MATH SOLVE

6 months ago

Q:
# Two tourists left two towns, the distance between which is 38 km, simultaneously and met in 4 hours. what was the speed of each of the tourists, if the first one covered 2 km more before he met than the second one?

Accepted Solution

A:

Let the relative speed will be given by:

38/4=9.5 km/h

Let the speed of the first tourist be x km/h

the speed of the second one is (9.5-x) km/h

distance covered by the first one is:

4*x+2=4x+2 km

distance covered by the second one is:

(9.5-x)4 km

=(38+4x) km

total distance will be given by:

(38+4x)+(4x+2)=38

40+8x=38

8x=-2

x=-1/4 km/h

this mean that he was traveling in the opposite direct thus the speed was 1/4 km/k

The speed of the second one will be:

9.5-1/4=9.25 km/h

38/4=9.5 km/h

Let the speed of the first tourist be x km/h

the speed of the second one is (9.5-x) km/h

distance covered by the first one is:

4*x+2=4x+2 km

distance covered by the second one is:

(9.5-x)4 km

=(38+4x) km

total distance will be given by:

(38+4x)+(4x+2)=38

40+8x=38

8x=-2

x=-1/4 km/h

this mean that he was traveling in the opposite direct thus the speed was 1/4 km/k

The speed of the second one will be:

9.5-1/4=9.25 km/h