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In a neighborhood consisting of 2,000 homes, 80 percent of

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Carcass
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In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   21 Dec 2017, 16:59
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In a neighborhood consisting of 2,000 homes, 80 percent of the homes are valued at $325,000 or less. Which of the following statements about the values of the homes in the neighborhood must be true?

Indicate \(all\) such statements.

A. The average (arithmetic mean) value is at most $325,000.
B. The median value is at most $325,000.
C. At most 400 homes have values greater than $325,000.

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B,C
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Popo
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   22 Dec 2017, 00:56
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So 0,8 * 2000 = 1600 values evaluated at at least 325 000
So 400 are greater than 325 000 C must be true

Since we have 1600 houses at 325 000 or less and 400 at more than 325 000
We cannot conclude that the median is 325 000. The 1000th and 1001 th values could be either less than 325 000 or equals 325000. So B is out.

The average is the number of houses * prices divided by the total of houses.
So it could be written as (<= 325 000)(1600) + (400)((>325000) /2000
Therefore A is out
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   22 Dec 2017, 21:52
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Popo wrote:
So 0,8 * 2000 = 1600 values evaluated at at least 325 000
So 400 are greater than 325 000 C must be true

Since we have 1600 houses at 325 000 or less and 400 at more than 325 000




I donot think your assumption is wright- can you plz explain how you assumed the remaing 400 houses have values more than 325000. As nothing is mentioned about the remaining 400 houses
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   26 Dec 2017, 06:58
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OA Added

Regards
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   15 Jan 2018, 18:29
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According to prompt 80% of h are at 325 or less, what if 1600h valued 324.9$ or 1600h valued 325$ other 400h valued 300$

Since question must be true, C cannot be correct answer
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   15 Jan 2018, 20:29
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yell2012prime wrote:
According to prompt 80% of h are at 325 or less, what if 1600h valued 324.9$ or 1600h valued 325$ other 400h valued 300$

Since question must be true, C cannot be correct answer


none of the other 400 houses can be valued at $300000 because those valuation is already included in less than or equal to 325000 other 400 houses must have valuation greater than 325000 as such c must be correct
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   16 Jan 2018, 07:28
sanjogsubedi wrote:
yell2012prime wrote:
According to prompt 80% of h are at 325 or less, what if 1600h valued 324.9$ or 1600h valued 325$ other 400h valued 300$

Since question must be true, C cannot be correct answer


none of the other 400 houses can be valued at $300000 because those valuation is already included in less than or equal to 325000 other 400 houses must have valuation greater than 325000 as such c must be correct



Yep, you are right :arrow: :oops:
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   23 Jan 2018, 10:00
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B and C both are correct

80% of 2000 -->1600 homes <=325000

Remaining homes= 2000-1600
=400

Median : (1000, 1001 Location values)/2
which will be atmost 325000
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   25 Jan 2018, 06:41
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pranab01 wrote:
Popo wrote:
So 0,8 * 2000 = 1600 values evaluated at at least 325 000
So 400 are greater than 325 000 C must be true

Since we have 1600 houses at 325 000 or less and 400 at more than 325 000




I donot think your assumption is wright- can you plz explain how you assumed the remaing 400 houses have values more than 325000. As nothing is mentioned about the remaining 400 houses


It is mentioned in the question. 80% are valued at 325k$ or less, therefore the remaining 20% (400 homes) must be valued at a higher price than 325k$.
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   24 Jul 2020, 06:36
Carcass wrote:
OA Added

Regards

how we can say that option B is must be true ???? we knew that 1600 houses at 325 000 or less , but we dont know about the 1000th and 1001th term maybe they are 325 000 or more less than 325 000 (like just 100,000 or less than it )
can you please explain it
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   24 Jul 2020, 09:10
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All are correct...A,B,C
A. at most means........all 1600 homes valued at 325k......then average at most will be same
B.Median as well follows the same scenario if all valued at 325k...
note : we are taking at most case here
C.400 at most should be greater than 325 k
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink] New post   08 Aug 2020, 13:30
saitejat wrote:
All are correct...A,B,C
A. at most means........all 1600 homes valued at 325k......then average at most will be same
B.Median as well follows the same scenario if all valued at 325k...
note : we are taking at most case here
C.400 at most should be greater than 325 k



The mean is not at most 325,000, b/c we can always have an outlier. For example a house worth 3 trillion dolllars would end up giving us a mean of more than 325,000 because the mean would explode just for that one house that is worth a crazy amount.
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Re: In a neighborhood consisting of 2,000 homes, 80 percent of [#permalink]
08 Aug 2020, 13:30
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