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

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**Solution:**

1. Let the speeds of the two trains be *S_1* and *S_2* (in m/s), and their lengths be *L_1* and *L_2* respectively.

2. When trains move in opposite directions, their relative speed is *S_1 + S_2*.
   - They cross each other in 14 seconds.
   - Total distance covered = sum of their lengths = *L_1 + L_2*.

Therefore, 
   $$L_1 + L_2 = (S_1 + S_2) \times 14$$

3. When trains move in the same direction, their relative speed is *|S_1 - S_2|*.
   - They cross each other in 3 minutes 2 seconds = 182 seconds.
   - Distance covered = *L_1 + L_2* (same as above).

Therefore, 
   $$L_1 + L_2 = |S_1 - S_2| \times 182$$

4. Equate the two expressions for *L_1 + L_2*:
   $$(S_1 + S_2) \times 14 = |S_1 - S_2| \times 182$$

5. Divide both sides by 14:
   $$S_1 + S_2 = |S_1 - S_2| \times 13$$

6. Assume *S_1 > S_2*, so:
   $$S_1 + S_2 = 13 (S_1 - S_2)$$

7. Expand and rearrange:
   $$S_1 + S_2 = 13 S_1 - 13 S_2$$
   $$S_1 + S_2 - 13 S_1 + 13 S_2 = 0$$
   $$-12 S_1 + 14 S_2 = 0$$

8. Rearranged:
   $$14 S_2 = 12 S_1$$
   $$\frac{S_1}{S_2} = \frac{14}{12} = \frac{7}{6}$$

9. The speed of the faster train is $ \frac{7}{6} $ times the speed of the slower train.

10. Percentage increase in speed:
    $$\left( \frac{S_1 - S_2}{S_2} \right) \times 100 = \left( \frac{7/6 - 1}{1} \right) \times 100 = \frac{1}{6} \times 100 = 16.67\%$$

**Answer:** 16.67\%

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