At a delivery store, seven packages have an average (arithme

Question

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

Aj85 · Accepted Answer

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.

abhi398 · Answer

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.

GMATmission · Answer

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).

hashjax · Answer

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!

nishtil · Answer

225*7 = 3X + 270*4
1575-1080 = 3X
495/3 = X
X= 165

nrgmat · Answer

I got B too..
Assuming all options on right of

270 is 270, then we get 
3x + 270*4 = 225*7 
3x = 1575-1080
x = 495/3
x= 165

samramandy · Answer

It is B. 
But I don't understand how come it's 700 level question.

BR
Mandeep

smthsteve6 · Answer

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

Petepie · Answer

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...

tyagigar · Answer

sum of first three is 495, in your case the sum 
(195+200+200) equals 595 .
kindly recheck

ravishankar1788 · Answer

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.

KarishmaB · Answer

For a discussion of Min-Max strategies, check out the following posts: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/01 ... -the-gmat/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/01 ... base-case/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/01 ... -extremes/

Bunuel · Answer

Good Min/Max Problems: DS: max-x-y-is-defined-as-the-maximum-of-x-and-y-and-min-x-y-128330.html for-any-integers-x-and-y-min-x-y-and-max-x-y-denote-the-94998.html each-person-on-a-committee-with-40-members-voted-for-exactly-129999.html PS: if-x-y-and-z-are-positive-integers-and-3x-4y-7z-then-140129.html coins-are-to-be-put-into-7-pockets-so-that-each-pocket-128587.html five-peices-of-wood-have-an-average-length-of-124-inches-and-123513.html in-a-class-of-30-students-2-students-did-not-borrow-any-129624.html if-the-population-of-city-a-is-increased-by-a-in-year-130222.html a-cyclist-travels-the-length-of-a-bike-path-that-is-116960.html four-carpenters-can-individually-complete-a-particular-task-133968.html if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html in-a-company-with-48-employees-some-part-time-and-some-full-132442.html of-the-200-employees-in-a-certain-company-25-percent-will-128189.html of-the-science-books-in-a-certain-supply-room-50-are-on-131100.html All DS Min/Max Problems to practice: search.php?search_id=tag&tag_id=42 All PS Min/Max Problems to practice: search.php?search_id=tag&tag_id=63

stonecold · Answer

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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At a delivery store, seven packages have an average (arithme

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