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Algebra Difficulty 2.7 Junior Find the answer

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?

Pick one

Solution

Let xx be the third number. It follows that the first number is 6x,6x, and the second number is x+40.x+40.
We have 6x+(x+40)+x=8x+40=96,6x+(x+40)+x=8x+40=96, from which x=7.x=7.
Therefore, the first number is 42,42, and the second number is 47.47. Their absolute value of the difference is ∣42−47∣=(E) 5.|42-47|=\boxed{\textbf{(E) } 5}.
~MRENTHUSIASM

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.