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Difference between revisions of "1991 AJHSME Problems/Problem 14"

(Created page with '==Problem== Several students are competing in a series of three races. A student earns <math>5</math> points for winning a race, <math>3</math> points for finishing second and …')
 
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There is only one way to get <math>13</math> points: <math>5+5+3</math>.  In this case, the largest score another person could get is <math>5+3+3=11</math>, so having <math>13</math> points guarantees having more points than any other person <math>\rightarrow \boxed{\text{D}}</math>.
 
There is only one way to get <math>13</math> points: <math>5+5+3</math>.  In this case, the largest score another person could get is <math>5+3+3=11</math>, so having <math>13</math> points guarantees having more points than any other person <math>\rightarrow \boxed{\text{D}}</math>.
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==Solution 2 ==
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If someone gets <math>11</math> points, the possible combinations are <math>5,5,1</math> and <math>5,3,3</math> If he gets <math>5,3,3</math> then someone else can be <math>5,3,3</math> which would not guarantee victory.
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If we have 13 points, the only way to make this is <math>5,5,3</math>. There is no way to get any number of points higher than this, so the answer is <math>\boxed{\text{D}}</math>.---stjwyl
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1991|num-b=13|num-a=15}}
 
{{AJHSME box|year=1991|num-b=13|num-a=15}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:33, 15 November 2022

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 $\rightarrow \boxed{\text{D}}$.

Solution 2

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 $\boxed{\text{D}}$.---stjwyl

See Also

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

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