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

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== See Also ==
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{{AMC12 box|year=2022|ab=A|num-b=1|num-a=3}}
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Revision as of 20:53, 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 $\boxed{(E) 5}$.

- phuang1024

See Also

2022 AMC 12A (ProblemsAnswer KeyResources)
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

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