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If y is a negative number greater than 8, is x greater than [#permalink]
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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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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 questions to practice: https://gmatclub.com/forum/ifxisapo ... 67734.html ( OG) https://gmatclub.com/forum/ifxisapo ... 28086.htmlhttps://gmatclub.com/forum/ifxisapo ... 31322.htmlhttps://gmatclub.com/forum/ifaisapo ... 05650.htmlhttps://gmatclub.com/forum/ifxandza ... 55651.htmlHope it helps.
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Re: 600 level question [#permalink]
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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: numberlineproblem22709.html#p719532Hope 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
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: numberlineproblem22709.html#p719532Hope 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 \(2x=8>8+y=84=12\)  answer to the question is YES; If \(y=2>8\), then \(x=8\) and \(2x=16<8+y=82=10\)  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 \(2x=8>8+y=84=12\)  answer to the question is YES; If \(y=2>8\), then \(x=8\) and \(2x=16<8+y=82=10\)  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, 16:23
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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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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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: numberlineproblem22709.html#p719532Hope 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: numberlineproblem22709.html#p719532Hope 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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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. = 86.5/2=7.25 and x> avg. But if x=7 and y=1 then avg. = 81/2=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. = 86.5/2=7.25 and x> avg. But if x=7 and y=1 then avg. = 81/2=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 [#permalink]
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16 Apr 2018, 08:28
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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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Re: If y is a negative number greater than –8 [#permalink]
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16 Apr 2018, 17:06
Given: 8 < y < 0 mean = \(\frac{y8}{2}\) Question asks if this is true: \(\frac{y8}{2}<x\)
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: \(\frac{y8}{2}<x\) > \(\frac{y8}{2}<4y\) > \(y8<8y\) > \(8<7y\) > \(\frac{8}{7}<y\) 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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Re: If y is a negative number greater than 8, is x greater than [#permalink]
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