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 !! :)
(1.) X*=X
(2.) (X - 1 ) * = (X - 2)
OA after some discussion !! :)
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
What is the value of x*?
Source: Beat The GMAT — Data Sufficiency |
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Location: Toronto
- Thanked: 1710 times
- GMAT Score:800
Let's actually look at the algebra.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 !!
(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.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
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
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.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
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
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.
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.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.
It is not wrong, but you have to consider the sign of x, x can be positive or negative.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?
In your case, you are only considering positive.
x(x-1) = x
then,
x-1 = +/- 1
which equals x = 0,2
78 clicks can change my life !
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.
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)
