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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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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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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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If 0<x<y, is yx < 0.00005 [#permalink]
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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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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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03 May 2016, 18:27
After viewing the responses, I think there's one potential twist where the GMAT creators could make this question harder.
Many posters have been saying that both (1) and (2) are obviously insufficient because they say nothing about the other variable. However, we do know that both x and y are greater than 0 and that y is greater than x. Were y to be less than 3/60,000, then (2) would be sufficient as x still has to be greater than 0 and therefore yx would be greater than 3/60,000.
Please let me know if I'm thinking about this right!



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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30 Oct 2017, 04:33
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. How do you know what sign the combined inequality takes when combining two inequalities with different signs?



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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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30 Oct 2017, 04:38
Edofarmer wrote: 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. How do you know what sign the combined inequality takes when combining two inequalities with different signs? ADDING/SUBTRACTING INEQUALITIES1. You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). 2. You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) ( take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Check for more the links below: Inequalities Made Easy!
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Re: If 0<x<y, is yx < 0.00005 [#permalink]
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30 Oct 2017, 09:56
Bunuel wrote: Edofarmer wrote: 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. How do you know what sign the combined inequality takes when combining two inequalities with different signs? ADDING/SUBTRACTING INEQUALITIES1. You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). 2. You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) ( take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Check for more the links below: Inequalities Made Easy!Thanks for the clarification




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