GMATPrep® Max/Min Statistics Problems
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Stacey is a GMAT Instructor living in Montreal. Click here to read more articles from Manhattan GMAT and to learn more about Manhattan GMAT's classes. |
This week, we’re 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 aren’t sure about the most efficient approach to these kinds of questions. We’ll tackle these two GMATPrep® questions this week in order to learn how to master max/min questions in general, and later this month I’ll 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. 
Let’s 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 we’re 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, they’re 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 isn’t a variable; we can’t 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 doesn’t 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 you’re sure you’ve 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. We’re asked to find the minimum possible population in the least populated district – or the smallest population that any one district could possibly have.
Let’s say that we’re 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 can’t have any more than that. If the smallest district has 500 people, then the largest district could have 550 people but can’t 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 don’t know the number of people in the smallest district, so let’s 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 can’t 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:
- figure out what variables are “in play” (what figures we can manipulate in the problem)
- 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)
- 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.
Meet the Top
abhasjha on November 11th, 2009 at 6:17 am
Thank you Stacey .Problems on minimization and maximization frequently appear on the G prep .Keep this in view I have collated few more problems that deal with the concept of minimization and maximization from the G prep itself .
They are :
Q1. for a certain race, 3 teams were allowed to enter 3 members each. A team earned 6-n points whenever one of its members finished in the nth place, where 1<=n<=5. There were no ties, disqualifications or withdrawals. if no team earned more than 6 points, what is the least possible score a team could have earned.
a) 0
b) 1
c) 2
d) 3
d) 4
Q2. . a certain meter records voltage between 0 and 10 volts, inclusive. if the average value of 3 recordings from the meter was 8 volts, what was the smallest possible recording in volts?
A) 2
B) 3
C) 4
D) 5
E) 6
Q3. Store S sold a total of 90 copies of a certain book during the seven days of last week, and it sold different numbers of copies on any two of the days. If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did Store S sell more than 11 copies on Friday?
(1) Last week Store S sold 8 copies of the book on Thursday.
(2) Last week Store S sold 38 copies of the book on Saturday.
Q4. Five pieces of wood have an average arithmetic mean length of 124 cms and a median length of 140 cms.What can be the maximum possible length in cms of the shortest piece of wood?
a.90
b.100
c.110
d.130
e.140
Q5. . Machine A produces pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?
a) 4
b) 4 and 2/3
c) 5 and 1/3
d) 6
e) 6 and ¼
Q6. In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest borrowed at least 3 books. If the average number of books per student was 2, what is the maximum number of books any single student could have borrowed?
A. 3
B. 5
C. 8
D. 13
E. 15
I am sure solving all these questions will help those who are preparing for GMAT .
Stacey on November 11th, 2009 at 7:44 am
abhasjha, this is great! I'm sure your fellow students appreciate the hard work you did to type these in.
Okay, guys, if you think you understand everything in the article, try these other problems out to test yourself!
Warning, though: if you haven't finished using GMATPrep, then you may be exposing yourself to problems that you will see when you take GMATPrep - so you may want to hold off if you are still planning to take GMATPrep.
J on November 11th, 2009 at 10:20 am
Are you planning to give the answers to these questions later so we can check whether we got the correct or not?
Thanks!
Stacey on November 11th, 2009 at 10:36 am
Great idea, J. abhasjha, when you get a chance, can you please post the OAs for each of the problems you posted?
If people would like to discuss solutions, etc, please don't do so here - the comments section isn't set up to make that an easy discussion, especially with multiple problems in abhasjha's comment. Start new threads (one for each problem) on the regular forums and, if you like, post links to those threads here. Then people can discuss each problem in depth over there.
(Though if you have questions on the original two problems, in the article, then please do post those questions here.)
Hesham on November 11th, 2009 at 11:00 am
Q4) A. Q5) A.
Hesham on November 11th, 2009 at 10:19 am
Answer to Q2) is 4 ?
Hesham on November 11th, 2009 at 10:25 am
Answer to Q3) is B. By strong guessing. Wish for a break down of a solution.
Hesham on November 11th, 2009 at 10:58 am
Q5) A
Hesham on November 11th, 2009 at 11:08 am
Can't solve Q1 ?????
Vinayak Parameshwaran on November 11th, 2009 at 12:33 pm
Answer to Q1 is Option (E) which is 4.
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.
Vinayak Parameshwaran on November 13th, 2009 at 8:50 am
Sorry the answer to Question 1 is Option (D) which is 3
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.
Milan on November 11th, 2009 at 1:57 pm
Hesham,
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.
Hesham on November 11th, 2009 at 4:36 pm
I am confused about the second question in the article. I can't picture the city with the highest population being 10% more than the city with the lowest population. I reality, is the only why to minimize the population of one city is to have same population in each of the other 10 cities ? Can each city have different population from one another, but remain no more than 10% of the other population ?
Stacey on November 16th, 2009 at 8:46 am
Hi, Hesham
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.
abhasjha on November 11th, 2009 at 10:09 pm
Answers to the questions in my previous posts are :
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.
Ram on November 11th, 2009 at 10:33 pm
Thanks abhasjha for the above questions.
MOreover u can post more spanning different topics..
Wud be of much help 2 many like me..
Thnx again..