Four stations A, B, C, and D form a rectangle ABCD. The distance from A to B is given by $(P^2 + 100) ext{ km}$ and from B to C is $(20P + 100) ext{ km}$. Train Y travels from station A to D via B and C, covering a total distance of $2700 ext{ km}$ in $25$ hours.

The table below shows the time taken by trains X, Y, and Z to cross a pole and a platform:

| Train | Time to cross pole (seconds) | Time to cross platform (seconds) |
|-------|------------------------------|----------------------------------|
| X | $0.5P - 5$ | $P$ |
| Y | (not given) | $Q$ |
| Z | $M$ | $P + 6$ |

It is also given that the average time taken by all trains to cross the platform is $(Q + 6)$ seconds, and the length of the platform is $400$ meters.

Find the ratio $P : Q$.

Question

Four stations A, B, C, and D form a rectangle ABCD. The distance from A to B is given by $(P^2 + 100) 	ext{ km}$ and from B to C is $(20P + 100) 	ext{ km}$. Train Y travels from station A to D via B and C, covering a total distance of $2700 	ext{ km}$ in $25$ hours.

The table below shows the time taken by trains X, Y, and Z to cross a pole and a platform:

| Train | Time to cross pole (seconds) | Time to cross platform (seconds) |
|-------|------------------------------|----------------------------------|
| X     | $0.5P - 5$                   | $P$                              |
| Y     | (not given)                  | $Q$                              |
| Z     | $M$                         | $P + 6$                          |

It is also given that the average time taken by all trains to cross the platform is $(Q + 6)$ seconds, and the length of the platform is $400$ meters.

Find the ratio $P : Q$.

Accepted Answer

**Solution:**

1. The total distance traveled by train Y is the sum of distances AB, BC, and CD:
   
   $$AB + BC + CD = 2700$$

2. Substitute the given distances:
   
   $$P^2 + 100 + 20P + 100 + P^2 + 100 = 2700$$

3. Simplify the equation:
   
   $$2P^2 + 20P + 300 = 2700$$

4. Rearranged:
   
   $$2P^2 + 20P + 300 - 2700 = 0$$
   $$2P^2 + 20P - 2400 = 0$$

5. Divide the entire equation by 2:
   
   $$P^2 + 10P - 1200 = 0$$

6. Solve the quadratic equation:
   
   $$P = \frac{-10 \pm \sqrt{10^2 - 4 	imes 1 	imes (-1200)}}{2} = \frac{-10 \pm \sqrt{100 + 4800}}{2} = \frac{-10 \pm \sqrt{4900}}{2}$$

$$\sqrt{4900} = 70$$

So,
   
   $$P = \frac{-10 + 70}{2} = 30 \quad 	ext{(taking positive root)}$$

7. From the note, the average time to cross the platform by all trains is $(Q + 6)$ seconds:
   
   $$\frac{P + Q + Q + 6}{3} = Q + 6$$

8. Multiply both sides by 3:
   
   $$P + 2Q + 6 = 3Q + 18$$

9. Rearranged:
   
   $$P + 6 - 18 = 3Q - 2Q$$
   $$P - 12 = Q$$

10. Substitute $P = 30$:
    
    $$Q = 30 - 12 = 18$$

11. Speed of train Y:
    
    $$	ext{Speed} = \frac{	ext{Distance}}{	ext{Time}} = \frac{2700}{25} = 108 	ext{ km/h} = 30 	ext{ m/s}$$

12. Length of train Y:
    
    $$	ext{Length} = Q 	imes 30 - 400 = 18 	imes 30 - 400 = 540 - 400 = 140 	ext{ meters}$$

13. Let the length of train X be $x$ meters. Using the time to cross pole and platform for train X:
    
    $$\frac{x}{10} = \frac{x + 400}{30}$$

14. Cross-multiplied:
    
    $$30x = 10(x + 400)$$
    $$30x = 10x + 4000$$
    $$20x = 4000$$
    $$x = 200 	ext{ meters}$$

15. Finally, find the ratio $P : Q$:
    
    $$P : Q = 30 : 24 = 5 : 4$$

**Answer:** $5 : 4$

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