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

Revision as of 23:02, 11 November 2022

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-==Problem==
+#redirect [[2022 AMC 12A Problems/Problem 2]]
-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?
-<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math>
-== Solution ==
-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>
-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
-== See Also ==
-{{AMC10 box|year=2022|ab=A|num-b=2|num-a=4}}
-{{MAA Notice}}