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IN the number expressed as O.xyz is O.xyz >2/3 1 [#permalink ]
06 Sep 2006, 07:23

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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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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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Answer: C
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.
Answer: C

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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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The thing here is to figure out the minimum value of â€œxâ€

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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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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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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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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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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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OA is c... good job everyone