ab+a+b=250
bc+b+c=300
ca+a+c=216
find a+b+c?

• 2(ab+bc+ca)+2(a+b+c)=766
  2(250-a-b+300-b-c+216-a-c)+2(a+b+c)=766
  1532-2(a+b+c)=766
  2(a+b+c)=766
  a+b+c=383
• 11 years agoHelpfull: Yes(21) No(21)
• I don't know what's the use of asking these types of questions in an aptitude test...
  Well, This problem can be solved in two methods (you are free to choose whichever you like...)
  Note:Firstly, these type of equations don't have any exact solutions. (Sadly,this one is also not one of those exceptional types!!)Hence, solution to this equation are only approximates...
  
  Method 1 :(The Conventional Way!!)
  ab + a + b = 250
  a(1+b) + b = 250
  a = (250−b)/(1+b)
  
  ac + a + c = 300
  c(1+a) + a = 300
  c = (300−a)/(1+a)=(300−(250−b)/(1+b))/(1+(250−b)/(1+b))
  = (300(1+b)−(250−b))/1+b+250−b
  = (300(1+b)−(250−b))/251
  =(50+301b)/251
  =(50+301b)/251
  
  bc + b + c = 216
  ((50+301b)b/251)+b+((50+301b)/251)=216
  (50 + 301b)b + 251b + 50 + 301b = 54216
  301b^2 +602b−54216 =0
  b=12.46
  
  a =(250−b)/(1+b)=(250−12.46)/(1+12.46) = 17.65
  c = (300−a)(1+a)=(300−17.65)/(1+17.65) = 15.14
  
  These values of a,b and c statisfies the given conditions,as given below
  ab + a + b = 17.65 *12.46 + 17.65 + 12.46 ≈ 250
  ac + a + c = 17.65 *15.14 + 17.65 + 15.14 ≈ 300.21
  bc + b + c = 12.46 *15.14 + 12.46 + 15.14 ≈ 216
  
  Hence,
  a + b + c = 17.65 + 12.46 + 15.14 = 45.25 [This may be one solution!! But remember there are many approximations which I have done here!!)
  
  Method 2: (Inequality approximation method)
  Take the three equations,
  ab+a+b=250.....(1)
  bc+b+c=300.....(2)
  ac+a+c=216.....(3)
  
  add them up, we get,
  2(a+b+c)+ab+ac+bc=766....(4)
  a+b+c= (766-(ab+ac+bc))/2....(5)
  
  So, we are getting some form here,but once again solving this equation is hard as we don't know the value of ab,bc,ac here.
  
  So, apply A.M.-G.M. Inequality (If you don't know..read them up in wikipedia!!)
  Take equation (1),apply AM-GM
  (ab+a+b)/3 >=(ab)^(2/3)
  put the value of ab+a+b=250.
  so, ab =< (+(250/3)^(1/2))^3 or (-(250/3)^(1/2))^3.........(6)
  Similarly for equation 2 and 3 we see that 300 and 216 are both divisible by 3,so we get whole numbers!!
  then,
  bc =< (+(100)^(1/2))^3 or (-(100)^(1/2))^3......(7)
  ac =< (+(72)^(1/2))^3 or (-(72)^(1/2))^3......(8)
  
  Using Equation 5 and 6,7,8. Plug in combination of values,you get the solutions for a+b+c.
  
  Please note: This kind of questions are purely options dependent. Try to nullify options.The second method is way faster than the first one...(according to me atleast!! but you are free to choose whatever you like!!)
  
  I would appreciate if anyone comes up with a more novel solution to this problem..
• 11 years agoHelpfull: Yes(4) No(28)
• ab is not a*b but a two digit no ab is a*10+b
  so there will be three eq
  11a+2b=250
  11b+2c=300
  11c+2a=216
  solving we get
  a=238/13
  b=316/13
  c=212/13
  
• 11 years agoHelpfull: Yes(4) No(0)
• 0.
  (ab+bc+ca)+2(a+b+c)=766
  (250-a-b+300-b-c+216-a-c)+2(a+b+c)=766
  
• 11 years agoHelpfull: Yes(3) No(0)
• is there any shortcut method to solve this one???? my method is also getting very lengthy..
• 11 years agoHelpfull: Yes(2) No(2)
• bloody soumik
• 11 years agoHelpfull: Yes(1) No(3)
• we know that
  A.M>=GM
  a+b/2>=sqrt(ab)
  or
  a+b=2*sqrt(ab)
  or
  ab+2*sqrt(ab)=250
  then
  ab=15.6(approx)
  bc=17.3(approx)
  ca=14.7(approx)
  or
  2*(a+b+c)=250-15.6+300-17.3+216-14.7
  or
  a+b+c=359.2(approx)
  is it rght???
• 11 years agoHelpfull: Yes(1) No(1)