Proffessor Ms.Rotunda and her predecessor,retired proffesor Dr.Calcus discoverd that the product of their ages is equal to the sum of their ages multiplied by 30,the age of their student who joined the faculty recently.Ms.Rotunda is 11 years older than Dr.calcus find the sum of ages of the three

• Given age of the student-30
  Let age of Ms.Rotunda & Dr. Calculus - x & y
  Now,condition->
  1st - xy=(x+y)*30
  2nd - x=y+11
  
  then,
  (y+11)(y)=(2y+11)*30
  => y^2+11y=60y+330
  => y^2-49y-330=0
  => y^2-55y+6y-330=0
  => y(y-55)+6(y-55)=0
  => (y+6)(y-55)=0
  => y=55(age can't be -ve)
  Dr. Calculus-> 55 years
  Ms. Rotunda-> 66 years
  Student -> 30 years
  
  Sum of the Age -> 151 years
• 10 years agoHelpfull: Yes(12) No(1)
• 1st condition: xy=(x+y)*30
  
  2nd condition: x=y+11
  
  Now,
  (y+11)(y)=(2y+11)*30
  => y^2+11y=60y+330
  => y^2-49y-330=0
  => y^2-55y+6y-330=0
  => y(y-55)+6(y-55)=0
  => (y+6)(y-55)=0
  => y=55 as the age can not be negative,
  
  Therefore the age of Dr. Calcus is: 55 years and that of Ms. Rotunda will be 66 years. I am confused about the illicit entry of their student in the question.
• 10 years agoHelpfull: Yes(8) No(1)
• Let age of Dr. calcus be x,
  age of Ms.Rotunda be x+11(given that age of ms Rotunda is 11 years more than Dr. calcus)
  ,and the age of the faculty who joined recently is 30(given)
  a/q
  x(x+11)=(x+x+11)30
  x^2+11x=60x+330
  x^2-49x-330=0
  (x-55)(x+6)=0
  x=55
  now,the sum of all the ages are 55+66+30=151 anss..
• 8 years agoHelpfull: Yes(1) No(0)
• plz giv me ans for this qtn?
• 10 years agoHelpfull: Yes(0) No(1)