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If a, b, x, and y are all positive, is a/b > x/y ? [#permalink]
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24 Apr 2017, 02:34
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If a, b, x, and y are all positive, is a/b > x/y ? [#permalink]
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24 Apr 2017, 03:20
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As a,b,x and y are positive, it makes our life easy while playing with inequalities. we have to find is : a/b>x/y > ay>bx > aybx > 0 S1. ay+1>bx > aybx>1 So, we cannot say whether aybx is always > 0 (for ex. aybx could be 0.5) Insufficient!S2. (ay/bx)^2 > (ay/bx) > ay/bx > 1 > ay>bx > aybx>0 Sufficient.Hence, answer should be B.
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Re: If a, b, x, and y are all positive, is a/b > x/y ? [#permalink]
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24 Apr 2017, 03:23
rewrite the question as (ay/bx)>1 statement 1 > divide the entire inequality by bx we get (ay/bx) + (1/bx) > 1 that is (positive number) + (positive Number) >1. This does not say anything about (ay/bx) as it can be 0.5 or 1.5 and so on This is not sufficient as we need to know whether (ay/bx)>1 Statement 2 (ay/bx)^2 > (ay/bx) >(ay/bx)^2  (ay/bx)>0 > (ay/bx)* ((ay/bx)1)>0 this means that either both the boldface terms are greater than zero or both are less than zero. as a,y,b and x are positive (ay/bx) cannot be less than zero. Hence both cannot be less than zero. Now both are greater than zero. which means, (ay/bx)> 0 and (ay/bx)  1 > 0 Remember that this is an AND condition. Both have to be true. therefore ay/bx >1 statement 2 alone is sufficient. So answer should be B
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Re: If a, b, x, and y are all positive, is a/b > x/y ? [#permalink]
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24 Apr 2017, 03:37
simplify the given stem
ay>bx aybx >0 ???? stat 1: aybx > 1,,not suff
stat 2 : divide the entire equation by ax/by
gives ax/by >1
ax>by axby >0.. suff
ans B



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Re: If a, b, x, and y are all positive, is a/b > x/y ? [#permalink]
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24 Apr 2017, 03:37
skothaka wrote: rewrite the question as (ay/bx)>1
statement 1 > divide the entire inequality by bx we get (ay/bx) + (1/bx) > 1 that is (positive number) + (positive Number) >1. This does not say anything about (ay/bx) as it can be 0.5 or 1.5 and so on
This is not sufficient as we need to know whether (ay/bx)>1
Statement 2 (ay/bx)^2 > (ay/bx) >(ay/bx)^2  (ay/bx)>0 >(ay/bx)*((ay/bx)1)>0
this means that either both the boldface terms are greater than zero or both are less than zero. as a,y,b and x are positive (ay/bx) cannot be less than zero. Hence both cannot be less than zero. Now both are greater than zero. which means, (ay/bx)> 0 and (ay/bx)  1 > 0
Remember that this is an AND condition. Both have to be true.
therefore ay/bx >1 statement 2 alone is sufficient. So answer should be B The solution provided for S1 is not complete. We cannot conclude anything with the equation (ay/bx) + (1/bx) > 1. See the above solution for a better explanation.
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Re: If a, b, x, and y are all positive, is a/b > x/y ? [#permalink]
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24 Apr 2017, 03:42
14101992 wrote: skothaka wrote: rewrite the question as (ay/bx)>1
statement 1 > divide the entire inequality by bx we get (ay/bx) + (1/bx) > 1 that is (positive number) + (positive Number) >1. This does not say anything about (ay/bx) as it can be 0.5 or 1.5 and so on
This is not sufficient as we need to know whether (ay/bx)>1
Statement 2 (ay/bx)^2 > (ay/bx) >(ay/bx)^2  (ay/bx)>0 >(ay/bx)*((ay/bx)1)>0
this means that either both the boldface terms are greater than zero or both are less than zero. as a,y,b and x are positive (ay/bx) cannot be less than zero. Hence both cannot be less than zero. Now both are greater than zero. which means, (ay/bx)> 0 and (ay/bx)  1 > 0
Remember that this is an AND condition. Both have to be true.
therefore ay/bx >1 statement 2 alone is sufficient. So answer should be B The solution provided for S1 is not complete. We cannot conclude anything with the equation (ay/bx) + (1/bx) > 1. See the above solution for a better explanation. The solution for S1 says that ay/bx could be anything between 0 and infinite. So S1 is not sufficient. That is enough to rule it out. Your solution is much simpler btw.
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Re: If a, b, x, and y are all positive, is a/b > x/y ?
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24 Apr 2017, 03:42






