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Is Y < Z ? 1. Y + Z = 1 2. Y^2 < Z^2 [#permalink]
 15 Jun 2005, 17:12
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Is Y < Z ?
1. Y + Z = 1
2. Y^2 < Z^2
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VP
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I pick E
Nothing says X and Y are integers.
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 18:08
WinWinMBA wrote: Is Y < Z ?
1. Y + Z = 1
2. Y^2 < Z^2
this one is a bit inolved
C.
1) insuff, even if X,Z >0 it can be clearly seen
2) insuff, obviously
1)+2) Z^2 > (1-Z)^2 => Z must be greater than 0.5, graph it to see it clearly, either as a line or two intersecting parabolas facing up.
so Z>0.5 and Y<0.5 sufficient
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 18:32
One more for C
Statement 1: y = 1-z
when z is negative y is not < z
when z is positive y may be < z
so not sufficient
Statement 2: Depends on whether y is negative or positive we will get 2 different answers, so not sufficient.
Combining these two: (1-z)^2 < y^2
when z is negative y is not < z
when z is between 0 and 1 y is not < z
when z is positive y is < z
sparky wrote: 1)+2) Z^2 > (1-Z)^2 => Z must be greater than 0.5, graph it to see it clearly, either as a line or two intersecting parabolas facing up.
so Z>0.5 and Y<0.5 sufficient
sparky, Just curious, how did you hone in on 0.5
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 18:54
rthothad wrote: sparky wrote: 1) + 2) Z^2 > (1-Z)^2 => Z must be greater than 0.5, graph it to see it clearly, either as a line or two intersecting parabolas facing up.
so Z>0.5 and Y<0.5 sufficient sparky, Just curious, how did you hone in on 0.5 i think, to hold the both equation, Z must be more than 0.5. i also agree with him/her. 
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Director
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Z^2 > (1-Z)^2 = 1 + Z^2 - 2Z
Z>0.5
sparky is a he by the way
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 19:18
rthothad wrote: HIMALAYA wrote: i think, to hold the both equation, Z must be more than 0.5. i also agree with him/her. Any calculations? WinWinMBA wrote: Is Y < Z ?
1. Y + Z = 1
2. Y^2 < Z^2
we know, from i, y and z could be anything, but their sum must be 1.
from ii, y and z could be both +ve and -ve.
from i and ii, z cannot be smaller than y because if z is smaller than y, the equation in i and ii do not hold.
so lets plug innnnnnnnnnnnnn......
if z = 1, y = 0=> possible
if z = 10, y = -9=>possible
if z = 0, y = 1 = > not possible because the eq z^2>y^2 doesnot satisy. so any value for z less than the y"s value doesnot satisfy the eqs.
so now lets plug in for some other values that are lowest for z:
if z = 0.9, y = .1, both equations are satisfied.
if z = 0.6, y = .4, both equations are satisfied.
if z = 0.51, y = .49, both equations are satisfied.
if z = 0.5, y = .5, both equations are not satisfied. so we cannot go for any value for z that is not more than 0.5 because sq of z must greater than the sq of y.
hope it is clear..............
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sparky wrote: Z^2 > (1-Z)^2 = 1 + Z^2 - 2Z
Z>0.5. sparky is a he by the way
thanks man.............
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Senior Manager
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Clearly each (1) and (2) alone are not sufficient.
Together we have
Y+Z=1
Y^2<Z^2
or
Y^2<(1-Y)^2
Y^2<Y^2 - 2Y +1
1-2Y>0 => 1>2Y => Y < 1/2
=> Z > 1/2
=> Z>Y
Answer is C
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1) Y and Z can be any value and Y can be > or < Z and both still add up to 1
2) Y^2-Z^2 < 0
(Y+Z)(Y-Z)<0
-Z < Y < Z
- not sufficient since dependent on sign of Y
Using 1 and 2,
if -z = -1 and y = 0, thn Y+Z = 1 and y<z
if -z = 0 and y=1, then Y+Z = 1 y>Z
I take E
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 19:30
HIMALAYA wrote: so lets plug innnnnnnnnnnnnn......
hope it is clear..............
HIMALAYA, Thanks for your time. Your thought process is similar to mine. plugging innnnnnnnnnnn. But I was curious to know of a less time consuming approach.
Sparky's approach looks a lot quicker than our plugging innnnnnnn.
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 19:38
rthothad wrote: HIMALAYA wrote: so lets plug innnnnnnnnnnnnn......
hope it is clear.............. HIMALAYA, Thanks for your time. Your thought process is similar to mine. plugging innnnnnnnnnnn. But I was curious to know of a less time consuming approach. Sparky's approach looks a lot quicker than our plugging innnnnnnn. yes, sparky's approach is indeed a lot faster. Nice one
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Director
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I just want to say something. When dealing with inequialities, try to solve them.
If you are plugging numbers on inequalities questions, you are pretty much doomed, unless you know what numbers to plug. But to know what numbers to plug, you need to have a general notion/sense how that particular inequality works which you might not have unless you have a lot of experince or are in a genious category.
Since it's all about signs try to get them in the following form
()*()/()*() >< 0
I bet you the things in brackets will correspond closely to what is given in 1) and 2)
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 20:00
ywilfred wrote: rthothad wrote: HIMALAYA wrote: so lets plug innnnnnnnnnnnnn......
hope it is clear.............. HIMALAYA, Thanks for your time. Your thought process is similar to mine. plugging innnnnnnnnnnn. But I was curious to know of a less time consuming approach. Sparky's approach looks a lot quicker than our plugging innnnnnnn. yes, sparky's approach is indeed a lot faster. Nice one ohhh yes, i agree with you guys. just trying to show how values of y and z are <0.5 and more than 0.5 respectively......... 
we can do in many ways, but, of course, we need a shortest one. this could be one of them:
from i, z=1-y ..............................1
from ii, y^2<z^2 .........................2
lets put the value of z in 1 to eq 2: y^2 <(1-y)^2
solving the eq ....... y^2 < 1-2y+y^2
2y<1
so y<1/2. if y<1/2 or 0.5, then z >1/2 0r 0.5.
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Re: DS: Value comparisons [#permalink]
15 Jun 2005, 21:12
WinWinMBA wrote: Is Y < Z ?
1. Y + Z = 1
2. Y^2 < Z^2
No confusion about 1 or 2 being insufficient.
Combined:
From 2:
(Y+Z)(Y-Z)<0
From 1:
Y+Z=1
=>Y-Z<0
=>Y<Z
Should be finished in 30 seconds.
_________________
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keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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Director
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The OA is C, HongHu's answer cannot be bettered !!!!
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Senior Manager
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HongHu, That's a kickass solution
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