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If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2

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If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2  [#permalink]

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New post 21 Dec 2018, 02:24
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  55% (hard)

Question Stats:

60% (01:48) correct 40% (01:34) wrong based on 88 sessions

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Re: If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2  [#permalink]

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New post 21 Dec 2018, 11:39
by analyzing the initial question, it is actually asking if x is (+ve) fraction less than 1 or not?

for statement 1: y/x > y,
if y is (+ve), so x is (+ve) fraction less than 1
if y is (-ve), so either x is (+ve) integer or fraction more than 1 or x is (-ve).

for statement 2: x^3 > x^2
it confirms that x must be (+ve) integer or fraction more than 1
which gives a definite No to the question.

so the answer is B IMO
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Re: If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2  [#permalink]

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New post 21 Dec 2018, 17:13
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Bunuel wrote:
If x is not equal to zero, is 1/x > 1 ?


(1) y/x > y

(2) x^3 > x^2


Question: start by figuring out what the question is really asking. You can't just multiply both sides of the inequality by x, unfortunately. The reason you can't do that is that x might be negative, in which case you'd have to flip the inequality sign. Since you don't know whether you need to do that or not, you can't simplify by multiplying by x.

Instead, use some logic to figure out what would make the answer 'yes' and what would make the answer 'no'. If x was negative, the answer would always be 'no', since 1/x would also be negative.

If x was positive, the answer would be 'no' if x was greater than (or equal to) 1, and 'yes' if x was less than 1.

Basically, the question is asking this:
- is x a positive number less than 1? Or, is it some other type of number?

Statement 2: I'm going to start here, because this statement only contains x, while the other statement adds a new variable. So, this one might be simpler.

If x^3 > x^2, then we can divide both sides by x^2. (We're allowed to do that, since x^2 is always positive, unlike x itself.)

That tells us that x > 1.

So, x isn't a positive number less than 1. The answer is a definite 'no', so the statement is sufficient.

Statement 1: y/x > y. As with the question, we can't just divide both sides by y, because y might be negative. Let's just try some numbers for y, since y could apparently be anything!

If y = 1, then we have 1/x > 1. That's the question we were asked in the first place! So, the answer would be 'yes'.

But if y = -1, we have -1/x > -1, which simplifies to 1/x < 1. The answer there would be 'no'.

Since the answer could be either 'yes' or 'no', this statement is insufficient.

The answer is B.
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If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2  [#permalink]

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New post 20 Jun 2019, 23:34
I had a doubt what if X=0.2
Then 1/X >X

Posted from my mobile device
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Re: If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2  [#permalink]

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New post 21 Jun 2019, 02:44
vinayyadav512 wrote:
I had a doubt what if X=0.2
Then 1/X >X

Posted from my mobile device


But that would violate the condition mentioned in the second statement:

x^3 > x^2
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Re: If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2  [#permalink]

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New post 21 Jun 2019, 12:36
Bunuel wrote:
If x is not equal to zero, is 1/x > 1 ?


(1) y/x > y

(2) x^3 > x^2



is 1/x > 1 ?
--> It's only possible when 0 < x < 1

(1) y/x > y
2 cases are possible

y = -ve & x > 1
Eg: -4/2 > -4 --> -2 > -4

y = +ve & 0<x<1
Eg: 2/0.5 > 2 --> 4 > 2

Insufficient

(2) x^3 > x^2

If x = -ve Eg: (-1)^3 > (-1)^2 --> -1 > 1 -- NOT Possible
If 0<x<1 Eg: (1/2)^3 > (1/2)^2 --> 0.125 > 0.25 -- NOT Possible
If x > 1 Eg: (2)^3 > (2)^2 --> 8 > 4 -- Possible

So, x > 1

Is 1/x > 1 ? Definitely NO

Sufficient

IMO Option B

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Re: If x is not equal to zero, is 1/x > 1 ? (1) y/x > y (2) x^3 > x^2   [#permalink] 21 Jun 2019, 12:36
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