Dealing with Very Hard Max/Min Statistics Problems

by , Nov 19, 2009

Last week, we tackled two GMATPrep questions; if you missed that article, go read it before continuing with this one. Make sure you try the two sample problems and take the time to master the concepts before you try the super-hard question below.

Okay, this sample problem is from our own archives. Set your timer for 2 minutes. and GO!

Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?

(A) 40

(B) 38

(C) 33

(D) 32

(E) 30

You have an answer, right? Even if you dont know what the answer is and you have to guess youve still picked an answer, right? If you havent, go pick an answer before you keep reading!

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

So, we have a set of 7 numbers. The average of those 7 numbers is 20; can we calculate anything from that? Yes the sum! The basic formula for an average is A = S/n, where A is the average, S is the sum, and n is the number of items. We know A and n, so plug the two numbers in to get 20*7 = 140 for the sum.

The problem is asking us to maximize one figure: the last (and, of course, highest) number in the set. If all 7 numbers have to add up to 140, and we want to make one number as large as possible, then what do we have to do to the remaining six numbers?

We have to minimize all 6 of the remaining numbers so, for the rest of the problem, we need to figure out how to make the other 6 numbers as small as possible.

Do we know anything about those 6 other numbers? We were told that the median is 20; what does that mean? Draw out some dashes on your scrap paper, one for each number in the set:

____ ____ ____ ____ ____ ____ ____

Now, how can we represent the fact that the median is 20? We have an odd number of terms. The median will be the middle term (the 4^th, in this case) and it will actually equal 20. So add that to your diagram, along with an x for the term were supposed to maximize:

____ ____ ____ _20_ ____ ____ __x__

The problem also gives us some info about the first term:

the smallest number in the set is 5 less than half the largest number

Hmm. We dont know what the largest number is, of course thats what the problem asks us to maximize! But were calling that largest number x so lets write the smallest term in terms of x: ()x 5. Add that to the diagram:

_()x 5_ ____ ____ _20_ ____ ____ __x__

Okay, now what whats our goal again? Oh, right, we want to minimize everything that isnt that last term, x. Okay, so what can we minimize? We cant change the 1^st term; thats going to be ()x 5 no matter what. And we cant change the fourth term; thats going to be 20 no matter what. What about the 2^nd, 3^rd, 5^th, and 6^th terms? What are the smallest possible values for each of those?

Lets start with the rules for writing out a bunch of numbers in order to show a median. When you write a set of numbers to find the median, the requirement is to write the numbers from smallest to largest. Lets say that we have to write these three numbers in order: 20, 14, 18. We would write them: 14, 18, 20. Moving to the right, each number is higher than the previous number. Moving to the left, each number is lower than the previous number.

Is that all? What if we had to write these three numbers in order: 20, 14, 20? Then, we would write: 14, 20, 20. Moving to the right, the second term, 20, is higher than the first term, 14, but the third term, 20, is equal to the second term, 20. So the full rule is not that the numbers have to increase as you move to the right or decrease as you move to the left. The rule is that, as you move to the right, the number has to be equal to or higher than the number to the left. Similarly, as you move to the left, the number has to be equal to or lower than the number to the right.

So, lets get back to our problem. Were trying to minimize the remaining slots. What is the smallest possible value for the 5^th term, keeping in mind that the number has to be equal to or higher than the 4^th term? The 4^th term is 20, so the smallest value for the 5^th term is also 20. For the same reason, the smallest value for the 6^th term is also 20.

_()x 5_ ____ ____ _20_ _20_ _20_ __x__

What about the 2^nd and 3^rd terms? The 2^nd term has to be equal to or higher than the 1^st term, and the first term is ()x 5. Therefore, the smallest possible value for the 2^nd term is equal to ()x 5. For the same reason, the smallest value for the 3^rd term is also ()x 5.

_()x 5_ _()x 5_ _()x 5_ _20_ _20_ _20_ __x__

Now we have representations for all seven terms: either real numbers or variable expressions. We know the seven terms add up to 140. Time to set up an equation and solve for x!

[()x 5] + [()x 5] + [()x 5] + 20 + 20 + 20 + x = 140

(3/2)x 15 + 60 + x = 140

(5/2)x = 95

x = 95(2/5)

x = 38

The correct answer is B.

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!)

Note: the key takeaways are the same as last week, when we did some lower-level max/min problems. The basic process doesn't change; we just have a bit more we need to know and a bit more we need to do on the very hard problem we did this week.