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What is the value of x*?

Expert replies

by him1985 » Tue Mar 27, 2012 11:16 pm
B is sufficient.

using 1) we find two values for x : X=0,2
using 2) we find single value for x: X=2

So only 2nd is sufficient, 1st is not sufficient.
Himanshu Chauhan
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by ronnie1985 » Thu Mar 29, 2012 10:13 am
(B) QED
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by MQ0451 » Mon Apr 16, 2012 9:45 am
short cut for the OA when solving 2.

n* = n(n-1)
(x-1)*=x-2
(x-1)(x-2)=x-2 <- don't distribute, just divide each side by x-2
x-1=1
x=2
Sufficient.
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by chris558 » Sun Sep 09, 2012 8:51 am
1) x(x-1)=x
x^2-x=x
x^2-2x=0
x(x-2)=0
x=0 or (x-2)=0 -> x=2
x=0 or 2, INSUFFICIENT

2) (x-1)(x-1-1)=x-2
(x-1)(x-2)=x-2
(x^2-3x+2)=x-2
x^2-4x+4=0
(x-2)(x-2)=0
x=2
SUFFICIENT

Answer is B
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by rajeshsinghgmat » Thu Mar 21, 2013 3:47 am
D in Answer.

x-1=1

x=2
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by Java_85 » Sun Oct 13, 2013 2:27 pm
B, Good question :)
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by swap-2 » Thu Jan 30, 2014 4:20 am
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 !! :)
OA- C
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by jaspreetsra » Thu Dec 04, 2014 1:48 pm
DATA SUFFICIENCY QUESTION:
For all integers n, n* = n(n - 1). What is the value of x*?

1. x* = x (not sufficient)
2. (x - 1 )* = (x - 2) (sufficient)
Explanation:
(x - 1 )* = (x-1)(x-1-1)=(x-1)(x-2)
From (2)
(x - 1 )* = (x - 2)
=>(x-1)(x-2) = (x - 2)
or
(x-1) = 1
x=2

So, B.
What is right answer?
Question is what is the value of x*?
I think answer should be C bcz from (2) we have the value of x. From (1),we have x*=x.
Is it right?
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by [email protected] » Thu Dec 04, 2014 3:15 pm
Hi jaspreetsra,

Fact 1 gives us 2 different answers while Fact 2 gives us just 1 answer.

Final Answer: B

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by smills » Fri Jan 30, 2015 7:33 am
Stupid question but I have to ask: Can someone please confirm what x* is? Is this the symbol for x squared??
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by DavidG@VeritasPrep » Fri Jan 30, 2015 7:44 am
Not a stupid question at all. x* isn't an official mathematical symbol that exists outside of this problem - the problem defines the symbol. Here it's defined as
x* = x(x-1). x(x-1) then expands to x^2 - x.

(But if you were given 2*, we'd know that 2* = 2(2-1). If we were given 3*, we'd know that 3* = 3(3-1) and so on.)

The important thing to note is that the GMAT will sometimes define a symbol within the context of a problem.
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by GMATGuruNY » Fri Jan 30, 2015 8:04 am
smills wrote:Stupid question but I have to ask: Can someone please confirm what x* is? Is this the symbol for x squared??
Sometime the GMAT will use an arbitrary symbol to represent a particular FUNCTION.
For example:
@x = (x)(x+5).
Here, @x indicates the operation to be performed upon any value of x:
(x)(x+5).
Thus:
@1 = (1)(1+5)
@2 = (2)(2+5)
@3 = (3)(3+5)
And so on.

In the problem posted above:
x* = (x)(x-1).
Implication:
1* = (1)(1-1)
2* = (2)(2-1)
3* = (3)(3-1)
And so on.
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by Matt@VeritasPrep » Mon Feb 09, 2015 2:06 am
smills wrote:Stupid question but I have to ask: Can someone please confirm what x* is? Is this the symbol for x squared??
This actually IS a mathematical symbol with a lot of different applications, most/all of which belong to math well out of the scope of the GMAT. (No, you aren't expected to know what a Hermitian adjoint is - the GMAT isn't that heinous!)

But none of the asterisk's fancy applications apply here: on the GMAT virtually any unfamiliar symbol (*, #, @, and even ^, which is an annoying internet way of indicating an exponent, but which doesn't properly mean that) doesn't have an inherent meaning and has to be defined. + and - mean what you think they do, but most other symbols are given a one-shot definition that only applies to that problem, and you aren't expected to know them before or remember them after.
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by nikhilgmat31 » Mon Aug 03, 2015 6:36 am
Statement 1 gives X as 0 or 2 so X* can be 0 or 2
Statement 2 gives X as 2 so x* is 2 always.

Answer is B.
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by pannalal » Thu Nov 30, 2017 4:15 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.
Assume for a second that 5 = 7 (although it is not correct). Now, if you are allowed to multiply both sides with 0, both sides shows 0. So, 5 becomes equal to 7, though we know that 5 is not equal to 7. Thus, in an equation, whenever you multiply or divide by any value, the necessary condition is you are not multiplying or dividing by 0. I hope, this is clear to you.
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