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If x and y are integers such that x<0<y, and z is non

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If x and y are integers such that x<0<y, and z is non  [#permalink]

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New post 24 Jan 2012, 09:49
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If x and y are integers such that x<0<y, and z is non negative integer then which of the following must be true?

A. \(x^2<y^2\)

B. \(x+y=0\)

C. \(xz<yz\)

D. \(xz=yz\)

E. \(\frac{x}{y}<z\)
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Re: PS-which of the following must be true  [#permalink]

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New post 24 Jan 2012, 09:59
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LM wrote:
If x and y are integers such that \(x<0<y\),and z is non negative integer then which of the following must be true?

A) \(x^2<y^2\)

B) \(x+y=0\)

C) \(xz<yz\)

D)\(xz=yz\)

E) \(\frac{x}{y}<z\)


Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

Given: \(x<0<y\) and \({0}\leq{z}\).

Evaluate each option:

A. \(x^2<y^2\) --> not necessarily true, for example: \(x=-2\) and \(y=1\);

B. \(x+y=0\) --> not necessarily true, for example: \(x=-2\) and \(y=1\);

C. \(xz<yz\) --> not necessarily true, if \(z=0\) then \(xz=yz=0\);

D. \(xz=yz\) --> not necessarily true, it's true only for \(z=0\);

E. \(\frac{x}{y}<z\) --> as \(x<0<y\) then \(\frac{x}{y}=\frac{negative}{positive}=negative<0\) and as \({0}\leq{z}\) then \(\frac{x}{y}<0\leq{z}\) --> always true.

Answer: E.
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Re: PS-which of the following must be true  [#permalink]

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New post 22 May 2012, 01:19
Bunuel wrote:
LM wrote:
If x and y are integers such that \(x<0<y\),and z is non negative integer then which of the following must be true?

A) \(x^2<y^2\)

B) \(x+y=0\)

C) \(xz<yz\)

D)\(xz=yz\)

E) \(\frac{x}{y}<z\)


Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

Given: \(x<0<y\) and \({0}\leq{z}\).

Evaluate each option:
A) \(x^2<y^2\) --> not necessarily true, for example: \(x=-2\) and \(y=1\);

B) \(x+y=0\) --> not necessarily true, for example: \(x=-2\) and \(y=1\);

C) \(xz<yz\) --> not necessarily true, if \(z=0\) then \(xy=yz=0\);

D)\(xz=yz\) --> not necessarily true, it's true only for \(z=0\);

E) \(\frac{x}{y}<z\) --> as \(x<0<y\) then \(\frac{x}{y}=\frac{negative}{positive}=negative<0\) and as \({0}\leq{z}\) then \(\frac{x}{y}<0\leq{z}\) --> always true.

Answer: E.



amazing ! couldn't figure out how option 3 was not necessarily true , forgot that non negative could mean that 0 is possible ,folks : non negative does not mean only positive integers , it could be 0 as well

Hypothetically speaking, Bunuel so if a question says, non positive numbers can we consider 0 as well , rather than only negative numbers.

Set of Non positive numbers { 0,-1,-5,-9 }
Set of Non negative numbers { 0,1, 4, 7, }

is this correct?
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Re: PS-which of the following must be true  [#permalink]

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New post 22 May 2012, 01:23
Joy111 wrote:
Bunuel wrote:
LM wrote:
If x and y are integers such that \(x<0<y\),and z is non negative integer then which of the following must be true?

A) \(x^2<y^2\)

B) \(x+y=0\)

C) \(xz<yz\)

D)\(xz=yz\)

E) \(\frac{x}{y}<z\)


Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

Given: \(x<0<y\) and \({0}\leq{z}\).

Evaluate each option:
A) \(x^2<y^2\) --> not necessarily true, for example: \(x=-2\) and \(y=1\);

B) \(x+y=0\) --> not necessarily true, for example: \(x=-2\) and \(y=1\);

C) \(xz<yz\) --> not necessarily true, if \(z=0\) then \(xy=yz=0\);

D)\(xz=yz\) --> not necessarily true, it's true only for \(z=0\);

E) \(\frac{x}{y}<z\) --> as \(x<0<y\) then \(\frac{x}{y}=\frac{negative}{positive}=negative<0\) and as \({0}\leq{z}\) then \(\frac{x}{y}<0\leq{z}\) --> always true.

Answer: E.



amazing ! couldn't figure out how option 3 was not necessarily true , forgot that non negative could mean that 0 is possible ,folks : non negative does not mean only positive integers , it could be 0 as well

Hypothetically speaking, Bunuel so if a question says, non positive numbers can we consider 0 as well , rather than only negative numbers.

Set of Non positive numbers { 0,-1,-5,-9 }
Set of Non negative numbers { 0,1, 4, 7, }

is this correct?


Yes, a set of non-positive numbers consists of zero and negative numbers.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: PS-which of the following must be true  [#permalink]

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New post 24 May 2012, 02:08
Bunuel wrote:
Yes, a set of non-positive numbers consists of zero and negative numbers.


Isn't that one of the first few things one gets to learn when trying to read the number system. Thank you bunnel for reminding everyone about it :)
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Re: If x and y are integers such that x<0<y, and z is non  [#permalink]

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New post 25 May 2012, 22:34
So 0 "Zero" is even
and can be part of both a NON Postive set and a NON negative set
Set of Non positive numbers { 0,-1,-5,-9 }
Set of Non negative numbers { 0,1, 4, 7, }
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Re: If x and y are integers such that x<0<y, and z is non  [#permalink]

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New post 04 Mar 2014, 10:42
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Re: If x and y are integers such that x<0<y, and z is non  [#permalink]

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New post 14 Sep 2016, 07:09
what a nice trap here!!!
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Re: If x and y are integers such that x<0<y, and z is non  [#permalink]

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Re: If x and y are integers such that x<0<y, and z is non &nbs [#permalink] 02 Sep 2018, 11:25
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