truck a = 10kg/min
truck b= 131/3kg/min
truck c = 5kg/min
if work simuntaneosuly
how min it take to load 2.4tons load

• clearly 2.4 tons=24oo kg
  A/q,truck is loaded simulaneously,
  thetrefore,in 1 mins,we can load=10+5+131/3=58.6kg
  so 2400kg can be loaded in=2400/58.6= 40.95 minutes
  
• 12 years agoHelpfull: Yes(29) No(1)
• a loaded in 1 min =10kg
  b loaded in 1 min=40/3kg
  c unloaded in 1 min=5kg
  so a,b,c are loaded in 1 min=55/3kg
  so 1 kg loaded by a,b,c in time =3/55min
  now, 2.4 tons loaded by a,b,c in time=(3/55)*2.4*1000=131min
  
• 12 years agoHelpfull: Yes(4) No(3)
• Approx 41 is answer...
  10+131/3+5=176/3=58.66
  so 58.66*41=2400
• 12 years agoHelpfull: Yes(3) No(0)
• Here it is mentioned clearly that one minute work of truck a ---10kg,one minute work of truck b--131/3,one minute work of truck c--5kg.
  let 'x' be the total time taken
  =>10x+(131/3)x+5x=2.4*10^3(since 10x-- denotes number of kg's transported)
  solving 58.667=2.4*10^3
  =>x=40.90minutes
• 12 years agoHelpfull: Yes(2) No(0)
• 450/11 MINS
• 12 years agoHelpfull: Yes(0) No(0)