Difference between revisions of "2022 AMC 10A Problems/Problem 3"
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math> | ||
− | == Solution == | + | == 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> | 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> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 == | ||
+ | Solve this using a system of equations. Let <math>x,y,</math> and <math>z</math> be the three numbers, respectively. We get three equations: | ||
+ | <cmath>\begin{align*} | ||
+ | x+y+z&=96, \\ | ||
+ | x&=6z, \\ | ||
+ | z&=y-40. | ||
+ | \end{align*}</cmath> | ||
+ | Rewriting the third equation gives us <math>y=z+40,</math> so we can substitute <math>x</math> as <math>6z</math> and <math>y</math> as <math>z+40.</math> | ||
+ | |||
+ | Therefore, we get | ||
+ | <cmath>\begin{align*} | ||
+ | 6z+(z+40)+z&=96 \\ | ||
+ | 8z+40&=96 \\ | ||
+ | 8z&=56 \\ | ||
+ | z&=7. | ||
+ | \end{align*}</cmath> | ||
+ | Substituting 7 in for <math>z</math> gives us <math>x=6z=6(7)=42</math> and <math>y=z+40=7+40=47.</math> | ||
+ | |||
+ | So, the answer is <math>|x-y|=|42-47|=\boxed{\textbf{(E) } 5}.</math> | ||
+ | |||
+ | ~alexdapog A-A | ||
+ | |||
+ | == Solution 3 == | ||
+ | In accordance with Solution 2, <cmath>y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}.</cmath> | ||
+ | '''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 | ||
== See Also == | == See Also == |
Latest revision as of 11:35, 24 March 2024
Contents
Problem
The sum of three numbers is The first number is
times the third number, and the third number is
less than the second number. What is the absolute value of the difference between the first and second numbers?
Solution 1
Let be the third number. It follows that the first number is
and the second number is
We have from which
Therefore, the first number is and the second number is
Their absolute value of the difference is
~MRENTHUSIASM
Solution 2
Solve this using a system of equations. Let and
be the three numbers, respectively. We get three equations:
Rewriting the third equation gives us
so we can substitute
as
and
as
Therefore, we get
Substituting 7 in for
gives us
and
So, the answer is
~alexdapog A-A
Solution 3
In accordance with Solution 2,
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
~Charles3829
Video Solution 3 (2 minutes)
https://youtu.be/7yAh4MtJ8a8?si=Xpc2h85yyqEMOnVb&t=322
~Math-x
Video Solution 4
See Also
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.