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# If 0<x<y, is y-x < 0.00005

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If 0<x<y, is y-x < 0.00005 [#permalink]

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05 Apr 2012, 09:26
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If 0<x<y, is y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000
[Reveal] Spoiler: OA

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Re: If 0<x<y, is y-x < 0.00005 [#permalink]

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05 Apr 2012, 16:46
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If 0<x<y, is y-x < 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 y-x<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): $$y-x<4-1$$ --> $$y-x<3$$. Sufficient.

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Re: If 0<x<y, is y-x < 0.00005 [#permalink]

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21 Feb 2013, 23:37
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DelSingh wrote:
If 0<x<y, is y-x < 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 y-x < 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 y-x < 1/15000 - 1/60,000
which is y-x < 1/20,000 i.e. y-x < 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 y-x < 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 y-x < 0.00005 [#permalink]

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05 Apr 2012, 17:57
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imhimanshu wrote:
If 0<x<y, is y-x < 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 y-x < 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 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:
$$y-x<\frac{1}{15,000}-\frac{1}{60,000}=\frac{1}{20,000}=0.00005$$

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Re: If 0<x<y, is y-x < 0.00005 [#permalink]

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22 Feb 2013, 00:20
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If 0<x<y, is y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000

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 y-x<0.00005, to try to disprove that we need to maximize y-x 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 y-x will always be less than 0.000048. Hence answer is (C)
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Re: If 0<x<y, is y-x < 0.00005 [#permalink]

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22 Feb 2013, 04:29
1
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DelSingh wrote:
yezz wrote:
DelSingh wrote:
If 0<x<y, is y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000

this ain't 700 or 600-700 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

x-y >-1/20,000... i.e. y-x<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)

y>1/15,000........2

y+ (-x) > 1/15,000 + (-1/60,000).................. simplify

y-x > 1/20,000
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Re: If 0<x<y, is y-x < 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 y-x < 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 y-x < 0.00005 [#permalink]

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21 Feb 2013, 17:01
[quote="DelSingh"]If 0<x<y, is y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000

this ain't 700 or 600-700 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

x-y >-1/20,000... i.e. y-x<1/20,000.....an then answer is a definite yes ...c
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Re: If 0<x<y, is y-x < 0.00005 [#permalink]

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21 Feb 2013, 17:53
yezz wrote:
DelSingh wrote:
If 0<x<y, is y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000

this ain't 700 or 600-700 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

x-y >-1/20,000... i.e. y-x<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 y-x < 0.00005 [#permalink]

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21 Feb 2013, 23:26
When you subtract 1 from 2, you get the value of y-x. However, since we know only one sided limits of these values, let's consider those values.

y-x=(1/15000)-(1/60,000)

Taking L.CM. y-x=1/20,000

y-x=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)

y-x=1/20,000-1/40,000=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 y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000

this ain't 700 or 600-700 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

x-y >-1/20,000... i.e. y-x<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 y-x < 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 y-x<0.00005? [#permalink]

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26 Mar 2013, 07:11
Since they are asking if "y-x<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:

y-x = 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 y-x < 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 y-x < 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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Re: If 0<x<y, is y-x < 0.00005 [#permalink]

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05 Dec 2014, 17:45
Bunuel wrote:
If 0<x<y, is y-x < 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 y-x<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): $$y-x<4-1$$ --> $$y-x<3$$. Sufficient.

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 y-x < 0.00005 [#permalink]

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27 Nov 2015, 10:50
Bunuel wrote:
If 0<x<y, is y-x < 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 y-x<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): $$y-x<4-1$$ --> $$y-x<3$$. Sufficient.

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 y-x < 0.00005 [#permalink]

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21 Dec 2015, 10:14
VeritasPrepKarishma wrote:
DelSingh wrote:
If 0<x<y, is y-x < 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 y-x < 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 y-x < 1/15000 - 1/60,000
which is y-x < 1/20,000 i.e. y-x < 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 y-x < 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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Re: If 0<x<y, is y-x < 0.00005   [#permalink] 21 Dec 2015, 10:14

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