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

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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

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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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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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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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: Last month 15 homes were sold in Town X. The average (arithmetic mean)  [#permalink]

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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? _________________ Math Expert V Joined: 02 Sep 2009 Posts: 58395 Re: Last month 15 homes were sold in Town X. The average (arithmetic mean) [#permalink] Show Tags 2 1 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.

More at: gmatprep-2010-statistics-92909.html#p715101
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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: Last month 15 homes were sold in Town X. The average (arithmetic mean)  [#permalink]

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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.
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Re: 15 homes in town X  [#permalink]

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why 3rd is wrong ? because if the median is 130000 , atleast a few elements shoulce be below it

pls help
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Re: 15 homes in town X  [#permalink]

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hirendhanak wrote:
why 3rd is wrong ? because if the median is 130000 , atleast a few elements shoulce be below it

pls help

Not necessarily.

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

So, for a set with odd number of elements the median is just the middle number: it's not necessary for any number of terms in a set to be more or less than the median. Consider the set {1, 1, 1,} --> median=1 or the set {1, 1, 2} --> median=1 no term is less than 1...

So if we take: $$x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=median$$, and $$x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173}$$ we can see that III is wrong with this scenario, so it's not always true.

Hope it's clear.
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Re: Mean / Median Question from GMAT Prep Test 1  [#permalink]

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Median means 7 prices were above the median and 7 below the median.

I. At least one of the homes was sold for more than $165,000. Total sales = 150K * 15 = 2250K Lets say all the seven homes above the median were sold for 165K Hence 165K * 7 + 130K * 8 = 1159K + 1040K = 2199K - Not high enough. Hence one home MUST have been above 165K A, D and E left. II. At least one of the homes was sold for more than$130,000 and less than $150,000. This is not true you can have the mean 150K even when 7 numbers are below 130K and 7 numbers are above 150K. I will not solve this scenario since this is intuitive. III. At least one of the homes was sold for less than$130,000.

This is not true since we can have 8 numbers at 130K and 7 numbers above 130K and have the mean 150K. I will not solve this scenario since this is intuitive.

D and E out. A remains.

CDM770234 wrote:
Can someone help me solve the following GMAT Prep Test 1 question:

23. Last month 15 homes were sold in town X. The average (Arithmetic Mean) price 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$130,000 and less than $150,000. III. At least one of the homes was sold for less than$130,000.

The answer choices are:

A. I only
B. II only
C. III only
D. I and II
E. I & III

I do not agree/understand the official answer! Retired Moderator Joined: 20 Dec 2010
Posts: 1572
Re: Last month 15 homes were sold in Town X. The average (arithmetic mean)  [#permalink]

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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 = 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.

Ans: "A"
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Re: Last month 15 homes were sold in Town X. The average (arithmetic mean)  [#permalink]

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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 Intern  Status: Waiting for Decisions Joined: 23 Dec 2012 Posts: 39 Location: India Sahil: Bansal GMAT 1: 570 Q49 V20 GMAT 2: 690 Q49 V34 GPA: 3 WE: Information Technology (Computer Software) Re: Last month 15 homes were sold in Town X. The average (arithmetic mean) [#permalink] Show Tags 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 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 Veritas Prep GMAT Instructor Joined: 11 Dec 2012 Posts: 312 Re: Last month 15 homes were sold in Town X. The average (arithmetic mean) [#permalink] Show Tags 2 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 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.

Hope this helps!
-Ron
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Re: Last month 15 homes were sold in Town X. The average (arithmetic mean)  [#permalink]

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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.

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

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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

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