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At a school-wide athletic fair, five students won a combined

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At a school-wide athletic fair, five students won a combined

by BTGmoderatorDC » Tue Feb 05, 2019 4:50 am

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At a school-wide athletic fair, five students won a combined total of 20 ribbons. If each of the five students won at least one ribbon and no two students won the same number of ribbons, what is the greatest number of ribbons that the student with the second-highest total could have won?

A. 5
B. 6
C. 7
D. 8
E. 9

OA B

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by [email protected] » Tue Feb 05, 2019 10:28 am
Hi All,

We're told that at a school-wide athletic fair, five students won a combined total of 20 ribbons, each of the five students won AT LEAST one ribbon and NO two students won the SAME number of ribbons. We're asked for the greatest number of ribbons that the student with the SECOND-HIGHEST total could have won. This is an example of a 'limit' question, but with a twist: we need to maximize the SECOND-HIGHEST total...

When a question asks for a largest or smallest possibility, we typically have to minimize or maximize (respectively) all of the other variables.

To start, we have to minimize the number of ribbons for the first three students. Since each student won AT LEAST one ribbon and no two students had the same number of ribbons, we would have to award those first three students with 1, 2 and 3 ribbons, respectively. That's 6 total ribbons, leaving 20 - 6 = 14 ribbons for the remaining two students.

We CANNOT award 7 and 7 though, since that would be the same number twice. Thus, we would have to award 6 and 8, meaning that the SECOND-HIGHEST possible number of ribbons would be 6.

Final Answer: B

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by Scott@TargetTestPrep » Wed Feb 06, 2019 6:33 pm
BTGmoderatorDC wrote:At a school-wide athletic fair, five students won a combined total of 20 ribbons. If each of the five students won at least one ribbon and no two students won the same number of ribbons, what is the greatest number of ribbons that the student with the second-highest total could have won?

A. 5
B. 6
C. 7
D. 8
E. 9

OA B

Source: Veritas Prep
To maximize the number of ribbons won by the top two students, we need to minimize the number of ribbons won by the remaining three students. Since each student won at least one ribbon and since no two students won the same number of ribbons, the minimum number of ribbons won by the lowest scoring three students is 1, 2 and 3.

Since the three lowest number of ribbons won was 1, 2, and 3 ribbons, there are a total of 14 ribbons left to be shared between the 2 people winning the two highest ribbon counts.

Note that no two students won the same number of ribbons; therefore, it is not possible that the top two students each won 7 ribbons. In order to make the second greatest number of ribbons won the largest, we need the greatest number of ribbons won to be equal 8, and thus the second greatest number is 6. Alternatively, it is also possible that the lowest number of ribbons won was 1,2 and 4 and the remaining two students won 6 and 7 ribbons, respectively.

Answer: B

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