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Mod x

by Mayur Sand » Wed Jul 22, 2009 12:15 pm
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3|&#8800; 0

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA C
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Re: Mod x

by El Cucu » Wed Jul 22, 2009 12:44 pm
Mayur Sand wrote:Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3|&#8800; 0

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA C
The questions asks whether x is between -1 and 1.
From 1) x could be 3 or 1/3 Insufficient
Form 2) x is not 3 Insufficient
Together x is 1/3 so C)
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by ogbeni » Wed Jul 22, 2009 1:18 pm
How do you get the 2nd value = 1/3 in statement 1?
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by El Cucu » Wed Jul 22, 2009 2:04 pm
ogbeni wrote:How do you get the 2nd value = 1/3 in statement 1?
-x-1=2x-2
1=3x
1/3= x
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by ogbeni » Wed Jul 22, 2009 2:06 pm
"-x-1=2x-2 "

If you negated the left half of the equation, why didn't you negate the absolute value on the right half?

When I consider both cases of the absolute value i get 3 :(
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by El Cucu » Wed Jul 22, 2009 2:11 pm
ogbeni wrote:"-x-1=2x-2 "

If you negated the left half of the equation, why didn't you negate the absolute value on the right half? (
You can´t as the equation will result in an absurdness. Ask Ian or Ron or some other guru if you need the theoretical explanation.
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by ogbeni » Thu Jul 23, 2009 10:10 am
anyone still care to explain why you cannot negate both sides in Statment 1?
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by raghavsarathy » Thu Jul 23, 2009 5:22 pm
ogbeni wrote:anyone still care to explain why you cannot negate both sides in Statment 1?
Hi

IMO whenever a modulus sign is seen , we can consider both positive and negative values. In this case since we have mod on both sides of the eqn , we can end up with 4 cases.

case 1 : (x+1) = 2(x-1) .. solve this to get x= 3

case 2 : Negative value for |x-1|
(x+1) = -2 (x-1).. solve this to get x = 1/3

case 3. Negative value for |x-1| and |x+1|
-(x+1) = -2(x-1)
This is essentially same as case 1. multiply both sides of the eqn with -1 and u end up with case 1

case 4. Negative value for |x+1|
-(x+1) = 2(x-1)
This is the same as case 2

So we end up with only 2 value of x.

So IMO we can consider negative values for both sides and still end up with same soln.


Ans - C
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