On the number line, the distance between x and y is greater : DS Archive
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# On the number line, the distance between x and y is greater

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On the number line, the distance between x and y is greater [#permalink]

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04 Feb 2009, 11:25
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100% (02:03) correct 0% (00:00) wrong based on 4 sessions

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On the number line, the distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?

1) xyz < 0
2) xy < 0
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04 Feb 2009, 12:54
bmiller0731 wrote:
On the number line, the distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?

1) xyz < 0
2) xy < 0

1: Either one could be -ve and rest two +ve or all could be -ve. so nsf.
2: If x is -ve, y is +ve and vice versa.

Togather: z has to be +ve and one of x and y is +ve and the other is -ve.

Suppose, x = -10, y = 5 and z = 6, no.
x = -10, z = 1, and y = 5. yes.

E.
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04 Feb 2009, 14:25
Anyway to set up the distance as an inequality |x-y|>|x-z| and solve algebraically?
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04 Feb 2009, 20:03
2
KUDOS
bmiller0731 wrote:
On the number line, the distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?

1) xyz < 0
2) xy < 0

each statement individually not sufficient

combined

xyz < 0 and xy < 0
--> z >0 and xy must be opposite signs

given condition |x-y|>|x-z|
y---0--z---x --> here z is lie between x and y and satisfies |x-y|>|x-z|

y---0---x---z --> here z is not lie between x and y satisies |x-y|>|x-z|

two possibilities
not sufficient

E.
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Re: DS: Number Line   [#permalink] 04 Feb 2009, 20:03
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