a is eating 27 chocolates in 60 min,b is eating 2 chocalates in 2 min, c is eating 7 chocalates in 20 min. inhow much time they can eat 120 chocalates

• A can eat 27 chocolates in 60 min
  B is eating 2 chocalates in 2 min. means 60 chocolates in 60 min
  C is eating 7 chocalates in 20 min. means 21 chocolates in 60 min
  in 60 mins, all 3 eat 108 chocolates.
  They can eat 120 chocalates in (120*60)/108 = 66.66667.
  After converting decimal value to seconds (i.e. 0.6667*60) it becomes around 66 minutes 40 seconds
  
  
• 12 years agoHelpfull: Yes(27) No(1)
• in 60 minutes A can eat 27 choclates
  in 1 minute A can eat 27/60 choclates
  in 2 minutes B can eat 2 choclates
  in 1 minutes B can eat 2/2 choclates
  in 20 minutes c can eat 7 choclates
  in 1 minutes c can eat 7/20 choclates
  so for a+b+c in 1 minute total no of choclates are (27/60+2/2+7/20)
  so a+b+c can eat 9/5 choclates in 1 minute
  so a+b+c can eat 1 choclate in 1/(9/5) minute
  so a+b+c can eat 120 choclates in (5/9)*120 minute
  =66.66667 minutes so arround 66 mins and 40 seconds
  
• 12 years agoHelpfull: Yes(12) No(0)
• 66mins 40secs.
  in 60 mins, all 3 eat 108 chocolates
  in next 6 mins, the total becomes 118.8
  n in next 40 secs they finish 120 chocs.
  Thus total is 66 mins n 40 secs.
• 12 years agoHelpfull: Yes(3) No(2)
• it can be simply solved by using unitary method
  27 cho.-60min
  60 cho.-60min
  21 cho.-60min
  total 108 cho. are over so for 120...by unitary method
  120*60/108=66.67
• 12 years agoHelpfull: Yes(2) No(1)
• a eats 27 choclates in 60 minutes
  So a will eat 1 choclate in 60/27 minutes
  b eats 2 choclates in 2 minutes
  b will eat 1choclate in 1 minute
  So b will eat 60 choclate in 60 minutes
  c eats 7 choclates in 20 minutes
  c will eat 1 choclate in 20/7 minutes
  So c will eat 21 choclate in 60 minutes
  So a,b and c will simultaneously eat 108(27+60+21)choclates in 60 minutes.
  As choclates will be given full (without breaking into pieces)to each of them.
  So out of remaining 12 choclates we will give 7 choclates to a, 7 choclates to b and 2 choclates to c which they will all finish in minimum time of 7 minutes.
  Thus they can eat 120 choclates in 67(60+7) minutes.
• 12 years agoHelpfull: Yes(1) No(0)
• Answer is 216 minutes
  
• 12 years agoHelpfull: Yes(0) No(10)
• i think it would be 81 minute nearly
• 12 years agoHelpfull: Yes(0) No(8)
• 27= 60 min
  60=60 min (1 per 1 min)
  21=60 min (7 X 3)
  totally 108 choco at 180 min
  to eat 120 =108+12
  ie
  180+12=192 min
• 12 years agoHelpfull: Yes(0) No(3)
• one minute work of
  a=27/60 = 9/20
  b=2/2 = 20/20
  c=7/20
  combined one day work of a, b and c = (9+20+7)/20
  =36/20
  =9/5
  therefore, time req. by them to eat 120 chocolates= (5/9)*120 min.
  = 4000 seconds.
  
• 12 years agoHelpfull: Yes(0) No(0)