April can run at a rate that is 2 miles per hour faster then donna's rate. one day, april gave donna a 25 minute head start on a run. if april passes donna five miles from the starting point, how fast is each runner running?

Let us say that:

s = speed of Donna

s + 2 = speed of April

Then we are also given that:

t + (25/60) = time of Donna

t = time of April

The distance taken by the two should be equal. We know that distance is the product of speed and time, so:

s * (t + 25/60) = (s + 2) t

We also know that:

(s + 2) * t = 5

s = 5/t – 2

So that:

(5/t – 2) * (t + 25/60) = 5

5 + 125/60t – 2t – 50/60 = 5

25/12t – 2t – 5/6 = 0

Multiply everything by t:

- 2t^2 – 5/6 t + 25/12 = 0

t^2 + 5/12 t = 25/24

Completing the square:

(t + 5/24)^2 = 25/24 + (5/24)^2

t + 5/24 = ± 1.04

t = -1.25 h, 0.83 h

Since time cannot be negative, therefore t = 0.83 h

So,

(s + 2) * t = 5

s + 2 = 5/0.83 = 6 miles per hour

s = 4 miles per hour

Answers:

speed of Donna = 4 miles / hr

speed of April = 6 miles / hr

a 10 percent increase