Last month 15 homes were sold in Town X. The average (arithmetic mean)

Question

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 $130,0000 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

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Bunuel · Accepted Answer

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 $130,0000 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

Note that we are asked which  MUST  be true.

Given:  {x_1}+{x_2}+...+{x_7}+({x_8=130})+{x_9}+...+{x_{15}}=15*150=2250

Let's start with the first one and try to make it false.

I. At least one of the homes was sold for more than $165,000.

Worst case scenario, when  x_{15}  has the least value (trying to make it less than 165), would be when  x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=max , and  x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}=min :
 8*130+7x=2250  -->  x_{min}\approx{173} .

So we got that I is always true: At least one of the homes was sold for more than $165,000 (as for the worst case scenario we got that least value of  x_{15}>165 ).

But if we take the scenario which we considered:  x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130 , and  x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173}  we can see that II and III with this scenario are false. So II and III are not always true.

Answer: A (I only).

Testluv · Answer

If we understand medians, use reasoning (instead of pure algebra), and have good technique, then we can answer this question very quickly.

Median just refers to the middle number in a sequence of ordered numbers. So, the median here: {3, 3, 3} is 3 even though all the numbers are 3.

We are told that the median is 130k. So, the 8th house sold for 130k. But the 1st through 7th houses may also have sold for 130k. Eliminate III; eliminate C and E.

We can also easily eliminate II. We could have 8 houses that sold for 130k or less; the rest can sell for well above 150k. Eliminate B and D.

The answer must be choice A, and there is no need to evaluate I.

Note that my approach was essentially the same as badgerboy's: start with the choices you can more easily prove untrue. The alternative is to start with the roman numerals that show up most frequently. Of course, another alternative is to use pure algebra, and for some people (I don't think many but some) that may well be more efficient.

Bunuel · Answer

We have:  x_1\leq{x_2}\leq{x_3}\leq{x_4}\leq{x_5}\leq{x_6}\leq{x_7}\leq{x_8=130}\leq{x_9}\leq{x_{10}}\leq{x_{11}}\leq{x_{12}}\leq{x_{13}}\leq{x_{14}}\leq{x_{15}} .

We are trying to make statement I false, which says:  At least one of the homes was sold for more than $165,000 . More than 165 can be terms from  x_9  to  x_{15} . Basically worst case scenario here means minimizing the value of  x_{15}  (finding the least possible value of  x_{15} ). How can we do that?

First we should maximize the values from  x_1  to  x_7  (by increasing/maximizing these terms, the lowest terms, we are decreasing/minimizing the highest terms). Their max values can be  130=x_8  (as  x_8  is the median value and the terms from  x_1  to  x_7  can not be more than this value).

Next: to minimize  x_{15}  we should make terms from  x_9  to  x_{15}  be the same.

As in solution the least possible value of  x_{15}  is  \approx{173} , thus values less then 165 are not possible. So at least one home was sold for more than $165.

Hope it's clear.

BrentGMATPrepNow · Answer

The key word in this question is  MUST 
So, if it's possible to create a scenario in which the statement is  not true , we can  eliminate it .

So, let's create a possible scenario and see which answer choices we can eliminate.

Aside : To make things simpler, let's  divide all of the prices by 1000 .

First, we'll use a nice rule that says:   sum of all values = (mean)(number of values)  
So, the sum of all 15 prices = ($150)(15) =   $2250  .

If the median is $130, then the middlemost value is  $130

So, one possible scenario is: 
130, 130, 130, 130, 130, 130, 130,  130 , 130, 130, 130, 130, 130, 130, 430

Aside : To find the last value (430), I took the sum of all 15 numbers (  2250  ) and subtracted (14)(130)

Notice that this scenario tells us that statements II and III need not be true.
Since answer choices B, C, D and E all include either II or III, we can eliminate them.

This leaves us with A, which must be the correct answer.

Cheers,
Brent

RELATED VIDEO 
 https://www.youtube.com/watch?v=Y9f_0QJU1ug

sidhu4u · Answer

Thanks for the quick response Bunuel. The answer is correct.

Great explanation, I'm just not able to digest how to attack such a problem in future. Also, the difficult part is how to identify the worst case scenario here? Why cant we take other different values instead of 8 values being 130 and the remaining 7 values at minimum?

ykaiim · Answer

We are discussing the worst scenario here. In addition, we are not given a lower limit and upper limit on the sale price of the homes.

What if some houses are below $130K?

Can someone explain more on this?

