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Re: 600 level question [#permalink]
02 Jun 2010, 04:32

4

This post received KUDOS

Expert's post

3

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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.

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.

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