I choose B

D

because from 1 we get x^2 = 0, therefore x has to be 0.

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.

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

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.

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?

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)

(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

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_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.

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.

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

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.

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)