Hi,
I had to open the image in another tab. I guess it would be better if you can upload it again.
Coming to the problem -
Let's draw the frequency distribution chart:
It should be arranged in increasing order of values and not frequencies.
Range----Frequency---Cumulative frequency
--0-9--------1--------1
10-19--------4--------5
20-29--------6--------11
30-39--------7--------18
40-49--------2--------20
Median will be average of 10th and 11th terms, both of which are in the range 20-29.
a)We need least possible number of seniors(items) with six pages from median. So, we place other items farther from median.
As the difference between 10th element and 11th element can be maximum 9, these two elements cannot be more than 5 pages within the average of them. So, in any case 10th and 11th elements have to be within 6 pages from median. There are many scenarios for this.I will be posting one case:
Let's consider the case:20,20,20,20,26,27. Median = 26.5 Only 26,27 are within 6 pages from median i.e. in the range(20.5,32.5). As we need least number within this range, we will choose the 7 values in the range 30-39 such that they are at least 33.
So, minimum is 2.
b)Now we employ similar strategy for maximizing
Choose all 4 items in the range 10-19 as 19
Choose all 6 items in the range 20-29 in such a way that median is 24.5. For instance, consider 24,24,24,24,24,25
Choose all 7 items in the range 30-39 as 30
As the median is 24.5, 4 items with values 19, 6 items with values(24,25) and 7 items with values 30 are within the 6 pages from median(24.5). Total = 4+6+7 = 17. This is the maximum we can get.

Btw, what is the source of this?

Thanks Frankenstein. Why did you not arrange the intervals in ascending order of frequency?

Hi,

I agree with Frankenstein on both parts.

I think he arranged the intervals in order of value rather than frequency because the median of a list is the element that lies in the middle when the terms are arranged by value (or, if there are even number of terms as there are here, it's the average of the two middle values).

Once you figure out that there are 20 terms, you can determine that the median is the average of the two middle terms (the 10th and 11th). Next, you must figure out in which interval those terms lie.

Since there are 5 terms total in 0-19, and then 6 terms in 20-29, we know that both term 10 and term 11 lie in 20-29.

As Frankenstein says, if your goal is to get as few terms as possible within 6 of the median, we should get all of the other terms out of the way of the median, by setting them to 20. It is inevitable that the 10th and 11th terms will be within 6 of the median, because:
1. the median is the average of those 2 terms
2. they can be at most (29-20) = 9 apart.
3. they can be at most 9/2 = 4.5 apart from their average. In other words, even if we made the 10th and 11th terms 20 and 29, the median would be 24.5 and they would both be within 6.

Once we confirm that we can get all of the other terms 6 away from these 2 middle terms, we can answer that 2 is the minimum.

On the other hand, if we make the median 25, and set all 10-19 terms at 19, and all 30-39 terms at either 30 or 31, we can get all of the terms in 10-39 within 6 of 25.

edge wrote:Thanks Frankenstein. Why did you not arrange the intervals in ascending order of frequency?

Hi,
Greg has already answered this. Anyway, I am incorporating this point as well in my earlier post for reference purpose. You haven't posted the source.I would like to know this because I feel it is a bit lengthy than we usually expect a GMAT question to be although we can save time from easier questions to solve such questions.

Frankenstein wrote:...You haven't posted the source...

Sorry about that; the source is MGMAT.

edge wrote:

Frankenstein wrote:...You haven't posted the source...

Sorry about that; the source is MGMAT.

Hey,
Okay Thanks..No probs