If y is a negative number greater than -8, is x greater than

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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Great explanation!! Kudos!!
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I still can't get option (2)...

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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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]

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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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
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Thank you guys..I will use it going forward...

'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.
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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!

'
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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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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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
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

Merging topics. Please refer to the discussion above.
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