# Statistical “Combo” Problems

*by*, Jan 20, 2012

Today were going to talk about statistics problems in which we have to combine knowledge of more than one concept. Try this GMATPrep problem first; set your timer for 2 minutes and go!

Last month 15 homes were sold in Town X.

The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?I. At least one of the homes was sold for more than $165,000.

II. At least one of the homes was sold for more than 130,000 and less than 150,000.

III. At least one of the homes was sold for less than $130,000.

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I and III

Sigh. I hate roman numeral questions. I have to do more work to solve the problem, and I never like that. :)

Okay. The question mentions both average and median. What are those two things again? Average, or mean, represents the sum of the sales prices divided by the number of houses sold, or *A* = *S*/*n*, where *A *is the average, *S* is the sum, and *n* is the number of items. In this case, they told us both *A* and* n*, so we can calculate *S*:

150,000 = *S* / 15

*S* = (150,000)(15) = 2,250,000

What else? The median is the middle number in a range of numbers written from smallest to largest. If the range consists of an even number of terms, then the median is the average of the two terms in the middle. In this case, were told that there are an odd number of homes, 15, so the median home is the 8^{th} home (if the sales prices are arranged in order from least to most expensive).

Quick check: how do you know that the median really is the 8^{th} home and not, say, the 7^{th}? What do you have to add to 8 to get to 15? Right: 8 + 7 = 15. Subtract that same number: 8 7 = 1. Our range runs from the 1^{st} home to the 15^{th} home, so 8 is indeed exactly in the middle.

Okay, so the cost of the 8^{th} home was $130,000. Homes 1 through 7 cost less than or equal to $130,000. Homes 9 through 15 cost more than or equal to $130,000.

Combining those two pieces of info, I realize that at least one home had to cost more than $130,000 otherwise, we couldnt get an average that is higher than the median. In fact, at least one home had to be more than $150,000 in order to get an average of $150k.

Thats interesting. $150k cant be the price of the most expensive home, because then the rest would be less than $150k, so we wouldnt average that number. I wonder what the bare minimum price is for the most expensive home?

Lets look at the statements. I dont want to calculate anything unless I have to, obviously! And check it out the first statement discusses a minimum for at least one of the homes; thats another way of saying: the most expensive home sold for more than $165,000.

Does that have to be true? Lets see. How would I minimize the cost of the most expensive home, house #15? All 15 houses together have to add up to $2,250,000. To minimize the cost of the most expensive, I want to maximize the cost of all of the others.

Houses 1 through 7 could each have cost $130,000 at most. House 8 was $130,000. Thats (130,000)(8) = $1,040,000. I still have $2,250,000 - $1,040,000 = $1,210,000 left to go. How can I maximize the values of the remaining 7 houses?

They could all equal the exact same thing! Lets take the remaining money, $1,210,000, and divide by 7: its something more than $170,000. (Note: I did long division there and, after I got the 17 to start, I stopped.) The statement says that at least one house cost more than $165,000. 170-something-thousand is more than 165,000, so I can already answer the question without knowing the exact number. Roman numeral I must be true.

Eliminate answers B and C. (Wondering why that always works? Let's say that the first 8 houses really are $130,000 and the next 7 are 170-something-thousand. How could the numbers change? The 8th has to be $130,000; that can't change. The 7 "below" that have to be $130k or less, so they can only go down. Houses 9 through 14 have to be equal to or less than the 15th house, so they can only go down unless the price for the 15th house goes up. If any house prices go down, then other houses have to be come more expensive to make up for that - so the 15th house can only become more expensive, not less expensive. Therefore, 170-something-thousand is the lowest possible price for the 15th house!)

Ugh. I hope the other two statements dont take as long to do. Lets start by seeing if we can accept or eliminate either of the other two statements based on what weve done so far.

Hey, excellent! We can eliminate statement II! In the example we just showed above, which is one possible set of prices for these houses, 8 of the houses sell for $130,000 and the other 7 sell for $170-something-thousand each. No houses in that set sell for between $130, and $150k, so this statement doesnt have to be true. Great! Eliminate answer D.

We can use the same reasoning for statement III. In the example we just tried, none of the homes sold for less than $130k, so this statement also does not have to be true. Eliminate answer E.

**The correct answer is A: I only.**

When we first start must be true roman numeral problems, we often just have to pick one of the statements and try to prove / disprove it. Often, it's easier to try to disprove, because we only need one case where the statement isn't true in order to eliminate it (which is the one really nice thing about "must be true" problems). If we get stuck, we leave that statement in and try a different statement instead.

**Key Takeaways for Statistics Combo Problems, and Roman Numerals:**

(1) If the problem mixes average (mean) and median, make sure you know what each term means *and* how they interrelate. What does it mean when the average is greater than the median? Less than the median? What might be true if the average equals the median? (Tell me in the comments and I'll tell you whether you're right!)

(2) Combo statistics problems lend themselves well to minimum / maximum type questions. Think about what you need to do to all of the other numbers in order to minimize or maximize a certain number. When all of the numbers have to add up to a certain fixed sum (as in Average problems), then minimizing one number will mean maximizing the others, and vice versa.

(3) On MUST be true questions, we have to find something that is always true, not just sometimes true. If possible, try to find one case in which the statement is false; then you know that you can eliminate that statement or answer. There *will* be traps revolving around things that are true sometimes but not *all* the time watch out for those traps!

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

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