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getzgetzu
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 01 Dec 2005, 00:32
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Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0



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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 Updated on: 01 Dec 2005, 09:31
Originally posted by SG1 on 01 Dec 2005, 08:47.
Last edited by SG1 on 01 Dec 2005, 09:31, edited 2 times in total.
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1. x1 = 3; x2=-3
---> insuff
2. x != 3
---> insuff

the answer is C
we discard x1 = 3
---> x = -3 < 1
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 01 Dec 2005, 09:09
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Ans = E.
Both statements not sufficient
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yb
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 07 Dec 2005, 15:31
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Can you please explain why it is E?
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 07 Dec 2005, 15:57
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C.

S1: x = 3, 1/3; insufficient
S2: any thing except 3; insufficient

together: x = 1/3 < 1
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 07 Dec 2005, 18:47
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C

1. X = 3 or -3
2. X<>3

Together 1. and 2. gives X = -3, |X| < 1
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| ? 0 07 Dec 2005, 19:36
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Bhai, how does statement 1 allow x to be -3 ?

(1) |x + 1| = 2|x - 1|

|-3+1| 2|-3-1|
2 8

The final answer is correct, but 1) allows x to be either 1 or 1/3, as duttsit explained.

Am i missing something?

Duttsit, Is there an intuitive way to derive the 2 possible values for x, or must we do it by plugging numbers? (I was able to get 3, but didn't realize 1/3 satisfied the equation too).

Thanks



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