GMATPrep® Max/Min Statistics Problems

by , Nov 11, 2009

This week, were going to tackle two GMATPrep questions, this time from the quant side of things. My students have been asking (really, complaining!) about maximize / minimize questions lately. A lot of students arent sure about the most efficient approach to these kinds of questions. Well tackle these two GMATPrep questions this week in order to learn how to master max/min questions in general, and later this month Ill give you a super hard one from our own archives just to see whether you learned the material as well as you thought you did. :)

Lets start with a sample problem. Set your timer for 2 minutes. and GO!

*Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

The most important thing to notice here is the word maximum. This one word is going to be the determining factor in how we set this problem up, right from the very beginning.

Most of the time, when were asked to maximize something, we will need to minimize the remaining variables in the problem. Conversely, if the problem asks us to minimize something, we will usually need to maximize the remaining variables. (There are times, though, when we will need to minimize some other variable in order to minimize the desired thing or maximize some other variable in order to maximize the desired thing so we do need to pay attention.)

This time, theyre asking us to maximize one figure: the lightest box.

If three items have an average weight of 7, then collectively, the three items have a weight of 7x3 = 21. If three items have a median weight of 9, then the middle of the three items is actually 9. This one isnt a variable; we cant change this number. The first (or lightest) of the three, therefore, has to be equal to or less than 9 (because it is to the left of the median). Check the answers quickly in this case, unfortunately, that information doesnt help us to eliminate any answers.

If the middle box is actually 9, then we can subtract that from 21 to get the combined weight for the other two boxes. 21 9 = 12. So the lightest and heaviest boxes have to add up to 12.

Now, do we want to minimize or maximize the weight of the heaviest box?

The heaviest box has to be equal to or greater than 9 (because it is to the right of the median). We want to maximize the weight of the lightest box, so we need to minimize the weight of the heaviest box. So, the smallest possible weight for the heaviest box is 9.

If the heaviest box is minimized to 9, and the heaviest and lightest add up to 12, then the maximum weight for the lightest box is 3. The correct answer is C.

Make sense? If youre sure youve got it, try this harder one. Set your timer for 2 minutes!

*A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have?

(A) 10,700

(B) 10,800

(C) 10,900

(D) 11,000

(E) 11,100

In this case, there are 11 voting districts, each with some number of people. Were asked to find the minimum possible population in the least populated district or the smallest population that any one district could possibly have.

Lets say that were going to minimize the population in District 1. Because all 11 districts have to add up to 132,000 people, we want to maximize the population in Districts 2 through 10. How can we do that? Now, we need more information from the problem:

no district is to have a population that is more than 10 percent greater than the population of any other district

So, if the smallest district has 100 people, then the largest district could have 110 people but cant have any more than that. If the smallest district has 500 people, then the largest district could have 550 people but cant have any more than that. How did we calculate those numbers? In each case, we take 10% of the original number and add that figure to the original number to give us our maximum.

In the given problem, we dont know the number of people in the smallest district, so lets call that x. If the smallest district is x, then calculate 10% and add that figure to x: x + 0.1x = 1.1x. So the largest district could be 1.1x but cant be any larger than that.

We want to maximize all of the 10 remaining districts, so we should assume that all 10 districts are equal to 1.1x. As a result, we have (1.1x)(10) = 11x people in the 10 maximized districts (Districts 2 through 10), as well as our original x people in the minimized district (District 1).

The problem told us that all 11 districts add up to 132,000, so write that out mathematically:

11x + x = 132,000

12x = 132,000

x = 11,000

The correct answer is D.

Key Takeaways for Max/Min Problems:

1. figure out what variables are in play (what figures we can manipulate in the problem)
2. figure out whether each variable needs to be maximized or minimized in order to achieve the desired outcome (the thing the problem asks us to do)
3. do the work (carefully, as always!)

* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.