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If 0<x<y, is yx < 0.00005 [#permalink]
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05 Apr 2012, 09:26
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If 0<x<y, is yx < 0.00005 (1) x>1/60,000 (2) y<1/15,000
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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If 0<x<y, is yx < 0.00005Notice that \(0.00005=\frac{5}{100,000}=\frac{3}{60,000}\), and \(\frac{1}{15,000}=\frac{4}{60,000}\). So, we can rewrite the question as: If 0<x<y, is yx<3(1) x>1 > if \(x=2\) and \(y=3\) then the answer is YES but if \(x=2\) and \(y=5\) the answer is NO. Not sufficient. (2) y<4 > if \(x=2\) and \(y=3\) then the answer is YES but if \(x=0.5\) and \(y=3.5\) the answer is NO. Not sufficient. (1)+(2) Remember we can subtract inequalities if their signs are in opposite directions > subtract (1) from (2): \(yx<41\) > \(yx<3\). Sufficient. Answer: C.
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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05 Apr 2012, 17:57
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imhimanshu wrote: If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000 1) NS  nothing about y 2) NS  nothing about x So it's between E and C Is yx < 1/20,000? LT = less than GT = Great than LT 1/15,000  GT 1/60,000 < 1/20,000. Multiply by 60,000 to simplify results in LT 4  GT 1 < 3? Test extremes  3.9  1.1 = 2.8 . YES ...sufficient. C



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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17 Jan 2013, 05:42
I have one question is this step possible for this question ( I have re written the equations in this way) 1/60000 < X Y< 4/60000 Add both equations and then subtract you reach back to the original question and can prove sufficiency.
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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21 Feb 2013, 17:01
[quote="DelSingh"]If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000
this ain't 700 or 600700 level question , it is way sub 600
anyways
the question is asking whether the difference between both +ve numbers x,y is very small ie. 5/100,000 = 1/20,000
obviously each alone is not suff
both
subtract 2 from 1
xy >1/20,000... i.e. yx<1/20,000.....an then answer is a definite yes ...c



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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21 Feb 2013, 17:53
yezz wrote: DelSingh wrote: If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000
this ain't 700 or 600700 level question , it is way sub 600
anyways
the question is asking whether the difference between both +ve numbers x,y is very small ie. 5/100,000 = 1/20,000
obviously each alone is not suff
both
subtract 2 from 1
xy >1/20,000... i.e. yx<1/20,000.....an then answer is a definite yes ...c I took off the difficulty, but GMAT Prep did rate this medium level. Anyway, I understand why the both statements are insufficient but I do not know how you combines them? What did you when you 'subtracted 2 from 1'?
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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21 Feb 2013, 23:26
When you subtract 1 from 2, you get the value of yx. However, since we know only one sided limits of these values, let's consider those values. yx=(1/15000)(1/60,000) Taking L.CM. yx=1/20,000 yx=0.00005 However, this just gives us the limit of the difference. Since y<1/15,000 and x>1/60,000, a bigger number on the L.H.S is being subtracted from a smaller number and hence, the actual difference will be less than 1/20,000. This is by applying concept. Let us test values for better understanding. For e.g. the value of y could be y=1/20,000(the greater the denominator, the smaller the number and hence y>1/15000) and x=1/40,000(by similar idea) yx=1/20,0001/40,000=1/40,000 1/40,000<1/20,000. Hence proved. Hope that helps! DelSingh wrote: yezz wrote: DelSingh wrote: If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000
this ain't 700 or 600700 level question , it is way sub 600
anyways
the question is asking whether the difference between both +ve numbers x,y is very small ie. 5/100,000 = 1/20,000
obviously each alone is not suff
both
subtract 2 from 1
xy >1/20,000... i.e. yx<1/20,000.....an then answer is a definite yes ...c I took off the difficulty, but GMAT Prep did rate this medium level. Anyway, I understand why the both statements are insufficient but I do not know how you combines them? What did you when you 'subtracted 2 from 1'?



