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What is the value of x*?

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Source: — Data Sufficiency |
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by TRANGZON » Sat Nov 21, 2009 8:13 am
I choose B
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by mehravikas » Thu Dec 03, 2009 7:29 pm
D

because from 1 we get x^2 = 0, therefore x has to be 0.
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by Stuart@KaplanGMAT » Thu Dec 03, 2009 8:49 pm
apoorva.srivastva wrote:For all integers n, n*=n (n-1). What is the value of x*?

(1.) X*=X
(2.) (X - 1 ) * = (X - 2)


OA after some discussion !! :)
Let's actually look at the algebra.

(1) x* = x

We also know that x* = x(x-1), so:

x(x-1) = x

x^2 - x = x

subtracting x from both sides:

x^2 - 2x = 0

x(x-2) = 0

x = 0 OR x = 2... insufficient.

(2) (x-1)* = (x-2)

We also know that (x-1)* = (x-1)(x-1-1) = (x-1)(x-2) = x^2 - 3x + 2, so:

x^2 - 3x + 2 = x - 2

subtracting x from both sides and adding 2 to both sides:

x^2 - 4x + 4 = 0

(x-2)(x-2) = 0

We have a perfect square, therefore x = 2... sufficient. Choose B.
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by thebigkats » Fri Jan 14, 2011 4:31 pm
I reached answer D, please help in why it is wrong:

STAT #1

x*=x
==> x(x-1)=x
==> (dividing both sides by x) (x-1) = 1
==> x = 2
SUFFICIENT

STAT #2
(x - 1 )* = (x - 2)
==> (x-1) (x-2) = (x-2)
==> (dividing both sides by (x-2)) (x-1) = 1
==> x = 2
SUFFICIENT

So why is it B and not D?

thanks
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by Zerks87 » Fri Jan 21, 2011 11:05 am
thebigkats wrote:I reached answer D, please help in why it is wrong:

STAT #1

x*=x
==> x(x-1)=x
==> (dividing both sides by x) (x-1) = 1
==> x = 2
SUFFICIENT

STAT #2
(x - 1 )* = (x - 2)
==> (x-1) (x-2) = (x-2)
==> (dividing both sides by (x-2)) (x-1) = 1
==> x = 2
SUFFICIENT

So why is it B and not D?

thanks
For statement 1 you have to expand the brackets before you can do anything, thus x(x-1) = x expands to x^2 - x = x which simplified to x^2 - 2x = 0 therefore x = 0 and x = 2, N.S.


Hope that helps.
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by diaca » Fri Jan 21, 2011 2:49 pm
So, always for an expression such as x*(x-1)=x, in which I have an x multiplying in both sides, Is it always wrong to cancel the x?
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by schumi_gmat » Sun Jan 23, 2011 10:26 am
diaca wrote:So, always for an expression such as x*(x-1)=x, in which I have an x multiplying in both sides, Is it always wrong to cancel the x?

To divide by x, x should not be equal to 0.

so in this case you cannot divide by x in condition (i)
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by rid » Mon Feb 28, 2011 9:54 am
(1) Insufficient
x=x(x-1) => x = 0 and/or x = 2

(2) Sufficient
(x-1)(x-2)=(x-2) => (x-2)(x-2)=0, x=2
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by madhan_dc » Mon Feb 28, 2011 10:08 am
As per one of the previous posts. I dont understand why we cannot divide each side by x in statement 1. Can anyone please explain this?

i divided both sides by x in statement 1 and got x =2

my answer was D.
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by diaca » Mon Feb 28, 2011 10:27 am
madhan_dc wrote:As per one of the previous posts. I dont understand why we cannot divide each side by x in statement 1. Can anyone please explain this?

i divided both sides by x in statement 1 and got x =2

my answer was D.
You cannot divide by X because we know nothing about X. We don't know if x equals zero. If the statement says that X doesn't equal zero, then you could divide by X.
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by rid » Mon Feb 28, 2011 10:30 am
madhan_dc wrote:As per one of the previous posts. I dont understand why we cannot divide each side by x in statement 1. Can anyone please explain this?

i divided both sides by x in statement 1 and got x =2

my answer was D.
Madhan: division by an unknown X may end up with a division by 0 (zero), which is undefined.
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by tetura84 » Mon Feb 28, 2011 10:52 am
diaca wrote:So, always for an expression such as x*(x-1)=x, in which I have an x multiplying in both sides, Is it always wrong to cancel the x?
It is not wrong, but you have to consider the sign of x, x can be positive or negative.
In your case, you are only considering positive.

x(x-1) = x
then,
x-1 = +/- 1
which equals x = 0,2
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by sushantgupta » Mon Jul 04, 2011 11:40 am
From satatement 1
x* = x(x-1) = x => x(x-2) = 0 hence x = 0 , 2
in sufficient

From statement 2
(x - 1 )* = (x - 2) => (x-1)(x-2) = (x-2) => (x-2)^2= 0
=> x = 2
sufficient.
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by prashant misra » Sun Nov 06, 2011 3:21 am
this was an easy one but i did some calculation mistake.finally i understood.i solved the first expression x*=x(x-1) and putting values as x=x(x-1) and second statement as (x-1)*=(x-1)(x-1-1)
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