# Thread: Problem Solving- Travel (Time is of the essence here...)

1. ## Problem Solving- Travel (Time is of the essence here...)

"Jim and Freda live 3 miles apart. They both leave their houses at the same time and walk to meet each other. Jim walks at 2.5 miles/hour and Freda walks at 3.5 miles/hour. How far will Jim walk before he meets Freda?

Please post a thorough response showing all the steps so I understand! Thank you in advance.

Here's a back-door solution . . .

Jim and Freda live 3 miles apart.
They both leave their houses at the same time and walk to meet each other.
Jim walks at 2.5 miles/hour and Freda walks at 3.5 miles/hour.
How far will Jim walk before he meets Freda?

They have a combined speed of: .$\displaystyle 2.5 + 3.5 \:=\:6$ mph.

To cover 3 miles at 6 mph, it will take them a half-hour.

In that time, Jim has covered: .$\displaystyle \tfrac{1}{2} \ \times 2\tfrac{1}{2} \:=\:1\tfrac{1}{4}$ miles.

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Assuming that we must use algebra . . .

We will use: .$\displaystyle \text{(Distance)} \:=\:\text{(Speed)} \times \text{}$

Let $\displaystyle t$ = time for them to meet (in hours).

Jim walks at 2.5 mph.
In $\displaystyle t$ hours, he walks: $\displaystyle 2.5t$ miles.

Freda walks at 3.5 mph.
In $\displaystyle t$ hours, she walks: $\displaystyle 3.5t$ miles.

Together, they cover 3 miles: .$\displaystyle 2.5t + 3.5t \:=\:3$

. . Then: .$\displaystyle 6t \:=\:3\quad\Rightarrow\quad t \:=\:\tfrac{3}{6} \:=\:\tfrac{1}{2}$ hour.

In that time, Jim walked: .$\displaystyle \tfrac{1}{2} \times 2.5 \:=\:1.25$ miles.