Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Now, the product of these two is 0. This means that at least one of them has to be 0. Either (y-1) = 0 or (5 - r) = 0 or both are 0. So, either y = 1 or r = 5 or both. Only if we know that r is not 5, then we can say that y must be 1. If r is 5, y may be 1 or may not be 1.

Stmnt 1: r^2 = 25 So r = +- 5 This statement tells us that r can be 5. If r = 5, y may or may not be 1. If r is not 5, y will be 1. Since we do not know whether r is 5 or not, we cannot say what the value of y is. Not sufficient.

Stmnt 2: r = 5 If r = 5, y may or may not be 1. Not sufficient.

Both together, r = 5. Again, not sufficient. Answer E _________________

Re: if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? (1) [#permalink]
13 May 2013, 02:30

1

This post received KUDOS

Expert's post

onedayill wrote:

Guys I'm bit lost here.

(y-1)[y+3-y+2] = r(y-1) (y-1)5 = r(y-1)

(y-1) is common here

so 5= r

which is what stmt 2 tells us.

I think i should be B.

Notice that we are asked to find the value of y not r. Also, I guess you reduced (y-1)5 = r(y-1) by y-1 which cannot be done here.

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) (y-1)5 = r(y-1) by y-1, you assume, with no ground for it, that y-1 does not equal to zero thus exclude a possible solution (notice that both y=1 AND r=5 satisfy the equation).

Hope it's clear.

Complete solution:

If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y?

Re: if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? (1) [#permalink]
07 Jan 2012, 14:31

Rephrasing the stem provides us:

(y-1)[y+3-y+2] = r(y-1) (y-1)5 = r(y-1) (y-1)(5-r) = 0 y = 1 or r = 5

1. r = +/- 5, if r = -5, y = 1 and if r = 5, y = anything. Insuff. 2. r = 5, which means y = anything. Insuff.

Combined, only thing common is r = 5, which still provides y = anything. Insuff.

E. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]
14 May 2013, 20:21

Hi Buneul, has my logic actually worked here? correct me if i am wrong...

Jasonammex wrote:

If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y?

(1) r^2=25 (2) r=5

From the given statement: (y-1)[Y+3-y+2]=r(y-1) Divide both sides by (y-1)

y+3-y+2=r r=5

Here I thought that in the given statement when Y does not exist it is impossible to calculate the value of Y. Thus both statements are not sufficient. Ans: E _________________

Do not forget to hit the Kudos button on your left if you find my post helpful.

Re: If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y? [#permalink]
15 May 2013, 00:53

Expert's post

atalpanditgmat wrote:

Hi Buneul, has my logic actually worked here? correct me if i am wrong...

Jasonammex wrote:

If (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y?

(1) r^2=25 (2) r=5

From the given statement: (y-1)[Y+3-y+2]=r(y-1) Divide both sides by (y-1)

y+3-y+2=r r=5

Here I thought that in the given statement when Y does not exist it is impossible to calculate the value of Y. Thus both statements are not sufficient. Ans: E

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) (y-1)5 = r(y-1) by y-1, you assume, with no ground for it, that y-1 does not equal to zero thus exclude a possible solution (notice that both y=1 AND r=5 satisfy the equation). _________________

HBS: Reimagining Capitalism: Business and Big Problems : Growing income inequality, poor or declining educational systems, unequal access to affordable health care and the fear of continuing economic distress...

I am not panicking. Nope, Not at all. But I am beginning to wonder what I was thinking when I decided to work full-time and plan my cross-continent relocation...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...