Ten years ago,the average of family of four members was 24 years.Three children having been born,the average age of family is same today.What are the present ages of children if two children are identically twins and differ by two years from the younger one?

• ten years ago,
  age of four members=24*4=96
  present total age of those four people=96+40=136
  present average of four people and three new children=24
  hence total sum=7*24=168
  sum of three children=168-136=32let younger member's age be 'z'
  z+(z-2)2=32
  =>z=12
  age of twins=10 each and younger member is of 12 years
• 10 years agoHelpfull: Yes(19) No(6)
• 24 * 7 = 168
  168-4(34)=168-136=32
  32 cant be the sum of the ages of children born within last ten years.
  according to given condition,10+10+8=28.
• 10 years agoHelpfull: Yes(6) No(2)