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2001 AMC 12 Problems/Problem 4: Difference between revisions

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<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 ==
== Solution 1==
Let <math>m</math> be the mean of the three numbers. Then the least of the numbers is <math>m-10</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
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>.
<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(10) = \boxed{(\text{D})30}</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 <imath>x</imath>, <imath>y</imath> and <imath>z</imath>. When we arrange them in ascending order then we can assume <imath>y</imath> is in the middle therefore <imath>y = 5</imath>.
 
We can also assume that the smallest number is <imath>x</imath> and the largest number of the three is <imath>z</imath>.
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 <imath>\frac{x+5+z}{3} = x + 10</imath> and simplifying we obtain <imath>z - 2x - 25 = 0</imath> doing so for the equation <imath>\frac{x+5+z}{3} = z - 15</imath> we obtain the equation <imath>x - 2z + 50 = 0</imath>
 
<cmath>x - 2z + 50 = 2x - 4z + 100</cmath>
 
when solve the above obtained equation and <imath>z - 2x - 25 = 0</imath> we obtain the values of <imath>z = 25</imath> and <imath>x = 0</imath>
 
Therefore the sum of the three numbers is <imath>25 + 5 + 0 = \boxed{\textbf{(D) }30}</imath>
 
==Video Solution by Daily Dose of Math==
 
https://youtu.be/t0UOhO9bJTo?si=MwvGzoT8Ya8uyx7n
 
~Thesmartgreekmathdude


== See Also ==
== See Also ==
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{{AMC10 box|year=2001|num-b=15|num-a=17}}
{{AMC10 box|year=2001|num-b=15|num-a=17}}
{{MAA Notice}}
{{MAA Notice}}
[[Category: Introductory Algebra Problems]]

Latest revision as of 19:30, 7 November 2025

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 $z$. 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 $25 + 5 + 0 = \boxed{\textbf{(D) }30}$

Video Solution by Daily Dose of Math

https://youtu.be/t0UOhO9bJTo?si=MwvGzoT8Ya8uyx7n

~Thesmartgreekmathdude

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

These problems are copyrighted © by the Mathematical Association of America.

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