Hi, the logic is as follows::
The mean price is 150000 for 15 houses. So the total sum = 2250000.
The median is 130000. Now median price imples that this house will be 8th in the series of houses when arranged in an ascendding order.
The worst case could be that all the first 8 houses got sold for 130000. Then the sum left for remaining 7 houses = 2250000 - 1040000 = 1210000.
This puts the mean price for the last 7 houses as = 1210000/7= 1710000.
This clearly confirms the price of at least one house to be more that 1650000.
GMATPrep Mean & Median
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Stuart@KaplanGMAT
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We know that there were 15 total.
The average is 150k.
The median is 130k.
So, the 8th (middle) term in the set is 130. Do we have to have a term below 130k?
No: a possible set would be 14 homes at 130k and 1 home at a really high price. Therefore, III isn't a MUST be true.
* * *
For 130 to be the median, the 8th home in the set must be 130. Let's look at two extreme examples:
14 homes at 130, 1 home above 130.
In this case, that one home above would have to be super expensive (to raise the average to 150), so (I) would be true.
8 homes at 130, 7 homes above 130.
In this case, we have the 8 homes on the low end dragging the average down by 8*20k (since 130k is 20k below 150k) = 160k. Therefore, the 7 homes above 130 must compensate and be a total of 160k above 150k, the average.
(By definition, for a set to average out to a number, there must be an equal amount below and above the average.)
So, if we spread that 160k out evenly among the 12 homes, each one would be roughly 23k above the average, or 173k each. If we don't spread it out evenly, then some homes will sell for more than 173k. Either way, we have at least one home selling for more than 165k and (I) is true.
At both extremes, (I) is true - therefore, it will always be true.
* * *
Let's look at II next:
Do we have to have a home between 130 and 150? Nope, we've already shown that we don't, with our extreme examples from examining (I). So, II doesn't have to be true.
Only I must be true: choose (A).
The average is 150k.
The median is 130k.
So, the 8th (middle) term in the set is 130. Do we have to have a term below 130k?
No: a possible set would be 14 homes at 130k and 1 home at a really high price. Therefore, III isn't a MUST be true.
* * *
For 130 to be the median, the 8th home in the set must be 130. Let's look at two extreme examples:
14 homes at 130, 1 home above 130.
In this case, that one home above would have to be super expensive (to raise the average to 150), so (I) would be true.
8 homes at 130, 7 homes above 130.
In this case, we have the 8 homes on the low end dragging the average down by 8*20k (since 130k is 20k below 150k) = 160k. Therefore, the 7 homes above 130 must compensate and be a total of 160k above 150k, the average.
(By definition, for a set to average out to a number, there must be an equal amount below and above the average.)
So, if we spread that 160k out evenly among the 12 homes, each one would be roughly 23k above the average, or 173k each. If we don't spread it out evenly, then some homes will sell for more than 173k. Either way, we have at least one home selling for more than 165k and (I) is true.
At both extremes, (I) is true - therefore, it will always be true.
* * *
Let's look at II next:
Do we have to have a home between 130 and 150? Nope, we've already shown that we don't, with our extreme examples from examining (I). So, II doesn't have to be true.
Only I must be true: choose (A).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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camitava
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smushkas,
Mean = 150000
Median = 130000
Items = 15
So if we arrange the cost of the houses in ascending order, cost of the 8th house will be = 130000. And the 7 items before the median entry will be less than 130000 or in worst case they are same with 130000. So thinking the worst condition, we can drop the stmt-3.
Now again taking the worst scenario, avg of the upper 7 items = ((150000 * 15) - (130000 * 8)) / 7 > 165000.
Now if we consider the cost distribution is uniform or something like that, - E.g.- cost of 7 houses : 129000 cost of 8th house : 130000 and avg. cost of upper 7 houses > 165000. So here we can say that we can drop the option - 2 as it can be a case that cost of each upper 7 houses > 165000 and no value between 130000 and 150000.
So this proves that stmt-1 is only the condition which MUST BE valid for any case. So IMO A
Mean = 150000
Median = 130000
Items = 15
So if we arrange the cost of the houses in ascending order, cost of the 8th house will be = 130000. And the 7 items before the median entry will be less than 130000 or in worst case they are same with 130000. So thinking the worst condition, we can drop the stmt-3.
Now again taking the worst scenario, avg of the upper 7 items = ((150000 * 15) - (130000 * 8)) / 7 > 165000.
Now if we consider the cost distribution is uniform or something like that, - E.g.- cost of 7 houses : 129000 cost of 8th house : 130000 and avg. cost of upper 7 houses > 165000. So here we can say that we can drop the option - 2 as it can be a case that cost of each upper 7 houses > 165000 and no value between 130000 and 150000.
So this proves that stmt-1 is only the condition which MUST BE valid for any case. So IMO A
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava