On the no line, the ditance between x and y is greater than the distance between x and z. DOes z lie between x and y on the no line?
1. xyz <0
2. xy <0
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- force5
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Can be
z x y
or
x z y
(1) says xyz are -ve which means either case 1 can be true only z -ve
or case 2 can be true with x -ve
or
all three can be negative. ( hence insufficient)
(2) xy -ve ( either one is negative) . which means x has to be negative. but still insuff)
combining- x is negative but y is not from stmnt2 hence the only possible scenario is the second one
x....z...y
Hence C.
z x y
or
x z y
(1) says xyz are -ve which means either case 1 can be true only z -ve
or case 2 can be true with x -ve
or
all three can be negative. ( hence insufficient)
(2) xy -ve ( either one is negative) . which means x has to be negative. but still insuff)
combining- x is negative but y is not from stmnt2 hence the only possible scenario is the second one
x....z...y
Hence C.
- HSPA
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try again force with below hint
Consider x=-1, z = -2 and y =5
x=-5, z= -6 and y = -1
Y is symmetric around X.
Consider x=-1, z = -2 and y =5
x=-5, z= -6 and y = -1
Y is symmetric around X.
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.
Second take: coming soon..
Regards,
HSPA.
- bubbliiiiiiii
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- manpsingh87
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here different cases are possible.HSPA wrote:On the no line, the ditance between x and y is greater than the distance between x and z. DOes z lie between x and y on the no line?
1. xyz <0
2. xy <0
1)xyz<0 x=+ve, y=+ve z=-ve, in this z won't lie between x and y. ( try diff. combinations).
x=+ve, y=-ve, z=+ve in this case also z won't lie between x and y.
x=-ve, y=+ve, z=+ve in this case z may lie between x and y.
hence 1 alone is not sufficient to answer the question.
2) xy<0, x=+ve and y=-ve, or x=-ve and y=+ve
now as no information is provided by z hence 2 alone is also not sufficient to answer the question.
combining 1 and 2.
x=+ve, y=-ve and z =-ve, z won't lie b/w x and y.
x=-ve, y=+ve, z=+ve z may lie between x and y.
as even after combining 1 and 2 we are not able to get the unique solution hence answer should be E.
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