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2022 AMC 12A Problem 2

Problem 2 of 25EasierAlgebra

The sum of three numbers is 96.96. The first number is 66 times the third number, and the third number is 4040 less than the second number. What is the absolute value of the difference between the first and second numbers?

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Solution

Let the third number be t.t. Then the first is 6t6t and the second is t+40.t+40. Their sum is 6t+(t+40)+t=8t+40=96,6t+(t+40)+t=8t+40=96, so t=7.t=7. The first number is 4242 and the second is 47,47, so the difference has absolute value 5.5. Thus, the correct answer is E.

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Concepts: system of equations · substitution

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.