Today’s GMAT tip comes from our friends at ManhattanGMAT. This article serves as a follow-up to a recent one on max/min questions. This time around, ManhattanGMAT instructor Stacey Koprince offers advice on an even more difficult sample GMATPrep® max/min question:
Recently, 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 don’t know what the answer is and you have to guess… you’ve still picked an answer, right? If you haven’t, go pick an answer before you keep reading!
As we discussed, 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 fourth, in this case) and it will actually equal 20. So add that to your diagram, along with an “x” for the term we’re 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 don’t know what the largest number is, of course – that’s what the problem asks us to maximize! But we’re calling that largest number “x” so let’s write the smallest term in terms of x: (½)x – 5. Add that to the diagram:
_(½)x – 5_ ____ ____ _20_ ____ ____ __ x __
Okay, now what – what’s our goal again? Oh, right, we want to minimize everything that isn’t that last term, “x.” Okay, so what can we minimize? We can’t change the first term; that’s going to be (½)x – 5 no matter what. And we can’t change the fourth term; that’s going to be 20 no matter what. What about the second, third, fifth, and sixth terms? What are the smallest possible values for each of those?
Let’s 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. Let’s 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, let’s get back to our problem. We’re trying to minimize the remaining slots. What is the smallest possible value for the fifth term, keeping in mind that the number has to be equal to or higher than the fourth term? The fourth term is 20, so the smallest value for the fifth term is also 20. For the same reason, the smallest value for the sixth term is also 20.
_(½) x – 5_ ____ ____ _20_ _20_ _20_ __ x __
What about the second and third terms? The second term has to be equal to or higher than the first term, and the first term is (½)x – 5. Therefore, the smallest possible value for the second term is also (½)x – 5. For the same reason, the smallest value for the third term is also (½)x – 5.
_(½) x – 5_ _(½) x – 5_ _(½) 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!
[(½)x – 5] + [(½)x – 5] + [(½)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 (same as before!):
(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 before, 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.
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