If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours
day, how much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?

• 26metric tonnes.
  Sol: We shall compare each item with the quantity of coal.
  Less engines, less coal consumed (direct)
  More working hours, more coal consumed (direct)
  If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.
  If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.
  Less rate of consumption, less coal consumed (direct).
  Therefore, number of engines 9:8 :: 24:x
  Working hours 8:13 :: 24:x
  Rate of consumption 1/3:1/4 :: 24:x.
  9*8*1/3*x= 8*13*1/4*24 or x= 26.
  Therefore, required consumption of coal 26 metric tonnes.
• 12 years agoHelpfull: Yes(14) No(5)
• 3 engines of former type for one hour consumes 3*24/(9*8)=1 ton
  i.e..4 engines of latter type consumes 1 ton for one hour
  hence 8 engines consumes 2 tons for one hour
  for 13 hours it is 13*2=26
• 11 years agoHelpfull: Yes(14) No(1)
• 26
  
  9engines consumes 24 metric ton for 8hrs
  9 engines for 1hr=3
  1 engine for 1hr=0.33
  
  3former engine = later 4
  1engine consumes for 1hr=0.33
  3 engine former/4 engine later in 1hr = 0.33*3=0.99
  8 engine in 13hrs = 0.99*2*13=26
• 9 years agoHelpfull: Yes(3) No(0)
• three engines of former type for one hour consumes 3*24/(9*8)=1 ton
  i.e.four engines of latter type consumes 1 ton for one hour
  hence 8 engines consumes 2 tons for one hour
  for 13 hours it is 13*2=twenty six.is that okay friends?
• 11 years agoHelpfull: Yes(2) No(2)
• consumption of one engine for 1 hr=x
  for 8 hrs it would be 8x
  for 9 engines 9*8*x=24 (given)
  hence x=1/3
  for latter type consumption of one engine for 1 hr=y
  nw it is given 4*y=3*x hence y=3x/4
  for 13 hrs consumption =13y
  for 8 engines= 8*13*y = 8*13*3*x/4
  put x=1/3 we get 26 metric tonnes
• 9 years agoHelpfull: Yes(1) No(1)
• 9 engines - 8 hr =24 ton coal
  1 engines -1 hr =24/(9*8)=1/3
  we have to find:-
  8 engine -13 hr=?
  given that:-
  3 engine for former(of 8 working hr engin)=4 *engine for latter(of 13 working hr engine)
  4* engine for latter in 1 hr=3* 1/3=>1
  so,
  8 engine -13 hr - 1 day= 2*13=26
  Ans
  26 ton
• 9 years agoHelpfull: Yes(0) No(0)