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Difficulty:
5%
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Question Stats:
92% (01:55) correct
8%
(02:32)
wrong
based on 591
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The 15 homes in a new development are each to be sold for one of three different prices so that the developer receives an average (arithmetic mean) of $200,000 per home. If 4 of the homes are to be sold for $170,000 each and 5 are to be sold for $200,000 each, what will be the selling price of each of the remaining 6 homes?
A. $200,000
B. $210,000
C. $215,000
D. $220,000
E. $230,000
A. $200,000
B. $210,000
C. $215,000
D. $220,000
E. $230,000
09 Nov 2012, 02:54
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bssys
Given that the average price of 15 of the homes is $200, the total price of these homes is 15*$200 = $3,000.
The total price of 4 of the homes is 4*$170 = $680;
The total price of 5 of the homes is 5*$200 = $1,000;
Therefore the total price of the remaining 6 of the homes is $3,000 - ($680 + $1,000) = $1,320. The average price per home is $1,320/6 = $220.
Answer: D.
Hope it's clear.
General Discussion
08 Nov 2012, 14:09
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This is a classic problem that during your gmat exam makes you out of balance, nervous because you deal with a lot of information.
Ok take control of the question. (I have structured my answer step by step, like a process of thoughts)
First of all we have 15 homes so odd number.
Second this 15 is grouped in 3 blocks or groups whatever you want to say.
Now is the best part: one group is 170 of 4 this is important to take in account the other 5 (strees on this) 200
Now look at the answer choices : A is not possible. out
Now is the moment to attack the question. The best strategy here is a weighted average strategy.
considering the average is more or less in the middle, it follows that the group of 6 is what we are looking for and is between 170 and 200 or after 200. We have 170 200 and 6*x and the average is 200: in some place we have to put 6*x
Now in the average what value has a major "weight" No the group pf 4 with 170 ( value 680) but the 5 houses with 200 (value 1000).
So the average tends to the latter or is after, otherwise we do not have 200 like average. In this kind of problem the last value (in increasing order as we can see) is never the answer because put our value to close at the end point of our weighted average.
So the answer must be D
I hope my explanation is useful
Ok take control of the question. (I have structured my answer step by step, like a process of thoughts)
First of all we have 15 homes so odd number.
Second this 15 is grouped in 3 blocks or groups whatever you want to say.
Now is the best part: one group is 170 of 4 this is important to take in account the other 5 (strees on this) 200
Now look at the answer choices : A is not possible. out
Now is the moment to attack the question. The best strategy here is a weighted average strategy.
considering the average is more or less in the middle, it follows that the group of 6 is what we are looking for and is between 170 and 200 or after 200. We have 170 200 and 6*x and the average is 200: in some place we have to put 6*x
Now in the average what value has a major "weight" No the group pf 4 with 170 ( value 680) but the 5 houses with 200 (value 1000).
So the average tends to the latter or is after, otherwise we do not have 200 like average. In this kind of problem the last value (in increasing order as we can see) is never the answer because put our value to close at the end point of our weighted average.
So the answer must be D
I hope my explanation is useful
