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Is lxl < 1? (1) lx + 1l = 2lx - 1l (2) lx - 3l ≠ 0

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Is lxl < 1? (1) lx + 1l = 2lx - 1l (2) lx - 3l ≠ 0 [#permalink] New post 04 Jun 2007, 19:19
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A
B
C
D
E

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Q5:
Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0


Q16:
If m, p, and t are positive integers and m < p < t, is the product mpt an even integer?
(1) t – p = p - m
(2) t – m = 16
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 [#permalink] New post 04 Jun 2007, 23:52
Q5:
Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0


(C) for me :)

lxl < 1 ?
<=> -1 < x < 1 ?

From (1)
lx + 1l = 2lx - 1l, what are the solutions of it?

o If x >=1, then:
lx + 1l = 2lx - 1l
<=> x+1 = 2*(x-1)
<=> x = 3 >>>>> 1 solution as x must be superior or equal to 1

o If -1 > x > 1, then:
lx + 1l = 2lx - 1l
<=> x+1 = 2*[-(x-1)]
<=> x+1 = -2*x + 2
<=> x = 1/3 >>>>> 1 solution as x must be between -1 and 1

Note that it's already insufficient, but let us continue to state later between (C) or (E)...

o If x =< -1, then:
lx + 1l = 2lx - 1l
<=> -(x+1) = 2*[-(x-1)]
<=> x+1 = 2*(x-1)
<=> x = 3 >>>>> no solution as x must be inferior or equal to -1

INSUFF.

From (2)
lx - 3l ≠ 0
<=> x < 3 or x > 3

INSUFF.

From (1) and (2)
With the system of sulution from 1, we have x = 3 or x = 1/3. Then, we have x that must be different from 3.

So, x = 1/3 (-1 < x < 1)

SUFF.
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 [#permalink] New post 06 Jun 2007, 10:35
Another method to solve

Stmt 1 lx + 1l = 2lx - 1l
(x + 1) ^2 = 4(x - 1) ^ 2
Solving for x we get 3 and 1/3

Stmt 2

x <> 3

Using 1 and 2 we see x = 1/3
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Re: 2 DS - Absolute / Number properties [#permalink] New post 06 Jun 2007, 15:21
jet1445 wrote:
Q5:
Is lxl < 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0

Q16:
If m, p, and t are positive integers and m < p < t, is the product mpt an even integer?
(1) t – p = p - m
(2) t – m = 16


Q5.
(1) Two cases: x+1 =2x-2 OR =-2x+2. In the 1st case, x=3; in the 2nd, x=1/3. Insuff.
(2) x<>3. Insuff.
(1&2) x has to be =1/3. Suff, then C.

Q16.
(1) Insuff. Take m, p, t = 1, 4, 7 and 1, 5, 9. In the 1st case mpt is even, in the 2nd case it´s odd.
(2) Insuff. Simple inspection.
(1&2) Insuff (take a similar example as in (1)), then E.
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 [#permalink] New post 10 Sep 2007, 18:36
for the first question, why cant we write:
x+1= 2x-2
x= 3
OR
-x-1=-2x+2
x=3

doesn't |x+1| = x+1 OR -x-1??
i guess i am not understanding how modulus works then.
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 [#permalink] New post 10 Sep 2007, 19:06
Question is really asking whether x a fraction.

St1:
|x+1| = 2|x-1|
If x = 3, |x+1| = 4, and 2|x-1| = 4 and |x| > 1
If x = 1/3, |x+1| = 4/3 and 2|x-1| = 4/3 and |x| < 1.
Insufficient.

St2:
X can be anything apart from 3. Insufficient.

Using st1 and st2:
x must be 1/3. sufficient.

ans C
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 [#permalink] New post 10 Sep 2007, 19:07
r019h wrote:
for the first question, why cant we write:
x+1= 2x-2
x= 3
OR
-x-1=-2x+2
x=3

doesn't |x+1| = x+1 OR -x-1??
i guess i am not understanding how modulus works then.


No, we only meddle with the signs on the RHS of the equation.
  [#permalink] 10 Sep 2007, 19:07
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