if (a+b)^2=25 and (a-b)^2=1
1)a is prime
2)modulus of a>2 and a is a natural number
a)
b)
c)
d)

• (a+b)^2=25 so (a+b)=+-5
  (a-b)^2=1 so (a-b)=+-1
  solve these equations
  a+b=+5,a-b=+1 then a=3,b=2
  a+b=+5,a-b=-1 then a=2,b=3
  a+b=-5,a-b=-1 then a=-3,b=-2
  a+b=-5,a-b=+1 then a=-2,b=-3
  so a=-2,-3,2,3
  1)a is prime is incorrect(negative no. cant be prime)
  2)modulus of a>2 and a is a natural number is incorrect(-2,-3 is not natural)
  both are incorrect statement
• 11 years agoHelpfull: Yes(121) No(6)
• here a can take 4 values i.e. 3, -3, 2, -2,,, on solving the equations with all the possibilities,,, now considering the frst case a is prime thus a can be either 2 or 3 as -ve numbers cant be prime so two value exists,,, but not giving the unique solution,,,, now considering the second case |a|>2 thus |-3|>2 and also |3|>2 thus again two values exists,,,, now considering both the statements a is prime and |a|>2 here we getting only one value is |3|>2 and also 3 is a prime number thus it can be answered using both
• 11 years agoHelpfull: Yes(8) No(2)
• (a+b)^2=25 and (a-b)^2=1
  by solving a=3,2,-3,-2
  1) a is prime then a= 2,3
  2) A) |a|> 2 i.e. a= 3,-3
  
  B) a is natural number i.e. a=2,3
  from A) AND B) a=3
  Statement 2) alone is sufficient
  
• 11 years agoHelpfull: Yes(8) No(0)
• either statement is sufficient
• 11 years agoHelpfull: Yes(4) No(9)
• a can be either -2, -3, 2 or 3
  So, none of the conclusion are true
  
  The possibilities for (a,b) are (2,3)(3,2)(-2,-3)(-3,-2)
  
  so, none of the conclusions are definete
  
  Answer is none follows
• 11 years agoHelpfull: Yes(2) No(4)
• @mahesh : -3,-2 are not natural no.
• 11 years agoHelpfull: Yes(2) No(0)
• since a can take negative values like -2,-3
  it cant be prime as well as natural.
  so both statements are incorrect
• 11 years agoHelpfull: Yes(1) No(0)
• by solving above two equation we will get four set of values
  (a=13,b=12),(a=12,b=13),(a=-12,b=-13),(a=-13,b=12)
  again a is prime no and a is also natural no so
  a can have only one value out of 4 ie a=13
  so b=12..........
• 11 years agoHelpfull: Yes(1) No(6)
• (a+b)^2=25========>a^2 + b^2 + 2ab=25...(1)
  (a=b)^2=1=========>a^2 + b^2 - 2ab=1....(2)
  solve above 2 eqn we get
  a^2+ b^2 = 13 in that eqn we put a=3 and b=2 get the value....and anssss c part
  
  
  
  
  
  
  
• 11 years agoHelpfull: Yes(1) No(0)
• a is prime
  
• 11 years agoHelpfull: Yes(0) No(2)
• a is prime
• 11 years agoHelpfull: Yes(0) No(2)
• a is a prime
• 11 years agoHelpfull: Yes(0) No(2)
• a is a prime number
• 11 years agoHelpfull: Yes(0) No(1)
• a is prime
• 11 years agoHelpfull: Yes(0) No(3)
• both are incorrect as a=2,-2,3,-3
  (which is nt a prime no. n a natural no.)
• 11 years agoHelpfull: Yes(0) No(0)
• both are sufficient
  from 1st statement we get a=1,3,-1,-3
  and in second statement it is given that modulus of a>2 so a=3 or-3 and it is also given that a is natural no. so a=3
• 10 years agoHelpfull: Yes(0) No(0)
• a is a prime
• 10 years agoHelpfull: Yes(0) No(1)
• The question is:
  f (a+b)^2=25 and (a-b)^2=1
  1)a is prime number.
  2) |a|>2 and a is an integer
• 5 years agoHelpfull: Yes(0) No(0)