|
|
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[2010 AMC 12B Problems/Problem 10]] |
| − | The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?
| |
| − | | |
| − | <math>\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}</math>
| |
| − | | |
| − | ==Solution==
| |
| − | We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives <math>100(100x)=99(50)+x \rightarrow 10000x=99(50)+x \rightarrow 9999x=99(50) \rightarrow 101x=50 \rightarrow x=\boxed{\text{(B)} \frac{50}{101}}</math>
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}}
| |
| − | {{MAA Notice}}
| |
