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# I'm having trouble with DS Yes/No questions.... I find that

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VP
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I'm having trouble with DS Yes/No questions.... I find that [#permalink]  23 Oct 2005, 21:48
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I'm having trouble with DS Yes/No questions.... I find that I always run out of time when trying to plug in different numbers. And more often than not, my answer is wrong. I've read the sticky post in this forum about picking smart numbers but these are still quite intimidating to me.

Below are some questions from OG.......

1. If x does not equal to -y, is (x-y)/(x+y) > 1?

(1) x>0
(2) y<0

2. Is lxl = y-z?

(1) x + y = z
(2) x<0

3. Is 1/(a-b)<b-a?

(1) a < b
(2) 1<la-bl

1. OA is E
2. OA is C
3. OA is A

Here is how I attacked the problems:

1. Is (x-y)/(x+y) > 1?

After simplifying, I get x-y > x+y, which got simplified to y < 0?

So (1) is obviously insufficient. And (2) is! But of course, as my luck would have it, the answer is E!

3. Is 1/(a-b)<b-a?

After simplifying, I get 1 < (b-a) (a-b)

From (1) we know that a-b<0 and b-a>0, from the equation above, 1 < -VE! I would've picked BCE right off the bat (as my luck would have it, I'm wrong again!) But if I left the equation the way it is, with the 1/(a-b)<b-a, I would've gotten it right; since 1/-VE < +VE.

It seems to me that simplifying is not always the easiest way, leaving them in their original format might be easier to comprehend. Or am I nuts? (as I think I am)

Last edited by TeHCM on 25 Oct 2005, 17:48, edited 1 time in total.
Senior Manager
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Here is my try:

1. If x does not equal to -y, is (x-y)/(x+y) > 1?

(1) x>0
(2) y<0

Clearly stmt 1 & 2 are insuff. Combining both,

x - y = +ve, however x + y can be +ve or -ve depending the which one is greater. Hence Insufficient.

2. Is lxl = y-z?

(1) x + y = z
(2) x<0

1) -x = y - z, But |x| would always give +ve value. Hence sufficient to answer the question.
2) Insufficient as we do not know y & z

3. Is 1/(a-b)<b-a?

(1) a < b
(2) 1<la-bl

1) provies a - b is -ve and b - a is +ve. Hence 1/-ve < +ve - Suff to answer.
2) |a-b| > 1, which gives a-b is -2 or 2, -3 or 3 and so on. Which is insufficient to answer.
Manager
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Re: Advices - Having trouble with Yes/No DS [#permalink]  23 Oct 2005, 22:12
TeHCM wrote:
I'm having trouble with DS Yes/No questions.... I find that I always run out of time when trying to plug in different numbers. And more often than not, my answer is wrong. I've read the sticky post in this forum about picking smart numbers but these are still quite intimidating to me.

Below are some questions from OG.......

1. If x does not equal to -y, is (x-y)/(x+y) > 1?

(1) x>0
(2) y<0

2. Is lxl = y-z?

(1) x + y = z
(2) x<0

3. Is 1/(a-b)<b-a?

(1) a < b
(2) 1<la-bl

I always find using algebra to be simpler than substituting numbers.

in question 1. is (x-y)/(x+y)>1 can be rewritten as is (x-y)>(x+y)? or is y<0?

x is immaterial, as it will cancel out in the equation.
statement I gives us x>0, insufficient.
statement II gives us y<0, which is sufficient.

question 2. is |x|=y-z
statement 1. x+y=z or x = -y+z or -x=y-z, therefore |x|=y-z. sufficient.
statement 2. x<0. which can translate to is y-z positive? this cannot be answered, thus insufficient.
VP
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Re: Advices - Having trouble with Yes/No DS [#permalink]  24 Oct 2005, 01:03
TeHCM wrote:

1. If x does not equal to -y, is (x-y)/(x+y) > 1?

(1) x>0
(2) y<0

2. Is lxl = y-z?

(1) x + y = z
(2) x<0

3. Is 1/(a-b)<b-a?

(1) a < b
(2) 1<la-bl

1. (x-y)/(x+y) > 1 ? means -2y/(x+y) > 0 ?
1) insuff and 2) insuff and 1)+2) as well b/c we dont know the values. E)...

2. Is lxl = y-z. basically the question is asking whether y-z is +ve ?

1)x + y = z means -x=y-z but we dont know x and so insuff
2) x is smaller than 0 but we dont know y and z
1)+2) we know that x is negative and y-z is x. hence, suff. C)...

