IF x and y are real then is x positive?
1)x^2m and y^2n have same sigh given both m and n are positive integers.
2)Both x^m and Y^n are positive, Given m and n are consecutive integers.
DATA SUFFICIENCY

• Both the statements are not suffient.,.
  1. x^2m will always be postive.
  2. if m and n are consecutive integers,we cant conclude that x is positive.
  for ex-
  x=-3 and m=2,n=3
  if this case -3 ^ 2 will be positive whereas we can see that its not positive
• 11 years agoHelpfull: Yes(32) No(3)
• Both r nt sufficient.first is of no use.in second eg.x is -2, m is 4, then xto power m is +, also if x is 2 we get the same rsult.how can v determine it den.y can b anything too.
• 11 years agoHelpfull: Yes(8) No(4)
• The 2nd statement alone is enough to answer this question and not both the statements alone because, (+x)^2m will also give a +ve number and (-x)^2m will also give a positive no. And how can you conclude whether x is +ve or _ve from the first statement. All you can conclude is X^2m is +ve and not x itself
• 11 years agoHelpfull: Yes(4) No(7)
• statement2 is sufficient to answer this question.
• 11 years agoHelpfull: Yes(4) No(3)
• acc to vidushi answer it can be concluded that both the statements are alone sufficient to answer the question.since any no to even power is always positive that 1 will hold true in every case
• 11 years agoHelpfull: Yes(3) No(2)
• since in 2 it is written x^m and y^n are positive
  means x^n is positive
  in only one condition when x is postive
• 11 years agoHelpfull: Yes(3) No(1)
• So can Glued can we conclude that statement 2 alone is sufficient to answer the question but not statement 1???
• 11 years agoHelpfull: Yes(3) No(0)
• both r not sufficient as if x=neg no the 1 cond equals x^2m which is square which will make it positive
  while if x is pos it too remains positive
  while in 2nd condition m is not told if it is even or odd if even we cant determine the no x is positive or negative
  so it is data insufficient
  
• 11 years agoHelpfull: Yes(3) No(2)
• let m=3 and n=4(consecutive nos. given)
  
  bcoz;x^m and y^n are +ve this implies x cant be negative;thus x is positive
• 11 years agoHelpfull: Yes(2) No(5)
• can't be say anything
  
  both statement is not sufficient
  
• 11 years agoHelpfull: Yes(1) No(0)
• I suppose so..
• 11 years agoHelpfull: Yes(0) No(0)
• jaspreet is right in this sense
• 11 years agoHelpfull: Yes(0) No(0)
• yahpp..its right..both are insufficient to answer.
• 9 years agoHelpfull: Yes(0) No(5)
• right @nitish nd @vidushi............only 2nd statement is sufficient to give answer
  nd we can not answer by 1st statement
• 9 years agoHelpfull: Yes(0) No(1)
• both are not sufficient as in second case (-2)^2 gives A +ve answer but x is not +ve....means x^m is positive but x is not
• 7 years agoHelpfull: Yes(0) No(0)