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At a delivery store, seven packages have an average (arithme
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20 Sep 2011, 20:17
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At a delivery store, seven packages have an average (arithmetic mean) weight of 225 pounds and a median weight of 270 pounds. What is the maximum possible weight, in pounds, of the lightest package? A. 25 B. 165 C. 195 D. 225 E. 270
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Re: Mean, Median
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20 Sep 2011, 20:43
Total weight of 7 packages = 225 * 7 = 1575 pounds.
In a set of 7 numbers, 4th element will be the middle one and hence median. weight of 4th element =270 pounds. to maximise weight of lightest package, minimize the weight of others. 4th number is 270, so let the 5th 6th and 7th number of the set also be 270 pounds which is least possible weight they can have. total weight of 4th, 5th , 6th and 7th numbers = 270 *4 = 1080 pounds.
so total weight remaining to be shared by 1st 2nd and 3rd number = 1575  1080= 495 pounds. let the 1st 2nd and 3rd element have the same value. so they can be 495/3 = 165 pounds each. so lightest package is 165 pounds in weight.




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Re: Mean, Median
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21 Sep 2011, 01:52
maheshsrini wrote: At a delivery store, seven packages have an average (arithmetic mean) weight of 225 pounds and a median weight of 270 pounds. What is the maximum possible weight, in pounds, of the lightest package?
a) 25 b)165 c)195 d)225 e)270 IMO B. You can eliminate options D and E with intuition. For average to be 225 and median to be 270, the lightest item has to be less than 225. If not, the average will be more than 225. Hence D and E are eliminated. Now since, option C looks closer to 225, test option B (165) first. I dint even test this case to come to answer. (I guess this is how you visualise the question).
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Re: Mean, Median
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21 Sep 2011, 04:10
for this type of question:
we know that median is 270 that means the 4th element in the set is 270. for maximum smallest integer we have to assume that the 5,6,7 will also be 270 and that smallest 3 element must be same.
hence, 3*x+270*4=225*7 hence x=165.



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Re: Mean, Median
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21 Sep 2011, 11:47
B.
225x 7 = 1575 total.
Lets say all options on right of 270 is 270.. therefore remaining three have to have total weight of: 1575  (270x4) = 495 495/3 = 165!



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Re: Mean, Median
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27 Sep 2011, 10:58
225*7 = 3X + 270*4 15751080 = 3X 495/3 = X X= 165
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Re: Mean, Median
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28 Sep 2011, 08:06
I got B too.. Assuming all options on right of
270 is 270, then we get 3x + 270*4 = 225*7 3x = 15751080 x = 495/3 x= 165



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Re: Mean, Median
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28 Sep 2011, 12:08
It is B. But I don't understand how come it's 700 level question.
BR Mandeep



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Re: Mean, Median
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05 Oct 2011, 10:42
Total number 7
mean 225 median 270 = 4th number = 270 to get a biggest first number the other numbers should be minimum = 4th , 5 th , 6 th and 7 th numbers are equal to median
= 45 more than mean that add ups 180
180 more than mean
the sum of first three numbers = 3*225  180 = 495
biggest first number = 165



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Re: At a delivery store, seven packages have an average (arithme
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22 Oct 2013, 08:26
Yes, but they do not have to be the same number, The smallest elements have to be smaller than the median and sum 495. So it would be correct to pick 195,200 and 200. can somebody explain why the smallest number has to be the same?? The question asks about maximum value of the smallest not of the three smallests...



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Re: At a delivery store, seven packages have an average (arithme
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07 May 2014, 00:56
Petepie wrote: Yes, but they do not have to be the same number, The smallest elements have to be smaller than the median and sum 495. So it would be correct to pick 195,200 and 200. can somebody explain why the smallest number has to be the same?? The question asks about maximum value of the smallest not of the three smallests... sum of first three is 495, in your case the sum (195+200+200) equals 595 . kindly recheck



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Re: At a delivery store, seven packages have an average (arithme
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08 May 2014, 13:37
B. 7 packages. Median is 4th package. Weight of median is 270.
to maximise weight of lightest package, minimize the weight of others. so the 5th 6th and 7th number of the set also be 270 pounds which is least possible weight they can have. total weight of 4th, 5th , 6th and 7th numbers = 270 *4 = 1080 pounds.
so total weight remaining to be shared by 1st 2nd and 3rd number = 1575  1080= 495 pounds. let the 1st 2nd and 3rd element have the same value. so they can be 495/3 = 165 pounds each. so lightest package is 165 pounds in weight.



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Re: At a delivery store, seven packages have an average (arithme
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08 May 2014, 22:46
maheshsrini wrote: At a delivery store, seven packages have an average (arithmetic mean) weight of 225 pounds and a median weight of 270 pounds. What is the maximum possible weight, in pounds, of the lightest package?
A. 25 B. 165 C. 195 D. 225 E. 270 For a discussion of MinMax strategies, check out the following posts: http://www.veritasprep.com/blog/2014/01 ... thegmat/http://www.veritasprep.com/blog/2014/01 ... basecase/http://www.veritasprep.com/blog/2014/01 ... extremes/
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Re: At a delivery store, seven packages have an average (arithme
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09 May 2014, 02:14



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At a delivery store, seven packages have an average (arithme
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15 Dec 2016, 15:08
Great Question this one. Here is my approach to this one =>
Let the packages be represented by => W1 W2 W3 W4 W5 W6 W7
Mean = 225
Mean = Sum/#
Hence Sum(7)=225*7=1575 Now median =270 #=7=> Odd Hence Median = 4th term = W4=270
Now to Maximise W1 we must minimise all others.
W1=W2=W3=W1 W4=W5=W6=W7=270
Hence 3W1+4*270=1575 3W1=495 W1=165 Pounds
Hence B
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