A thief flees City A in a car towards City B on a stretch of straight road, 300km long, at a speed of 60kmph. In 15 min, a police party X leaves City A to chase the thief at 65kmph. If a police party Y were to leave the city B at the same time as police party X leaving the city to catch the thief at 60kmph. Then who will catch first and by how much time

• Ans. Police party Y will catch 37.5 minutes earlier than Police party X.
  
  When police party X from city A leaves the thief has covered 15 Km. So police party have to cover 15 km relative to thief at a speed of 5 Kmph (65-60) as they are moving in the same direction. So they will catch in 15/3= 3 hours.
  When the police party Y will leav city B they will be 285 Km away from thief. As thief and police will be moving in opposite directions they will be approaching each other at a speed of 120 kmph(60+60). And police party Y leaving city B will catch the thief in 285/120 i.e 2 hrs. 22.5 minutes
  So police party Y will catch first by 37.5 minutes.
• 12 years agoHelpfull: Yes(4) No(0)
• When police party X from city A leaves the thief has covered 15 Km. So police party X have to cover 15 km relative to thief at a speed of 5 Kmph (65-60) as they are moving in the same direction. So they will catch in 15kmph/3kmph= 3 hours.
  When the police party Y will leave city B they will be 285 Km away from thief. As thief and police will be moving in opposite directions they will be approaching each other at a speed of 120 kmph(60+60). And police party Y leaving city B will catch the thief in 285/120 i.e 2 hrs. 22 minutes 30 seconds.
  
  Thus police party Y will catch the thief first in 2 hrs. 22 minutes 30 seconds.
• 11 years agoHelpfull: Yes(0) No(0)