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Re: should we always solve DS q's with taking sample numbers ? [#permalink]

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19 Jul 2008, 23:08

1

This post received KUDOS

1) Tells us 2x-2y =1. This tells us definitely that x > y however both x and y can greater, eqaul to or less than 0. So A and D out.

2) x/y > 1 This tells us x > y if y is +ve (which means x is +ve as well) Or x < y is y is -ve (which means x is -ve as well). Again two possible values so not possible.

Combining we have x > y from 1 and 2 says x > y only if x and y are positive. So Answer C.

However having solved that, I will concur that it is pretty trick problem and for most of the people picking number is better strategy.

I would do the problem exactly as abhijit did above. To this question:

"should we always solve DS q's with taking sample numbers ?"

I would say no, not 'always'. I'd only try to solve DS questions with sample numbers if you cannot solve the problem conceptually or algebraically. It is often the case that if you don't pick a sufficient variety of sample numbers, you will reach an incorrect conclusion, and therefore a wrong answer.
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Re: should we always solve DS q's with taking sample numbers ? [#permalink]

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19 Jul 2008, 23:47

some people get it and some people dont but i feel that if you have two variablee inequality DS problem, plotting the conditions on xy plane is the best method.. only thing is, dont try this in exam without practise

question : x,y > 0 ... target area is Q1

statement 1 : line passes through Q1, Q2 and Q3 ... not suff statement 2 : green area ... in Q1 and Q3 .. not suff

combine, line segament in blue, alwasy in Q1 ...suff ..answer

Re: should we always solve DS q's with taking sample numbers ? [#permalink]

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20 Jul 2008, 00:33

abhaypratapsingh wrote:

Let take one more example : On number line , the distance between x and y is greater than distance between x and z. does z lie between x and y ?

1) xyz <0 2) xy<0

Now this resolves to is |X-Y| > |X-Z| ?

lets try solve algebracally and by sample data ... and see how much time each takes ...

is it E

statement 1 ... clearly not suff ... x,y,z all can be -ve and you never know z was on which side of x statement 2 ... not suff ... doesn tell anything about Z

combine : ...z is Positive and one of x or y is negative ....

if y is negative .... both x and z are positive and z can be on either side of x ... not suff

Re: should we always solve DS q's with taking sample numbers ? [#permalink]

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30 Jul 2008, 16:27

durgesh79 wrote:

some people get it and some people dont but i feel that if you have two variablee inequality DS problem, plotting the conditions on xy plane is the best method.. only thing is, dont try this in exam without practise

question : x,y > 0 ... target area is Q1

statement 1 : line passes through Q1, Q2 and Q3 ... not suff statement 2 : green area ... in Q1 and Q3 .. not suff

combine, line segament in blue, alwasy in Q1 ...suff ..answer

Attachment:

DSQ3.JPG

Isn't the green area colored incorrectly? If x/y > 1, that means that x>y. The graph should than be colored such that everything to the right of the line y=x is green. Am I missing something?

Isn't the green area colored incorrectly? If x/y > 1, that means that x>y. The graph should than be colored such that everything to the right of the line y=x is green. Am I missing something?

If x/y > 1, that means that:

x > y if y > 0 x < y if y < 0

Remember, we need to reverse the inequality if we multiply both sides by a negative number, and y can certainly be negative.
_________________

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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: should we always solve DS q's with taking sample numbers ? [#permalink]

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30 Jul 2008, 18:52

IanStewart wrote:

mrblack wrote:

Isn't the green area colored incorrectly? If x/y > 1, that means that x>y. The graph should than be colored such that everything to the right of the line y=x is green. Am I missing something?

If x/y > 1, that means that:

x > y if y > 0 x < y if y < 0

Remember, we need to reverse the inequality if we multiply both sides by a negative number, and y can certainly be negative.