Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Last month 15 homes were sold in Town X. The average [#permalink]

Show Tags

19 Apr 2010, 06:05

14

This post received KUDOS

54

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

53% (00:59) correct
47% (01:03) wrong based on 1007 sessions

HideShow timer Statistics

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

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

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.

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?
_________________

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?

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.

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?
_________________

Want to improve your CR: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html

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.

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.

Plugin to check the options might well consume time in above questions. If there is some way in which we can deduce the question in terms of equation and then verify (just like we do in inequality, co-ordinate geometry), then it might consume less time. I have tried to deduce the data in terms of equation but in the end, I did the same to plugin data to satisfy above conditions. If someone can deduce the data into equations and solve it without plugin, then it will be less time consuming.

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.

Last month 15 homes were sold in Town X. The average sale price of the houses was $150,000 and the median sales price was $130,000. Which must be true? At least 1 house sold for > $165,000 At least 1 house sold for > $130,000 but < $150,000 At least 1 home sold for < $130,000 I II III I and II I and III

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.

median (130) is lower than average (150). Thus the first 8 has at best 130 and rests must have more average greater than 172. For this one of the 7 values must be greater than 165.

total 150*15=2250 the first 8s = 130*8= 1040 the rest 7s = 2250-1040= 1210 1210/7= > 172, for which one value must be >165.

Re: Last month 15 homes were sold in Town X. The average [#permalink]

Show Tags

26 May 2012, 10:03

sidhu4u wrote:

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

Simplest way is by elimination, if we take III it may be possible but a series can have 13000,x,x.....Median....>130,000 so "Must" makes it false, so is the second one, so even without calcs we can say answer is 'A'

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

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).

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???

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

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.

Re: Last month 15 homes were sold in Town X. The average [#permalink]

Show Tags

19 Nov 2013, 13:38

Hi Moderators, Here is my approach. Since median is 130K there 8 values which are bringing down the average by at least 20, so the average for the remaining 7 values must go up by more than 20 i.e. 150+ (20+). As such 'I' has to be true. Based on this understanding we can eliminate II and III. Does it look good? Thanks!
_________________

Please contact me for super inexpensive quality private tutoring

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Re: Last month 15 homes were sold in Town X. The average [#permalink]

Show Tags

08 Nov 2014, 04:41

Bunuel wrote:

sidhu4u wrote:

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?

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.

Absolutely Nailed it. Extremely Clear and Concise Explanation for a generic Approach which people usually tend to miss out.

Re: Last month 15 homes were sold in Town X. The average [#permalink]

Show Tags

23 Aug 2015, 06:22

For all of the people like me who are not as intelligent as Bunuel , I solved II and III first and assumed I by POE. During review, when I did solve I to understand the concept, I made x_1-x_8 to be 130 and x_9-x_15 to be 165 and found out that it was less than the 150 mean, so we know that least one of the houses had to be higher than 165.

For all of the people like me who are not as intelligent as Bunuel , I solved II and III first and assumed I by POE. During review, when I did solve I to understand the concept, I made x_1-x_8 to be 130 and x_9-x_15 to be 165 and found out that it was less than the 150 mean, so we know that least one of the houses had to be higher than 165.

The trick to all these questions is to recognize the fact that in order to maximize 1 entity, you need to minimize the others and in order to minimize 1 entity you need to maximize the others. This is what Bunuel has done in his post above.

Additionally, average = number of terms * sum of the terms

The above 2 tips should provide you the correct answer.
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...