Asked in TCS 08-03-2014 off campus test

a^2*b^2+a^2+b^2=259
b^2*c^2+b^2+c^2=1299
c^2*a^2+c^2+a^2=499
then find a+b+c=?

• a^2*b^2+a^2+b^2+1= (a^2+1)(b^2+1)
  by using above one we can write
  a^2*b^2+a^2+b^2+1=260..(1)
  b^2*c^2+b^2+c^2+1=1300..(2)
  c^2*a^2+c^2+a^2+1=500..(3)
  by doing (1)/(2) we will get a^2+1/c^2+1 = 1/5..(4)
  by doing (4)*(3) we will get (a^2+1)^2=10,
  by solving a=3, b=5,c=7
  so a+b+c=15
  
• 10 years agoHelpfull: Yes(23) No(4)
• In this type question you can choose a option that is most probable from the options.I just choose the option b) i.e.15 and break it into a=3,b=5,c=7 than i put these values in these 3 equations and these values are satisfying the all of these equations so answer is
  (a+b+c)=15.
  Remember one thing always if you have min 3 to 5 mins than only you can apply this method ...
• 10 years agoHelpfull: Yes(3) No(1)
• Hint (x^2-1)(y^2-1)= (xy)^2 + x^2 + y^2 +1
  so we get
  a^2*b^2+a^2+b^2=259 => (a^2-1)(b^2-1)= 259+1=260 ... (1)
  b^2*c^2+b^2+c^2=1299 => (b^2-1)(c^2-1)= 1299+1 = 1300 ... (2)
  c^2*a^2+c^2+a^2=499 => (c^2-1)(a^2-1)= 499+1 = 500 ... (3)
  
  Divide (1)/(2) we get (a^2-1)/(c^2-1)=1/5 ....(4)
  Multiply (4) * (3) and solve for a
  divide (4) / (3) and solve to get c
  and den solve for b
  Then you'll get a+b+c value
  Remember... if it maybe a dummy question so you may not find the correct answer in the options
• 10 years agoHelpfull: Yes(2) No(5)
• put get a=3,b=5,c=7
  so a+b+c=15
• 10 years agoHelpfull: Yes(1) No(4)