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Algebraic expression.

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bhumika.k.shah Legendary Member Default Avatar
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Algebraic expression.

Post 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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ajith Legendary Member
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Post 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

_________________
Always borrow money from a pessimist, he doesn't expect to be paid back.

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bhumika.k.shah Legendary Member Default Avatar
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Post Sat Feb 06, 2010 3:22 am
Yup did it the same way . Very Happy

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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harsh.champ Legendary Member
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Post 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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bhumika.k.shah Legendary Member Default Avatar
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Post 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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bhumika.k.shah Legendary Member Default Avatar
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Post 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

Smile

Answer E

Hope it helps Smile

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harsh.champ Legendary Member
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Post Sat Feb 06, 2010 3:03 am
[quote="harsh.champ"]
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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bhumika.k.shah Legendary Member Default Avatar
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Post Sat Feb 06, 2010 3:05 am
Yes!

keep it simple !

Smile

[quote="harsh.champ"]
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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