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| Problem | | #redirect [[2022 AMC 10A Problems/Problem 3]] |
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| 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?
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| Solution
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| 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>. Using the given sum of the numbers, we obtain the equation <math>(6x)+(x+40)+x = 96</math>, which solves <math>x=7</math>.
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| The first number is <math>6x = 6(7) = 42</math>, and the second number is <math>x+40 = 7+40 = 47</math>, and the difference between the two is <math>\boxed{(E) 5}</math>.
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| - phuang1024
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