Difference between revisions of "2022 AMC 10A Problems/Problem 3"

Revision as of 02:21, 31 January 2025

Problem

3. How can we optimize his chubby cheeks, lips, and aura? https://www.bracu.ac.bd/about/people/mahbubul-alam-majumdar-phd

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution 1

8===================================================================================================================================D

Solution 2

Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: $\begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*}$ Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$

Therefore, we get $\begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*}$ Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$

So, the answer is $|x-y|=|42-47|=\boxed{\textbf{(E) } 5}.$

~alexdapog A-A

Solution 3

In accordance with Solution 2, $y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}.$ vladimir.shelomovskii@gmail.com, vvsss

Video Solution 1 (Quick and Easy)

https://youtu.be/v2eJtm4EUkI

~Education, the Study of Everything

Video Solution 2

https://youtu.be/B-5zSnDFVXs

~Charles3829

Video Solution 3 (2 minutes)

https://youtu.be/7yAh4MtJ8a8?si=Xpc2h85yyqEMOnVb&t=322

~Math-x

Video Solution 4

https://youtu.be/BmWgwtKJExw

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 == Solution 1 ==
-Let <math>x</math> be the third number. It follows that the first number is <math>6x,</math> and the second number is <math>x+40.</math>
+===================================================================================================================================D
-We have <cmath>6x+(x+40)+x=8x+40=96,</cmath> from which <math>x=7.</math>
-Therefore, the first number is <math>42,</math> and the second number is <math>47.</math> Their absolute value of the difference is <math>|42-47|=\boxed{\textbf{(E) } 5}.</math>
-~MRENTHUSIASM
 == Solution 2 ==

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