Lennon has 6 hours to spend in Ha Ha Tonka State Park. He plans to drive around the park at an average speed of 20 miles per hour, looking for a good trail to hike. Once he finds a trail he likes, he will spend the remainder of his time hiking it. He hopes to travel more than 60 miles total while in the park. If he hikes at an average speed of \(1.5\) miles per hour, which of the following systems of inequalities can be solved for the number of hours Lennon spends driving, \(d\), and the number of hours he spends hiking, \(h\), while he is at the park? A) \(1.5 h+20 d>60\) \(h+d \leq 6\) B) \(1.5 h+20 d>60\) \(h+d \geq 6\) C) \(1.5 h+20 d<60\) \(h+d \geq 360\) D) \(20 h+1.5 d>6\) \(h+d \leq 60\)

Short Answer

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The short answer is: Option A: \(1.5 h+20 d>60\) \(h+d \leq 6\)

Step by step solution

01

\(h + d \leq 6\). It represents the sum of hours spent driving and hiking. Now, we need to find an inequality for the distance constraint. #Step 2: Distance constraint# Lennon wants to travel more than 60 miles total in the park. While driving, he covers 20 miles per hour, so when driving the distance covered is \(20d\). While hiking, he covers 1.5 miles per hour, so when hiking the distance covered is \(1.5h\). The total distance is the sum of driving and hiking distances:

\(1.5h + 20d > 60\). It represents that the total distance traveled should be greater than 60 miles. #Step 3: Compare the options with the derived inequalities# Now, we will compare the derived inequalities with the given options, looking for a match. The derived inequalities are: 1. \(1.5h + 20d > 60\) 2. \(h + d \leq 6\) Option A: 1. \(1.5 h+20 d>60\) 2. \(h+d \leq 6\) Option B: 1. \(1.5 h+20 d>60\) 2. \(h+d \geq 6\) Option C: 1. \(1.5 h+20 d < 60\) 2. \(h+d \geq 360\) Option D: 1. \(20 h+1.5 d>6\) 2. \(h+d \leq 60\) Comparing the derived inequalities with the options, we see that Option A matches the derived inequalities. Therefore, the correct choice is:
02

Answer

Option A: \(1.5 h+20 d>60\) \(h+d \leq 6\)

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