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:

Re: 600 level question [#permalink]
02 Jun 2010, 04:32

4

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

shekar123 wrote:

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?

(1) On the number line, x is closer to -8 than it is to y. (2) x = 4y

Given: \(-8<y<0\). Q: is x greater than the average of -8 and x? Or: is \(x>\frac{-8+y}{2}\)? --> \(2x>-8+y\)?

-----{-8}-----{average}-----{y} (average of y and -8 is halfway between y and -8).

(1) On the number line, x is closer to -8 than it is to y.

Now, as \(x\) is closer to \(-8\) than it (\(x\)) is to \(y\), then \(x\) is either in the green area, so less than average OR in the red area, so also less than average. Answer to the question is NO.

Sufficient.

(2) \(x=4y\) --> is \(2x>-8+y\)? --> is \(8y>-8+y\)? --> is \(y>-\frac{8}{7}\)? We don't now that. Not sufficient. (we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)

Re: 600 level question [#permalink]
02 Jun 2010, 14:19

Bunuel wrote:

shekar123 wrote:

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?

(1) On the number line, x is closer to -8 than it is to y. (2) x = 4y

Given: \(-8<y<0\). Q: is x greater than the average of -8 and x? Or: is \(x>\frac{-8+y}{2}\)? --> \(2x>-8+y\)?

-----{-8}-----{average}-----{y} (average of y and -8 is halfway between y and -8).

(1) On the number line, x is closer to -8 than it is to y.

Now, as \(x\) is closer to \(-8\) than it (\(x\)) is to \(y\), then \(x\) is either in the green area, so less than average OR in the red area, so also less than average. Answer to the question is NO.

Sufficient.

(2) \(x=4y\) --> is \(2x>-8+y\)? --> is \(8y>-8+y\)? --> is \(y>-\frac{8}{7}\)? We don't now that. Not sufficient. (we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)

Re: 600 level question [#permalink]
02 Jun 2010, 14:52

2

This post received KUDOS

Expert's post

shekar123 wrote:

Bunuel wrote:

shekar123 wrote:

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?

(1) On the number line, x is closer to -8 than it is to y. (2) x = 4y

Given: \(-8<y<0\). Q: is x greater than the average of -8 and x? Or: is \(x>\frac{-8+y}{2}\)? --> \(2x>-8+y\)?

-----{-8}-----{average}-----{y} (average of y and -8 is halfway between y and -8).

(1) On the number line, x is closer to -8 than it is to y.

Now, as \(x\) is closer to \(-8\) than it (\(x\)) is to \(y\), then \(x\) is either in the green area, so less than average OR in the red area, so also less than average. Answer to the question is NO.

Sufficient.

(2) \(x=4y\) --> is \(2x>-8+y\)? --> is \(8y>-8+y\)? --> is \(y>-\frac{8}{7}\)? We don't now that. Not sufficient. (we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)

Based on the above 2 informations can we answer whether: \(2x>-8+y\)? No. If \(y=-1>-8\), then \(x=-4\) and \(2x=-8>-8+y=-8-4=-12\) - answer to the question is YES; If \(y=-2>-8\), then \(x=-8\) and \(2x=-16<-8+y=-8-2=-10\) - answer to the question is NO.

Two different answers to the question is \(2x>-8+y\)?

Re: 600 level question [#permalink]
02 Jun 2010, 20:59

great explanation.

Statement:(2) \(x=4y\)

Based on the above 2 informations can we answer whether: \(2x>-8+y\)? No. If \(y=-1>-8\), then \(x=-4\) and \(2x=-8>-8+y=-8-4=-12\) - answer to the question is YES; If \(y=-2>-8\), then \(x=-8\) and \(2x=-16<-8+y=-8-2=-10\) - answer to the question is NO.

Two different answers to the question is \(2x>-8+y\)? great explanation Hope it's clear.[/quote]

Re: Number line question [#permalink]
19 Sep 2010, 15:24

zisis wrote:

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?

(1) On the number line, x is closer to -8 than it is to y.

(2) x = 4y

I just substituted numbers in each of the choices and arrived at A. But it took me 3:05 mins to solve this...I am always taking extra time in solving such problems...ANyone please suggest a short cut in solving such type of problems...Always when I substitute values in these kind of problems I take a minimum of 3 minutes.. Any suggestion is greatly appreciated.

