Is Y

Question

Is Y < Z ?

1. Y + Z = 1
2. Y^2 < Z^2

Folaa3 · Answer

I pick E

Nothing says X and Y are integers.

sparky · Answer

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

rthothad · Answer

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, Just curious, how did you hone in on 0.5

HIMALAYA · Answer

i think, to hold the both equation, Z must be more than 0.5. i also agree with him/her. :wink:

sparky · Answer

Z^2 > (1-Z)^2 = 1 + Z^2 - 2Z

Z>0.5

sparky is a he by the way

HIMALAYA · Answer

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..............

HIMALAYA · Answer

thanks man............. :wink:

AJB77 · Answer

Clearly each (1) and (2) alone are not sufficient. Together we have Y+Z=1 Y^20 => 1>2Y => Y < 1/2 => Z > 1/2 => Z>Y Answer is C

ywilfred · Answer

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

rthothad · Answer

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.

ywilfred · Answer

yes, sparky's approach is indeed a lot faster. Nice one :wink:

sparky · Answer

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)

HIMALAYA · Answer

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......... 8-)

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^21/2 0r 0.5.

HongHu · Answer

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

WinWinMBA · Answer

The OA is C, HongHu's answer cannot be bettered !!!!

AJB77 · Answer

HongHu, That's a kickass solution

Is Y < Z ? 1. Y + Z = 1 2. Y^2 < Z^2

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