Difference between revisions of "2016 AMC 10A Problems/Problem 7"
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<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math> | <math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math> | ||
| − | == Solution == | + | == Solution 1 == |
Since <math>x</math> is the mean, | Since <math>x</math> is the mean, | ||
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Therefore, <math>7x=540+x</math>, so <math>x=\boxed{\textbf{(D) }90}.</math> | Therefore, <math>7x=540+x</math>, so <math>x=\boxed{\textbf{(D) }90}.</math> | ||
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| + | Note: if the mean of a set is in the set, it can be discarded and the mean of the remaining numbers will be the same. This means that if using the mean in your reasoning, you can just take the mean of the other 6 numbers, and it'll solve it marginally faster. -[[User:Integralarefun|Integralarefun]] ([[User talk:Integralarefun|talk]]) 18:19, 27 September 2023 (EDT) | ||
==Solution 2== | ==Solution 2== | ||
| − | Note that <math>x</math> must be the median so it must equal either <math>60</math> or <math>90</math>. You can see that the mean is also <math>x</math>, and by intuition <math>x</math> should be the greater one. <math>x=\ | + | Note that <math>x</math> must be the median so it must equal either <math>60</math> or <math>90</math>. You can see that the mean is also <math>x</math>, and by intuition <math>x</math> should be the greater one. <math>x=\boxed{\textbf{(D) }90}.</math> |
| + | ~bjc | ||
==Check== | ==Check== | ||
Order the list: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> must be either <math>60</math> or <math>90</math> because it is both the median and the mode of the set. Thus <math>90</math> is correct. | Order the list: <math>\{40,50,60,90,100,200\}</math>. <math>x</math> must be either <math>60</math> or <math>90</math> because it is both the median and the mode of the set. Thus <math>90</math> is correct. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/qDqe8pztaJQ | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
Latest revision as of 17:19, 27 September 2023
Contents
Problem
The mean, median, and mode of the 
data values 
are all equal to 
. What is the value of ?
Solution 1
Since is the mean,
Therefore, 
, so 
Note: if the mean of a set is in the set, it can be discarded and the mean of the remaining numbers will be the same. This means that if using the mean in your reasoning, you can just take the mean of the other 6 numbers, and it'll solve it marginally faster. -Integralarefun (talk) 18:19, 27 September 2023 (EDT)
Solution 2
Note that must be the median so it must equal either 
or 
. You can see that the mean is also , and by intuition should be the greater one.
~bjc
Check
Order the list: 
. must be either or because it is both the median and the mode of the set. Thus is correct.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/XXX4_oBHuGk?t=163
~IceMatrix
~savannahsolver
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AMC 10 Problems and Solutions | ||
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