Re: Number line question [#permalink]
19 Sep 2010, 15:28

vigneshpandi wrote:

I just substituted numbers in each of the choices and arrived at A. But it took me 3:05 mins to solve this...I am always taking extra time in solving such problems...ANyone please suggest a short cut in solving such type of problems...Always when I substitute values in these kind of problems I take a minimum of 3 minutes.. Any suggestion is greatly appreciated.

Choice (A)

In this question, if you draw a number line, most of the answer will just come to you as obvious _________________

Re: 600 level question [#permalink]
17 Jan 2011, 09:40

'A' can't be the correct answer -

given --> -8<y<0

so , when y =-1 , [-1+(-8)] /2=-4.5 when y=-7, [-7+(-8)]/2= -7.5

now if x is closer to -8 means , x<-4.5

but there is no connection between values of X and Y [ie, value of x is independent of Y] so, if x=-4.6 then , x>-7.5 [where -7.5 is one of the values of average of y and -8] if x=-7.8 then , x<-7.5 [where -7.5 is one of the values of average of y and -8]

but if x=4y is also considered then x will always be greater than average of y and -8

Re: 600 level question [#permalink]
17 Jan 2011, 09:52

Expert's post

reg wrote:

'A' can't be the correct answer -

given --> -8<y<0

so , when y =-1 , [-1+(-8)] /2=-4.5 when y=-7, [-7+(-8)]/2= -7.5

now if x is closer to -8 means , x<-4.5

but there is no connection between values of X and Y [ie, value of x is independent of Y] so, if x=-4.6 then , x>-7.5 [where -7.5 is one of the values of average of y and -8] if x=-7.8 then , x<-7.5 [where -7.5 is one of the values of average of y and -8]

but if x=4y is also considered then x will always be greater than average of y and -8

so correct ans must be C .

please correct me if i'm wrong

OA (official answer) for this question is A. not C.

If you consider y=-7 so that the average of -8 and y to be -7.5 then x=-4.6 is not a proper value for x as (1) says that x is closer to -8 than it (x) is to y.

Refer to the correct solutions above. _________________

Re: If y is a negative number greater than -8, is x greater than [#permalink]
01 Nov 2014, 14:59

Bunuel wrote:

shekar123 wrote:

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?

(1) On the number line, x is closer to -8 than it is to y. (2) x = 4y

Given: \(-8<y<0\). Q: is x greater than the average of -8 and x? Or: is \(x>\frac{-8+y}{2}\)? --> \(2x>-8+y\)?

-----{-8}-----{average}-----{y} (average of y and -8 is halfway between y and -8).

(1) On the number line, x is closer to -8 than it is to y.

Now, as \(x\) is closer to \(-8\) than it (\(x\)) is to \(y\), then \(x\) is either in the green area, so less than average OR in the red area, so also less than average. Answer to the question is NO.

Sufficient.

(2) \(x=4y\) --> is \(2x>-8+y\)? --> is \(8y>-8+y\)? --> is \(y>-\frac{8}{7}\)? We don't now that. Not sufficient. (we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)

What exactly do you mean by "(we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)"

Re: If y is a negative number greater than -8, is x greater than [#permalink]
02 Nov 2014, 05:28

Expert's post

russ9 wrote:

Bunuel wrote:

shekar123 wrote:

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?

(1) On the number line, x is closer to -8 than it is to y. (2) x = 4y

Given: \(-8<y<0\). Q: is x greater than the average of -8 and x? Or: is \(x>\frac{-8+y}{2}\)? --> \(2x>-8+y\)?

-----{-8}-----{average}-----{y} (average of y and -8 is halfway between y and -8).

(1) On the number line, x is closer to -8 than it is to y.

Now, as \(x\) is closer to \(-8\) than it (\(x\)) is to \(y\), then \(x\) is either in the green area, so less than average OR in the red area, so also less than average. Answer to the question is NO.

Sufficient.

(2) \(x=4y\) --> is \(2x>-8+y\)? --> is \(8y>-8+y\)? --> is \(y>-\frac{8}{7}\)? We don't now that. Not sufficient. (we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)

What exactly do you mean by "(we've gotten that if \(0>y>-\frac{8}{7}\) (for instance if \(y=-1\)), then the answer to the question is YES, but if \(y\leq{-\frac{8}{7}}\) (for instance if \(y=-2\)), then the answer to the question is NO.)"

I can follow everything minus that. Thanks!

' For (2) we know that \(x=4y\). After substituting this into the question the question becomes "is \(y>-\frac{8}{7}\)?" We cannot asnwer this, so the statement is insufficient.

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

I’ll start off with a quote from another blog post I’ve written : “not all great communicators are great leaders, but all great leaders are great communicators.” Being...