A man can row $10 ext{ km/h}$ in still water. When the river is running at a speed of $4.5 ext{ km/h}$, it takes him $2$ hours to row to a place and come back to the initial point. How far is the place (in km) rounded off to two decimal places?

Question

A man can row $10 	ext{ km/h}$ in still water. When the river is running at a speed of $4.5 	ext{ km/h}$, it takes him $2$ hours to row to a place and come back to the initial point. How far is the place (in km) rounded off to two decimal places?

Accepted Answer

**Solution:**

1. Define variables:
   - Speed of boat in still water, $v_b = 10 	ext{ km/h}$
   - Speed of current, $v_c = 4.5 	ext{ km/h}$
   - Distance to the place, $d$ km
   - Total time for round trip, $T = 2$ hours

2. Calculate downstream and upstream speeds:
   - Downstream speed: $v_{downstream} = v_b + v_c = 10 + 4.5 = 14.5 	ext{ km/h}$
   - Upstream speed: $v_{upstream} = v_b - v_c = 10 - 4.5 = 5.5 	ext{ km/h}$

3. Set up the total time equation:
   $$
   T = \frac{d}{v_{downstream}} + \frac{d}{v_{upstream}} = \frac{d}{14.5} + \frac{d}{5.5} = 2
   $$

4. Combine the fractions:
   $$
   2 = d \left( \frac{1}{14.5} + \frac{1}{5.5} ight) = d 	imes \frac{5.5 + 14.5}{14.5 	imes 5.5} = d 	imes \frac{20}{79.75}
   $$

5. Solve for $d$:
   $$
   2 = \frac{20d}{79.75} \implies 2 	imes 79.75 = 20d \implies 159.5 = 20d \implies d = \frac{159.5}{20} = 7.975
   $$

6. Round off to two decimal places:
   $$
   d \approx 7.98 	ext{ km}
   $$

**Answer:** The distance to the place is approximately $7.98$ km.

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