Difference between revisions of "2010 AMC 12B Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\times50=4,950</math>. With 99 terms, the average is <math>\frac{99\times50}{99}</math>, which equals 50. Since the average is <math>100x</math>, we set set <math>100x</math> equal to <math>50</math> and solve for <math>x=\frac{1}{2}</math>. Thus, the answer is <math>\boxed{\text{C}}</math>. | + | We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\times50=4,950</math>. With 99 terms, the average is <math>\frac{99\times50}{99}</math>, which equals 50. Since the average is <math>100x</math>, we set set <math>100x</math> equal to <math>50</math> and solve for <math>x</math> = <math>\frac{1}{2}</math>. Thus, the answer is <math>\boxed{\text{C}}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | ||
{{stub}} | {{stub}} | ||
Revision as of 19:58, 5 February 2011
Problem 10
The average of the numbers 
and 
is 
. What is ?
Solution
We first sum the first 
numbers: 
. With 99 terms, the average is 
, which equals 50. Since the average is , we set set equal to 
and solve for = 
. Thus, the answer is 
.
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 | |
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