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shekar123
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If y is a negative number greater than -8, is x greater than 01 Jun 2010, 20:39
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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
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If y is a negative number greater than -8, is x greater than 02 Jun 2010, 05:32
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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
Show more

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.)

Answer: A.

Similar questions to practice:
https://gmatclub.com/forum/if-x-is-a-po ... 67734.html (OG)
https://gmatclub.com/forum/if-x-is-a-po ... 28086.html
https://gmatclub.com/forum/if-x-is-a-po ... 31322.html
https://gmatclub.com/forum/if-a-is-a-po ... 05650.html
https://gmatclub.com/forum/if-x-and-z-a ... 55651.html

Hope it helps.
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If y is a negative number greater than -8, is x greater than 02 Jun 2010, 07:46
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Great explanation!! Kudos!!
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If y is a negative number greater than -8, is x greater than 02 Jun 2010, 15:19
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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.)

Answer: A.

Similar question: number-line-problem-22709.html#p719532

Hope it helps.
Show more


I still can't get option (2)...:(
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If y is a negative number greater than -8, is x greater than 02 Jun 2010, 15:52
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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.)

Answer: A.

Similar question: number-line-problem-22709.html#p719532

Hope it helps.


I still can't get option (2)...:(
Show more

Stem: \(-8<y<0\).

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

Hope it's clear.
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If y is a negative number greater than -8, is x greater than 02 Jun 2010, 21:59
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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]
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If y is a negative number greater than -8, is x greater than 19 Sep 2010, 16:23
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zisis
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
Show more

A is enough, if x is closer to -8 than y and y>-8 then x will be less than the average of -8 and y (answering NO to the question)

B is not enough. Eg. y=-0.5, x=-2, avg=-4.25 x is greater; y=-6, x=-24, avg=-16 x is lesser

Hence, ans is A
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If y is a negative number greater than -8, is x greater than 19 Sep 2010, 16:24
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zisis
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
Show more

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)
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If y is a negative number greater than -8, is x greater than 19 Sep 2010, 16:28
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vigneshpandi

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)
Show more

In this question, if you draw a number line, most of the answer will just come to you as obvious
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If y is a negative number greater than -8, is x greater than 19 Sep 2010, 16:53
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Thank you guys..I will use it going forward...
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If y is a negative number greater than -8, is x greater than 17 Jan 2011, 10:40
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'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
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If y is a negative number greater than -8, is x greater than 17 Jan 2011, 10:52
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reg
'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
Show more

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.
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If y is a negative number greater than -8, is x greater than 01 Nov 2014, 15:59
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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.)

Answer: A.

Similar question: number-line-problem-22709.html#p719532

Hope it helps.
Show more

Hi Bunuel,

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!
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If y is a negative number greater than -8, is x greater than 02 Nov 2014, 06:28
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shekar123
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.)

Answer: A.

Similar question: number-line-problem-22709.html#p719532

Hope it helps.

Hi Bunuel,

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!
Show more
'
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.

Hope it's clear.
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If y is a negative number greater than -8, is x greater than 11 Oct 2016, 05:45
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'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.

.
Show more

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
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If y is a negative number greater than -8, is x greater than 11 Oct 2016, 06:04
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reg
'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.

Please suggest me.

Thanks
Show more

How is x=-7 closer to -8 than x=-7 is to y=-6.5?
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If y is a negative number greater than -8, is x greater than 16 Apr 2018, 17:06
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Given: -8 < y < 0
mean = \(\frac{y-8}{2}\)
Question asks if this is true: \(\frac{y-8}{2}<x\)

Statement 1
Drawing a number line helped me process this info.

<-------(-8)-----(mean)------(y)---------0--------->

The mean is the halfway point between -8 and y. If x is closer to -8 than to y, then x lies to the left of the mean on the number line. So x is not greater than the mean.
Sufficient

Statement 2
x=4y
So question becomes is this true: \(\frac{y-8}{2}<x\) --> \(\frac{y-8}{2}<4y\) --> \(y-8<8y\) --> \(-8<7y\) --> \(-\frac{8}{7}<y\)
Is y greater than -8/7? We dont know. y can be -7 and the answer is NO. or y can be -1 and the answer is YES.
Not sufficient

Answer: A
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If y is a negative number greater than -8, is x greater than 16 Apr 2025, 05:55
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