Speed of a boat in still water is 12 kmph and speed of stream is 'x' kmph. If in traveling 270 km upstream boat takes $66 \frac{2}{3}\%$ more time than traveling 270 km downstream, then find the value of 'x'.

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**Solution:**

1. Let the speed of the boat in still water be $12 	ext{ kmph}$ and speed of the stream be $x 	ext{ kmph}$.

2. Speed upstream = $(12 - x) 	ext{ kmph}$

3. Speed downstream = $(12 + x) 	ext{ kmph}$

4. Distance traveled in both directions = $270 	ext{ km}$

5. Time taken upstream = $\frac{270}{12 - x}$

6. Time taken downstream = $\frac{270}{12 + x}$

7. Given that time taken upstream is $66 \frac{2}{3}\% = \frac{2}{3}$ more than time taken downstream, so:

$$\frac{270}{12 - x} = \left(1 + \frac{2}{3}ight) 	imes \frac{270}{12 + x} = \frac{5}{3} 	imes \frac{270}{12 + x}$$

8. Simplify the equation:

$$\frac{270}{12 - x} = \frac{5}{3} 	imes \frac{270}{12 + x}$$

Multiply both sides by $(12 - x)(12 + x)$:

$$270 (12 + x) = \frac{5}{3} 	imes 270 (12 - x)$$

9. Divide both sides by 270:

$$(12 + x) = \frac{5}{3} (12 - x)$$

10. Multiply both sides by 3:

$$3(12 + x) = 5(12 - x)$$

$$36 + 3x = 60 - 5x$$

11. Bring like terms together:

$$3x + 5x = 60 - 36$$

$$8x = 24$$

$$x = 3$$

**Answer:** $3 	ext{ kmph}$

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