Bunuel · Answer

Note that we are asked which MUST be true. Given: {x_1}+{x_2}+...+{x_7}+{x_8=130}+{x_9}+...+{x_{15}}=15*150=2250 Let's start with the first one and try to make it false. I. At least one of the homes was sold for more than $165,000. Worst case scenario, when x_{15} has the least value (trying to make it less than 165), would be when x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=max , and x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}=min : 8*130+7x=2250 --> x_{min}\approx{173} . So we got that I is always true: At least one of the homes was sold for more than $165,000 (as for the worst case scenario we got that least value of x_{15}>165 ). But if we take the scenario which we considered: x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130 , and x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173} we can see that II and III with this scenario are false. So II and III are not always true. Answer: I only. More at: gmatprep-2010-statistics-92909.html#p715101

RGM · Answer

This is my take on this.

Consider the case that the first 8 houses are priced at 130K. The total sale amount is 2250K, and with 8 houses at 130K (1040K), you're left with 1210K for the 7 remaining houses. Dividing this amount by 7 would give you the least prices that you could price these 7 houses. With approx. 172K per house, a house priced above 165K must be a part of the portfolio.

From the above, one can see that number 2 is not necessarily correct. One can have 8 houses at 130K and the rest at the quoted price above.

The third - well, we can satisfy the conditions and not have a house priced below 130K.

hirendhanak · Answer

why 3rd is wrong ? because if the median is 130000 , atleast a few elements shoulce be below it

pls help

Bunuel · Answer

Not necessarily.

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

So, for a set with odd number of elements the median is just the middle number: it's not necessary for any number of terms in a set to be more or less than the median. Consider the set {1, 1, 1,} --> median=1 or the set {1, 1, 2} --> median=1 no term is less than 1...

So if we take:  x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=median , and  x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173}  we can see that III is wrong with this scenario, so it's not always true.

Hope it's clear.

gmat1220 · Answer

Median means 7 prices were above the median and 7 below the median.

I. At least one of the homes was sold for more than $165,000. 
Total sales = 150K * 15 = 2250K

Lets say all the seven homes above the median were sold for 165K

Hence 
165K * 7 + 130K * 8 = 1159K + 1040K = 2199K - Not high enough. Hence one home MUST have been above 165K

A, D and E left.

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

This is not true you can have the mean 150K even when 7 numbers are below 130K and 7 numbers are above 150K. I will not solve this scenario since this is intuitive.

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

This is not true since we can have 8 numbers at 130K and 7 numbers above 130K and have the mean 150K. I will not solve this scenario since this is intuitive.

D and E out. A remains.

fluke · Answer

Let's work with the numbers dividing everything by 1000. Total value of the sale = 15*150=2250 Median sale = $130 If we arrange the sale price in ascending order the 8th value would be 130 I. Let's try to prove it wrong. x,x,x,x,x,x,x,130,y,y,y,y,y,y,Z To make Z<=165 x should be as big as possible. Let's make them all 130*8=1040 Let's make all y as 165 165*6=990 Z = 2250-990-1040=2250-2030=220>165. Apparently; it's true. II. At least 1 house sold for > $130,000 but < $150,000 Not true. Same example as st1. 130,130,130,130,130,130,130,130,165,165,165,165,165,165,225 Not true. III. At least 1 home sold for < $130,000 Not true. Same example as st1. 130,130,130,130,130,130,130,130,165,165,165,165,165,165,225 Not true. Not true. Ans: "A"

bsahil · Answer

Hi Bunuel,

Why have we not considered the case that some terms from 1-7 are less than 130, 8th term is 130 and then again terms from 9-15 some are between 130- 150 some even above 165.. In that case all three statements will be true...
What am i missing here???

Sahil

VeritasPrepRon · Answer

Hi Sahil, I have left the only quote from Bunuel's explanation that is needed to answer your question. We are looking for something that  must  be true, not that can be true.

If something can be true, then come up with any fairy tale explanation you want. Maybe 1 house was 1,000,000 and the others were all 5$. If 14 of the houses are sold for exactly 130,000 but one is sold for 505,000$. This satisfies the situation, but II) and III) do not happen. Thus, we have found a situation where II and III do not occur, and we can eliminate them.

If I'm looking for something that must be true, I cannot find a single example that will negate it. Try as you will with number I), no scenario will satisfy the conditions and not have a house sold for more than 165,000$. The mathematics of the question guarantee it.

Hope this helps!
-Ron

ScottTargetTestPrep · Answer

We are given that 15 homes were sold in Town X last month. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. We can start by reducing $150,000 and $130,000 by dividing each number by 1,000.
 
We now have that the mean sale price of the homes was $150 and the median sale price was $130. Let’s now analyze the statements to determine which MUST be true.
 
I. At least one of the homes was sold for more than $165,000.
 
(We can reinterpret Roman numeral I as: At least one of the homes was sold for more than $165.)
 
