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If x > 0, is x/y > x ?

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If x > 0, is x/y > x ? [#permalink]

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08 Oct 2009, 13:27
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If x > 0, is x/y > x ?

(1) 0 < y < 1
(2) x > 1
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Re: x>0 [#permalink]

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08 Oct 2009, 13:52
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amitgovin wrote:
if x>0, is (x/y)>x ?

1) 0<y<1

2) x>1

I got the answer second time around but can someone please give me a good explanation that doesn't use substituting numbers?

1.) x > 0, 0<y<1

then x/y will always be greater than x as y is +ve and also less than 1.

Even if x < 1, (x/y) will be greater than x.

So sufficient.

2.) x>1 .

This does not give any info about y. So insufficient.

So A.
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Re: x>0 [#permalink]

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08 Oct 2009, 13:54
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amitgovin wrote:
if x>0, is (x/y)>x ?

1) 0<y<1

2) x>1

I got the answer second time around but can someone please give me a good explanation that doesn't use substituting numbers?

Q is if x>0 is(x/y)>x --> x(1-y)/y>0

(1) x>0 (given), 0<y<1 --> both denominator and nominator >0 so x(1-y)/y>0 Sufficient

(2) x>1 x(1-y)/y can be <0 and >0 depending on y. 0<y<1 it's positive, for all other values of y it's negative. Not sufficient.

A.
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Re: x>0 [#permalink]

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08 Oct 2009, 16:40
Hi Bunuel!

How did you so quickly come to this conclusion that "for all other values of 'y' the equation 'x(1-y)/y' will be '-ve'?

Also, how can I strengthen my inequalitites knowledge (for GMAT).

Please explain in detail.

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Re: x>0 [#permalink]

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08 Oct 2009, 17:13
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Syed wrote:
Hi Bunuel!

How did you so quickly come to this conclusion that "for all other values of 'y' the equation 'x(1-y)/y' will be '-ve'?

Also, how can I strengthen my inequalitites knowledge (for GMAT).

Please explain in detail.

OK, first of all get rid of x, it's positive thus won't affect anything in the case of sign. We have (1-y)/y: even not doing any deep inquires it's obvious that we have an inequality with y and knowing nothing about it, so we can not conclude whether (1-y)/y positive or not. But if we just for practice want to determine when inequality (1-y)/y>0 holds true we can do the following:

We have 1-y and y, thus we have two check points 1 and 0 (check points y-1=0 --> y=1 and y=0). We should check three cases for (1-y)/y:

1. y<0 --> denominator y is negative, nominator is positive 1-negative=1+positive=positive, so (1-y)/y negative (positive/negative=negative)

2. 0<y<1 --> denominator y is positive, nominator also positive 1-positive number less than 1=positive, so (1-y)/y positive (positive/positive)

3. y>1 --> denominator y is positive, nominator is negative 1-positive number more than 1=negative, so (1-y)/y negative (negative/positive)

We have that (1-y)/y (and thus x(1-y)/y) is positive when y is in range (0;1) and negative when y<0 or y>1.

So statement (2) gives us two scenarios for x(1-y)/y, hence not sufficient.

Hope now it's clear.
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Re: x>0 [#permalink]

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08 Oct 2009, 17:42
Indeed! It is clear. Thanks again!

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Re: x>0 [#permalink]

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09 Oct 2009, 11:51
Quick question - couldn't we simply rewrite the question to determine y?

$$X/y>x$$ could be simply rewritten as $$x/x>y$$and therefore is $$1>y$$?

1- Tells us y is less than 1
2 - Tell us nothing about Y

Therefore A?

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Re: x>0 [#permalink]

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09 Oct 2009, 12:25
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hogann wrote:
Quick question - couldn't we simply rewrite the question to determine y?

$$X/y>x$$ could be simply rewritten as $$x/x>y$$and therefore is $$1>y$$?

