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 ???
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Algebraic expression.
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bhumika.k.shah
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harsh.champ
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IMO ,the answer is C.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 ???
-
bhumika.k.shah
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No its not C
harsh.champ wrote:IMO ,the answer is C.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 ???
-
bhumika.k.shah
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- Joined: Sun Dec 27, 2009 12:28 am
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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
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
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harsh.champ
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harsh.champ wrote:Ok,I get it.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 ???
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.
Last edited by harsh.champ on Sat Feb 06, 2010 3:07 am, edited 1 time in total.
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bhumika.k.shah
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Yes!
keep it simple !
keep it simple !
harsh.champ wrote:harsh.champ wrote:Ok,I get it.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 ???
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.
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ajith
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aθc = 0bhumika.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 ???
(a-c)/(a+c) =0
=>
a-c =0
a=c
E
Always borrow money from a pessimist, he doesn't expect to be paid back.
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bhumika.k.shah
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Yup did it the same way . 
ajith wrote:aθc = 0bhumika.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 ???
(a-c)/(a+c) =0
=>
a-c =0
a=c
E
