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# IN the number expressed as O.xyz is O.xyz >2/3 1

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IN the number expressed as O.xyz is O.xyz >2/3 1 [#permalink]

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06 Sep 2006, 08:23
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IN the number expressed as O.xyz
is O.xyz >2/3

1 x+y>13

2 x+z>14
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06 Sep 2006, 09:11
Yes/No question.
A,D and B are out

Must be b/n C and E.

I'm getting C --> 0.689
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Last edited by X & Y on 06 Sep 2006, 09:31, edited 1 time in total.
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06 Sep 2006, 09:31
One more for C
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06 Sep 2006, 09:43

S1: x + y > 13

@ x = 5, y = 9, # = 0.59z < 2/3
@ x = 7, y = 7, # = 0.77z > 2/3

Not sufficient.

S2: x + z > 14
@ x = 6, z= 9, # = 0.6y9
If y = 0, 0.609 < 2/3
If y = 7, 0.679 > 2/3
Not sufficient.

S1 & S2: x+y > 13, x+z > 14
@ x = 6, y=8, z = 9, #=0.689 > 2/3
@ x = 7, y=7, z =8, # = 0.778 > 2/3

Sufficient.

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06 Sep 2006, 09:44
X & Y wrote:
Yes/No question.
A,D and B are out

Must be b/n C and E.

I'm getting C --> 0.689

How did you arrive at the asnwer?
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06 Sep 2006, 10:16
The thing here is to figure out the minimum value of â€œxâ€
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06 Sep 2006, 11:01
C

0.xyz > 0.666666666????

St1: for 0.95z YES, for 0.59z NO : INSUFF

St2: for 0.6y9 may be YES or may be NO, for 0.9y6 YES: INSUFF

Together:
Taking worst case:
For 0.689 YES. Hence SUFF.
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06 Sep 2006, 11:40
IN the number expressed as O.xyz
is O.xyz >2/3

1 x+y>13

2 x+z>14

I have another approach for this

0.xyz > 2/3 = 3(xyz)/3000 > 2000/3000

thus now we are comparing 3(xyz) to 2000 is 3(xyz) > 2000

from both statments worst case scenario is x =6 thus y at least is equall 8

and this is enough to deduct that 3(680) > 2000 whatever the value of z is

Thus 3(xyz)>2000 thus O.xyz >2/3
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06 Sep 2006, 23:45
Why is it necessary that y , z should be a single integer.

x+y>13 cant it ALSO mean that x=2 and y=12
x+z>14 cant it ALSO mean that x=2 and z=14
So, the number is 0.21214 , which is less than 2/3.

Am i missing something?
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06 Sep 2006, 23:59
dilbert wrote:
Why is it necessary that y , z should be a single integer.

x+y>13 cant it ALSO mean that x=2 and y=12
x+z>14 cant it ALSO mean that x=2 and z=14
So, the number is 0.21214 , which is less than 2/3.

Am i missing something?

x, y, z are single integers..that is a valid assumption...
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08 Sep 2006, 13:47
I get C...

first we dont know if XYZ are distinct digits...

0.XYZ > 0.75?

(1) X+Y > 13..

well x=6,y=9=15 > 13 .69 < .69
x=9, y=6 > 13... .96>.75
x=9 y=9, .99 >.75

(2)

x+z>14; x=6, z=8 works
or x=8, z=6 works..

if x=8 Y=9, 0.89 which is greater than 0.75
or x=6, z=8, 0.698 <.75

if x=9, y=9, z=9
.999 >.75 All over the place...

THIS IS STRAIGHT C
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09 Sep 2006, 14:52
OA is c... good job everyone
09 Sep 2006, 14:52
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