Is X^2 divisible by 8?
(a)If X^3 divisible by 16.
(b)X divisible by 18.

• From (a)
  as x^3 is divisible by 16, hence x^3 is even and so x must be even, therefore we can write x^3/16=(2*2*2*something)/16 or (8*something)/16 or (something)/2 which is equal to zero as x^3 is divisible by 16. Therefore x is atleast a multiple of 4. Hence it is clear that x^2 is divisible by 8.
  From (b)
  x is divisible by 18, we can only guess x is an even number, nothing else. x^2 may or may not be divisible by 8.
  So answer is (a) alone is sufficient
• 8 years agoHelpfull: Yes(13) No(0)
• a) for ex take 4 or 8 as x and verify both a and b only a is true
• 8 years agoHelpfull: Yes(8) No(0)
• To proof: x^2=a*8(multiple of 8)
  
  checking i). x^3=b*16
  take cube root and then square both sides
  x^2=2^(8/3) * b^(2/3)
  which is not a multiple of 8 certainly.
  
  checking ii). x=c*18=2*3^2*c
  square both sides
  x^2=4*3^4*c^2
  which is not always a multiple of 8 .for it c should be even.
  taking both
  x^3/x= x^2=8*(b/9c)=8*k
  
  BOTH statements (1) and (2)
  TOGETHER are sufficient to
  answer the question asked,
  but NEITHER statement
  ALONE is sufficient to answer
  the question asked
• 8 years agoHelpfull: Yes(2) No(9)
• Only a) is correct:
  Ex: Take x=4,
  if x^3 =64, IS divisible by 16;
  than x^2= 16, Is divisible by 8;
  
  So a alone is sufficient...!
• 8 years agoHelpfull: Yes(2) No(2)
• Both of the statement are sufficient to answer the question.
  (Note: it is not given that X is an integer)
  So X can be irrational as well
  
  In the first case we can have a number like 2*(2)^(1/3).
  
  But when we combine both the statements, we get that X^2 must be divisible by 8.
  
  Had it been given that X is an integer then statement 1 would have been enough.
• 8 years agoHelpfull: Yes(0) No(3)
• Both statement must to ans
• 8 years agoHelpfull: Yes(0) No(0)