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

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

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New post 13 Apr 2018, 03:00
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Question Stats:

65% (01:02) correct 35% (01:09) wrong based on 71 sessions

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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

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

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New post 13 Apr 2018, 04:00
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


The number of ribbons that will give the second highest is arranged as follow:
1, 2, 3, 6, 8.

Therefore option "B" = 6 is the second highest and the answer.
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Re: At a school-wide athletic fair, five students won a combined total of  [#permalink]

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New post 28 Apr 2018, 17:24
First 3 would get 1,2 and 3 ribbons

14 ribbon remain
Can't be equal so second highest will get 6 and highest one will get 8

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

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New post 31 Aug 2018, 01:13
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 &nbs [#permalink] 31 Aug 2018, 01:13
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