For any two numbers we define an operation $ yielding another number, X $ Y such that following condition holds:
• X $ X = 0 for all X
• X $ (Y $ Z) = X $ Y + Z
Find the Value of 2012 $ 0 + 2012 $ 1912
Options
o 2112
o 100
o 5936
o Can not be determined

• here $ represent the - operator so X-X=0 first condition
  x$(y$z)=x-(y-z)=x-y+z
  it can be written as x$y+z
  2012-0+2012-1912=2112 so ans is 2112
• 12 years agoHelpfull: Yes(96) No(1)
• 2112 just replace $ with -
• 12 years agoHelpfull: Yes(4) No(2)
• when you take $ as + , you will be getting the answer as 5936 . As they have given in the question that for all values of x, so x can also be zero . if x is 0 , X+X=0 .
  
• 10 years agoHelpfull: Yes(2) No(1)
• operator so X-X=0 first condition
  x$(y$z)=x-(y-z)=x-y+z
  it can be written as x$y+z
  2012-0+2012-1912=2112 so ans is 2112
• 10 years agoHelpfull: Yes(2) No(0)
• Replace $ to -
  You will get the ans
  
• 12 years agoHelpfull: Yes(1) No(0)
• x$x=0
  so,x-x=0
  so, 2012-0+2012-1912=2112
• 9 years agoHelpfull: Yes(1) No(0)
• we can take only x-x=0, x+x=0 this condition is true only possible if x=0.
• 9 years agoHelpfull: Yes(0) No(2)
• if we take $='+'
  then value will be (2012+0+2012+1912=5936)
  when we take $='-'
  then value will be (2012-0+2012-1912=2112)
  i think both ans r correct
  
• 8 years agoHelpfull: Yes(0) No(1)
• X $ X = 0 ie: X - X = 0
  X$(Y$Z) = X$Y+Z ie: X - (Y - Z)
  X - Y + Z = X$Y+Z
  2012$0+2012$1912 = 2012 - 0 + 2012 - 1912 = 2112
• 8 years agoHelpfull: Yes(0) No(0)