MATH SOLVE

9 months ago

Q:
# 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?

Accepted Solution

A:

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.

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.