All of the districts add up to a static number (132,000), so if we want to minimize the population of one district, then we have to maximize the populations of all of the other districts.

How we maximize is dependent upon whatever restrictions they give us. In this case, we only have one restriction (the largest is no more than 10% higher than the smallest). Because we want to maximize all 10 of the other districts, we want to make them all 10% higher than the smallest district.

It may help to understand this if we use smaller numbers and easier concepts. Let's say we have 10 districts and we're told that the largest district cannot be more than 5 higher than the smallest district. If the smallest district is 10, then the largest would have to be 15, right? What is the *maximum* total population that we could have, given that the smallest is 10 and the largest could be as high as 15? Well, I can't change the smallest - that's 10. Now, ask yourself: what's the largest that I can make each of the remaining 9?

Is it possible for each of the remaining 9 to be 15? Yes. Is it possible for any to be larger than that? No. Is it possible for any to be smaller than that? Yes - but then we wouldn't be *maximizing* the total possible number. So the *maximum* possible is for all of the remaining 9 to be 15.

So you can have some cities be smaller than 15 - but then you wouldn't be maximizing the possible number.

A B C
4 5 3
2 1

For the least possible score for one team you need to maximize the scores in the others.

1.d .2.c. 3. b. 4.b. 5.a.6.d

In case you want to explanation to these problems you can search google . These problems have been discusse on forums such as beat the gmat , manhaatan, urch an GMAT club .

Guys all the best for your gmat exams.

Q2. Yes, it is 4.

Q4. is not A, it is B. Avg of 5 pieces of wood is 124 and median = 140. 124 * 5 = 620cm of wood total, so for the smallest to be max, all of the pieces of wood greater than the mean should be minimized, so that occurs at 140.

So you have 3 pieces of wood at 140, total of 420 and the other 2 are unknown, but you should know that to maximize the smallest piece, the second to smallest piece should be minimized. So from your total of 620, you have 420 from the biggest 3, leaving 200 for the smallest 2, or to maximize the smallest, you get 100. Choice B.

Here is how I arrived at the solution.

Let the teams be A, B and C. with team members [a1, a2, a,3] [b1, b2, b3] and [c1, c2, c3] respectively.

So now I have 9 people participating in a race.

Now take a look at the conditions:

Each point is given by 6-n where 1<=n<=5. This follows that there are no points for the 6th, 7th, 8th and 9th positions.

6-n would mean 5 points for the first position (6-1), 4 points for the second (6-2) and so on till the last (6-5).

Please look at the constraints mentioned in the problem, that no team gets more than 6 points.

A B C
5 4 2
1 3

You can now arrange the points in such a way that the conditions mentioned in the problem are satisfied. In each case the team with the least score has 4 points.

To solve this within 2 mins on the GMAT you really don't need the garbage that I have typed above. You just need to figure out the possible outcomes fit them to satisfy the conditions mentioned in the question.

Hope this helps.