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Re: There are five homes. If the median price of a home is $200,000 [#permalink]
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itisSheldon wrote:
There are five homes. If the median price of a home is $200,000, is the range of all prices greater than $80,000?

1) The average price of the five homes is $240,000.
2) Three of five homes have the same price.


Strategy


Draw it out:



_ _ 200 _ _

Range > 80?

1) Stmt 1
Avg = 240
Sum = 240 * 5 = 1200

Consider extremes:

Case 1: Max Range


1 1 200 _ _ => Yes (Can see it easily. No math needed)

Case 2: Min Range


200 200 200 x x
600 + 2x = 1200
x = 300
Range = 300 - 200 = 100 Yes

Hence, statement is sufficient.

2) Stmt 2
200 200 200 x 300 => Yes
200 200 200 x 250 => No

Hence, stmt 2 is not sufficient.

Answer: A
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There are five homes. If the median price of a home is $200,000 [#permalink]
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amanvermagmat wrote:
itisSheldon wrote:
There are five homes. If the median price of a home is $200,000, is the range of all prices greater than $80,000?

1) The average price of the five homes is $240,000.
2) Three of five homes have the same price.


Statement 1:
Average price of 5 homes = $240,000. This means total price of 5 homes = 5*240000 = $1,200,000.
To keep our range minimum, we should try to keep smallest and largest values (out of these 5) as close as possible.
Our smallest value can go up to 200,000 (since thats the median), and in that case, first 3 values will be 200,000 only.

So let the price of first 3 homes be $200,000 each. This makes it $600,000. Remaining price of 2 costlier homes = 1,200,000 - 600,000 = $600,000
In this case, costliest room will get the smallest value if we assume price of 4th and 5th homes to be equal, i.e., $300,000 each.

So here we have accomplished the purpose of keeping the smallest and largest value as close as possible. And the range in this case = price of highest - price of lowest = 300,000 - 200,000 = $100,000. So when here only the range is more than $80,000 then in any other case it will definitely be more than $80,000. Sufficient.

Statement 2:
Which three homes have same price we don't know. And also this doesn't help in determining the range. So not sufficient.

Hence A answer



I think even B is sufficient.... working for option B.


even though we dont know which 2 houses have same price, we have to make either of the choices.. as there are only two choices available if we keep house in ascending order of their value.


choice I : 1, 2 & 3 have same price
Choice II : 2, 3 & 4 have same price



Choice I : 1 2 & 3 have same price, since the median is known to us 200000. price of 1, 2 & 3 totals to 600,000 and rest 3 & 4 have a value of 600,000.

now whatever value we give to 4 & 5 range is going to be more than 80,000.

Choice II : 2 3 & 4 have same price, again as we know median is 200000. price of 2 3 & 4 totals to 600,000 and rest 1 & 5 have a value of 600,000.

now whatever value we give to 1 & 5 range is going to be more than 80,000.

Hence under both choices we get same answer that is range of all prices IS greater than $80,000. so even B is sufficient statement.


pls help me, if m on the wrong track !!
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Re: There are five homes. If the median price of a home is $200,000 [#permalink]
amanvermagmat wrote:
itisSheldon wrote:
There are five homes. If the median price of a home is $200,000, is the range of all prices greater than $80,000?

1) The average price of the five homes is $240,000.
2) Three of five homes have the same price.


Statement 1:
Average price of 5 homes = $240,000. This means total price of 5 homes = 5*240000 = $1,200,000.
To keep our range minimum, we should try to keep smallest and largest values (out of these 5) as close as possible.
Our smallest value can go up to 200,000 (since thats the median), and in that case, first 3 values will be 200,000 only.

[/b]


Hi, how do you know to find the minimum of the range in the first place? Isn't that also true if the maximum of range is smaller than $80k => range is always smaller than $80K => SUFFICIENT??
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Re: There are five homes. If the median price of a home is $200,000 [#permalink]
akara2500 wrote:
amanvermagmat wrote:
itisSheldon wrote:
There are five homes. If the median price of a home is $200,000, is the range of all prices greater than $80,000?

1) The average price of the five homes is $240,000.
2) Three of five homes have the same price.


Statement 1:
Average price of 5 homes = $240,000. This means total price of 5 homes = 5*240000 = $1,200,000.
To keep our range minimum, we should try to keep smallest and largest values (out of these 5) as close as possible.
Our smallest value can go up to 200,000 (since thats the median), and in that case, first 3 values will be 200,000 only.

[/b]


Hi, how do you know to find the minimum of the range in the first place? Isn't that also true if the maximum of range is smaller than $80k => range is always smaller than $80K => SUFFICIENT??
Tks


Hello

Yes, if MAX of range is smaller than $80k; => range is always smaller than $80k.
Similarly, if MIN of range is greater than $80k; => range is always greater than $80k. And thats what I have shown here.
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Re: There are five homes. If the median price of a home is $200,000 [#permalink]
This is how I thought about it.

There are five homes. If the median price of a home is $200,000, is the range of all prices greater than $80,000?

1) The average price of the five homes is $240,000.
Since the average is 240K, choose the scenario in which the price of a,b,c are maximized (b/c the smaller they are, the wider the gap becomes between the min and max price. It doesn't matter what d is so long as it's greater than the median.
a = 200K
b = 200K
c = 200K
d = 240K

Evidently, we have a 120K shortfall (a, b, c are each 40k smaller than the average)
This means e >= 360K
360 - 200K = 160

Sufficient

2) Three of five homes have the same price.
Use the same test cases above, you can manipulate the range by changing the max value of the last house (there's no upper limit).
Insufficient.

A
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Re: There are five homes. If the median price of a home is $200,000 [#permalink]
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Re: There are five homes. If the median price of a home is $200,000 [#permalink]
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