Assume that the rate of consumption of coal by a locomotive engine varies as the square of its speed and is 1000 kg per hour when the speed is 60 kmph, when in motion. If coal costs the railway company Rs.15 per 100 kg and if the other expenses of running the train is Rs.12 per hour, find a formula for the cost in paise per kilometre when the speed is S km per hour.
options:
(1200/S)+(25S^2/6)
(1200/S)+(5S^2/18)
(1200/S)+(25S/6)
none

• R = K*S^2
  put R=1000, S=60
  1000 = k*60^2 => K = 5/18
  
  R = (5/18)*S^2
  
  Total Cost(C) = Coal cost + Other costs =
  C = 12*100/S + (5/18 *S^2)/S *15
  C = (1200/S) + (25S/6)
• 10 years agoHelpfull: Yes(57) No(9)
• Let R be the rate of consumption.
  Let S be the speed of the engine.
  So,by the problem,
  R=K*S^2
  where K is an arbitrary constant.
  When R=1000 kg per hr,S=60 km per hr,
  1000=K*60^2
  or,K=5/18.
  Therefore,R=5/18*S^2
  Again,cost of 100 kg=Rs.15
  Cost of 1 kg coal=Rs.15/100
  cost of R i.e, 5/18*S^2=Rs.(5/18*S^2*15/100)
  =(5/18*S^2*15) paise
  Cost of coal for S km=(5/18*S^2*15) paise
  Cost of coal per km=(5/18*S^2*15/S)
  Again,when it is S km,other cost=Rs.12=1200 paise
  Other cost per km=(1200/S) paise
  Therefore,total cost in paise per km
  =(1200/S)+(5/18*S^2*15/S)
  =(1200/S)+(25S/6)
  Hence,3rd option is the right answer.
• 9 years agoHelpfull: Yes(24) No(0)
• C( in paise) = 12*100 + (5/18)*S^2* 15 [for 1 kg coal, 15/100 *100 paise]
  
  C(paise/ km) = 12*100/S + (5/18)*S^2* 15 / S [divide by S]
• 10 years agoHelpfull: Yes(16) No(2)
• Ans: 3rd option: (1200/S)+25S/6;
  verification:
  
  soln: consuption= (1000*15)/100 =>150
  total consumption=150 + extra 12 rupees= 162rupees per 60 km
  Rs.per Km= 162/60 =>2.7 rupees per km => 270 paise per km which satisfy the option..
• 10 years agoHelpfull: Yes(11) No(4)
• @RAKESH
  pls explain this step. i dont get this
  C = 12*100/S + (5/18 *S^2)/S *15
• 10 years agoHelpfull: Yes(5) No(2)
• Abhay pandey....y (1000*15)/100 can u pls explain
  
• 10 years agoHelpfull: Yes(0) No(0)
• hello rakesh
  y 12*100/s
• 10 years agoHelpfull: Yes(0) No(0)
• @ABHAY PANDEY
  How is satisfy the 3rd option
• 6 years agoHelpfull: Yes(0) No(0)