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Birthday109
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Last month 15 homes were sold in Town X. The average (arithmetic mean) 19 Nov 2024, 21:35
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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.
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Last month 15 homes were sold in Town X. The average (arithmetic mean) 20 Nov 2024, 00:29
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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.
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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.
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Last month 15 homes were sold in Town X. The average (arithmetic mean) 04 Jun 2025, 02:29
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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
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Last month 15 homes were sold in Town X. The average (arithmetic mean) 23 May 2026, 04:55
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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?
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Last month 15 homes were sold in Town X. The average (arithmetic mean) 16 Aug 2026, 01:33
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Quick algebraic approach for Statement I using deviations from the mean:

Median = 130K, so the 8th term is 130K. To minimize the upper 7 terms, maximize the first 8 by setting all of them to 130K.

Their total shortfall from the 150K mean is:
8(150k−130k)=160K

So the upper 7 terms must compensate by 160K to staify the mean value of 150k. Their average gain must be:
160k/7 = ~22.8

Value for each home above median = 150k+22.8 = 172.8K

Since 172.9K > 165K, at least one home must have sold for more than 165K.
Hence, Statement I must be true.

Note: If any of the first 8 terms are below 130K, the required average of the upper 7 homes only increases.
sidhu4u
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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