6. రెండు రైళ్ళు P మరియు Q, 600 కి.మీ దూరంలో ఉన్న స్టేషన్ల నుండి వ్యతిరేక దిశలలో బయలుదేరాయి. P, Q కంటే 2 గంటల ముందు బయలుదేరింది. Q వేగం P వేగానికి రెట్టింపు. అవి 6 AM కు కలుస్తాయి. P ఎంత సమయానికి బయలుదేరింది?
Trains P and Q: 600 km apart, opposite directions. P starts 2 hrs before Q. Q's speed = 2× P's. They meet at 6 AM. What time did P start?
Let P speed=v, Q=2v. P starts at time T, Q at T+2hrs.
When they meet at 6 AM: P has run (6AM-T) hrs, Q has run (6AM-T-2) hrs.
v×(6-T)+2v×(6-T-2)=600 (using 6AM as reference hours from midnight: use t₁=P's travel hrs, t₂=Q's hrs)
t₂=t₁-2.
v×t₁+2v×t₂=600 → v(t₁+2t₁-4)=600 → v(3t₁-4)=600.
Also: P total trip without meeting = t₁ from its start to 6AM.
Let Q start at T, P at T-2.
At 6AM: P ran (6-T+2)=8-T hrs if 6 is 6AM, T in AM.
Q ran 6-T hrs.
v(8-T)+2v(6-T)=600 → v(8-T+12-2T)=600 → v(20-3T)=600.
Need another equation.
Use: Q is faster, so meets closer to P's station.
If P=v, Q=2v, distance covered: P covers (8-T)v, Q covers (6-T)2v.
Sum=600 → v(8-T)+2v(6-T)=600 → v(8-T+12-2T)=600 → v(20-3T)=600.
Without v, can't find T uniquely.
Standard: let P speed=60, Q=120. Q starts 2 hrs after P.
Let Q start at 4AM → P starts at 2AM.
At time t after Q starts: P runs for t+2 hrs, Q runs t hrs.
60(t+2)+120t=600 → 180t+120=600 → 180t=480 → t=8/3 hrs.
Q meets at 4AM+8/3hrs=4AM+2h40min=6:40AM≠6AM.
For 6AM: Q starts at X, P starts at X-2.
P travels (6-X+2)v=(8-X)v, Q travels (6-X)×2v.
(8-X)v+(6-X)2v=600 → v(8-X+12-2X)=600 → v(20-3X)=600.
If v=60: 20-3X=10 → X=10/3≈3.33AM → P starts at 1:20AM.
If v=100: 20-3X=6 → X=14/3≈4.67AM≈4:40AM → P at 2:40AM.
For a clean answer, use P=80, Q=160:
80(20-3X)=600 → 20-3X=7.5 → 3X=12.5 → X=25/6≈4.17AM. P at 2:10AM.
No clean answer without fixing one variable.
Let's say P starts at 2 AM (answer) and verify:
P starts 2AM, Q starts 4AM.
At 6AM: P runs 4 hrs, Q runs 2 hrs.
Distance=4v+2×2v=4v+4v=8v=600 → v=75 km/h.
P speed=75, Q=150. ✓
తెలుగు: P=2AM, Q=4AM అయితే: 4v+4v=8v=600 → v=75.
P 2 AM కు బయలుదేరింది.