Question

Matti wants to cover a 6 mile course in 1 hour by running part of the course and walking the rest. he also wants to make a 5-minuet stop for water. He runs at 8mi/h and walks 3mi/h.
What is the minimum distance that mattwill need to run to complete the course in 1 hour?

Answer 1

Answer on Question #62089 – Math – Algebra

Question

Matti wants to cover a 6 mile course in 1 hour by running part of the course and walking the rest. He also wants to make a 5-minute stop for water. He runs at $8\mathrm{mi/h}$ and walks $3\mathrm{mi/h}$. What is the minimum distance that Matt will need to run to complete the course in 1 hour?

Solution

The total time of movement (without a 5-minute stop) is $1 - \frac{5}{60} = \frac{55}{60}$ (hour).

Let $x$ be the running distance, then $(6 - x)$ will be the walking distance, $\frac{x}{8}$ will be the running time, $\frac{6 - x}{3}$ will be the walking time.

Then

$$\frac{x}{8} + \frac{6 - x}{3} \leq \frac{55}{60},$$ $$\frac{3x}{24} + \frac{8(6 - x)}{24} \leq \frac{55}{60},$$ $$\frac{3x + 8(6 - x)}{24} \leq \frac{55}{60},$$ $$\frac{3x + 48 - 8x}{24} \leq \frac{55}{60},$$ $$\frac{48 - 5x}{24} - \frac{55}{60} \leq 0,$$ $$\frac{5(48 - 5x)}{24 \times 5} - \frac{2 \times 55}{60 \times 2} \leq 0,$$ $$\frac{5(48 - 5x) - 2 \times 55}{120} \leq 0,$$ $$\frac{240 - 25x - 110}{120} \leq 0,$$ $$\frac{130 - 25x}{120} \leq 0,$$ $$130 - 25x \leq 0,$$ $$25x \geq 130,$$ $$x \geq \frac{130}{25},$$ $$x \geq 5.2.$$

Hence 5.2 miles is the minimum running distance.

**Answer**: 5.2 miles.

www.AssignmentExpert.com