Two trains cross each other in 14 seconds when they are moving in opposite directions, and when they are moving in the same direction they cross each other in 3 minutes 2 seconds. Find by what percent the speed of the faster train is more than the speed of the slower train.

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**Solution:**

1. Let the sum of the lengths of both trains be $L$.

2. Let the speeds of the two trains be $x$ (faster) and $y$ (slower).

3. When trains move in opposite directions, their relative speed is $x + y$.
   - Time taken to cross each other = 14 seconds.
   - So, $\frac{L}{x + y} = 14$  (Equation 1)

4. When trains move in the same direction, their relative speed is $x - y$.
   - Time taken to cross each other = 3 minutes 2 seconds = $3 	imes 60 + 2 = 182$ seconds.
   - So, $\frac{L}{x - y} = 182$  (Equation 2)

5. From Equation 1 and 2:
   $$\frac{L}{x - y} = 182, \quad \frac{L}{x + y} = 14$$

6. Equate $L$ from both equations:
   $$14(x + y) = 182(x - y)$$

7. Simplify:
   $$14x + 14y = 182x - 182y$$
   $$14x - 182x = -182y - 14y$$
   $$-168x = -196y$$
   $$168x = 196y$$

8. Divide both sides by 28:
   $$6x = 7y$$

9. So,
   $$x = \frac{7}{6} y$$

10. The percentage by which the faster train's speed is more than the slower train's speed:
    $$\frac{x - y}{y} 	imes 100 = \left(\frac{\frac{7}{6}y - y}{y}ight) 	imes 100 = \frac{1}{6} 	imes 100 = 16.67\%$$

**Answer:** 16.67\%

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