A fruit-seller sells half of his fruits plus 4 oranges; again he sells half of what remains plus 4 oranges; and so again a third time. His store is now exhausted; How many did he start with?

• let f is the total no. of fruit then in first time he sold=f/2+4 and remaining are f/2-4
  second time he sold (f/2-4)/2+4 and remaining are [(f/2-4)/2]/2-4;
  so total fruits are first time +second time+third time=total
  (f/2+4)+(f/2-4)/2+4+[(f/2-4)/2]/2-4=f
  after solving
  f=56.
• 11 years agoHelpfull: Yes(2) No(1)
• the oranges are also a part of all fruits.
  so everytym we take half of remaining fruits, we have to subtract the extra four oranges given.
  therefore, x/2 +4 + (x-8)/4 +4 + (x-24)/8 +4 = x..
  solving, 8x=7x+96-40
  so, x=56... solved...
• 11 years agoHelpfull: Yes(1) No(2)
• let the total fruits is x.
  so after first time remaining fruits is = x/2 - 4
  = (x-8)/2
  
  after second time remaining fruits is = ((x-8)/2)/2 -4
  = (x-8)/4 - 4
  = (x-8-16)/4
  = (x-24)/4
  
  after third time remaining fruits is = ((x-24)/4)/2 - 4
  = (x-24)/8 - 4
  = (x-24-32)/8
  = (x-56)/8
  
  and given that after third time store is exhausted so
  
  (x-56)/8 = 0
  so ans is x=56
  
  
  
• 11 years agoHelpfull: Yes(1) No(0)