1991 AJHSME Problems/Problem 14

Problem

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?

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

Solution

There are two ways for a student to get $11$ : $5+5+1$ and $5+3+3$ . Clearly if someone gets one of these combinations someone else could get the other, so we are not guaranteed the most points with $11$ .

There is only one way to get $13$ points: $5+5+3$ . In this case, the largest score another person could get is $5+3+3=11$ , so having $13$ points guarantees having more points than any other person $\rightarrow \boxed{\text{D}}$ .

If someone gets $11$ points, the possible combinations are $5,5,1$ and $5,3,3$ If he gets $5,3,3$ then someone else can be $5,3,3$ which would not guarantee victory. If we have 13 points, the only way to make this is $5,5,3$ . There is no way to get any number of points higher than this, so the answer is $\boxed{\text{D}}$ .

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1991 AJHSME Problems/Problem 14

Problem

Solution

See Also