The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago. Four years ago, Jennifer, who is a year younger than Margaret, was 3/2 times as old as Abigail. The sum of the ages of Jennifer and Margaret two years, hence, will be:

• 
  let the present age of Abigail=a
  let the present age of Jennifer=j
  let the present age of Margaret=m
  The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago
  that means ,
  (a+j+m)-9=52
  => a+j+m=61------------------(1)
  
  Four years ago, Jennifer, who is a year younger than Margaret,was 3/2 times as old as Abigail
  means,
  m-j = 1 ----------------------(2)
  j - 4 = 3/2(a-4) -----------------------------(3)
  from no 3 equation we can get a = (2j + 4)/3
  now put the value of a in no 1 equation
  and we can get 5j + 3m = 179------------------------(4)
  now from no 2 equation m=1+j
  put ithis into no 4 equation
  and we get j = 22
  and so from equation 2 we can get m = 23
  so The sum of the ages of Jennifer and Margaret two years, hence, will be= 22+23+4 = 49
  
• 7 years agoHelpfull: Yes(5) No(0)
• ans=44
  
  solution:
  Abigail =a
  Jennifer =j
  Margaret=m
  then
  The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago.
  
  (a+j+m)-3=52
  present age is a+j+m=55
  
  Four years ago, Jennifer, who is a year younger than Margaret
  j-4=m+1 =>j-m=5
  
  was 3/2 times as old as Abigail.
  j-4=(3/2)a => -(3/2)a+j=4
  
  after solving a,m,j
  a=13
  j=23.5
  m=18.5
  The sum of the ages of Jennifer and Margaret two years, hence, will be: j+m+2
  =23.5+18.5+2
  =44(ans)
  
• 8 years agoHelpfull: Yes(2) No(4)
• answer is 49
  
• 7 years agoHelpfull: Yes(1) No(1)
• DHILEEP BRO, BUT THE ANSWER IS NOT MATCHING AS YOU SOLVED,THE OPTIONS ARE 51,47,48,49.
• 8 years agoHelpfull: Yes(0) No(2)
• 27
  x+3/2x+3/2x-1=55
  x=8
  m+j=27
  
  
  
• 7 years agoHelpfull: Yes(0) No(1)
• ans=44
  
  solution:
  Abigail =a
  Jennifer =j
  Margaret=m
  then
  The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago.
  
  (a-3 + j-3 + m-3)=52
  present age is a+j+m=61
  
  Four years ago, Jennifer, who is a year younger than Margaret
  j-4=(m-4)+1 =>j-m=1
  
  was 3/2 times as old as Abigail.
  j-4=(3/2) (a-4) => 3a-2j=4
  
  after solving a,m,j
  a=16.5
  j=22.75
  m=21.75
  The sum of the ages of Jennifer and Margaret two years, hence, will be: j+m+2
  =22.75+21.75+2
  =46.5(ans)
• 6 years agoHelpfull: Yes(0) No(1)