Average and Median Sale Price

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Average and Median Sale Price

by nakul_anand » Sat Nov 28, 2009 8:29 am
Last month, 15 homes were sold in town X. The average (Arithmetic Mean) sale price was $150000 and the median sale price was $130000. Which of the following is true?

(I) At least 1 home was sold for more than $165000
(II) At least 1 home was sold for more than $130000 but less than $150000.
(III) At least 1 home was sold for less than $130000


(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

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by papgust » Sat Nov 28, 2009 8:01 pm
IMO B

I. Not necessary to have at least 1 home sold for more than $165000 since median is $130000 and mean is $150000, it is possible to keep the highest home value to be under $165000.
III. Again not necessary to have at least 1 home sold for less than $130000 since it is possible to have the least home value of $130000 and still have Mean price as $150000 and the median price as $130000.

Only option left now is II (So, B)

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by ssuarezo » Mon Nov 30, 2009 12:49 pm
nakul_anand wrote:Last month, 15 homes were sold in town X. The average (Arithmetic Mean) sale price was $150000 and the median sale price was $130000. Which of the following is true?

(I) At least 1 home was sold for more than $165000
(II) At least 1 home was sold for more than $130000 but less than $150000.
(III) At least 1 home was sold for less than $130000


(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
IMHO: E

III: 130 is the median, and 150 is the average, we need values > 130, in order to get the 150 average, but if so, we need lower values than 130, so it's the median. am I right?

I: Since the median is 130 and 150 is the average, we need at least one of 170 to get the average.

II: values between 130 and 150 do nothing to the avg and the median.

Please Nakul, what's the OA?

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by ace_gre » Mon Nov 30, 2009 3:46 pm
Hi,

Given average of 15 homes is 150K. Total sale price = 15*150K=2250K.
Given median of 15 homes, n is odd,so arranging in ascending order sale price of 8th home is 130K.
This can mean that price of first 7 homes can be <=130k.
Lets assume 130k to be the price upto 8 homes. Then 130*8=1040K.So the total sale price of the last 7 homes is 2250-1040k=1210k. So this has to be split among the last 7 homes to make the average 150k.

This automatically makes I true. If I were to be false, say last 7 homes were <=165K, then 165*7=1155K and is less than 1210K.

Hence atleast one house has to be >165K.

II does not have to be true. It could be that first 8 homes were sold for 130K, 6 homes for 150K and 1 home could have been sold for 310K. average is still 150K and median is 130K

III Not necessarily true since only average and mean are given. Case discussed in II will negate this statement.

IMO A must be true. II and III could be true but not necessarily true.

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by sunnyjohn » Mon Nov 30, 2009 5:49 pm
IMO:A

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by nakul_anand » Mon Nov 30, 2009 7:14 pm
Official Answer - A

I agree with ace_gre's explaination.

Tricky question though!!!!

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by djkvakin » Thu Dec 03, 2009 8:59 am
The main question ' which IS true' is poorly formulated IMHO. I hope real GMAT questions are more precise: which of the following MUST be true/Could be true. In this case the the difference between Must be vs. Could be is crucial.

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by tallazndood » Sun Feb 14, 2010 10:51 pm
djkvakin wrote:The main question ' which IS true' is poorly formulated IMHO. I hope real GMAT questions are more precise: which of the following MUST be true/Could be true. In this case the the difference between Must be vs. Could be is crucial.
For the record, this is a retired official GMAT question. The actual wording should be "must be true" instead of "is true".

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by Anaira Mitch » Sat Dec 10, 2016 10:15 pm
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.

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by Jeff@TargetTestPrep » Tue Dec 13, 2016 4:48 pm
nakul_anand wrote:Last month, 15 homes were sold in town X. The average (Arithmetic Mean) sale price was $150000 and the median sale price was $130000. Which of the following is true?

(I) At least 1 home was sold for more than $165000
(II) At least 1 home was sold for more than $130000 but less than $150000.
(III) At least 1 home was sold for less than $130000


(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

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by Brent@GMATPrepNow » Tue Dec 13, 2016 7:41 pm
nakul_anand wrote:Last month, 15 homes were sold in town X. The average (Arithmetic Mean) sale price was $150000 and the median sale price was $130000. Which of the following is true?

(I) At least 1 home was sold for more than $165000
(II) At least 1 home was sold for more than $130000 but less than $150000.
(III) At least 1 home was sold for less than $130000


(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. As in, which of the following MUST be true. So, if it's possible that one scenario is not true, we can eliminate it.

So, let's looks at one 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, 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 II and III need not be true (since our scenario does not conform to either one).

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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by tomkruz » Thu Dec 15, 2016 4:46 am
well looks like mine is wrong,

gotta to calculate for another time equation right ?

thanks so much

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