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

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

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12 Aug 2017, 11:40
himanshukamra2711 wrote:
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
2250-1934=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.

Sorry, but I was not able to decipher what you were trying to say...
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Re: Last month 15 homes were sold in Town X. The average  [#permalink]

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13 Aug 2017, 00:55
Bunuel wrote:
himanshukamra2711 wrote:
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
2250-1934=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.

Sorry, but I was not able to decipher what you were trying to say...

Option 2.
If i take value of house as 149000 for 6 house(149*6+130*8=1934),so now the total is 1934000 and we need to have the total as 225000 so the last 7th house can have the value as 316000.Moreover,if the minimum value of the house is 173000 then how can 165000 be the value of 6 houses.(I am not able to come to the conclusion that if minimum is 173k then atleast one house is of 165k,and if this is so then why not 149k)
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Re: Last month 15 homes were sold in Town X. The average  [#permalink]

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11 Sep 2017, 10:57
Bunuel wrote:
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: http://gmatclub.com/forum/gmatprep-2010 ... ml#p715101 Hi Bunuel For stmt 1, why cant we have a scenario where houses 1 to 8 are 130 each, house 9 is 160, house 10 is 186 and remaining being the balance average i.e. around 173? Manager Joined: 30 Jul 2014 Posts: 137 GPA: 3.72 Re: Last month 15 homes were sold in Town X. The average [#permalink] ### Show Tags 13 Sep 2017, 03:29 I solved for statement 1,2, and 3 correctly, but while evaluating the options forgot what I deduced - marked E, when the solution could be option A. It seems that it is a problem of attention span - I don't know how to solve such psychological problems, such as this one, lack of focus etc... Could anyone help. _________________ A lot needs to be learned from all of you. Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3896 Location: United States (CA) Re: Last month 15 homes were sold in Town X. The average [#permalink] ### Show Tags 08 Dec 2017, 11:57 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 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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions CEO Joined: 12 Sep 2015 Posts: 3024 Location: Canada Re: Last month 15 homes were sold in Town X. The average [#permalink] ### Show Tags 20 Apr 2018, 15:05 Top Contributor 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 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 FROM OUR COURSE _________________ Brent Hanneson – GMATPrepNow.com Sign up for our free Question of the Day emails Director Joined: 09 Mar 2016 Posts: 946 Last month 15 homes were sold in Town X. The average [#permalink] ### Show Tags 18 Jun 2018, 06:37 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 hey pushpitkc, here is my reasoning to the problem above, I think I am on the right track but couldn't figure out how to check to be sure that at least one of the homes was sold for more than$165,000.

$$y$$ is house and $$130$$ is median = $$x$$, average is $$150$$

so we have following:

130,130,130,130,130,130,130,173 etc
$$y$$, $$y$$, $$y$$, $$y$$ ,$$y$$, $$y$$, $$y$$, $$X$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$

say each house costs 150 but since since median is 130 the first 7 houses cant cost more than 130. so we need to reduce the cost of 7 houses by at least 20.

from each of the first 7 houses we get extra $$20 *7 = 140 +20(median) = 160$$ so this is the money that need to be distributed over the next 7 houses (that cost at least 150 each) after median.

$$160/7 = 23$$ Approx ---- > hence $$150 +23 = 173$$
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Re: Last month 15 homes were sold in Town X. The average  [#permalink]

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18 Jun 2018, 11:00
1
dave13 wrote:
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

hey pushpitkc,

here is my reasoning to the problem above, I think I am on the right track but couldn't figure out how to check to be sure that at least one of the homes was sold for more than $165,000. $$y$$ is house and $$130$$ is median = $$x$$, average is $$150$$ so we have following: 130,130,130,130,130,130,130,173 etc $$y$$, $$y$$, $$y$$, $$y$$ ,$$y$$, $$y$$, $$y$$, $$X$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$ say each house costs 150 but since since median is 130 the first 7 houses cant cost more than 130. so we need to reduce the cost of 7 houses by at least 20. from each of the first 7 houses we get extra $$20 *7 = 140 +20(median) = 160$$ so this is the money that need to be distributed over the next 7 houses (that cost at least 150 each) after median. $$160/7 = 23$$ Approx ---- > hence $$150 +23 = 173$$ Hey dave13 First of all, the reasoning is correct. But it is not necessary that each house after the median has a price greater than$165000. There is a possibility that each &
every house except one can have a selling price of less than $165000 As you rightly pointed, the$160000 needs to be distributed among the remaining
7 houses

