TheAnuja55 wrote:Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?
This is a MAX/MIN problem.
To MINIMIZE the number of people not assigned to a group, we must MAXIMIZE the number of people who ARE assigned.
Each group is composed of the following ratio:
A : B : C = 1:2:3.
Since each group must include 3 of the 40 people from division C, the maximum number of people from Division C who could be assigned is 13:
13*3 = 39.
Multiplying A:B:C = 1:2:3 by 13, we get:
A : B : C = 13(1:2:3) = 13:26:39.
Thus:
Of the 20 people from Division A, the maximum who could be assigned = 13, the minimum not assigned = 20-13 = 7.
Of the 30 people from Division B, the maximum who could be assigned = 26, the minimum not assigned = 30-26 = 4.
Of the 40 people from Division C, the maximum who could be assigned = 39, the minimum not assigned = 40-39 = 1.
Minimum not assigned from the 3 divisions = 7+4+1 = 12.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
