P, Q and R are three towns located on a river flowing at a constant rate. Town Q is situated equidistant from towns P and R. A person rows from P to Q and back in 10 hours. The same person can row from P to R in 4 hours. Determine the ratio of the person's rowing speed in still water to the speed of the river current.

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Let's define our variables:
- Let x km be the distance from P to Q (also from Q to R since Q is equidistant)
- Let a km/hr be the speed of the boat in still water
- Let b km/hr be the speed of the river current

For the journey from P to Q and back:
- Downstream speed (P to Q) = a + b km/hr
- Upstream speed (Q to P) = a - b km/hr
- Time taken = $\frac{x}{a+b} + \frac{x}{a-b} = 10$ hours...(i)

For the journey from P to R:
- Since Q is equidistant from P and R, the distance from P to R is 2x km
- Time taken = $\frac{2x}{a+b} = 4$ hours
- This gives us x = 2(a+b)...(ii)

Substituting (ii) into equation (i):
$\frac{2(a+b)}{a+b} + \frac{2(a+b)}{a-b} = 10$

Simplifying:
$2 + \frac{2(a+b)}{a-b} = 10$

$\frac{2(a+b)}{a-b} = 8$

Multiplying both sides by (a-b):
$2(a+b) = 8(a-b)$

Expanding:
$2a + 2b = 8a - 8b$

Rearranging:
$2a - 8a = -8b - 2b$
$-6a = -10b$
$\frac{a}{b} = \frac{10}{6} = \frac{5}{3}$

Therefore, the ratio of the speed in still water to the speed of the river current is **5:3**.

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