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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21 Feb 2013, 23:37
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DelSingh wrote: If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000
Source: GMAT Prep question pack 1 There are two ways to deal with it. Method 1: Is yx < 0.00005? We can see that both statements alone are not sufficient. (1) x>1/60,000 (2) y<1/15,000 We know that we can add inequalities when they have the same sign ie. a < b c < d then, a+c < b+d Also, when we multiply an inequality by 1, the inequality sign flips. x>1/60,000 implies x < 1/60,000 You can add these two inequalities: x < 1/60,000 and y<1/15,000 to get yx < 1/15000  1/60,000 which is yx < 1/20,000 i.e. yx < 0.00005 Another method is to see this on the number line. Draw a number line to understand this. 0<x<y implies that x and y are both positive and x is to the left of y on the number line. Is yx < 0.00005 means is the distance between x and y less than .00005? (1) x>1/60,000 means x lies to the right of 1/60,000 (2) y<1/15,000 means y lies to the left of 4/60,000 So the distance between them must be less than 4/60,000  1/60,000 = 3/60,000 = .00005
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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22 Feb 2013, 00:20
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If 0<x<y, is yx < 0.00005 (1) x>1/60,000 (2) y<1/15,000 Solution: (Answer is C)What do we know?X is positive and Y is greater than X. What do we need to know?Is Y is less than 0.00005 + X? Whenever you face a Data Sufficiency question asking Yes, No. Simply substitute and try to disprove the statement. Statement(1): X is greater than 1/60,000 = 0.00001666 Which does not tell any relation between X and Y Hence it is insufficient. Statement (2) is also insufficient as it only tells that Y is less than 0.000066 (It is very important to know the importance of converting fractions to percentage) If we combine both the statements, we get that X is greater than 0.000016 and Y is less than 0.000066 Now the question is asking us that yx<0.00005, to try to disprove that we need to maximize yx and for that let us get the maximum value of y and minimum value of x. Let us say y = 0.000065 and x = 0.000017 So the maximum difference is = 0.000065  0.000017 = 0.000048 Hence combining both the statements we can say that yx will always be less than 0.000048. Hence answer is (C)
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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22 Feb 2013, 04:29
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DelSingh wrote: yezz wrote: DelSingh wrote: If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000
this ain't 700 or 600700 level question , it is way sub 600
anyways
the question is asking whether the difference between both +ve numbers x,y is very small ie. 5/100,000 = 1/20,000
obviously each alone is not suff
both
subtract 2 from 1
xy >1/20,000... i.e. yx<1/20,000.....an then answer is a definite yes ...c I took off the difficulty, but GMAT Prep did rate this medium level. Anyway, I understand why the both statements are insufficient but I do not know how you combines them? What did you when you 'subtracted 2 from 1'? for 2 ineq to subtract they have to be with opposit direction , one of them is bigger than and 2nd is smaller than and what u do is keep the sign ( direction in terms of bigger than or smaller than) of the ineq from which u subtract the 2nd .... Another way of seeing it is as follows if we subtract 1 from 2 is like flipping the sign of 1 and adding it to the 2nd , thus x>1/60,000 becomes x<1/60,000...............1 after changing direction ( flipping the sign) now add 2 to 1 y>1/15,000........2 y+ (x) > 1/15,000 + (1/60,000).................. simplify yx > 1/20,000



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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02 Mar 2013, 11:07
I want to follow this posting.
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Re: If 0<x<y, is yx<0.00005? [#permalink]
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26 Mar 2013, 07:11
Since they are asking if "yx<0.00005"; first I will try if the statement 2 is enough : Is y<0.00005 or y<1/20000 ?
y = 1/15000 (bigger that 1/20000) ; therefore it´s not enough,
I will resolve both sides of the equation now:
yx = 1/15000  1/60000 = 3/60000 = 1/20000 0.00005 = 5/100000 = 1 /20000
Therefore you need both statements. Answer C



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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30 Mar 2013, 16:02
In relation to this question, I have a basic scientific notation question, how would you write the sci notation of 1 / 20,000?



