Difference between revisions of "2011 AMC 8 Problems/Problem 21"
(→Video Solution 2) |
|||
| (12 intermediate revisions by 9 users not shown) | |||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | If at least half the guesses are | + | If at least half the guesses are too low, then Norb's age must be greater than <math>36.</math> |
| − | 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 | + | 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> |
| − | Lastly, | + | Lastly, Norb's age is a prime number so the answer must be <math>\boxed{\textbf{(C)}\ 37}</math> |
| + | |||
| + | ==Solution 2 (Alternative approach)== | ||
| + | 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>. | ||
| + | |||
| + | == 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== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=20|num-a=22}} | {{AMC8 box|year=2011|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:31, 15 July 2024
Contents
Problem
Students guess that Norb's age is 
, and 
. 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?
Solution
If at least half the guesses are too low, then Norb's age must be greater than 
If two of the guesses are off by one, then his age is in between two guesses whose difference is 
It could be 
or 
but because his age is greater than 
it can only be 
or 
Lastly, Norb's age is a prime number so the answer must be 
Solution 2 (Alternative approach)
Since two guesses are off by one, we know that both 
and 
are in the list where 
is the age of Norb. Now, we know that and are 
and 
, and 
, and 
and 
and . From these values, we know that must be 
, 
, and . Since half of the guesses are too low, 
and are all too low so we can eliminate all numbers in our list lesser than or equal to . Therefore, our list has only left so the answer is .
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3886
~ pi_is_3.14159265
Video Solution by WhyMath
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
