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In a certain district , the minimum of the monthly rainfall

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In a certain district , the minimum of the monthly rainfall [#permalink] New post 14 Nov 2004, 13:18
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In a certain district , the minimum of the monthly rainfall is 0.2, max is 12.2 and the average of 12 months is 3.5. Which of the following is the greatest possible value of the median?

A. 2.4
B 3.5
C 4.2
D 4.8
E 6.2
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 [#permalink] New post 14 Nov 2004, 16:22
D) 4.8
Median is b/w 6-7 month
Total rain: 12*3.5 = 42
Minimize 5 lowest rain months with .2 each month you get 5*.2 = 1
Most rainy month is 12.2
Total is 13.2
To have max median, spread 42-13.2 = 28.8 over remaining 6 months of rain: 28.8/6 = 4.8 --> max median value
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 [#permalink] New post 15 Nov 2004, 00:07
You are right Paul. Thanks a lot :!:
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 [#permalink] New post 15 Nov 2004, 00:28
Could you please explain why you minimized it to "5 lowest rain months" ?
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 [#permalink] New post 15 Nov 2004, 12:08
Paul, that was very useful. I gues the key is greatest possible value for the median. That can be obtained if the first 5 months had the lowest rainfall!
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 [#permalink] New post 15 Nov 2004, 12:47
In order to have the highest median value, the average of the 6 and 7 month, you have to make it such that the first 5 months are at a minimum(0.2) so as to keep the most for the median value when you reach 6th and 7th month.
For the values above the 7th month, you have to minimize them so as to keep, save, the highest possible score for the 7th month. However, you know that the highest month is 12.2 so get that one out of the way as I did and you will have the remainder (28.8) which you can now spread over the remaining 6 middle months (6,7,8,9,10,11). This way, you will maximize the median which is b/w 6th and 7th month

From lowest to highest raining month:
1: 0.2
2: 0.2
3: 0.2
4: 0.2
5: 0.2
6: 4.8
7: 4.8

8: 4.8
9: 4.8
10: 4.8
11: 4.8
12: 12.2
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Paul

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 [#permalink] New post 15 Nov 2004, 12:53
Got it, thanks Paul :idea:
  [#permalink] 15 Nov 2004, 12:53
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