A man covers a certain distance between his house and office. If his average speed is 20 km/hr, he is late by 10 min and if his average speed is 50 km/hr, he reaches 20 min early. Find the distance between his house and office?
A. 50 km
B. 50/3 km
C. 55 km
D. 56/7 km

• =(20*50/30)*(30/60)
  =50/3
  
• 8 years agoHelpfull: Yes(2) No(0)
• shor cut:
  take lcm of 50 and 20 ...................100
  10 min late ........20 min early................ total 30 min
  100km.................3hr=180min(100/50=2,100/20=5.................difference 5-2=3hr)
  30min
  as 180/6=30
  hence 100/6=50/3 (ans)
• 8 years agoHelpfull: Yes(0) No(0)
• ANS
  
  =(Products of speed/difference in speed) * Difference in Time
  
  = ((20*50)/30)*(30/60)
  
  =50/3
  
  So, Ans is B
• 4 years agoHelpfull: Yes(0) No(0)