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Is yx -z > x - yz? (1) y < 1 (2) x > 1

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Is yx -z > x - yz? (1) y < 1 (2) x > 1  [#permalink]

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New post 03 Mar 2019, 07:49


  45% (medium)

Question Stats:

60% (01:24) correct 40% (01:48) wrong based on 15 sessions

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Is yx -z > x - yz?

(1) y < 1

(2) x > 1


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Joined: 31 Oct 2013
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Concentration: Accounting, Finance
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Re: Is yx -z > x - yz? (1) y < 1 (2) x > 1  [#permalink]

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New post 03 Mar 2019, 07:58
rohan2345 wrote:
Is yx-z>x-yz?

(1) y<1

(2) x>1

yx - z > x - yz

yx - x > z - yz

x(y - 1) > z( 1 - y)

Important thing is that this question has 3 variables. we don't have any idea about their sign .

Statement 1: y<1. No information about x and z. NOT sufficient.

Statement 2: x>1. NO information about y and z.

Combining both statement :

Still No idea about z.

Since all these variables are in multiplication relationship, without knowing the sign of these variables , it's impossbile to answer this question.

The best answer is E.
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Re: Is yx -z > x - yz? (1) y < 1 (2) x > 1  [#permalink]

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New post 05 Mar 2019, 07:02
Top Contributor
Hi Selim,

Inequalities questions become particularly tricky if you do not have a structure to solve them. People tend to struggle on this topic because they go with a process of plugging in, rather than choosing to analyze. Plugging in values should always be the last mode of attack on any Inequality or Absolute Value question.

The process on analysis works best when you keep the following points in mind while solving inequalities

1. Always breakdown the question stem

The key to solving Inequality questions is to always breakdown the question stem and rephrase the data sufficiency.

2. Always simplify the inequality by keeping the RHS as '0' and maintain the LHS as a product or a division of values by factoring out common terms

This makes the analysis process a lot easier. Let us consider xy < 0 and x + y < 0. If xy < 0, then x and y both have different signs. This is very straightforward and easy to analyze. The same can be said about x/y < 0. But when we look at x + y < 0, there are three cases that apply here

1. x and y both are negative
2. x is positive and y is negative, and the magnitude of y is greater than the magnitude of x
3. x is negative and y is positive, and the magnitude of x is greater than the magnitude of y

So clearly when we have an addition or a subtraction, in addition to the signs we need to worry about the magnitudes as well. This is not the case with a multiplication and division.

2. Always factor out even powered terms (x^2, x^4....) as the sign of these terms will always be either 0 or positive, so they will have no impact on the sign of the entire quantity given.

Keep these three points in mind as hygiene factors while solving questions based on Inequalities.

Now let us consider the question stem given.

Is yx - z > x - yz -----> Taking terms to the LHS.

Is yx - z - x + yz > 0 ------> yx + yz - z - x > 0 -----> y(x + z) -1(z + x) -----> (x + z)(y - 1) > 0

Statement 1 : y < 1 -----> y - 1 < 0

We do not know the sign of x + z, x + z can either be positive or negative. Insufficient.

Statement 2 : x > 1

This does not tell you anything about y and z. Insufficient.

Combined the statements are again insufficient since we do not have any information about z.

Answer : E

- CrackVerbal Prep Team

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Re: Is yx -z > x - yz? (1) y < 1 (2) x > 1   [#permalink] 05 Mar 2019, 07:02
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