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Algebraic expression.

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by bhumika.k.shah » Sat Feb 06, 2010 2:26 am
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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by harsh.champ » Sat Feb 06, 2010 2:52 am
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.
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by bhumika.k.shah » Sat Feb 06, 2010 2:54 am
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.
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by bhumika.k.shah » Sat Feb 06, 2010 2:57 am
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 :)
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componendo and dividendo corollary

by harsh.champ » Sat Feb 06, 2010 3:03 am
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.
Last edited by harsh.champ on Sat Feb 06, 2010 3:07 am, edited 1 time in total.
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by bhumika.k.shah » Sat Feb 06, 2010 3:05 am
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.
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by ajith » Sat Feb 06, 2010 3:18 am
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
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by bhumika.k.shah » Sat Feb 06, 2010 3:22 am
Yup did it the same way . :D
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
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