Difference between revisions of "2022 AMC 12A Problems/Problem 2"

Revision as of 20:15, 11 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?

Solution

Let $x$ be the third number. It follows that the first number is $6x$ and the second number is $x+40$ . Using the given sum of the numbers, we obtain the equation $(6x)+(x+40)+x = 96$ , which solves $x=7$ .

The first number is $6x = 6(7) = 42$ , and the second number is $x+40 = 7+40 = 47$ , and the difference between the two is $\boxed{(E) 5}$ .

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Difference between revisions of "2022 AMC 12A Problems/Problem 2"

Revision as of 20:15, 11 November 2022