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Is y < z ?

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Is y < z ? [#permalink] New post 07 Aug 2014, 10:37
Expert's post
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

54% (01:48) correct 46% (00:52) wrong based on 50 sessions
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Re: Is y < z ? [#permalink] New post 07 Aug 2014, 23:51
(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] New post 08 Aug 2014, 02:55
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] New post 09 Aug 2014, 06:17
Bunuel can you explain it more clearly plz
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Re: Is y < z ? [#permalink] New post 09 Aug 2014, 07:07
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] New post 12 Aug 2014, 06:27
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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.

Answer: C.

Hope it's clear.
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Re: Is y < z ?   [#permalink] 12 Aug 2014, 06:27
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