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The participants in a race consisted of 3 teams with 3

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The participants in a race consisted of 3 teams with 3

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The participants in a race consisted of 3 teams with 3 runners on each team. A team was awarded 6 -n points if one of its runners finished in nth place, where 1 <= n <= 5. If all of the runners finished the race and if there were no ties, was each team awarded at least one point?

(1) No team was awarded more than a total of 6 points.
(2) No pair of teammates finished in consecutive places among the top five places.

OA A

Source: Princeton Review

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BTGmoderatorDC wrote:
The participants in a race consisted of 3 teams with 3 runners on each team. A team was awarded 6 -n points if one of its runners finished in nth place, where 1 <= n <= 5. If all of the runners finished the race and if there were no ties, was each team awarded at least one point?

(1) No team was awarded more than a total of 6 points.
(2) No pair of teammates finished in consecutive places among the top five places.

OA A

Source: Princeton Review
Given there are 6 teams with 3 3 runners on each team, we have 3*3 = 9 runners.

Again given, the points awarded equal to 6 - n, where n = position of the runner in the race; and 1 ≤ n ≥ 5, only the first five positions will be awarded points.

We know that all of the runners finished the race and there were no ties.

Question: Was each team awarded at least one point?

Say the three runners of the teams are represented by T1, T2, and T3.

Let's take each statement one by one.

(1) No team was awarded more than a total of 6 points.

In order to test the question, we must think of a scenario such that none of the runners of a team wins. So, all the five positions must be won by the other two teams. The number of points accrued by winning all five positions equals to 5 + 4 + 3 + 2 + 1 = 15.

However, as per Statement 1, the maximum points a team can earn is 6, thus, the maximum points two teams can earn is 6 + 6 = 12. Thus, the third team must have earned at least 15 - 12 = 3 points.

The answer is yes, each team was awarded at least one point. Sufficient.

(2) No pair of teammates finished in consecutive places among the top five places.

Say the first five positions are lined up as given below.

Case 1: T1, T2, T1, T2, T1.

T3 team runners come after 5th position, earning them no points. The answer is no, each team was NOT awarded at least one point.

Case 2: T1, T2, T1, T2, T3.

A T3 team runner comes at 5th position, earning it 1 point. The answer is yes, each team was awarded at least one point.

No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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Manhattan Review GMAT Prep

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BTGmoderatorDC wrote:
The participants in a race consisted of 3 teams with 3 runners on each team. A team was awarded 6 -n points if one of its runners finished in nth place, where 1 <= n <= 5. If all of the runners finished the race and if there were no ties, was each team awarded at least one point?

(1) No team was awarded more than a total of 6 points.
(2) No pair of teammates finished in consecutive places among the top five places.

Source: Princeton Review
\[A,B,C\,\,{\text{teams}}\,\,\left( {3\,\,{\text{people}}\,\,{\text{in}}\,\,{\text{each}}} \right)\]
\[?\,\,:\,\,{\text{all}}\,\,{\text{teams}}\,\,{\text{got}}\,\,{\text{points}}\]
\[\begin{array}{*{20}{l}}
{{\text{Place}}}&{{\text{Points}}} \\
1&5 \\
2&4 \\
3&3 \\
4&2 \\
5&1
\end{array}\]
\[15\,\,{\text{points}}\,\,{\text{given}}\,\,{\text{in}}\,\,{\text{total}}\,\,\,\left( * \right)\]
\[\left( 1 \right)\,\sum {{\text{any}}\,{\text{two}}\,\,{\text{teams}}} \,\,\, \leqslant \,\,12\,\,{\text{points}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]
\[\left( 2 \right)\,\,\,\left\{ \begin{gathered}
\hfill \,{\text{Take}}\,\,A = \left\{ {1,3,5} \right\}\,\,{\text{and}}\,\,B = \left\{ {2,4} \right\}\,\,\,\left( {{\text{places}}} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \\
\hfill \,{\text{Take}}\,\,A = \left\{ {1,3} \right\}\,,\,B = \left\{ {2,4} \right\}\,\,{\text{and}}\,\,C = \left\{ 5 \right\}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \\
\end{gathered} \right.\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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