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 "2001 AMC 12 Problems/Problem 4"
(→‎Solution 2)
 
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #4]] and [[2001 AMC 10 Problems|2001 AMC 10 #16]]}}
 +
 
== Problem ==
 
== Problem ==
 
The mean of three numbers is <math>10</math> more than the least of the numbers and <math>15</math>
 
The mean of three numbers is <math>10</math> more than the least of the numbers and <math>15</math>
Line 4: Line 6:
 
sum?
 
sum?
  
−
<math>\text{(A)}\ 5\qquad \text{(B)}\ 20\qquad \text{(C)}\ 25\qquad \text{(D)}\ 30\qquad \text{(E)}\ 36</math>
+
<math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 36</math>
 +
 
 +
== Solution 1==
 +
Let <math>m</math> be the mean of the three numbers. Then the least of the numbers is <math>m-10</math>
 +
and the greatest is <math>m + 15</math>. The middle of the three numbers is the median, <math>5</math>. So
 +
<math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which can be solved to get <math>m=10</math>.
 +
Hence, the sum of the three numbers is <math>3\cdot 10 = \boxed{\textbf{(D) }30}</math>.
 +
 
 +
==Solution 2==
 +
Say the three numbers are <math>x</math>, <math>y</math> and <math>z</math>. When we arrange them in ascending order then we can assume <math>y</math> is in the middle therefore <math>y = 5</math>.
 +
 
 +
We can also assume that the smallest number is <math>x</math> and the largest number of the three is <math>y</math>.
 +
Therefore,
 +
 
 +
<cmath>\frac{x+y+z}{3} = x + 10 = z - 15</cmath>
 +
<cmath>\frac{x+5+z}{3} = x + 10 = z - 15</cmath>
 +
 
 +
Taking up the first equation <math>\frac{x+5+z}{3} = x + 10</math> and simplifying we obtain <math>z - 2x - 25 = 0</math> doing so for the equation <math>\frac{x+5+z}{3} = z - 15</math> we obtain the equation <math>x - 2z + 50 = 0</math>
 +
 
 +
<cmath>x - 2z + 50 = 2x - 4z + 100</cmath>
 +
 
 +
when solve the above obtained equation and <math>z - 2x - 25 = 0</math> we obtain the values of <math>z = 25</math> and <math>x = 0</math>
  
−
== Solution ==
+
Therefore the sum of the three numbers is <math>25 + 5 + 0 = \boxed{\textbf{(D) }30}</math>
−
Let <math>m</math> be the mean of the three numbers. Then the least of the numbers is <math>m − 10</math>
 
−
and the greatest is <math>m + 15</math>. The middle of the three numbers is the median, 5. So
 
−
<math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which implies that <math>m=10</math>.
 
−
Hence, the sum of the three numbers is <math>3(10) = 30</math>, and the answer is <math>\text{(D)}</math>.
 
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2001|num-b=3|num-a=5}}
 
{{AMC12 box|year=2001|num-b=3|num-a=5}}
 +
{{AMC10 box|year=2001|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 21:53, 10 February 2024

The following problem is from both the 2001 AMC 12 #4 and 2001 AMC 10 #16, so both problems redirect to this page.

Problem

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 36$

Solution 1

Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$. The middle of the three numbers is the median, $5$. So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$, which can be solved to get $m=10$. Hence, the sum of the three numbers is $3\cdot 10 = \boxed{\textbf{(D) }30}$.

Solution 2

Say the three numbers are $x$, $y$ and $z$. When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$.

We can also assume that the smallest number is $x$ and the largest number of the three is $y$. Therefore,

\[\frac{x+y+z}{3} = x + 10 = z - 15\] \[\frac{x+5+z}{3} = x + 10 = z - 15\]

Taking up the first equation $\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$

\[x - 2z + 50 = 2x - 4z + 100\]

when solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$

Therefore the sum of the three numbers is $25 + 5 + 0 = \boxed{\textbf{(D) }30}$

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=2001_AMC_12_Problems/Problem_4&oldid=215202"