us to compare MEDIANS, because the data (as Rich pointed out) might cluster to one end of the range or the other. The only exception would be if the ranges don't OVERLAP at all. If one class's range was 20 to 40 and the other's was 50 to 80, then clearly the median of the second is higher. If the ranges overlap, though, (and we don't know anything else about whether the sets are evenly spaced, etc), then we don't know. Insufficient.

In the 2nd statement, if we can pair each term in Dr. Adam's class with a higher corresponding term in Dr. Brown's class, then the middle term in Dr. Adam's class must be lower than the middle term in Dr. Brown's class. Sufficient.