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Last month 15 homes were sold in Town X. The average
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19 Apr 2010, 06:05
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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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Last month 15 homes were sold in Town X. The average
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19 Apr 2010, 06:44
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).
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Re: GMATPrep 2010  Statistics
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03 Jul 2010, 17:27
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.




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Re: GMATPrep 2010  Statistics
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19 Apr 2010, 07:04
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?
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Re: GMATPrep 2010  Statistics
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19 Apr 2010, 07:48
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.
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Re: PS : Sale Price of home
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24 May 2010, 08:04
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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Re: PS : Sale Price of home
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24 May 2010, 08:14
ykaiim wrote: 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? 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: gmatprep2010statistics92909.html#p715101
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Re: Statistics from GMATPrep
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03 Jul 2010, 08:51
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.



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Re: Statistics from GMATPrep
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03 Jul 2010, 10:20
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, coordinate 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.



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Re: Average
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23 Mar 2011, 00:51
Onell wrote: 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 = 22509901040=22502030=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"
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Re: Average
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23 Mar 2011, 08:17
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 = 22501040= 1210 1210/7= > 172, for which one value must be >165. Ans A
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Re: Statistics
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01 May 2011, 17:11
15 homes can have the following values 8 Homes having value equal to $130k 3 Homes having value equal to $150k 4 Homes having value equal to $190K With above values median will be $130k and average will be $150k Statement II and III become not correct/true. So A. I only no need of checking Statement I



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Re: Last month 15 homes were sold in Town X. The average
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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'



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Re: GMATPrep 2010  Statistics
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20 Aug 2013, 09:04
Bunuel 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
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). Hi Bunuel, Why have we not considered the case that some terms from 17 are less than 130, 8th term is 130 and then again terms from 915 some are between 130 150 some even above 165.. In that case all three statements will be true... What am i missing here??? Sahil



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Re: GMATPrep 2010  Statistics
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20 Aug 2013, 09:22
bsahil wrote: Bunuel wrote: Note that we are asked which MUST be true.
Hi Bunuel, Why have we not considered the case that some terms from 17 are less than 130, 8th term is 130 and then again terms from 915 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. Hope this helps! Ron
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Re: Last month 15 homes were sold in Town X. The average
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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!
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Re: Last month 15 homes were sold in Town X. The average
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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.



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Re: Last month 15 homes were sold in Town X. The average
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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_1x_8 to be 130 and x_9x_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.



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Re: Last month 15 homes were sold in Town X. The average
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23 Aug 2015, 06:26
TooLong150 wrote: 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_1x_8 to be 130 and x_9x_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.



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Last month 15 homes were sold in Town X. The average
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12 Aug 2017, 11:19
Bunuel 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
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). what if in case 2 i take 149k as the value of 6 properties,making the total as 149*6+1040=1934 22501934=316k,which can be the 7th or last property value in order to maintain the average,and if in case the min value is 173k then how 165k is taken as the value of 6 properties.




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