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AMC8 1991 Test Paper
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Sample Question 14:

Several students are competing in a series of three races. A student earns \(5\) points for winning a race, \(3\) points for finishing second and \(1\) point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?

\(\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15\)




Answer Keys

Question 14: D


Solutions

Question 14
Step 1: Consider the point tally for several students. Note that two combinations can result in a score of 11 points: this can happen with \(5+5+1\) and \(5+3+3\).

Step 2: Observe that if any one student achieves a total of 11 points using either combination, another student could also end up with 11 points by using the other possible combination. Therefore, even with 11 points, the student's victory is not guaranteed.

Step 3: To ensure victory, consider other numbers of points a student can earn. Note that there is only one way to get 13 points: \(5+5+3\).

Step 4: In the scenario where a student gets 13 points, the highest score that another student can achieve is 11 points (using either combination: \(5+5+1\) or \(5+3+3\)).

Step 5: Therefore, earning 13 points guarantees the achievement of more than any other student, no matter how the points are distributed among the remaining students. This makes 13 the smallest number of points that guarantees victory.

So, the answer is (D) 13.