Collectively, the five children in the Kramer family have 30

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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

The OA is B.

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by GMATGuruNY » Sat Aug 18, 2018 3:08 am
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.
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by Jeff@TargetTestPrep » Mon Aug 20, 2018 10:39 am
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
We can let the three children who have the least number of trophies have 1, 2, and 3 trophies, respectively, since no child can have the same number of trophies as any other sibling. Thus, these three children have a total of 1 + 2 + 3 = 6 trophies.

So we are left with a total of 30 - 6 = 24 trophies for the two children with the greatest number of trophies. Since we want the second-most number of trophies to be a large as possible, we will ensure that the greatest number of trophies is as small as possible while still retaining its role as the greatest number. We see that a 12-12 split almost works, but the two numbers are not allowed to be the same value.

Thus, we can let the child with the greatest number of trophies have 13 trophies, and hence, the child with the second most trophies has 11.

Answer: B

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