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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.)

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.)

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 \(2x>-8+y\)?

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]

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.

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
_________________

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 , 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]

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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.)"

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]

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24 Dec 2015, 06:20

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Re: If y is a negative number greater than -8, is x greater than [#permalink]

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11 Oct 2016, 04:45

Bunuel wrote:

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.

.

how could Ans be A??

if x=-7 and y=-6.5 then Avg. = -8-6.5/2=-7.25 and x> avg. But if x=-7 and y=-1 then avg. = -8-1/2=-4.5 thus x<avg.

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.

.

how could Ans be A??

if x=-7 and y=-6.5 then Avg. = -8-6.5/2=-7.25 and x> avg. But if x=-7 and y=-1 then avg. = -8-1/2=-4.5 thus x<avg.

Please suggest me.

Thanks

How is x=-7 closer to -8 than x=-7 is to y=-6.5?
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