Doomtown is 300 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9 a.m. Mr. Archer leaves Sagebrush for Joshua. At 1 p.m. Mr. Sassoon leaves Doomtown for Joshua. If Mr. Archer travels at an average speed 20 mph faster than Mr. Sassoon and they each reach Joshua at 4 p.m., how fast is each traveling?

Refer to the diagram shown below.

Let the distance that Sassoon travels is d miles.
Let his average speed be x mph.
The time of travel is from 1 p.m. to 4 p.m., which is 3 hours.
Therefore 
3x = d                            (1)

The distance that Archer travels is (d + 300) miles.
The time of travel is from 9 a.m. to 4 p.m., which is 7 hours.
The average traveling speed is (x + 20) mph, therefore
7(x + 20) = d + 300     
That is,
7x + 140 = d + 300
7x = d + 160                  (2)

Subtract (1) from (2).
7x - 3x = d + 160 - d
4x = 160
x = 40 mph  (Sassoon's average speed)
x+20 = 60 mph (Archer's average speed)
From (1), obtain d = 120 mi

Answer:
Archer's speed is 60 mph.
Sassoon's speed is 40 mph.