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

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.

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.

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).

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!

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

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

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

BR
Mandeep

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

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

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.

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/
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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
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if-the-population-of-city-a-is-increased-by-a-in-year-130222.html
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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


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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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