Two trains, A and B, start simultaneously from station X towards station Y. Train A travels the first 60 km at its original speed, then reduces its speed by 25% and reaches station Y at 10 A.M. Train B travels the first 90 km at its original speed, then reduces its speed by 25% and reaches station Y at 9:45 A.M. If both trains had maintained their original speeds throughout the journey, they would have arrived at station Y at 8:30 A.M.

What is the sum of the time taken by train A and train B to cover the distance between station X and station Y if both trains travel the entire distance at their original speeds?

Question

Two trains, A and B, start simultaneously from station X towards station Y. Train A travels the first 60 km at its original speed, then reduces its speed by 25% and reaches station Y at 10 A.M. Train B travels the first 90 km at its original speed, then reduces its speed by 25% and reaches station Y at 9:45 A.M. If both trains had maintained their original speeds throughout the journey, they would have arrived at station Y at 8:30 A.M.

What is the sum of the time taken by train A and train B to cover the distance between station X and station Y if both trains travel the entire distance at their original speeds?

Accepted Answer

**Solution:**

1. Let the original speed of train A be $S_A$ and train B be $S_B$.

2. Let the total distance from station X to Y be $D$ km.

3. Both trains start at the same time.

4. Train A travels 60 km at speed $S_A$, then reduces speed by 25%, so new speed is $0.75 S_A$ for the remaining distance $(D - 60)$ km.

5. Train B travels 90 km at speed $S_B$, then reduces speed by 25%, so new speed is $0.75 S_B$ for the remaining distance $(D - 90)$ km.

6. Given that train A reaches at 10:00 A.M. and train B at 9:45 A.M.

7. If both trains had traveled the entire distance at their original speeds, they would have arrived at 8:30 A.M.

8. Let the start time be $T$.

9. Time taken by train A with speed reduction:
   $$t_A = \frac{60}{S_A} + \frac{D - 60}{0.75 S_A} = \frac{60}{S_A} + \frac{4}{3} 	imes \frac{D - 60}{S_A} = \frac{60 + \frac{4}{3}(D - 60)}{S_A}$$

10. Time taken by train B with speed reduction:
    $$t_B = \frac{90}{S_B} + \frac{D - 90}{0.75 S_B} = \frac{90}{S_B} + \frac{4}{3} 	imes \frac{D - 90}{S_B} = \frac{90 + \frac{4}{3}(D - 90)}{S_B}$$

11. Time taken by train A at original speed for full distance:
    $$T_A = \frac{D}{S_A}$$

12. Time taken by train B at original speed for full distance:
    $$T_B = \frac{D}{S_B}$$

13. According to the problem, the trains start at the same time $T$ and:
    - Train A arrives at 10:00 A.M., so
      $$T + t_A = 10:00$$
    - Train B arrives at 9:45 A.M., so
      $$T + t_B = 9:45$$
    - Both would arrive at 8:30 A.M. if they traveled at original speeds, so
      $$T + T_A = 8:30$$
      $$T + T_B = 8:30$$

14. From the last two equations:
    $$T_A = T_B$$
    This implies
    $$\frac{D}{S_A} = \frac{D}{S_B} \Rightarrow S_A = S_B$$

15. Let original speed of both trains be $S$.

16. Using the arrival times:
    $$T + t_A = 10:00$$
    $$T + t_B = 9:45$$
    $$T + T_A = 8:30$$

17. Subtracting the third from the first:
    $$t_A - T_A = 1.5 	ext{ hours}$$

18. Subtracting the third from the second:
    $$t_B - T_B = 1.25 	ext{ hours}$$

19. Calculate $t_A - T_A$:
    $$t_A - T_A = \left( \frac{60}{S} + \frac{D - 60}{0.75 S} ight) - \frac{D}{S} = \frac{60}{S} + \frac{4}{3} 	imes \frac{D - 60}{S} - \frac{D}{S}$$
    $$= \frac{60 + \frac{4}{3}(D - 60) - D}{S} = \frac{60 + \frac{4D}{3} - 80 - D}{S} = \frac{\frac{1}{3}D - 20}{S}$$

20. Set equal to 1.5 hours:
    $$\frac{\frac{1}{3}D - 20}{S} = 1.5 \Rightarrow \frac{1}{3}D - 20 = 1.5 S$$

21. Similarly, calculate $t_B - T_B$:
    $$t_B - T_B = \left( \frac{90}{S} + \frac{D - 90}{0.75 S} ight) - \frac{D}{S} = \frac{90}{S} + \frac{4}{3} 	imes \frac{D - 90}{S} - \frac{D}{S}$$
    $$= \frac{90 + \frac{4}{3}D - 120 - D}{S} = \frac{\frac{1}{3}D - 30}{S}$$

22. Set equal to 1.25 hours:
    $$\frac{\frac{1}{3}D - 30}{S} = 1.25 \Rightarrow \frac{1}{3}D - 30 = 1.25 S$$

23. From equations (20) and (22):
    $$\frac{1}{3}D - 20 = 1.5 S$$
    $$\frac{1}{3}D - 30 = 1.25 S$$

24. Subtract second from first:
    $$10 = 0.25 S \Rightarrow S = 40$$

25. Substitute $S=40$ into equation (20):
    $$\frac{1}{3}D - 20 = 1.5 	imes 40 = 60$$
    $$\frac{1}{3}D = 80 \Rightarrow D = 240 	ext{ km}$$

26. Time taken by each train at original speed:
    $$T_A = T_B = \frac{D}{S} = \frac{240}{40} = 6 	ext{ hours}$$

27. Sum of times:
    $$6 + 6 = 12 	ext{ hours}$$

**Answer:** 12 hours

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