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"

(→‎Solution 2)
(→‎Solution 4)
 
(10 intermediate revisions by 6 users not shown)
Line 8: Line 8:
  
 
==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>n</math> be the original number of elements in the set and we let <math>m</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,
 
<cmath>13=m+2n</cmath>
 
<cmath>13=m+2n</cmath>
 
<cmath>14=2m+n</cmath>
 
<cmath>14=2m+n</cmath>
Line 14: Line 14:
  
 
==Solution 3==
 
==Solution 3==
−
Warning:This solution will rarley ever work however seeing that you can so easily plug and chug in probem 25 it is funny to see this
+
Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in problem 25 it is funny to see this.
  
−
Plug and chug random numbers with the answer choices we can start with <math>4</math> numbers. You see that if you have 4 5s and you add 15 the resulting mean will be 7 we can verify this with math
+
Plug and chug random numbers with the answer choices, starting with the choice of <math>4</math> numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math
 
<cmath>\frac{5+5+5+5+15}{5}=7</cmath>
 
<cmath>\frac{5+5+5+5+15}{5}=7</cmath>
 
adding in 1 to the set you result in the mean to be 6.
 
adding in 1 to the set you result in the mean to be 6.
 
<cmath>\frac{5+5+5+5+15+1}{6}=6</cmath>
 
<cmath>\frac{5+5+5+5+15+1}{6}=6</cmath>
 
Thus we conclude that 4 is the correct choice or <math>\boxed{\textbf{(A)}}</math>
 
Thus we conclude that 4 is the correct choice or <math>\boxed{\textbf{(A)}}</math>
 +
 +
 +
Video solution by Canadamath
 +
https://www.youtube.com/watch?v=hYq8GgJ0it4
 +
 +
==Solution 4==
 +
 +
Let <math>S</math> be the sum of the numbers. Let <math>x</math> be the number of integers, and the mean be <math>y</math>. We have the equation <math>\frac{S}{x} = y</math>. We also have <math>\frac{S+15}{x+1} = y+2</math> and <math>\frac{S+16}{x+2} = y+1</math>. From the first equation, we have <math>S = xy</math>. We can multiply both sides by the denominators of the second and third equations to yield <math>S+15 = xy+2x+y+2</math> and <math>S+16 = xy+x+2y+2</math>. Since <math>S = xy</math>, we can substitute that into the equations and subtract by <math>xy</math> to get <math>15 = 2x+y+2</math> or <math>13 = 2x+y</math>. We also get <math>x+2y = 14</math>, and adding the two equations up gives <math>3x+3y = 27</math>, or <math>x+y = 9</math>. Subtracting <math>x+y=9</math> from <math>2x+y=13</math> gives <math>x = 4</math> and <math>y = 5</math>.
 +
 +
We can check that this is the correct answer by having <math>4, 9, 6,</math> and <math>1</math> as our numbers. The sum is 20, and the mean is 5. Adding 15 gives that the sum is 35, and the number of integers is 5, so the mean is 2, an increase of 2. Adding 1 to the integers gives that the sum is 36, and the number of numbers is 6, for a mean of 6, a decrease of 1.
 +
 +
We can also prove that since the sum is 20, the number of integers is 4, and the mean is 5, we have <math>\frac{20}{4} = 5</math>, which is true. The second equation gives <math>\frac{35}{5} = 7 = 5+2</math>, which satisfies the condition. We also have <math>\frac{36}{6} = 6 = 7-1</math>, which also satisfies the condition.
 +
 +
Therefore, since we correctly solved it and checked our work multiple ways, we can be confident that the answer is <math>\textbf{(A)}</math>.
 +
 +
== Video Solution ==
 +
https://www.youtube.com/watch?v=u8zi0QikJK0  ~David
  
 
== 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}}

Latest revision as of 18:11, 17 September 2023

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 $n$ be the original number of elements in the set and we let $m$ 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 $n=\boxed{\textbf{(A)}4}$.

Solution 3

Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in problem 25 it is funny to see this.

Plug and chug random numbers with the answer choices, starting with the choice of $4$ numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math \[\frac{5+5+5+5+15}{5}=7\] adding in 1 to the set you result in the mean to be 6. \[\frac{5+5+5+5+15+1}{6}=6\] Thus we conclude that 4 is the correct choice or $\boxed{\textbf{(A)}}$


Video solution by Canadamath https://www.youtube.com/watch?v=hYq8GgJ0it4

Solution 4

Let $S$ be the sum of the numbers. Let $x$ be the number of integers, and the mean be $y$. We have the equation $\frac{S}{x} = y$. We also have $\frac{S+15}{x+1} = y+2$ and $\frac{S+16}{x+2} = y+1$. From the first equation, we have $S = xy$. We can multiply both sides by the denominators of the second and third equations to yield $S+15 = xy+2x+y+2$ and $S+16 = xy+x+2y+2$. Since $S = xy$, we can substitute that into the equations and subtract by $xy$ to get $15 = 2x+y+2$ or $13 = 2x+y$. We also get $x+2y = 14$, and adding the two equations up gives $3x+3y = 27$, or $x+y = 9$. Subtracting $x+y=9$ from $2x+y=13$ gives $x = 4$ and $y = 5$.

We can check that this is the correct answer by having $4, 9, 6,$ and $1$ as our numbers. The sum is 20, and the mean is 5. Adding 15 gives that the sum is 35, and the number of integers is 5, so the mean is 2, an increase of 2. Adding 1 to the integers gives that the sum is 36, and the number of numbers is 6, for a mean of 6, a decrease of 1.

We can also prove that since the sum is 20, the number of integers is 4, and the mean is 5, we have $\frac{20}{4} = 5$, which is true. The second equation gives $\frac{35}{5} = 7 = 5+2$, which satisfies the condition. We also have $\frac{36}{6} = 6 = 7-1$, which also satisfies the condition.

Therefore, since we correctly solved it and checked our work multiple ways, we can be confident that the answer is $\textbf{(A)}$.

Video Solution

https://www.youtube.com/watch?v=u8zi0QikJK0 ~David

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=198326"