Roger was told to think of a number, add 20 , then divide the result by 3 . The answer was Jower than Roger's original number. Deteruine the range of possible numbers Roger may have originally chosen.

Let's imagine Roger's original number is represented by \( x \). Following the steps: when Roger adds 20 to \( x \), he gets \( x + 20 \). Dividing that by 3 gives us \( \frac{x + 20}{3} \). According to the problem, this result is lower than his original number, which gives us the inequality: \[ \frac{x + 20}{3} < x \] To solve this inequality, we can multiply both sides by 3 to eliminate the fraction (just remember that this will not change the direction of the inequality): \[ x + 20 < 3x \] Now we solve for \( x \): \[ 20 < 3x - x \] \[ 20 < 2x \] \[ 10 < x \] So, \( x > 10 \). Now, let's check the implications of our analysis. If we further analyze it, let's consider the scenario when \( x \) is equal to 10. If he chose 10, then: \[ \frac{10 + 20}{3} = \frac{30}{3} = 10 \] Since the answer is not lower, Roger must have chosen a number greater than 10. Thus, the original number must be in the range: \[ \text{Range: } (10, \infty) \]