swerve wrote:Collectively, the five children in the Kramer family have 30 trophies. If each child has at least one trophy and no two children have the same number of trophies, what is the greatest number of trophies that the child with the second-highest number of trophies could have?
A. 10
B. 11
C. 12
D. 13
E. 14
In ascending order, let the 5 trophy values be a, b, c, d and e.
To maximize the value of d -- the second highest value -- we must MINIMIZE the values of a, b and c.
Since every child must have a different number of trophies, the least possible values for a, b and c are a=1, b=2, and c=3.
Since there are a total of 30 trophies, d+e = 30 - (1+2+3) = 24.
Since d and e must be distinct positive integers with a sum of 24, d=11 and e=13.
Thus, the greatest possible value of d -- the second-highest number of trophies -- is 11.
The correct answer is
B.
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