1- Tells us y is less than 1
2 - Tell us nothing about Y

Therefore A?

x>0, (x/y)>x can be simplified this way x(1-y)/y>0 --> as x>0 it doesn't affect the sign of x(1-y)/y>0, so we can get rid of it --> (1-y)/y>0. This is maximum we can to do before considering the statements.

We can NO WAY rewrite x>0, (x/y)>x as "x/x>y --> 1>y" Because at this stage we don't know the sign of y.
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Re: x>0 [#permalink]

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09 Oct 2009, 12:40
Bunuel wrote:

We can NO WAY rewrite x>0, (x/y)>x as "x/x>y --> 1>y" Because at this stage we don't know the sign of y.

Exactly which is where the conditions 1 and 2 come in. Since we don't know the sign we are now asking.

Is 1 > y?

Condition 1 says yes
Condition 2 mentions nothing about Y

Is 1 > y is much easier to answer than is (x/y)>y

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Re: x>0 [#permalink]

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09 Oct 2009, 13:06
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hogann wrote:
Bunuel wrote:

We can NO WAY rewrite x>0, (x/y)>x as "x/x>y --> 1>y" Because at this stage we don't know the sign of y.

Exactly which is where the conditions 1 and 2 come in. Since we don't know the sign we are now asking.

Is 1 > y?

Condition 1 says yes
Condition 2 mentions nothing about Y

Is 1 > y is much easier to answer than is (x/y)>y

Not so. Let's consider another example:

Is 1/y>1?

(1) y<0.5
(2) y<0.3

According to your logic you would rewrite the statement "is 1/y>1?" as "is y<1"?

Afterwards you consider the statements:

(1) y<0.5 according to your logic you would say: yes y<0.5<1 so sufficient;
(2) y<0.3 according to your logic you would say: yes y<0.3<1 so sufficient;

And thus you would answer D, both are sufficient.

BUT it's WRONG: answer to my rearranged question is E not D

In our initial question it just happened to be that the statements given didn't revealed the mistake you've made in simplification, BUT generally your way is not wright. You can not multiply inequality by the variable not knowing the sign of it.
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Re: x>0 [#permalink]

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09 Oct 2009, 13:26
Bunuel wrote:

You can not multiply inequality by the variable not knowing the sign of it.

I get this rule now - the simplification only worked because y could not be negative (as indicated in condition 1)

If y was negative the sign would be flipped.

Thanks for your help!

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Re: If x > 0, is x/y > x ? [#permalink]

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considering statement 1, we can say that y>0 and a fraction. As we now know the sign of y we can safe cross multiply. Therefore, the x/y > x evaluates to --> x>xy. If y is a fraction, x is obviously a greater that xy because anything multiplied with fraction is lower than that anything i.e. x. So st1 is sufficient

considering 2 - there is no information given about y so we cannot evaluate and prove x/y > x

Therefore, answer is A

Hope it helped

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Re: If x > 0, is x/y > x ? [#permalink]

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Hi Bunuel, guys,

I have gone through everyone'e explanations and probably have found a easier solution. What do you guys suggest?

If x > 0

We can rephrase x/y > x ---> 1/y > 1 ( since x is positive just dividing x on both sides)

Hence the question is : 1/y > 1

Statement 1:

0 < y < 1

since y is a proper fraction, therefore ---> 1/proper fraction > 1

Statement 2:

x > 1 --- > no mention of x ---> insufficient

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Re: If x > 0, is x/y > x ? [#permalink]

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16 Apr 2014, 06:54
irda wrote:
Hi Bunuel, guys,

I have gone through everyone'e explanations and probably have found a easier solution. What do you guys suggest?
If x > 0
We can rephrase x/y > x ---> 1/y > 1 ( since x is positive just dividing x on both sides)
Hence the question is : 1/y > 1
Statement 1:
0 < y < 1
since y is a proper fraction, therefore ---> 1/proper fraction > 1
Statement 2:
x > 1 --- > no mention of x ---> insufficient

I did the same. Is it justified? Experts please comment.

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Re: If x > 0, is x/y > x ? [#permalink]

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