7+1*($130000) + 6*$164000 + x = 15*$150000 ->$1040000 + $984000 + x =$2250000
-> x = $226000($2250000 - $2024000) Here, only one house will have a sale price which is greater than$165000

Hope this helps you!
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Re: Last month 15 homes were sold in Town X. The average  [#permalink]

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19 Jun 2018, 08:28
pushpitkc wrote:
dave13 wrote:
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

hey pushpitkc,

here is my reasoning to the problem above, I think I am on the right track but couldn't figure out how to check to be sure that at least one of the homes was sold for more than $165,000. $$y$$ is house and $$130$$ is median = $$x$$, average is $$150$$ so we have following: 130,130,130,130,130,130,130,173 etc $$y$$, $$y$$, $$y$$, $$y$$ ,$$y$$, $$y$$, $$y$$, $$X$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$, $$y$$ say each house costs 150 but since since median is 130 the first 7 houses cant cost more than 130. so we need to reduce the cost of 7 houses by at least 20. from each of the first 7 houses we get extra $$20 *7 = 140 +20(median) = 160$$ so this is the money that need to be distributed over the next 7 houses (that cost at least 150 each) after median. $$160/7 = 23$$ Approx ---- > hence $$150 +23 = 173$$ Hey dave13 First of all, the reasoning is correct. But it is not necessary that each house after the median has a price greater than$165000. There is a possibility that each &
every house except one can have a selling price of less than $165000 As you rightly pointed, the$160000 needs to be distributed among the remaining
7 houses

7+1*($130000) + 6*$164000 + x = 15*$150000 ->$1040000 + $984000 + x =$2250000
-> x = $226000($2250000 - $2024000) Here, only one house will have a sale price which is greater than$165000

Hope this helps you!

hello pushpitkc

many thanks for explanation i just have two questions

Why are you here (6*$164000) multiplying by 164000 and not by 173000 And second qestion how based on this x =$226000($2250000 -$2024000) you conclude that only one house will have a sale price which is greater than $165000 thank you Senior PS Moderator Joined: 26 Feb 2016 Posts: 3191 Location: India GPA: 3.12 Last month 15 homes were sold in Town X. The average [#permalink] ### Show Tags 19 Jun 2018, 09:12 dave13 wrote: pushpitkc wrote: Hey dave13 First of all, the reasoning is correct. But it is not necessary that each house after the median has a price greater than$165000. There is a possibility that each &
every house except one can have a selling price of less than $165000 As you rightly pointed, the$160000 needs to be distributed among the remaining
7 houses

7+1*($130000) + 6*$164000 + x = 15*$150000 ->$1040000 + $984000 + x =$2250000
-> x = $226000($2250000 - $2024000) Here, only one house will have a sale price which is greater than$165000

Hope this helps you!

hello pushpitkc

many thanks for explanation i just have two questions

Why are you here (6*$164000) multiplying by 164000 and not by 173000 And second qestion how based on this x =$226000($2250000 -$2024000) you conclude that only one house will have a sale price which is greater than $165000 thank you Hey dave13 The reason I am able to say that only one house has the price greater than$165000 is
because I am assuming that 6 houses(which cost greater than the median) cost less than
$165000 each and we can conclude that only 1 house costs more than$165000

7+1*($130000) + 6*$164000 + x = 15*$150000 -> x =$2250000 - $2024000 =$226000

Now, 8 houses cost $130000 each, 6 houses cost$164000 and 1 house costs $226000(which is greater than$165000).

Hope this clears your confusion!
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