Click to see a similar question. ---------- Four packages have an average weight of 10.5 pounds. What is the minimum possible weight of the heaviest package in pounds if the median is 10 pounds?
A. 10 B. 10.5 C. 11 D. 12 E. 13
(D) Let us denote the weights of the packages in pounds by v, x, y, z naming from the lightest one to the heaviest one. The median is 10 pounds. Therefore (x + y) / 2 = 10. x + y = 20
The average is 10.5 pounds. Therefore (v + x + y + z) / 4 = 10.5. v + (x + y) + z = 42 v + 20 + z = 42 v + z = 22
The weight v must be no greater than 10, since 10 is the median. Therefore the minimum possible weight of the heaviest package is 22 – 10 = 12 pounds (all the other packages would weigh 10 pounds in this case).
The correct answer is (D). ----------
I would think the answer could also be 13 as the lightest package should be less than 10 (in this case could be 9)?
what is your definition of median? I thought that median of numbers 1, 2, 3 , 4 is any number in interval [2, 3]. With this definition packages of weights 10, 10, 11, 11 have median weight 10 (not only 10, but any number of interval [10, 11]) and mean 10.5, so the correct answer would be C. Is it only matter of definition or did I overlook something?
Is it only matter of definition or did I overlook something?
GMAT is based on school math. Median is defined as the middle element of an arranged set, if the number of elements in the set is odd. It is defined as the average of the two middle elements of the arranged set, if the number of elements in the set is even.
So the median of a fixed set is always a single number. Use only this definition in GMAT.
Users browsing this forum: Yahoo [Bot] and 6 guests
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.