Averages and Median in Office-experts pls help

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Five offices have an average of 8 people per office and a median of 7 people per office, and none of the offices are vacant. What is the maximum number of people who can be in the largest office?

A. 23 B. 24 C. 25 D. 26 E. 27
although OA is 24, I think 23 will also fit this scenario

Pls help...
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by GmatMathPro » Fri Aug 03, 2012 11:41 am
mathewmithun wrote:Five offices have an average of 8 people per office and a median of 7 people per office, and none of the offices are vacant. What is the maximum number of people who can be in the largest office?

A. 23 B. 24 C. 25 D. 26 E. 27
although OA is 24, I think 23 will also fit this scenario

Pls help...
There are 5 offices with an average of 8 people per office, so there are 5*8=40 people. To maximize the number of people in one office, we need to minimize the number of people in the other four offices while staying within the restrictions stated in the problem.

1. We know the median is 7, so the list of number of people in the offices, in ascending order, looks like this : ___, ____, 7, ____, ____.

2. We know that no offices are vacant, so there must be at least 1 person in each office. Nothing in the problem states that there is a different number of people in each office, so both of the numbers to the left of 7 could be 1: 1, 1, 7, _____, _____.

3. The fourth number of the list must be greater than or equal to 7 because 7 is the median. The smallest possible value in this range is 7: 1, 1, 7, 7, _____.

4. So far, we have 1+1+7+7=16 people. We know the office has 40 people total, so the largest office has [spoiler]40-16=24 people[/spoiler]
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by GMATGuruNY » Fri Aug 03, 2012 11:48 am
mathewmithun wrote:Five offices have an average of 8 people per office and a median of 7 people per office, and none of the offices are vacant. What is the maximum number of people who can be in the largest office?

A. 23 B. 24 C. 25 D. 26 E. 27
although OA is 24, I think 23 will also fit this scenario

Pls help...
Let the five offices, in ascending order, be ABCDE.
This is a MAX/MIN problem.
To MAXIMIZE E, we need to MINIMIZE A, B, C, and D.

The total number of people = (number of offices)(average per office) = 5*8 = 40.

Since the median = 7, the 5 offices are as follows:
A...B...7...D...E

Since no office can be empty, the minimum possible value for A and B = 1:
1...1...7...D...E

Since D cannot be less than the median, the minimum possible value for D = 7:
1...1...7...7...E

Thus, the maximum possible value for E = 40 - (1+1+7+7) = 24.

The correct answer is B.
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by mathewmithun » Fri Aug 03, 2012 12:09 pm
thanks, I somehow overlooked "maximum". B is the answer

GMATGuruNY, GmatMathPro and other experts,

I am just surprised by the ease with which you approach quants and DS. Although theory and practice goes a long way in building confidence to face difficult quant/DS, I am still unable to face such question (after practicing quant and ds for a very long time), especially during GMAT exam.
One mistake i constantly make while approaching any quant question is that I try to find some recognizable pattern between the question and the ones I practiced before, and I waste a lot of time in that loop. I think I get in that loop because of lack of confidence to face the question head on. If some of you could give tips on how to face questions head on, the right way (not theory or the usual GMAT specific steps), it could really help me face quant section.

thanks
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