Question

two bicyclists started biking toward each other when they were 88 miles apart. three hours later there were 4 miles left between them. how fast was each bicyclist going if one of them was 4 mph slower than the other?

Answer 1

Alright, lets get started.

Suppose first bicyclist speed is = x mph

As given in question, second one is 4 mph slower than first, hence

Speed of second bicyclist is = (x-4) mph

They both travels 3 hrs so,

distance covered by first bicyclist :

$distance = speed* time$

$d1 = 3x$

distance covered by second bicyclist :

$d2= 3 (x-4)$

4 miles were still left after 3 hrs of driving.

It means they covered 88 - 4 = 84 miles together

$d1 + d2 = 84$

$3x + 3 (x-4) = 84$

$3x + 3x - 12 = 84$

Adding 12 in both sides

$6x - 12 + 12 = 84 + 12$

$6x = 96$

Dividing 6 in both sides

$\frac{6x}{6}= \frac{96}{6}$

$x = 16$

Means speed of first bicyclist = 16 mph

Speed of second bicyclist = $16 - 4 = 12$ mph

Hence speed of both bicyclist are 16 mph and 12 mph. : Answer

Hope it will help :)

Answer 2

29.3 mph because the first bicyclist went 88 miles while the second went only 84 at 28 mph