An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ???

bhumika.k.shah wrote:An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ??? IMO ,the answer is C.

No its not C harsh.champ wrote:bhumika.k.shah wrote:An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ??? IMO ,the answer is C.

Again another problem that looks difficult but actually is not . a # c = a-c / a+c Now they have given , a # c = 0 therefore, a-c / a+c = o Multiplying both sides with a + c we get a-c = 0 therefore , c = a Answer E Hope it helps

harsh.champ wrote:bhumika.k.shah wrote:An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ??? Ok,I get it. now, a ≠ -c,so a+c ≠ 0,so addition sign ruled out. also a+b ≠ 0 , From the question i can figure out that it is componendo and dividendo, hence a-c/a+c = 0 => a=c. Hence,answer would be E.

Yes! keep it simple ! harsh.champ wrote:harsh.champ wrote:bhumika.k.shah wrote:An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ??? Ok,I get it. now, a ≠ -c,so a+c ≠ 0,so addition sign ruled out. also a+b ≠ 0 , From the question i can figure out that it is componendo and dividendo, hence a-c/a+c = 0 => a=c. Hence,answer would be E.

bhumika.k.shah wrote:An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ??? aθc = 0 (a-c)/(a+c) =0 => a-c =0 a=c E

Yup did it the same way . ajith wrote:bhumika.k.shah wrote:An operation θ is defined by the equation a θ b =a – b / a + b , for all numbers a and b such that a ≠ –b. If a ≠ –c and a θ c = 0, then c = (A) –a (B) –1/a (C) 0 (D)1/a (E) a How to start solving this sum ??? aθc = 0 (a-c)/(a+c) =0 => a-c =0 a=c E