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Re: 600 level question [#permalink]
02 Jun 2010, 04:32
5
This post received KUDOS
Expert's post
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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, 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, 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.
Re: If y is a negative number greater than -8, is x greater than [#permalink]
24 Dec 2015, 06:20
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