A stationary engine has enough fuel to move 12 hours when its tank is 4/5 full, how much
hours will it run when the tank is 1/3 full

• when engine is (4/5) full,it moves for 12 hr
  so, when tank is full, it will move = 12/(4/5) = 15 hr
  
  So, when it is 1/3 full, it will move = 15*(1/3)= 5 hours
• 10 years agoHelpfull: Yes(46) No(0)
• Ans: 5hrs
  fuel and time is directly propotional
  4/5fuel=12 hrs
  1fuel=12*5/4=15hrs
  1/3 fuel= 15/3= 5hrs
  
• 10 years agoHelpfull: Yes(6) No(0)
• 12----->4/5
  x------->1/3
  
  12/x = (4/5)/(1/3) =>12*(1/3)=x*(4/5)
  =>x=5hrs
• 10 years agoHelpfull: Yes(3) No(0)
• ans is 5hr..
  if 4/5 takes ---->12hr
  1/3 takes----->
  12*(1/3)/4/5=5
• 10 years agoHelpfull: Yes(2) No(0)
• when engine is (4/5) full,it moves for 12 hr
  so, when tank is full, it will move = 12/(4/5) = 15 hr
  
  So, when it is 1/3 full, it will move = 15*(1/3)= 5 hours
• 10 years agoHelpfull: Yes(2) No(0)
• Ans. 5 hours
  
  4/5 full tank has fuel to move engine for 12 hours
  full tank has fuel to move engine for 12*5/4 hours
  1/3 full tank has fuel to move engine for 12*5*1/(4*3) hours = 5 hours
• 10 years agoHelpfull: Yes(0) No(0)
• 12 hours in 4/5th of tank , so the engine will run for 15 hours in full tank.
  for 1/3rd of tank, the engine will run for 5 hours.
• 10 years agoHelpfull: Yes(0) No(0)
• when tank is 4/5 full, engine goes for 12 hrs
  1 full, engine goes for 12*5/4 hrs
  1/3 full, engine goes for 12*5/4*1/3=5hrs
• 10 years agoHelpfull: Yes(0) No(0)
• 5 hours is correct
  
• 10 years agoHelpfull: Yes(0) No(0)
• it will take 5 hours
• 10 years agoHelpfull: Yes(0) No(0)
• 4/5 ---->12 hrs
  1/3----->x hrs
  
  x*4/5=1/3*12
  
  i.e x=5 hrs
  
• 10 years agoHelpfull: Yes(0) No(0)
• ans is 5 hours
  solve it by unitary method.
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• 4/5:1/3::12:x
  12*x=12*5
  then x=5
• 10 years agoHelpfull: Yes(0) No(0)
• 5 hrs is the answer
• 9 years agoHelpfull: Yes(0) No(0)