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Algebraic expression.

This topic has 7 member replies
bhumika.k.shah Legendary Member Default Avatar
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Algebraic expression.

Post Sat Feb 06, 2010 2:26 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    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.

    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.

    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

    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

    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.

    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.

    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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