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Magic: 4 = 3

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Magic: 4 = 3

by aneesh.kg » Sat May 12, 2012 11:46 pm
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.
Aneesh Bangia
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Source: — Quantitative Reasoning |

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by Brent@GMATPrepNow » Sun May 13, 2012 6:52 am
I'm going to say . . . sorcery :-)

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by eagleeye » Sun May 13, 2012 11:52 pm
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".
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by neelgandham » Mon May 14, 2012 12:14 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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by hey_thr67 » Mon May 14, 2012 4:48 am
yes, since a+b-c =0 , So, actually it is 4*0=3*0
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by Aanderson » Mon May 14, 2012 5:40 am
lol - here is another one

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

taking root on both sides:

-2 = 2 !!!!!

0 = 4 !!!!
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by hey_thr67 » Mon May 21, 2012 9:26 am
You were to come up with some sample usage for this trick Aneesh ...
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by aneesh.kg » Mon May 21, 2012 9:38 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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by neelgandham » Mon May 21, 2012 12:59 pm
1] What is the value of x?
(1) x^2 = 2x
x*(x-2) = 0
x = 0 or x = 2. Insufficient!
(2) x is an even integer
x can be any even integer. Insufficient!
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?
(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.
(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.
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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