A train runs between four stations A, B, C, and D on Monday and Tuesday. The average speed of the train for the entire journey on Monday is $50 ext{ km/h}$, and on Tuesday it is $62.5 ext{ km/h}$. On Monday, the train takes one hour less to cover the distance between A and B compared to Tuesday. The time taken to travel from B to C is the same on both days. However, the train takes three hours more to cover the distance between C and D on Monday than on Tuesday. The distance between A and B is $40\%$ less than the distance between C and D. The distance between B and C is $50\%$ more than the distance between A and B. Note that the train does not stop at any station between A and D.

What is the total time taken by the train to cover the distance from A to D on Tuesday?

Question

A train runs between four stations A, B, C, and D on Monday and Tuesday. The average speed of the train for the entire journey on Monday is $50 	ext{ km/h}$, and on Tuesday it is $62.5 	ext{ km/h}$. On Monday, the train takes one hour less to cover the distance between A and B compared to Tuesday. The time taken to travel from B to C is the same on both days. However, the train takes three hours more to cover the distance between C and D on Monday than on Tuesday. The distance between A and B is $40\%$ less than the distance between C and D. The distance between B and C is $50\%$ more than the distance between A and B. Note that the train does not stop at any station between A and D.

What is the total time taken by the train to cover the distance from A to D on Tuesday?

Accepted Answer

**Solution:**

1. Let the distance between C and D be $x$ km.

2. Then, the distance between A and B is $40\%$ less than $x$, so:
   $$AB = x - 0.4x = 0.6x$$

3. The distance between B and C is $50\%$ more than AB, so:
   $$BC = 0.6x + 0.5 	imes 0.6x = 0.6x + 0.3x = 0.9x$$

4. Total distance from A to D:
   $$AD = AB + BC + CD = 0.6x + 0.9x + x = 2.5x$$

5. Let the time taken on Tuesday to cover AB, BC, and CD be $t$, $u$, and $v$ hours respectively.

6. Given:
   - On Monday, time to cover AB is one hour less than Tuesday: Monday's AB time = $t - 1$
   - Time to cover BC is the same on both days: Monday's BC time = $u$
   - On Monday, time to cover CD is three hours more than Tuesday: Monday's CD time = $v + 3$

7. Average speed on Monday is $50$ km/h, so total time on Monday:
   $$T_{Mon} = (t - 1) + u + (v + 3) = t + u + v + 2$$

8. Average speed on Tuesday is $62.5$ km/h, so total time on Tuesday:
   $$T_{Tue} = t + u + v$$

9. Using the formula for average speed:
   $$	ext{Speed} = \frac{	ext{Distance}}{	ext{Time}}$$

For Monday:
   $$50 = \frac{2.5x}{t + u + v + 2}$$

For Tuesday:
   $$62.5 = \frac{2.5x}{t + u + v}$$

10. From Tuesday's equation:
    $$t + u + v = \frac{2.5x}{62.5} = \frac{2.5x}{62.5} = 0.04x$$

11. From Monday's equation:
    $$t + u + v + 2 = \frac{2.5x}{50} = 0.05x$$

12. Subtracting the two equations:
    $$(t + u + v + 2) - (t + u + v) = 0.05x - 0.04x$$
    $$2 = 0.01x$$
    $$x = \frac{2}{0.01} = 200 	ext{ km}$$

13. Now, total time on Tuesday:
    $$t + u + v = 0.04x = 0.04 	imes 200 = 8 	ext{ hours}$$

**Answer:** 8 hours

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