The driver of a maruti car driving at a speed of 68km/hr located a bus 40m ahead of him. After 10sec the bus is 60m behind. The speed of the bus is:
a) 32km/hr
b) 30km/hr
c) 25km/hr
d)38km/hr

• 
  Speed of the Car = 68 km/h = 18.89 m/s;
  Distance travelled in 10 sec = 188.89 metres;
  
  Let,
  
  Initial position of car is 'A'
  Final position of car is 'D'
  Initial position of bus is 'B'
  Final position of bus is 'C' ;
  
  A, B, C D are on one straight line;
  
  AD is the distance travelled by the car in 10 sec;
  BC is the distance travelled by the bus in 10 sec;
  
  AB = 40 metres;
  CD = 60 metres;
  AD = 188.89 metres;
  
  AB + BC + CD = AD --- --- --- --- --- --- --- (1);
  
  So, it follows that
  
  BC = AD - AB - CD = 188.89 - 40 - 60 = 88.89 metres;
  
  BC = 88.89 metres;
  
  BC is the distance travelled by the bus in 10 sec;
  
  So, Speed of the bus is 88.89/10 = 8.889 m/s = 32 km/h
• 8 years agoHelpfull: Yes(0) No(0)
• total distance covered here 60+40=100m, time given =10sec, so to cross this distance , speed should be=dis/time >>100/10
  s=10m/s=10*18/5=36km/h..
  car speed is 68km
  68-36=32 bus speed
• 5 years agoHelpfull: Yes(0) No(0)