There are 3 societies A, B, C having some tractors each. A Gives B and C as many tractors as they already have. After some days B gives A and C as many tractors as they have. After some days C gives A and B as many tractors as they have. Finally each has 24 tractors. What is the original Number of tractors each had in the beginning? Ans. A ?39. B? 21. C? 12.

• Let us take in the reverse order,
  
  all A, B & C each has 24 remaining at the end.
  
  A B C
  24 24 24
  
  before C gave to A & B the had
  A B C
  12 12 24+12+12=48
  
  now before b gave A & C the same amount of tractor they already had
  
  A B C
  6 12+6+24=42 24
  
  Now befor A gave B & C, they must had
  
  A B C
  6+21+12=39 21 12
  
  A=39, B=21, C=12
• 10 years agoHelpfull: Yes(21) No(1)
• A B C
  24 24 24 (at the end)
  12 12 24+12+12
  6 12+6+24 24+12+12
  6+12+6+12+3 12+6+3 12
• 11 years agoHelpfull: Yes(1) No(4)
• before starting t solve this question i would like to say you should have t know about backtracking technique so,
  
  last
  A=24 B=24 c=24
  okay
  now just see last c option means last person who give tractors okay
  so last one is c
  after c gave tractors how much he have
  so divide a/2=12
  b/2=12
  now add
  C=12+12+24;
  C=48
  now sales method on B
  a/2=6
  c/2=24
  B=6+12+24=42;
  now again use this method
  for A
  b=42/2=21
  c=24/2=12
  A=12+21+6=39
  
  now initial they
  have
  A=39 B=21,C=12
  
  hope you get your answers please like and share
• 6 years agoHelpfull: Yes(1) No(0)