Difference between revisions of "2001 AMC 12 Problems/Problem 4"

Latest revision as of 20:45, 15 July 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 $25 + 5 + 0 = \boxed{\textbf{(D) }30}$

Video Solution by Daily Dose of Math

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

~Thesmartgreekmathdude

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.

@@ Line 30: / Line 30: @@
 Therefore the sum of the three numbers is <math>25 + 5 + 0 = \boxed{\textbf{(D) }30}</math>
+==Video Solution by Daily Dose of Math==
+https://youtu.be/t0UOhO9bJTo?si=MwvGzoT8Ya8uyx7n
+~Thesmartgreekmathdude
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