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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
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Bunuel 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


Difficulty Level: 600

We have to maximize the second highest ribbon winner so we must minimize other 3 winners, as each of the 5 students have won at least one ribbon and no two student won same number of ribbons so we can minimize 3 winners as

Lowest Winner won = 1 ribbon
Second winner= 2 ribbons
third winner= 3 ribbons

Now we have to maximize other two winner to get our answer

1+2+3= 6
20-6= 14

we have 14 ribbons left as no two students won same number of ribbons we can calculate second highest as 6 ribbons winner and highest winner have won 7 ribbons

Hence Answer: B
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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
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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:

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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
Hi there!

How did everyone know to award the first three students 1, 2, 3 ribbons?

Why was it not 2, 3, 4, 5, 6 ribbons respectively for the five students? This also equal 20 total ribbons...

Please help explain why you solved it like you did. How do I know when to use the same technique you used to solve similar questions?

Thank you in advance!
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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
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Hi jojo95,

When a question asks us to find the largest or smallest possible value that a variable, we typically have to minimize or maximize (respectively) all of the other variables involved. Giving the first three students 2, 3 and 4 ribbons would NOT be minimizing those possibilities though - in this situation, you end up giving out 9 total ribbons (instead of the 6 total ribbons when the first three students receive 1, 2 and 3 ribbons, respectively). By extension, you would have fewer total ribbons for the last two students.

In your example, the second-highest number is 5... but that's not actually the highest value that is possible for that student.

GMAT assassins aren't born, they're made,
Rich
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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
EMPOWERgmatRichC wrote:
Hi jojo95,

When a question asks us to find the largest or smallest possible value that a variable, we typically have to minimize or maximize (respectively) all of the other variables involved. Giving the first three students 2, 3 and 4 ribbons would NOT be minimizing those possibilities though - in this situation, you end up giving out 9 total ribbons (instead of the 6 total ribbons when the first three students receive 1, 2 and 3 ribbons, respectively). By extension, you would have fewer total ribbons for the last two students.

In your example, the second-highest number is 5... but that's not actually the highest value that is possible for that student.

GMAT assassins aren't born, they're made,
Rich


This helps so much! Thank you for explaining!
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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
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Bunuel 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


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 numbers of ribbons won were 1, 2, and 3 ribbons, there is 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 to 8, and thus the second greatest number is 6.

Answer: B
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Re: At a school-wide athletic fair, five students won a combined total of [#permalink]
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