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 "2010 AMC 10B Problems/Problem 4"

(Created page with 'The average of two numbers, <math>a</math> and <math>b</math>, is defined as <math>\frac{a+b}{2}</math>. Thus the average of <math>x</math> and <math>x^2</math> would be <math>\f…')
 
m
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
For a real number <math>x</math>, define <math>\heartsuit(x)</math> to be the average of <math>x</math> and <math>x^2</math>. What is <math>\heartsuit(1)+\heartsuit(2)+\heartsuit(3)</math>?
 +
 +
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20</math>
 +
 +
==Solution==
 
The average of two numbers, <math>a</math> and <math>b</math>, is defined as <math>\frac{a+b}{2}</math>. Thus the average of <math>x</math> and <math>x^2</math> would be <math>\frac{x(x+1)}{2}</math>. With that said,  we need to find the sum when we plug, <math>1</math>, <math>2</math> and <math>3</math> into that equation. So:
 
The average of two numbers, <math>a</math> and <math>b</math>, is defined as <math>\frac{a+b}{2}</math>. Thus the average of <math>x</math> and <math>x^2</math> would be <math>\frac{x(x+1)}{2}</math>. With that said,  we need to find the sum when we plug, <math>1</math>, <math>2</math> and <math>3</math> into that equation. So:
  
  
−
<math>\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\mathrm{(C)} \text{ or } 10}</math>.
+
<cmath>\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}</cmath>
 +
 
 +
==Video Solution==
 +
https://youtu.be/uAc9VHtRRPg?t=209
 +
 
 +
~IceMatrix
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2010|ab=B|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Revision as of 02:58, 26 September 2020

Problem

For a real number $x$, define $\heartsuit(x)$ to be the average of $x$ and $x^2$. What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$

Solution

The average of two numbers, $a$ and $b$, is defined as $\frac{a+b}{2}$. Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$. With that said, we need to find the sum when we plug, $1$, $2$ and $3$ into that equation. So:


\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}\]

Video Solution

https://youtu.be/uAc9VHtRRPg?t=209

~IceMatrix

See Also

2010 AMC 10B (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 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=2010_AMC_10B_Problems/Problem_4&oldid=134108"