3. Is 1/(a-b)<b-a? means 0>(a-b)^2+a-b ?

1) a-b<0 suff b/c (a-b)^2 is always +ve
2) insuff b/c |a-b| can be -4 or 4 or any number smaller or bigger than -1 or 1
A)...
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Director
Joined: 10 Oct 2005
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I dont agree with Christoph

Christoph wrote -
2. Is lxl = y-z. basically the question is asking whether y-z is +ve ?

1)x + y = z means -x=y-z but we dont know x and so insuff
2) x is smaller than 0 but we dont know y and z
1)+2) we know that x is negative and y-z is x. hence, suff. C)...

For any y = |x| , y = +x or -x. Both are always true. So if we have an equation y = x we can say that y = |x|. so 1 is sufficient.

BTW what is OA.
VP
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Re: Advices - Having trouble with Yes/No DS [#permalink]  24 Oct 2005, 08:51
chets wrote:

I always find using algebra to be simpler than substituting numbers.

in question 1. is (x-y)/(x+y)>1 can be rewritten as is (x-y)>(x+y)? or is y<0?

x is immaterial, as it will cancel out in the equation.
statement I gives us x>0, insufficient.
statement II gives us y<0, which is sufficient.

question 2. is |x|=y-z
statement 1. x+y=z or x = -y+z or -x=y-z, therefore |x|=y-z. sufficient.
statement 2. x<0. which can translate to is y-z positive? this cannot be answered, thus insufficient.

Very good explanation. Thanks. For (1) I was gonna chose C as I missed cancelling x from both side. Dont know how to remember these small yet important things.

Also agree that it is always better to use algebra to simplify before picking numbers (if any).
Manager
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Quote:
I always find using algebra to be simpler than substituting numbers.

in question 1. is (x-y)/(x+y)>1 can be rewritten as is (x-y)>(x+y)? or is y<0?

x is immaterial, as it will cancel out in the equation.
statement I gives us x>0, insufficient.
statement II gives us y<0, which is sufficient.

This doesn't hold up when you plug in numbers.

If x=0 and y=-1:
(0+1)/(0-1)=-1 < 1

If x=100 and y=-1:
(100+1)/(100-1) > 1

Thus, statement 2 by itself is insufficient.

If x=1 and y=-1:
(1+1)/(1-1) = undefined

Thus, statements 1 and 2 together are insufficient.

The correct answer should be E.
Manager
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BumblebeeMan wrote:
Quote:
I always find using algebra to be simpler than substituting numbers.

in question 1. is (x-y)/(x+y)>1 can be rewritten as is (x-y)>(x+y)? or is y<0?

x is immaterial, as it will cancel out in the equation.
statement I gives us x>0, insufficient.
statement II gives us y<0, which is sufficient.

This doesn't hold up when you plug in numbers.

If x=0 and y=-1:
(0+1)/(0-1)=-1 < 1

If x=100 and y=-1:
(100+1)/(100-1) > 1

Thus, statement 2 by itself is insufficient.

If x=1 and y=-1:
(1+1)/(1-1) = undefined

Thus, statements 1 and 2 together are insufficient.

The correct answer should be E.

the question stem says x is not equal to -y. in your example with x=1 and y =-1, x is equal to -y.

Manager
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chets wrote:
BumblebeeMan wrote:
Quote:
I always find using algebra to be simpler than substituting numbers.

in question 1. is (x-y)/(x+y)>1 can be rewritten as is (x-y)>(x+y)? or is y<0?

x is immaterial, as it will cancel out in the equation.
statement I gives us x>0, insufficient.
statement II gives us y<0, which is sufficient.

This doesn't hold up when you plug in numbers.

If x=0 and y=-1:
(0+1)/(0-1)=-1 < 1

If x=100 and y=-1:
(100+1)/(100-1) > 1

Thus, statement 2 by itself is insufficient.

If x=1 and y=-1:
(1+1)/(1-1) = undefined

Thus, statements 1 and 2 together are insufficient.

The correct answer should be E.

the question stem says x is not equal to -y. in your example with x=1 and y =-1, x is equal to -y.

The algebraic explanation is logical. I also agree that the x=1 , y=-1 example is not in line with the question stem..... but if I plug in another set of numbers, I still get insufficient. Am I doing something wrong?

x= 4, y=-2

(4+2)/(4-2) = 6/2 >1 = 3>1 YES

x= -4, y=-2

(-4+2)/(-4-2) = -2/-6 = 1/3>1 NO
Manager
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This doesn't hold up when you plug in numbers.

If x=0 and y=-1:
(0+1)/(0-1)=-1 < 1

If x=100 and y=-1:
(100+1)/(100-1) > 1

Thus, statement 2 by itself is insufficient.

If x=1 and y=-1:
(1+1)/(1-1) = undefined

Thus, statements 1 and 2 together are insufficient.

