Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 10B Problems/Problem 25"

m (helps clarify what you have to go through)
(→‎Solution)
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 lot of algebra, the solution is found to be <math>y= \boxed{\textbf{(A)}\ 4} </math>.
+
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>\dfrac{x+15}{y+1}=\dfrac{x}{y}+2</math> and <math>\dfrac{x+15+1}{y+1+1}=\dfrac{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>.
 +
==Solution 2==
 +
We let <math>m</math> be the number of elements in the set and we let <math>n</math> be the average of the terms of the original list. Then we have <math>mn</math> is the sum of all the elements of the list. So we have two equations: <cmath>mn+15=(m+2)(n+1)=mn+m+2n+2</cmath> and <cmath>mn+16=(m+1)(n+2)=mn+2m+n+2.</cmath>
 +
Simplifying both equations and we get,
 +
\begin{align*}
 +
13&=m+2n\\
 +
14&=2m+n
 +
\end{align*}
 +
Solving for <math>m</math> and <math>n</math>, we get <math>m=5</math> and <math>n=\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 23:39, 29 December 2017

Problem

When $15$ is appended to a list of integers, the mean is increased by $2$. When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$. How many integers were in the original list?

$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$

Solution

Let $x$ be the sum of the integers and $y$ be the number of elements in the list. Then we get the equations $\dfrac{x+15}{y+1}=\dfrac{x}{y}+2$ and $\dfrac{x+15+1}{y+1+1}=\dfrac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1$. With a lot of algebra, the solution is found to be $y= \boxed{\textbf{(A)}\ 4}$.

Solution 2

We let $m$ be the number of elements in the set and we let $n$ be the average of the terms of the original list. Then we have $mn$ is the sum of all the elements of the list. So we have two equations: \[mn+15=(m+2)(n+1)=mn+m+2n+2\] and \[mn+16=(m+1)(n+2)=mn+2m+n+2.\] Simplifying both equations and we get, \begin{align*} 13&=m+2n\\ 14&=2m+n \end{align*} Solving for $m$ and $n$, we get $m=5$ and $n=\boxed{\textbf{(A)}4}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_25&oldid=89255"