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

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#redirect [[2022 AMC 12A Problems/Problem 2]]
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==Problem==
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The sum of three numbers is <math>96.</math> The first number is <math>6</math> times the third number, and the third number is <math>40</math> less than the second number. What is the absolute value of the difference between the first and second numbers?
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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>
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== Solution ==
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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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We have <cmath>6x+(x+40)+x=8x+40=96,</cmath> from which <math>x=7.</math>
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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>
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~MRENTHUSIASM
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== See Also ==
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{{AMC10 box|year=2022|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 00:04, 12 November 2022

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

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

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