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# Is y < z ?

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Is y < z ? [#permalink]

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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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08 Aug 2014, 00:51
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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

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Re: Is y < z ? [#permalink]

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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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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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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.

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Re: Is y < z ? [#permalink]

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12 Aug 2014, 07:27
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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.

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

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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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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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06 Jun 2016, 11:36
AbdurRakib wrote:
Is y < z ?

(1) y + z = 1

(2) $$y^2$$ < $$z^2$$

Merging topics. Please refer to the discussion above.
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Re: Is y < z ? [#permalink]

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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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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

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