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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30 Mar 2013, 16:24
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jmuduke08 wrote: In relation to this question, I have a basic scientific notation question, how would you write the sci notation of 1 / 20,000? \(\frac{1}{20000}=\frac{1}{2}*\frac{1}{10000}=0.5*10^^4=5*10^^5\)
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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30 Mar 2013, 16:26
Zarrolou wrote: jmuduke08 wrote: In relation to this question, I have a basic scientific notation question, how would you write the sci notation of 1 / 20,000? \(\frac{1}{20000}=\frac{1}{2}*\frac{1}{10000}=0.5*10^^4\) ahh thank you, I was multiplying .5 by 10,000 instead of 1/10,000 and knew it wasnt possible



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Re: If 0 < x < y , is y  x < 0.00005 ? [#permalink]
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27 Apr 2013, 02:58
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(1) Insufficient. We know nothing about \(y\). (2) Insufficient. We know nothing about \(x\). (1)+(2) Sufficient. We know that \(y<\frac{1}{15,000}\) and \(x<\frac{1}{60,000}\). If we add this two inequalities we will get: \(yx<\frac{1}{15,000}\frac{1}{60,000}=\frac{1}{20,000}=0.00005\) The correct answer is C.
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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05 Dec 2014, 17:45
Bunuel wrote: If 0<x<y, is yx < 0.00005
Notice that \(0.00005=\frac{5}{100,000}=\frac{3}{60,000}\), and \(\frac{1}{15,000}=\frac{4}{60,000}\).
So, we can rewrite the question as:
If 0<x<y, is yx<3
(1) x>1 > if \(x=2\) and \(y=3\) then the answer is YES but if \(x=2\) and \(y=5\) the answer is NO. Not sufficient. (2) y<4 > if \(x=2\) and \(y=3\) then the answer is YES but if \(x=0.5\) and \(y=3.5\) the answer is NO. Not sufficient.
(1)+(2) Remember we can subtract inequalities if their signs are in opposite directions > subtract (1) from (2): \(yx<41\) > \(yx<3\). Sufficient.
Answer: C. Hi Bunuel, This is great. I actually went the long division route and it took quite some time. Can you suggest similar problems where we manipulate fractions/decimals as such? I clicked on the tab on the top right but it just let me to regular inequalities problems. Thanks,



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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27 Nov 2015, 10:50
Bunuel wrote: If 0<x<y, is yx < 0.00005
Notice that \(0.00005=\frac{5}{100,000}=\frac{3}{60,000}\), and \(\frac{1}{15,000}=\frac{4}{60,000}\).
So, we can rewrite the question as:
If 0<x<y, is yx<3
(1) x>1 > if \(x=2\) and \(y=3\) then the answer is YES but if \(x=2\) and \(y=5\) the answer is NO. Not sufficient. (2) y<4 > if \(x=2\) and \(y=3\) then the answer is YES but if \(x=0.5\) and \(y=3.5\) the answer is NO. Not sufficient.
(1)+(2) Remember we can subtract inequalities if their signs are in opposite directions > subtract (1) from (2): \(yx<41\) > \(yx<3\). Sufficient.
Answer: C. In the beginning of your explanation, how do you get 1/15,000 and 4/60,000 ?



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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21 Dec 2015, 10:14
VeritasPrepKarishma wrote: DelSingh wrote: If 0<x<y, is yx < 0.00005
(1) x>1/60,000 (2) y<1/15,000
Source: GMAT Prep question pack 1 There are two ways to deal with it. Method 1: Is yx < 0.00005? We can see that both statements alone are not sufficient. (1) x>1/60,000 (2) y<1/15,000 We know that we can add inequalities when they have the same sign ie. a < b c < d then, a+c < b+d Also, when we multiply an inequality by 1, the inequality sign flips. x>1/60,000 implies x < 1/60,000 You can add these two inequalities: x < 1/60,000 and y<1/15,000 to get yx < 1/15000  1/60,000 which is yx < 1/20,000 i.e. yx < 0.00005 Another method is to see this on the number line. Draw a number line to understand this. 0<x<y implies that x and y are both positive and x is to the left of y on the number line. Is yx < 0.00005 means is the distance between x and y less than .00005? (1) x>1/60,000 means x lies to the right of 1/60,000 (2) y<1/15,000 means y lies to the left of 4/60,000 So the distance between them must be less than 4/60,000  1/60,000 = 3/60,000 = .00005 well explained. You got KUDO for this.
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