Hi All!
Received a PM asking me to respond:
sanju09 wrote:Class A and Class B took the same test. For class A, the median score is 80, and the average (arithmetic mean) score is 82; for class B, the median score is 78, and the average (arithmetic mean) score is 74. Combining the two classes A and B together, is the average (arithmetic mean) of the combination greater than its median?
(1) Class A has 37 students and Class B has 40 students.
(2) Class A and Class B together have 77 students.
This is actually a pretty tough question; one in which we will need to use number sense (rather than actually testing numbers). I will show you some example data sets just to drive the concept home, but this would NOT be the method I would encourage you to use on the test - it took me a bit of thinking and number crunching to come up with sets that would prove the point!
Okay, so the question tells us about 2 classrooms' scores: their means and medians. Once we combine the classes, we are then asked if the combined mean is greater than the combined median.
Let's take a look at the (somewhat) easier statement...
Statement 2: Class A and Class B together have 77 students.
- If I only know the combined numbers of students, there is actually very little I know about the combined class. In order to get the combined average (this is really just a weighted average) I would need exact numbers for each of the classes, and there is absolutely nothing I can tell about the median because I know NOTHING about the makeup of the actual individual test scores.
To provide an example, I picked much smaller numbers for my classes (again, only to show an example) = 11 total.
Example (1) Let's say that 3 students came from Class A, and 8 from Class B:
Class A = {75,
80,91}
Class B = {60,60,60,
78,78,85,85,86}
Combined: {60,60,60,75,78,
78,80,85,85,86,91}
Mean=76.2, Median=78
The median is GREATER than the Mean
Example (2) Let's say that 7 students came from Class A and 4 from Class B:
Class A = {78,78,78,80,85,85,86}
Class B = {62,
78,78,78}
Combined: {62,78,78,78,78,
78,78,80,85,85,86}
Mean=79.1, Median=78
The median is LESS than the Mean.
We have 2 different answers, with 2 different solutions. NOT Sufficient!
Statement 1: Class A has 37 students and Class B has 40 students.
- So this statement actually gives us a bit more information. We know the weighted average of the combined class now: (37*82 + 40*74)/77 = 77.84.
We also know that Class B has a few more students than class A but they are roughly equal (not sure if this will matter yet but worth noting as our 2 picked sets of numbers above flip-flopped this relationship and might not have had the numbers so close). And finally, we can see that C will NOT be an answer choice (because everything from statement 2 can be figured out from statement 1, so it won't add any info - we're down to A or E as options)!!
So let's start introducing some theory. We know that the mean is 77.84. Let's think about trying to MAKE the median BELOW and ABOVE that number (trying to prove insufficiency).
General Background:
With the combined class of 77, there will be 38 grades <= the median, and 38 grades >= the median. (for odd sets, the median is the actual middle number - here it will be entry #39)
For Class A, there will be 18 grades <= the median (80) and 18 grades >= the median. (again, odd set, it will be entry #19)
For Class B, there will be 20 grades <= the median (78) and 20 grades >= the median, but the AVERAGE of the 20th & 21st terms has to be the median (for even sets, the median is the average of the 2 middle numbers).
Test 1:
So let's say we make EVERY number below the median of Class A, less than 70 (and then pick balancing values for the scores above). So we would have 18 grades under 70, 1-80, then a bunch of grades above 86 (to keep it simple with my plan for Class B).
For Class B, lets make the 2 middle terms very different so they have to average out to the median. Make every number below the median (entries 1-19) below 70. Then makeentry #20=70 and entry #21=86 (average out to 78). And finally, make the other 19 entries greater than or equal to 86.
Together, we have 18+19=37 grades below 70, so:
Entries #1-37 are below 70
Entry #38 = 70 (from B, entry #20)
Entry #39 = 80 (from A)
Entry #40 = 86 (from B, entry #21)
Entries #41-77 are above 86
The median=80 and is GREATER than the average of 77.84
Test 2:
Now we're trying to find a SMALLER Median that 77.84. Is that going to be possible? Let's try for the opposite of what we did - rather than make the 2 middle terms of B different (one below 80 and one above), lets try to keep them both below. We can make them the same!
So again, let's say we make EVERY number below the median of Class A, less than 70 (and then pick balancing values for the scores above). So we would have 18 grades under 70, 1-80, then a bunch of grades above 80.
Now, for Class B, lets make the 2 middle terms the same. Again make every number below the median (entries 1-19) below 70, just to keep it simple. Then make entry #20=78 and entry #21=78. And finally, to keep trying to include as many small numbers as we can, lets make all the 19 entries above the median equal to 78. This is actually fine because the average for Class B is BELOW the median, so we want more weight at the bottom!
Again, we have 18+19=37 grades below 70, so:
Entries #1-37 are below 70
Entry #38 = 78 (from B, entry #20)
Entry #39 = 78 (from B, entry #21)
19 more entries = 78
Entry #59 = 80 (from median of A)
Entries #60-77 are above 80
Notice that when we tried to get a median as small as possible, we ended up on the median from B this time.
And that is going to always be the case. If we steer the 2 middle numbers of set B far apart (one above 80 and one below), the median will end up resting on the 80. If we move them back together (both under 80) then the median will end up resting on the greater of the 2 = a number between 78 and 80.
So the smallest median we can get is 78, and the largest 80 - but everything in between is GREATER than the mean. SUFFICIENT!
The answer is A.
**Overall Lesson**
This is a TERRIBLY tough problem and if you're seeing this on the GMAT, you're doing VERY well. Just take note of how much information statement 1 gives us - we know the relative numbers, so we can (1) calculate the exact weighted average, and (2) we can start to see the relative sizes of each group and try to understand the possible locations for the new median.
I hope this clears up the confusion!
Whit
