Last visit was: 07 Jun 2026, 18:02 It is currently 07 Jun 2026, 18:02
Home
Messages
MBA Fair
Tests
Schools
Events
Close
GMAT Club Daily Prep
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
User avatar
Birthday109
User avatar
Joined: 27 Jul 2024
Last visit: 15 Dec 2024
Posts: 4
Given Kudos: 5
Location: New Zealand
GMAT Focus 1: 575 Q79 V81 DI76
GMAT Focus 1: 575 Q79 V81 DI76
Posts: 4
Kudos: 0
Last month 15 homes were sold in Town X. The average (arithmetic mean) 19 Nov 2024, 21:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Why should we make terms from x_9 to x_15 the same? Our purpose is to minimise x_15 so wouldn't we want to maximise x_9 to x_14? But then again, we don't get given a maximum value that the house price could be.
Bunuel
sidhu4u
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.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 07 Jun 2026
Posts: 111,137
Own Kudos:
819,406
Given Kudos: 106,717
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 111,137
Kudos: 819,406
Last month 15 homes were sold in Town X. The average (arithmetic mean) 20 Nov 2024, 00:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Birthday109
Why should we make terms from x_9 to x_15 the same? Our purpose is to minimise x_15 so wouldn't we want to maximise x_9 to x_14? But then again, we don't get given a maximum value that the house price could be.
Bunuel
sidhu4u
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.
Show more

The key reason for making terms from x_9 to x_15 the same is to minimize x_15, as explained in the solution. The goal is to create the "worst-case scenario" where x_15 is as low as possible while satisfying the constraints of the problem. By equalizing these terms, you ensure that x_15 reaches its minimum value under the given conditions.
User avatar
BinodBhai
User avatar
Joined: 19 Feb 2025
Last visit: 15 Oct 2025
Posts: 106
Own Kudos:
31
Given Kudos: 252
Posts: 106
Kudos: 31
Last month 15 homes were sold in Town X. The average (arithmetic mean) 04 Jun 2025, 02:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sum = 15 x 150k = 2,250,000 (2.25 Mn)
If out of 15, 14 homes valued at 130k = 14x130k = 1,820,000 (1.82 Mn)

Diff = 2.25 Mn - 1.82 Mn = 430k

Now analysing statements
I. At least one of the homes was sold for more than $165,000. - Yes (as calculated above)
II. At least one of the homes was sold for more than $130,0000 and less than $150,000 - not necessarily, as seen above
III. At least one of the homes was sold for less than $130,000. - not necessarily, as seen above

Hence answer is A
User avatar
KTH95
User avatar
Joined: 12 Jun 2024
Last visit: 06 Jun 2026
Posts: 1
Own Kudos:
5
Given Kudos: 315
Products:
GMAT Club Tests Forum Quiz
Posts: 1
Kudos: 5
Last month 15 homes were sold in Town X. The average (arithmetic mean) 23 May 2026, 04:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
It is a must be true question, since there is a scenario where st 2 and st 3 are true, there is also a scenario where they aren't. so for a must be true question they do not qualify. In such questions i try to find a scenario where it is not true to simplify the process
ykaiim
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?
Show more
   1   2 
Moderator:
Bunuel
Math Expert
111137 posts