The correct answer should be E.[/quote]

the question stem says x is not equal to -y. in your example with x=1 and y =-1, x is equal to -y.

The algebraic explanation is logical. I also agree that the x=1 , y=-1 example is not in line with the question stem..... but if I plug in another set of numbers, I still get insufficient. Am I doing something wrong?

x= 4, y=-2

(4+2)/(4-2) = 6/2 >1 = 3>1 YES

x= -4, y=-2

(-4+2)/(-4-2) = -2/-6 = 1/3>1 NO
Senior Manager
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TeHCM - you're not alone. I posted a similar topic last night and received some good advice...

http://www.gmatclub.com/phpbb/viewtopic.php?t=21933

I think it all has to do with plugging in all the right numbers, in the right equations, in the right amount of time...with the right amount of beer.

For number 1, I like the algebraic approach Chet mentioned. Without it, I picked C. With it, I picked B.

BTW - what does "VE" stand for?
Manager
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ValleyBall1 wrote:
TeHCM - you're not alone. I posted a similar topic last night and received some good advice...

http://www.gmatclub.com/phpbb/viewtopic.php?t=21933

I think it all has to do with plugging in all the right numbers, in the right equations, in the right amount of time...with the right amount of beer.

For number 1, I like the algebraic approach Chet mentioned. Without it, I picked C. With it, I picked B.

BTW - what does "VE" stand for?

mhm was wondering that as well... but now I think -ve = negative and +ve = positive :D
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Re: Advices - Having trouble with Yes/No DS [#permalink]  25 Oct 2005, 17:32
chets wrote:
I always find using algebra to be simpler than substituting numbers.

Yes, this is VERY important. You should almost always try to work the algebra out for this kind of questions and just use number picking as a way to verify or explore. It'll also be eaiser to pick the right numbers if you've work the algebra out.
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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

VP
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1. OA is E
2. OA is C
3. OA is A

Here is how I attacked the problems:

1. Is (x-y)/(x+y) > 1?

After simplifying, I get x-y > x+y, which got simplified to y < 0?

So (1) is obviously insufficient. And (2) is! But of course, as my luck would have it, the answer is E!

3. Is 1/(a-b)<b-a?

After simplifying, I get 1 < (b-a) (a-b)

From (1) we know that a-b<0 and b-a>0, from the equation above, 1 < -VE! I would've picked BCE right off the bat (as my luck would have it, I'm wrong again!) But if I left the equation the way it is, with the 1/(a-b)<b-a, I would've gotten it right; since 1/-VE < +VE.

It seems to me that simplifying is not always the easiest way, leaving them in their original format might be easier to comprehend. Or am I nuts? (as I think I am)
Senior Manager
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I guess whether you rephrase & simplify or leave the statement as is depends on the statement and what you're solving for. I'm guessing this is a skilled art that I have not even come close to mastering.
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Re: Advices - Having trouble with Yes/No DS [#permalink]  25 Oct 2005, 18:19
CHETS and ALL: Be careful when you expand algebric expression especially when you have inequalities....

(x-y)/(x+y)>1 ?

to simplify you got (x-y)>(x+y) <---this is wrong...

without knowing if (x+y) is positive or negative you cannot do this....

so One you need to know is, is X>y? is X>0? is |x|>|y|...

without this info you cannot expand the equation as you did...be careful on the exam this is a trap that ETS expects you to take...

chets wrote:
TeHCM wrote:
I'm having trouble with DS Yes/No questions.... I find that I always run out of time when trying to plug in different numbers. And more often than not, my answer is wrong. I've read the sticky post in this forum about picking smart numbers but these are still quite intimidating to me.

Below are some questions from OG.......

1. If x does not equal to -y, is (x-y)/(x+y) > 1?

(1) x>0
(2) y<0

2. Is lxl = y-z?

(1) x + y = z
(2) x<0

3. Is 1/(a-b)<b-a?

(1) a < b
(2) 1<la-bl

I always find using algebra to be simpler than substituting numbers.

in question 1. is (x-y)/(x+y)>1 can be rewritten as is (x-y)>(x+y)? or is y<0?

x is immaterial, as it will cancel out in the equation.
statement I gives us x>0, insufficient.
statement II gives us y<0, which is sufficient.

question 2. is |x|=y-z
statement 1. x+y=z or x = -y+z or -x=y-z, therefore |x|=y-z. sufficient.
statement 2. x<0. which can translate to is y-z positive? this cannot be answered, thus insufficient.
VP
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fresinha12, thank you! I know where I did wrong now!

Without knowing if (x+y) is pos. or neg, I cannot simplify for sure without leaving the signs unchanged. I could sleep better now.
Senior Manager
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Sleep is good.

Good point fresinha!
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