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

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<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math>
 
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math>
  
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== Solution ==
+
== Solution 1==
 
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>.
 
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==
 
==Solution 2==
 
We let <math>m</math> be the original number of elements in the set and we let <math>n</math> be the original 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,
 
We let <math>m</math> be the original number of elements in the set and we let <math>n</math> be the original 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,

Revision as of 19:50, 22 October 2018

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 1

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 original number of elements in the set and we let $n$ be the original 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, \[13=m+2n\] \[14=2m+n\] Solving for $m$ and $n$, we get $m=5$ and $n=\boxed{\textbf{(A)}4}$.

See also

2002 AMC 10B (Problems • Answer Key • Resources)
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