If p=2q then q=r*r , if p-odd then q is even, whether r is even or odd ?
a) First condition is sufficient
b) Second condition is sufficient
c) Both are sufficient
d) Both are not sufficient

• d)both are not sufficient
  
• 10 years agoHelpfull: Yes(33) No(0)
• there is no way for p to be odd.p=even and so q=odd.
  when q must be odd then r must be odd.
  ANS:c
• 10 years agoHelpfull: Yes(6) No(2)
• b
  Because 2* any no. is even then P is even
  r*r their r any no.
  because of any no. multi. is even no.
  
• 10 years agoHelpfull: Yes(5) No(6)
• 1)p=2q then q=r*r
  2)p-odd then q is even
  consider r=3(odd),the r^2=9(answer is odd)
  consider r=4(even),the r^2=16(answer is even)
  so Both are sufficient
• 9 years agoHelpfull: Yes(4) No(2)
• if q=r*r and p=2q p is even in any case r even or odd
  
  so only first condition sufficient
• 10 years agoHelpfull: Yes(3) No(6)
• cleary both are not sufficient as if we take r as even then q will be even and p will also be even at the same time.hence can't be true.
• 10 years agoHelpfull: Yes(3) No(0)
• both are not sufficient
  
• 10 years agoHelpfull: Yes(2) No(1)
• d)both are not sufficient
  as p is anyway even but q and r may b even or odd both.
• 10 years agoHelpfull: Yes(2) No(1)
• C option is the best because there is no way for p to be odd.p=even and therefore q=odd.
  when q must be odd then r must be odd. :)
  
• 10 years agoHelpfull: Yes(1) No(2)
• d)clearly both are not sufficient.
  if we multiply any number(whether odd or even)from 2, we always get an even number .
• 10 years agoHelpfull: Yes(1) No(0)
• b is sufficient.
  
  q=r*r
  r=sqrt(q)
  so if q is even then square root of q is also even
• 9 years agoHelpfull: Yes(0) No(1)
• C option is the best because there is no way for p to be odd.p=even and therefore q=odd.
  when q must be odd then r must be odd. :)
  
• 9 years agoHelpfull: Yes(0) No(1)
• the second condition is sufficient
• 4 years agoHelpfull: Yes(0) No(0)
• both are suffecient
• 4 years agoHelpfull: Yes(0) No(0)
• both are sufficient
• 7 Months agoHelpfull: Yes(0) No(0)