Difference between revisions of "2022 AMC 12A Problems/Problem 2"
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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? | 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? | ||
| − | Solution | + | ==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>. | 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>. | ||
Revision as of 20:16, 11 November 2022
Problem
The sum of three numbers is 
. The first number is 
times the third number, and the third number is 
less than the second number. What is the absolute value of the difference between the first and second numbers?
Solution
Let 
be the third number. It follows that the first number is 
and the second number is 
. Using the given sum of the numbers, we obtain the equation 
, which solves 
.
The first number is 
, and the second number is 
, and the difference between the two is 
.
- phuang1024
