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Is y < z ?

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Is y < z ? [#permalink]

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New post 07 Aug 2014, 11:37
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Is y < z ?

(1) y + z = 1

(2) y^2 < z^2
[Reveal] Spoiler: OA

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Re: Is y < z ? [#permalink]

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(1) For both \(Y=0.3, Z=0.7\) and \(Y=0.7, Z=0.3\) we have \(Y+Z=1\). Insufficient

(1) For both \(Y=0.3, Z=0.7\) or \(Y=0.3, Z=-0.7\) we have \(Y^2 < Z^2\). Insufficient

(1)+(2) From Statement (1) \(Z=1-Y\), and put in Statement (2):
\(Y^2 < (1-Y)^2\)
\(Y^2 < 1-2Y+Y^2\)
\(2Y< 1\)
\(Y<1/2\)
Therefore, \(-Y>-1/2\) and \(1-Y>1-1/2\) and \(Z=1-Y>1/2\). Hence, \(Y<Z\). Sufficient

The correct answer is C.
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Re: Is y < z ? [#permalink]

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New post 08 Aug 2014, 03:55
Statement 1 : y+z = 1 , This can be possible in different cases :
Case 1 : 0<y<1 and 0<z<1
Case 2 : y>=1 and as Z = 1 - y so z<=0
Case 3 : Z>=1 and as y = 1-z so y<=0
As we can't conclude whether y>z or y<z, statement 1 is not sufficient

Statement 2 : y^2 < z^2 , again which is possible in many cases
case 1 : z>y>0
case 2 : z>0>y and z>|y|
case 3 : y>0>z and y<|z|
case 4 : 0>y>z
As we can't conclude whether y>z or y<z, statement 2 is also not sufficient

By combining two statements and observing all the cases from both statements we can eliminate case 4 and case 3 from statement 2 as they violate the condition y + z = 1 from statement 1. So from the remaining cases which satisfy both statements we can clearly observe z>y .
Hence we can say yes y<z.

Therefore Choice [C] is the correct answer.
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Re: Is y < z ? [#permalink]

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New post 09 Aug 2014, 07:17
Bunuel can you explain it more clearly plz
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Re: Is y < z ? [#permalink]

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New post 09 Aug 2014, 08:07
Gnpth wrote:
Is \(Y<Z\)?

1) \(Y+Z=1\)

2) \(Y^2 < Z^2\)



Ans: C

Statement1 : Y & Z can take any values , so Definitely it cannot give the answer

Statement 2 : If Y = 5 & Z = 10 Then Y < Z
But if Y= 0.5 & Z = 0.2 Then y >Z

1+2 :
\(Y^2 < Z^2\)

\(Y^2 - Z^2 < 0\)

\((Y-Z)(Y+Z)<0\)

From statement 1 : Y+Z =1

Putting the value \(Y-Z< 0\)

\(Y < Z\)

For any positive value of y+z this will be true.

Answer C
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Re: Is y < z ? [#permalink]

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Is y < z ?

(1) y + z = 1. The sum of two numbers is 1. Obviously from this info we cannot say which one is bigger. Not sufficient.

(2) y^2 < z^2 --> take the square root from both sides (we can safely do that since we know that both sides are non-negative): |y| < |z| --> z is farther from 0 than y is. Not sufficient.

(1)+(2) From (2): (y - z)(y + z) < 0. This implies that y - z and y + z have opposite signs. Since we know from (1) that y + z is positive, then y - z must be negative: y - z < 0 --> y < z. Sufficient.

Answer: C.

Hope it's clear.
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Re: Is y < z ? [#permalink]

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New post 31 Mar 2016, 18:45
Gnpth wrote:
Is y < z ?

(1) y + z = 1

(2) y^2 < z^2


1. y=-3, z=4 => y+z=1 or y=4, z=-3 =y+z=1. so 1 alone is insufficient.
2. y^2 < z^2
y=2, z=-3 -> y>z
or
y=-2, z=3 -> y<z
2 alone not sufficient.

1+2
2 can rewrite:
y^2 - z^2 < 0
(y+z)(y-z)<0
we know y+z=1
so y-z<0
y<z

sufficient.

C
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Is y < z ? [#permalink]

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New post 06 Jun 2016, 11:09
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Is y < z ?

(1) y + z = 1

(2) \(y^2\) < \(z^2\)
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Re: Is y < z ? [#permalink]

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New post 06 Jun 2016, 11:36
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Re: Is y < z ? [#permalink]

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New post 06 Jun 2016, 11:39
AbdurRakib wrote:
Is y < z ?

(1) y + z = 1

(2) \(y^2\) < \(z^2\)



1) y + z = 1: Scenario 1 - y= 0.3, z = 0.7, y + z = 1 and y < z; Scenario 2 - when y = 4, z = -3, y > z. Hence, not sufficient.

2) \(y^2\) < \(z^2\), From this statement, we can only that absolute value of y is greater than absolute value of z. y =-4 and z = 5, and \(y^2\) = 16 and \(z^2\) = 25, \(y^2\) < \(z^2\), but z > y. Not sufficient.

Consider both statements together.
when both y and z are positive, y< z as \(y^2\) < \(z^2\).
when only 1 of y and z are positive, y + z = 1 -> y = 1 -z . If y is negative, z has to be positive, so y < z
when only 1 of y and z are positive, y + z = 1 -> y = 1 -z , if y is positive, z has to negative. If y = 4, z = -3. If y = 5, z = -4. If y=6, z=-5. As you can see |y| > |z|, but from statement 2, \(y^2\) < \(z^2\) or |y| < |z|. So, we can never have a scenario where y is positive and z is negative.
Note that y and z both cannot be negative as y + z = 1.
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Re: Is y < z ? [#permalink]

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New post 06 Jun 2016, 11:39
AbdurRakib wrote:
Is y < z ?

(1) y + z = 1

(2) \(y^2\) < \(z^2\)



(1) y + z = 1

If y = 5
z=-4

then y>z

if y = 0
Z= 1
z>y

Not sufficient

\(y^2\) < \(z^2\)

if y = 4 z = 5 then yes
if y= -4 z= -5 then no
if y= -4 z=5 then yes
if y= 4 z= -5 then no
if y=0 z= 1 then yes

Combining both statements

y+z= 1 and \(y^2\) < \(z^2\)
only if y= -4 z=5 or if y=0 z= 1 is possible

hence z>y

C is the answer
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Re: Is y < z ?   [#permalink] 06 Jun 2016, 11:39
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