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Magic: 4 = 3

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aneesh.kg GMAT Destroyer!
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Magic: 4 = 3 Post Sun May 13, 2012 12:46 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Let me show you some magic here.

    Say it's given that
    a + b = c

    Since a = 4a - 3a, b = 4b - 3b, c = 4c - 3c
    Step 1: (4a - 3a) + (4b - 3b) = (4c - 3c)

    Bringing all the 4s on LHS and 3s on the RHS
    Step 2: 4a + 4b - 4c = 3a + 3b - 3c

    Taking 4 common on LHS and 3 on RHS
    Step 3: 4(a + b - c) = 3(a + b - c)

    Cancelling off (a + b - c) on both the sides
    Step 4: 4 = 3


    Woah! Where did I just trick you?

    Spend some time thinking about this. Convince yourself, and then we will discuss a very important concept tested often by the GMAT.

    Dear experts,
    Please just let the others try.
    Thanks.

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    Brent@GMATPrepNow GMAT Instructor
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    Post Sun May 13, 2012 7:52 am
    I'm going to say . . . sorcery Smile

    Cheers,
    Brent

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    Post Mon May 14, 2012 12:52 am
    In step 4, cancelling off is the trick since cancelling off requires division on the two sides by a non zero number, and in the example a+b-c = 0 since a+b=c. Division by 0, gives indeterminable values and "leads to" the crazy "result".

    neelgandham Community Manager
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    Post Mon May 14, 2012 1:14 am
    I concur with Brent - A great magic trick.

    Another one!
    1*0 = 0*0
    1 = 0 (Cancelling 0's from both sides)
    Is it?

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    Post Mon May 14, 2012 5:48 am
    yes, since a+b-c =0 , So, actually it is 4*0=3*0

    Aanderson Just gettin' started! Default Avatar
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    Post Mon May 14, 2012 6:40 am
    lol - here is another one

    (-2)^2 = (2)^2

    taking root on both sides:

    -2 = 2 !!!!!

    0 = 4 !!!!

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    Post Mon May 21, 2012 10:26 am
    You were to come up with some sample usage for this trick Aneesh ...

    aneesh.kg GMAT Destroyer!
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    Post Mon May 21, 2012 10:38 am
    Hi,

    Thanks for reminding me about it.
    Brent, Anil and others: Thanks for resisting the temptation of giving away the trick.
    Aanderson: Your trick was quite amusing, to say the least.
    eagleeye, hey_thr67: you were absolutely correct. We cannot cancel off 0s from LHS and RHS.

    Ok now, let's get down to business.

    Let's start with a few basic DS problems.

    1] What is the value of x?

    (1) x^2 = 2x
    (2) x is an even integer

    2] Is |x| > 1?

    (1) x^3 > x
    (2) |x| = x

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    Post Mon May 21, 2012 1:59 pm
    1] What is the value of x?

    Quote:
    (1) x^2 = 2x
    x*(x-2) = 0
    x = 0 or x = 2. Insufficient!
    Quote:
    (2) x is an even integer
    x can be any even integer. Insufficient!
    Quote:
    From 1 and 2
    x can be 0 or 2. Insufficient

    IMO E

    2] Is |x| > 1?
    The question can be rephrased to Is x>1 or x<-1?
    Quote:
    (1) x^3 > x
    x^3 - x > 0.
    x*(x^2 -1) > 0.
    x*(x-1)*(x+1)>0.
    Let us now check the values which satisfy the inequation.
    If x<-1, then x*(x-1)*(x+1)<0.
    If -1<x<0, then x*(x-1)*(x+1)>0.
    If 0<x<1, then x*(x-1)*(x+1)<0.
    If x>1 then x*(x-1)*(x+1)>0.
    So, If x*(x-1)*(x+1)>0, then -1<x<0 or x>1. Since x can be any number between -1 and 0 or any number greater than 1, statement 1 is insufficient to answer the question.
    Quote:
    (2) |x| = x
    Implies x>0
    Since x can be any value between 0 and 1 and any number greater than 1, statement 2 is insufficient to answer the question.
    Quote:
    1+2
    Intersection of -1<x<0 or x>1 and x>0 is x>1. So 1 + 2 combined is sufficient to answer the question.
    IMO C

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