Difference between revisions of "2002 AMC 10B Problems/Problem 25"
LighteningAB (talk | contribs) m (helps clarify what you have to go through) |
|||
| Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
| − | Let <math>x</math> be the sum of the integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\frac{x+15}{y+1}=\frac{x}{y}+2</math> and <math>\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a | + | Let <math>x</math> be the sum of the integers and <math>y</math> be the number of elements in the list. Then we get the equations <math>\frac{x+15}{y+1}=\frac{x}{y}+2</math> and <math>\frac{x+15+1}{y+1+1}=\frac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1</math>. With a lot of algebra, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>. |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=B|num-b=24|after=Last problem}} | {{AMC10 box|year=2002|ab=B|num-b=24|after=Last problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:05, 17 January 2016
Problem
When 
is appended to a list of integers, the mean is increased by 
. When 
is appended to the enlarged list, the mean of the enlarged list is decreased by . How many integers were in the original list?
Solution
Let 
be the sum of the integers and 
be the number of elements in the list. Then we get the equations 
and 
. With a lot of algebra, the solution is found to be 
.
See also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last problem | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
