Two buses leave towns 268 miles apart at the same time and travel toward each other. One bus travels 10 miles per hour slower than the other. If they meet in 2 hours, what is the rate of each bus?

Rate of first bus       = 72 mph

Rate of second bus  = 62 mph

Step-by-step explanation:

let first bust travels at x mph

then,  2nd but will travel at ( x - 10) mph

distance travelled by first bus in 2 hours = 2x

distance travelled by 2nd but in 2 hours = 2( x- 10) = 2x - 20

distance between the two is 268 miles

so we can write

2x + 2x - 20 = 268

4x = 288

x = 288 / 4 = 72 mph

rate of first bus = 72 mph

and second bus = 72 - 10 = 62 mph

speed of slower bus is 62 mph.

Speed of faster bus is 72 mph

Step-by-step explanation:

Let x represent the the speed of the slower bus.

One bus travels 10 miles per hour slower than the other. It means that the speed of the faster bus is (x + 10) mph

Distance = speed × time

The 2 buses meet in 2 hours.

Distance travelled by the slower bus in 2 hours is 2 × x = 2x

Distance travelled by the faster bus in 2 hours is

2(x + 10)

By the time they meet, they would have travelled 268 miles. It means that

2x + 2(x + 10) = 268

2x + 2x + 20 = 268

4x + 20 = 268

4x = 268 - 20

4x = 248

x = 248/4

x = 62 mph

Speed of the faster bus is

x + 10 = 62 + 10 = 72 mph

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