Question

The vagabonds were 200 miles apart at 2 p.M and were headed toward each other if they met at 6 p.M and one was traveling 10 miles per hour faster than the other what was the speed of each vagabond

Answer 1

First vagabond: 30 miles per hour

Second vagabond: 20 miles per hour

Explanation:

Let's call the speed of the first vagabond A, and the speed of the second vagabond B.

They were heading toward each other, so the relative speed is the sum of both speeds: A+B

The total number of hours spent until they meet is 6 - 2 = 4 hours, so we have the first equation:

(A+B) * 4 = 200

A + B = 50

One vagabond was traveling 10 miles faster than the other, so we have the second equation:

A = B + 10

Using the value of A from the second equation in the first one, we have:

B + 10 + B = 50

2B = 40

B = 20 miles per hour

Now, from the second equation, we find the value of A:

A = 20 + 10 = 30 miles per hour

Answer 2

The answer is 200

Explanation:

To solve the question given, let us recall the following,

Let r4=200 miles apart

r=50 mph combined speed

Where

2f+10=50

2f=40

f=20

f+10=30

or we can express it in another way,

which is

f x4+(f+10)x 4=200

f x4+f x4+40=200

8f+40=200

Then

8f is =160

f=20

f+10=30

Therefore,

4x 20+4 x 30=200

80+120=200

200=200