Question

A journey $192$ $km$ from town $A$ to town $B$ take $2$ hrs more by an ordinary passenger train than a dupe train . If the speed of the frighten train is $16$ $km/hr$ more. Find the speed of the fate is the passenger train .

Answer 1

• Let the speed of the average passenger train be 'x' km/hr.
  
• So, the time taken by the passenger train = 192/x.
  
• So, the average speed of the super fast train = (x + 16) km/hr
  
• So, the time taken by the super fast train = 192/(x + 16).
  
• Given : Time taken by the passenger train to travel 192 km - time taken by super fast train to cover the distance of 192 km = 2 hours.
  
• Therefore,
  
• 192/x - 192/(x + 16) = 2 Taking L.C.M.. we get
  
• (192x + 3072 + 192x)/(x)(x+16) = 2
  
• 3072/x(x + 16) = 2 After cross multiplying, we get
  
• 3072 = 2x² + 32x
  
• 2x² + 32x - 3072 = 0
  
• Dividing it by 2, we get
  
• x² + 16x - 1536 = 0
  
• x² + 48x - 32x - 1536 = 0
  
• x(x + 48) - 32(x + 48) = 0
  
• (x + 48) (x - 32) = 0
  
• x = - 48 or x = 32
  
• The speed of the train cannot be negative, therefore the speed of the passenger train is 32 km/hr
  

And the speed of the super fast train is 32 + 16 = 48 km/hr