To determine whether the above statement MUST be true, let’s see if we can find a scenario in which none of the homes is priced at more than $165. Furthermore, since the median price is $130, let’s create a scenario in which the eighth value of the 15 values (i.e., the middle number) is 130, the first seven values are all 130 and the last seven values are all 165. That is:
 
130, 130, 130, 130, 130, 130, 130, 130, 165, 165, 165, 165, 165, 165, 165, 165
 
If this is the case, then the sum would be 130 x 8 + 165 x 7 = 1,040 + 1,155 = 2,195. However, since the average of price of the homes is $150, the sum of the prices of the homes is 150 x 15 = 2,250. This means that at least one of the numbers in the list above has to be changed to a number greater than 165 to make up the difference between 2,195 and 2,250. (For example, since the difference is 55, we can change the last number 165 to 220 to make up this difference.)
 
So Roman numeral I is correct.
 
II. At least one of the homes was sold for more than $130,000 and less than $150,000.
 
(We can reinterpret II as: At least one of the homes was sold for more than $130 and less than $150.)
 
In the analysis of Roman numeral I, we showed that none of the homes had to be priced between $130 and $165. Roman numeral II is not correct.
 
III. At least one of the homes was sold for less than $130,000.
 
(We can reinterpret III as: At least one of the homes was sold for less than $130.)
 
In the analysis of Roman numeral I, we showed that none of the homes had to be priced less than $130. Roman numeral III is not correct.
 
Answer: A

KarishmaB · Answer

15 homes were sold and median is 130k. This means if we put them all in increasing order, the 8th home was sold for 130k. So the first seven homes were sold for a maximum of 130k and could be below that amount.

Since mean is 150k, the other 7 sales must make up for the deficit of 8 * 20k = 160k. So if all 7 were sold for the same price, they would have been sold for 160k/7 = 23k more than 150k i.e. for about 173k each.

So one possibility is
130k, 130k, ... 130k, 173k, 173k, ... 173k

I. At least one of the homes was sold for more than $165,000.

Correct. To make up for 160k, each of the 7 homes must be sold for 173k. Here we are minimising the highest sale value. So at least one home must be sold for more than $165k.

II. At least one of the homes was sold for more than $130,0000 and less than $150,000

Not necessary. In our example, we saw that we don't need to have a sale value in between this range.

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

Not necessary, again as we see from our example.

Answer (A)

adityaganjoo · Answer

Total houses = 15 Mean SP = $150k Median SP = $130K Statement I : At least one of the homes was sold for more than $165,000 Let us assume that all the houses were sold at the highest value possible, and see if we can still reach $165k Maximum possible SP for first 8 houses = Median = $130k Maximum possible SP for next 7 houses = x 130k(8) + 7x = 150k(15) => x=(2250k - 1040k)/7 => x=1210k/7 = 174.something k >165k Therefore, even if we maximise the SP of all houses, in order to distribute the total value among all, SP for some houses will definitely cross 165k Statement II At least one of the homes was sold for more than $130,0000 and less than $150,000 Let us assume that cheapest 7 houses were sold at $1 each. Next 7 at median value of $130k each and the most expensive one at x So, 7(1) + 130k(7) + x = 150k(15) Solving, we get: x = 1340k < 165k Therefore, Statement II is not definitely true. Statement III At least one of the homes was sold for less than $130,000. This is easy, since Median SP < Mean SP. Suppose, all houses were sold at median SP, mean would be equal to the median. But mean is larger than median. One case could be: 14 houses were sold at 130K and 15th at somewhat higher, such that the mean became 150k. Thus, no house was sold at less than 130k. Obviously, cases where a few houses were sold below 130k are possible, but we cannot conclude that the Statement III is definitely true. Therefore, III is out Therefore, only Statement I is 'always true'. Thus, the correct answer is (A)

Vardaan24 · Answer

Hi, is there a solution showing how to answer this question without a calculator? I believe the quant section does not provide access to a calculator anymore and the solutions provided thus far require a calculator. Thanks!

nishantswaft · Answer

Hey don't get worried by the large amount of data and steps required for this question. If you logically follow the question you can see that it can be done in an easier way without using a calculator.

we are asked which  MUST  be true.  
 Median just refers to the middle number in a sequence of ordered numbers.

We have to try to find a scenario in which statements are proven false.

Sum = average * number of values 
 So, the sum of all 15 prices = ($150)(15) = $2250

since we know that median is $130k

Lets say all the seven homes above the median were sold for 165K

Hence 
 165K * 7 + 130K * 8 = 1159K + 1040K = 2199K this is not equal to $2250 one home MUST have been above 165K

Now if you look at answer choices, the options are in such a way that only A is correct here.

Hope you find this helpful.

Last month 15 homes were sold in Town X. The average (arithmetic mean)

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