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Is y < x+z/2

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Source: — Data Sufficiency |
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by thephoenix » Fri Apr 09, 2010 8:12 pm
JGaynor wrote:From gmatprep:

Is y<x+ z/2?

1) y-x< z -y
2) z-y>z -x/z

OA: [spoiler]C....I'm lost as to why it is C, any help will be appreciated[/spoiler]

Thank you,

Jason
is the q correct as s1 is suff in my opinion
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by reply2spg » Fri Apr 09, 2010 9:04 pm
I think this question is missing brackets. However, as per the current question I got answer as A
JGaynor wrote:From gmatprep:

Is y<x+ z/2?

1) y-x< z -y
2) z-y>z -x/z

OA: [spoiler]C....I'm lost as to why it is C, any help will be appreciated[/spoiler]

Thank you,

Jason
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by sanju09 » Sat Apr 10, 2010 2:36 am
JGaynor wrote:From gmatprep:

Is y<x+ z/2?

1) y-x< z -y
2) z-y>z -x/z

OA: [spoiler]C....I'm lost as to why it is C, any help will be appreciated[/spoiler]

Thank you,

Jason
Please let me reword

Is y < ½ (x + z)?

(1) y - x < z - y

(2) z - y > (z - x)/z

Just as to take the question for granted and verify C.

(1) This info amounts to 2 y < x + z, or y < ½ (x + z). C cannot be the OA now.

(2) This info amounts to y < z - 1 + (x/z), which gives no inspiration whatsoever to answer the core question.

Hence [spoiler]A[/spoiler] (take the question for granted)

Now please do not let me reword

Is y < x + (z/2)?

(1) y - x < z - y

(2) z - y > z - (x/z)


(1) This info amounts to 2 y < x + z, or y < ½ (x + z). C could be the OA so far, why not? Reasoning being, y < ½ (x + z) is not guaranteeing that whether or not y < x + (z/2), take different set of integers for the variables and get the different opinion about the same, if you don't believe me; though it's nowhere given in the question that the used variables are integers, but if this assumption is sufficient to negate something on test, then the assumption is good. Assumptions or the self made restrictions on the real test are bad only when it make you believe something as correct even when there's much scope in verification still pending.

(2) This info amounts to y < x/z, which alone gives no muse at all to answer the nucleus query, simply because we know nothing about the character of the variables used in place of the anonymous real numbers.

Now undertake the bull with both horns in hands

y < ½ (x + z)

y < x/z

Add them, resolve and get y < x + ¼ z + ½ (x/z) - ¾ x, and the heart question reduces to yet another question

Is ½ (x/z) - ¾ x ≥ ¼ z?

With no statements to follow, alas!

[spoiler]E[/spoiler]

Please understand the significance of an incontrovertible post on forums as this. Or at least consider editing the post once anything fishy is found in there. It normally happens...
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by JGaynor » Sat Apr 10, 2010 7:55 am
Sorry for the confusion....yes there should be brackets...

Is y<(x+z)/2

1) y-x < z-y
2) z-y > (z-x)/2

I checked again, and I chose 'A' for the answer in my GmatPrep and it was wrong- the OA is C.

Possibly an error on GmatPrep??? - anyone else have any idea..???

Jason
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by liferocks » Sat Apr 10, 2010 9:14 am
JGaynor wrote:Sorry for the confusion....yes there should be brackets...

Is y<(x+z)/2

1) y-x < z-y
2) z-y > (z-x)/2

I checked again, and I chose 'A' for the answer in my GmatPrep and it was wrong- the OA is C.

Possibly an error on GmatPrep??? - anyone else have any idea..???

Jason
if this is the correct version of question shouldn't the ans be D ie either of the questions alone is sufficient?
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by Ashish8 » Sat Apr 10, 2010 9:17 am
I got D. Can someone verify my math.

Is y<(x+z)/2.
= Is 2y<x + z ?

1. y - x < z - y
= Add y to both sides --> 2y - x < z
= Add x to both sides -- > 2y < x + z

1 is sufficient

2. z - y > (z - x)/2
= Multiply both sides by 2 --> 2(z - y) > z - x --> 2z - 2y > z - x
= Add both sides by -2z --> -2y > -z - x
= Mulitply both sides by -1 (flip inequality sign) --> 2y < z + x

2 is sufficient
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by gmatassistance » Sun Apr 11, 2010 1:06 pm
Ashish 8 - you are correct, the ans should be D - both statements alone are sufficient

JGaynor - can you please double check the answer?
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by gmatmachoman » Mon Apr 12, 2010 1:10 am
gmatassistance wrote:Ashish 8 - you are correct, the ans should be D - both statements alone are sufficient

JGaynor - can you please double check the answer?
D should be the answer!!
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by Cdawg » Mon Apr 12, 2010 9:42 am
Ashish8 wrote:I got D. Can someone verify my math.

Is y<(x+z)/2.
= Is 2y<x + z ?

1. y - x < z - y
= Add y to both sides --> 2y - x < z
= Add x to both sides -- > 2y < x + z

1 is sufficient

2. z - y > (z - x)/2
= Multiply both sides by 2 --> 2(z - y) > z - x --> 2z - 2y > z - x
= Add both sides by -2z --> -2y > -z - x
= Mulitply both sides by -1 (flip inequality sign) --> 2y < z + x

2 is sufficient
I checked the calc. - no error IMO => D is right
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