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Difference between revisions of "2011 AMC 8 Problems/Problem 21"
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==Problem==
 
Students guess that Norb's age is <math>24, 28, 30, 32, 36, 38, 41, 44, 47</math>, and <math>49</math>. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
 
Students guess that Norb's age is <math>24, 28, 30, 32, 36, 38, 41, 44, 47</math>, and <math>49</math>. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
  
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==Solution==
 
==Solution==
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If at least half the guesses are two low, then his age must be greater than <math>36.</math>
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If at least half the guesses are too low, then Norb's age must be greater than <math>36.</math>
  
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If two of the guesses are off by one, then his age is in between two guesses whose difference is <math>2.</math> It could <math>31,37,</math> or <math>48,</math> but because it is greater than <math>36</math> it can only be <math>37</math> or <math>48.</math>
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If two of the guesses are off by one, then his age is in between two guesses whose difference is <math>2.</math> It could be <math>31,37,</math> or <math>48,</math> but because his age is greater than <math>36</math> it can only be <math>37</math> or <math>48.</math>
  
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Lastly, his age is a prime number so the answer must be <math>\boxed{\textbf{(C)}\ 37}</math>
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Lastly, Norb's age is a prime number so the answer must be <math>\boxed{\textbf{(C)}\ 37}</math>
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==Solution 2 (Alternative approach)==
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Since two guesses are off by one, we know that both <math>x+1</math> and <math>x-1</math> are in the list where <math>x</math> is the age of Norb. Now, we know that <math>x+1</math> and <math>x-1</math> are <math>28</math> and <math>30</math>, <math>30</math> and <math>32</math>, <math>36</math> and <math>38</math> and <math>47</math> and <math>49</math>. From these values, we know that <math>x</math> must be <math>29</math>, <math>31</math>, and <math>37</math>. Since half of the guesses are too low, <math>24, 28, 30, 32,</math> and <math>36</math> are all too low so we can eliminate all numbers in our list lesser than or equal to <math>36</math>. Therefore, our list has only <math>37</math> left so the answer is <math>\boxed{\textbf{(C)}\ 37}</math>.
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== Video Solution by OmegaLearn==
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https://youtu.be/HISL2-N5NVg?t=3886
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~ pi_is_3.14159265
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==Video Solution by WhyMath==
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https://youtu.be/i6BmM5W_Vm8
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=20|num-a=22}}
 
{{AMC8 box|year=2011|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 10:31, 15 July 2024

Problem

Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$

Solution

If at least half the guesses are too low, then Norb's age must be greater than $36.$

If two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48.$

Lastly, Norb's age is a prime number so the answer must be $\boxed{\textbf{(C)}\ 37}$

Solution 2 (Alternative approach)

Since two guesses are off by one, we know that both $x+1$ and $x-1$ are in the list where $x$ is the age of Norb. Now, we know that $x+1$ and $x-1$ are $28$ and $30$, $30$ and $32$, $36$ and $38$ and $47$ and $49$. From these values, we know that $x$ must be $29$, $31$, and $37$. Since half of the guesses are too low, $24, 28, 30, 32,$ and $36$ are all too low so we can eliminate all numbers in our list lesser than or equal to $36$. Therefore, our list has only $37$ left so the answer is $\boxed{\textbf{(C)}\ 37}$.

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=3886

~ pi_is_3.14159265

Video Solution by WhyMath

https://youtu.be/i6BmM5W_Vm8

~savannahsolver

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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