2022 AMC 10A Problems/Problem 3

Revision as of 18:03, 25 December 2022 by Charles3829 (talk | contribs) (Video Solution 1 (Quick and Easy))

Problem

The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?

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

Solution

Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$

We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$

Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{\textbf{(E) } 5}.$

~MRENTHUSIASM

Solution 2

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

Therefore, we get \[6z+(z+40)+z=96\] \[8z+40=96\] \[8z=56\] \[z=7\]

Substituting 7 in for $z$ gives us \[x=6z=6(7)=42\] and \[y=z+40=7+40=47\]

So $|x-y|=|42-47|=\boxed{\textbf{(E) } 5}$

~alexdapog A-A

Video Solution 1 (Quick and Easy)

https://youtu.be/v2eJtm4EUkI

~Education, the Study of Everything

Video Solution 2

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

~Charles3829

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png