A boat goes 28 km downstream and while returning covers only 75\% of the distance covered downstream. If the boat takes 3 hours more to cover upstream than downstream, then find the speed of the boat in still water (km/hr) if the speed of the stream is $\frac{5}{9}$ m/sec?

Question

Accepted Answer

**Solution:**

1. Convert speed of stream from m/sec to km/hr:
   $$	ext{Speed of stream} = \frac{5}{9} 	imes \frac{18}{5} = 2 	ext{ km/hr}$$

2. Let the speed of the boat in still water be $x$ km/hr.

3. Given:
   - Downstream distance = 28 km
   - Upstream distance = 75\% 	imes 28 = 21 	ext{ km}$
   - Time difference between upstream and downstream = 3 hours

4. Speed downstream = $(x + 2)$ km/hr
   Speed upstream = $(x - 2)$ km/hr

5. Time taken downstream:
   $$t_{down} = \frac{28}{x + 2}$$

6. Time taken upstream:
   $$t_{up} = \frac{21}{x - 2}$$

7. According to the problem:
   $$t_{up} - t_{down} = 3$$

Substitute values:
   $$\frac{21}{x - 2} - \frac{28}{x + 2} = 3$$

8. Multiply both sides by $(x - 2)(x + 2)$:
   $$21(x + 2) - 28(x - 2) = 3(x - 2)(x + 2)$$

9. Expand:
   $$21x + 42 - 28x + 56 = 3(x^2 - 4)$$

Simplify left side:
   $$-7x + 98 = 3x^2 - 12$$

10. Rearrange:
    $$3x^2 + 7x - 110 = 0$$

11. Solve quadratic equation:
    Using quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=3$, $b=7$, $c=-110$:

$$\Delta = 7^2 - 4 	imes 3 	imes (-110) = 49 + 1320 = 1369$$

$$\sqrt{1369} = 37$$

$$x = \frac{-7 \pm 37}{6}$$

Possible values:
    - $x = \frac{30}{6} = 5$
    - $x = \frac{-44}{6} = -7.33$ (not possible as speed cannot be negative)

12. Therefore, speed of boat in still water is $5$ km/hr.

**Answer:** 5 km/hr

Answer

Answer

Answer