At a reception, one-third of the guests departed at a certain time.Later two-fifths of the guests departed. Even later two-thirds of the remaining guests departed. if six people were left, how many were originally present at the perty?

• Let Original Member be x
  First One third guest departed i.e x/3
  Remaining guest x-(x/3) = 2x/3
  Now from Remaining (2x/3) two-fifths departed(4x/15)
  i.e. Nw Remaining guest will be (2x/3 - 4x/15 = 2x/5)
  Now from Remaining (2x/5) two-thirds departed(4x/15)
  Nw remaining (2x/5 - 4x/15)= 2x/15
  which is equal to 6 (according to question)
  2x/15=6 which gives x=45
• 10 years agoHelpfull: Yes(37) No(0)
• MR goyal how can u say that , "Later two-fifths of the guests departed." in this statement remaining people left.
  It may be 2/5th of whole.
• 10 years agoHelpfull: Yes(4) No(1)
• answer is 45 people were originally present. Solve considering initial number of people to be 'x'.
• 10 years agoHelpfull: Yes(3) No(0)
• 45 people were present when party begins..
• 10 years agoHelpfull: Yes(3) No(3)
• x is the number of guests
  then 1/3(3/5(2/3(x)))=6
  x=45
• 10 years agoHelpfull: Yes(2) No(0)
• @gaurav kumar :1/3 is less than 2/5. so it is 2/5 of remaining guests only..
• 10 years agoHelpfull: Yes(1) No(0)
• can u pls xplain
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• 45
  x-x/3=(2/3)x
  (2/3)x-(4/15)x=(2/5)x
  (2/5)x-(4/15)x=(2/15)x
  (2/15)x=6
  x=45
• 6 years agoHelpfull: Yes(0) No(0)