Tackling Max/Min Statistics on the GMAT - Part 1
Blast from the past! I first discussed the problems in this series way back in 2009. Im reviving the series now because too many people just arent comfortable handling the weird maximize/minimize problem variations that the GMAT sometimes tosses at us.
In this installment, were going to tackle two GMATPrep questions. Next time, Ill give you a super hard one from our own archivesjust to see whether you learned the material as well as you thought you did. :)
Heres your first GMATPrep problem. Go for it!
*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
When you see the word maximum (or a synonym), sit up and take notice. This one word is going to be the determining factor in setting up this problem efficiently right from the beginning. (The word minimum or a synonym would also apply.)
When youre asked to maximize (or minimize) one thing, you are going to have one or more decision points throughout the problem in which you are going to have to maximize or minimize some other variables. Good decisions at these points will ultimately lead to the desired maximum (or minimum) quantity.
This time, they want to maximize the lightest box. Step back from the problem a sec and picture three boxes sitting in front of you. Youre about to ship them off to a friend. Wrap your head around the dilemma: if you want to maximize the lightest box, what should you do to the other two boxes?
Note also that the problem provides some constraints. There are three boxes and the median weight is 9 kg. No variability there: the middle box must weigh 9 kg.
The three items also have an average weight of 7. The total weight, then, must be (7)(3) = 21 kg.
Subtract the middle box from the total to get the combined weight of the heaviest and lightest boxes: 21 9 = 12 kg.
The heaviest box has to be equal to or greater than 9 (because it is to the right of the median). Likewise, the lightest box has to be equal to or smaller than 9. In order to maximize the weight of the lightest box, what should you do to the heaviest box?
Minimize the weight of the heaviest box in order to maximize the weight of the lightest box. The smallest possible weight for the heaviest box is 9.
If the heaviest box is minimized to 9, and the heaviest and lightest must add up to 12, then the maximum weight for the lightest box is 3.
The correct answer is (C).
Make sense? If youve got it, try this harder GMATPrep problem. 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
Hmm. There are 11 voting districts, each with some number of people. Were asked to find the minimum possible population in the least populated districtthat is, the smallest population that any one district could possibly have.
Lets say that District 1 has the minimum population. Because all 11 districts have to add up to 132,000 people, youd need to maximize the population in Districts 2 through 10. How? Now, you 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 up to 10% more, or 110 people, but it cant have any more than that. If the smallest district has 500 people, then the largest district could have up to 550 people but thats it.
How can you use that to figure out how to split up the 132,000 people?
In the given problem, the number of people in the smallest district is unknown, 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. The largest district could be 1.1x but cant be any larger than that.
Since you need to maximize the 10 remaining districts, set all 10 districts equal to 1.1x. As a result, there are (1.1x)(10) = 11x people in the 10 maximized districts (Districts 2 through 10), as well as the original x people in the minimized district (District 1).
The problem indicated 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).
Practice this process with any max/min problems youve seen recently and join me next time, when well tackle a super hard problem.
Key Takeaways for Max/Min Problems:
(1) Figure out what variables are in play: what can you manipulate in the problem? Some of those variables will need to be maximized and some minimized in order to get to the desired answer. Figure out which is which at each step along the way.
(2) Did you make a mistakemaximize when you should have minimized or vice versa? Go through the logic again, step by step, to figure out where you were led astray and why you should have done the opposite of what you did. (This is a good process in general whenever you make a mistake: figure out why you made the mistake you made, as well as how to do the work correctly next time.)
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.