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

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

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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 [#permalink]

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02 Jun 2010, 05:32
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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.)

Answer: A.

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

Hope it helps.
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Re: 600 level question [#permalink]

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02 Jun 2010, 07:46
Great explanation!! Kudos!!
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Re: 600 level question [#permalink]

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02 Jun 2010, 15:19
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.)

Answer: A.

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

Hope it helps.

I still can't get option (2)...

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Re: 600 level question [#permalink]

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02 Jun 2010, 15:52
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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.)

Answer: A.

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

Hope it helps.

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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Re: 600 level question [#permalink]

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02 Jun 2010, 21:59
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]

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Re: Number line question [#permalink]

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19 Sep 2010, 16:19
A is enough... Basically it is asking if x is closer to y or -8?, A gives us the answer...

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Re: Number line question [#permalink]

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19 Sep 2010, 16:23
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

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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Re: Number line question [#permalink]

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19 Sep 2010, 16:24
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.

Choice (A)
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Re: Number line question [#permalink]

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19 Sep 2010, 16:28
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
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Re: 600 level question [#permalink]

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19 Sep 2010, 16:53
Thank you guys..I will use it going forward...
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Re: 600 level question [#permalink]

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14 Oct 2010, 19:49
A

1) x is closer to -8 than y so x is towards -8 away from mid-point (=avg). so x is < avg ==> SUFF

2) x = 4y

y=-7 => x=-28 --> no
y=1 => x=4 ---> yes
no and yes ==> INSUFF
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Re: 600 level question [#permalink]

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17 Jan 2011, 10:40
'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

so correct ans must be C .

please correct me if i'm wrong

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Re: 600 level question [#permalink]

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17 Jan 2011, 10:52
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.
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Re: If y is a negative number greater than -8, is x greater than [#permalink]

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31 Aug 2013, 06:22
Bumping for review and further discussion.
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Re: If y is a negative number greater than -8, is x greater than [#permalink]

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01 Nov 2014, 15:59
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.)

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!

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

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02 Nov 2014, 06:28
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.)

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!

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

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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, 05:45
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.

Please suggest me.

Thanks

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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, 06:04
rohit8865 wrote:
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.

Please suggest me.

Thanks

How is x=-7 closer to -8 than x=-7 is to y=-6.5?
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Re: If y is a negative number greater than -8, is x greater than   [#permalink] 11 Oct 2016, 06:04
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