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Joined: 02 Sep 2009
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62% (00:52) correct 38% (01:05) wrong based on 154 sessions
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Re M1133
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15 Sep 2014, 23:45



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Re: M1133
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21 May 2015, 15:00
Hi Bunel, I might have gone bonkers over here. Can you explain why B is not correct? As all the numbers are positive, zy = y x will certainly give z>x. Thanks D



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22 May 2015, 03:56



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23 May 2015, 05:08
Quote: z  y = y  x is the same as x + z = 2y. How does this imply that z > x? Thanks for your response!! I could be missing something fundamental, but when I see "z" as any positive number(e.g.8) and subtract "y" any other positive number (e.g. 6) > "86 = 2" Now, in order to obtain 2 on the RHS(yx), I have y, which is "6" (as assumed previously)  (x) some positive number (6x). Now that number,x, is "4" in this case, and therefore less than 8, which is z. I know when you simplify the equation to x+z=2y , it doesn't result into anything concrete, but if we keep the equation in asis form, it implies z>x. As I mentioned earlier, I could be missing something fundamental, therefore apologies in advance if my question is too pedantic. Thanks a lot! D



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23 May 2015, 05:44
Ronin30 wrote: Quote: z  y = y  x is the same as x + z = 2y. How does this imply that z > x? Thanks for your response!! I could be missing something fundamental, but when I see "z" as any positive number(e.g.8) and subtract "y" any other positive number (e.g. 6) > "86 = 2" Now, in order to obtain 2 on the RHS(yx), I have y, which is "6" (as assumed previously)  (x) some positive number (6x). Now that number,x, is "4" in this case, and therefore less than 8, which is z. I know when you simplify the equation to x+z=2y , it doesn't result into anything concrete, but if we keep the equation in asis form, it implies z>x. As I mentioned earlier, I could be missing something fundamental, therefore apologies in advance if my question is too pedantic. Thanks a lot! D hi, you are wrongly assuming that z>y... what if z = 6 and y=8... zy=68=2.. therefore yx==2.... 8x=2.. so x=10.. which is greater than z...
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Re: M1133
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23 May 2015, 12:38
Makes complete sense.. As i thought, fundamental is amiss... Thanks a lot!! D



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19 Aug 2015, 03:30
For this question best is to use some smart numbers i think
for statement 1 let x= 5 y = 100 z= x = 5 > since given
so lets check : is 5% of 100 > 100% of 5   no they are equal.  this will hold true for any positive number for x and y and z
so sufficient
Statement 2: first simplify  the given zx=yx becomes x+z =2y Again smart numbers  choose numbers that satisfy the above condition Let y =5.  so x and z should be 2 numbers that add up to 2y here it is 10
so for example x can be 7 and z can be 3
lets check. is 7% of 5 > 5% of 3? > yes.. greater percent of a bigger number > smaller percent of a smaller number
So lets see if we can get an opposite ans what if x = 3 and z = 7 ( they still add up to 10 which is 2y) now is 3%of 5 > 5% of 7? No it is not...
So this statement is not suff
So



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Re: M1133
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15 Feb 2016, 14:45
How is this question any different from M2421 ? I had noticed this earlier as well  a lot of questions are just simple repeats.



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15 Feb 2016, 23:34



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16 Feb 2016, 09:57
Bunuel,
I can give you a couple of examples. M1428 and M2220. They are very almost identical. Also, D0145 has a twin brother somewhere. I had already solved that earlier, and saw it again yesterday in one of my Quizzes.



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15 Jul 2016, 23:14
Bunuel, Please help.
second statemnt yx=zy
From this doesnt it form that no.s x, y and z are in Aritmetic progression. Thus, x%of y is always coming less from y%of x.
I am unable to think of any combination where it is going to fail.



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28 Oct 2017, 04:27
S1  Given x = z. Lets assume x=10 y =100 and z =10 this will given me x% of y = y% of z. Can you please explain where i am going wrong here?



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08 Jun 2018, 14:20
gbshrey wrote: S1  Given x = z. Lets assume x=10 y =100 and z =10 this will given me x% of y = y% of z. Can you please explain where i am going wrong here? Your question doesn't make sense. What you wrote is exactly what the question is asking so where is the confusion?
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17 Nov 2018, 01:06
Hi chetan2uFacing problem with B Stem Is y(xz)>0 From B 2y=x+z. Substituting in stem> (x+z)(xz)>0 x+z is positive... xz must be positive :O Not able to figure out my mistake
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