Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
Absolute value
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|x| < 1?
(1) |x+1| = 2|x-1|
|a| = a (if a >=0)
|a| = -a (if a<0)
Therefore,
Either x+1 = 2(x-1)
=> x=3
Or
x+1 = -2(x-1)
x+1 = -2x +2
x=1/3
Not Sufficient
(2) |x-3| != 0
If we set |x-3| = 3, we get x = 6, 0
Not Sufficient
Combining both statements
(1) => x = 3, or 1/3
(2) This statement eliminates the possibility that x = 3, leaving us with x = 1/3
Therefore |x| < 1
Sufficient
C.
(1) |x+1| = 2|x-1|
|a| = a (if a >=0)
|a| = -a (if a<0)
Therefore,
Either x+1 = 2(x-1)
=> x=3
Or
x+1 = -2(x-1)
x+1 = -2x +2
x=1/3
Not Sufficient
(2) |x-3| != 0
If we set |x-3| = 3, we get x = 6, 0
Not Sufficient
Combining both statements
(1) => x = 3, or 1/3
(2) This statement eliminates the possibility that x = 3, leaving us with x = 1/3
Therefore |x| < 1
Sufficient
C.
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I was just trying to find a situation in which statement given will yield different results. If we can find such a situation, then the question cannot be answered on the basis of the information given.
Since we were only told that |x-3| !=0, we can set |x-3| = any non zero number. I picked 3 because I realized numbers >=3 will yield different results. Try |x-3| = 4,5,6...
Since we were only told that |x-3| !=0, we can set |x-3| = any non zero number. I picked 3 because I realized numbers >=3 will yield different results. Try |x-3| = 4,5,6...
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Statement (1)beater wrote:Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
CaseI
|x + 1| = 2|x - 1|
-x - 1 = -2x + 2
2x - x = 2+1
x=3
CaseII
|x + 1| = 2|x - 1|
x+1 = 2x - 2
3 = x
x = 3
CaseIII
|x + 1| = 2|x - 1|
x+1 = -2x + 2
x=1/3
CaseIV
|x + 1| = 2|x - 1|
-x-1 = 2x-2
x=1/3
therefore x=3 or 1/3. Insufficient.
Statement (2)
|x - 3| ≠ 0
This simply implies x cannot be 3.
Combining (1) & (2)
x=1/3. Sufficient.
Thus C is the answer.
This is a very long method, which I wouldnt recommend on the actual test, since it is not only very time consuming but also risky in terms of silly mistakes.
Try and go through Ian's posts. He has a very good number line approach for solving questions like this one.
Hope this helps.
CASE IV is not valid. This situation will never arrive since the range of value of x can be in this case.Morgoth wrote:Statement (1)beater wrote:Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
CaseI
|x + 1| = 2|x - 1|
-x - 1 = -2x + 2
2x - x = 2+1
x=3
CaseII
|x + 1| = 2|x - 1|
x+1 = 2x - 2
3 = x
x = 3
CaseIII
|x + 1| = 2|x - 1|
x+1 = -2x + 2
x=1/3
CaseIV
|x + 1| = 2|x - 1|
-x-1 = 2x-2
x=1/3
therefore x=3 or 1/3. Insufficient.
Statement (2)
|x - 3| ≠ 0
This simply implies x cannot be 3.
Combining (1) & (2)
x=1/3. Sufficient.
Thus C is the answer.
This is a very long method, which I wouldnt recommend on the actual test, since it is not only very time consuming but also risky in terms of silly mistakes.
Try and go through Ian's posts. He has a very good number line approach for solving questions like this one.
Hope this helps.
x > 1
-1 < x < 1
x < -1
so, for case IV it cannot be that |x+1| => - (x+1) and 2|x-1| => 2(x-1), because if x < -1, then 2|x-1| => -2(x-1)
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Agreed! nice spot.nitin86 wrote: CASE IV is not valid. This situation will never arrive since the range of value of x can be in this case.
x > 1
-1 < x < 1
x < -1
so, for case IV it cannot be that |x+1| => - (x+1) and 2|x-1| => 2(x-1), because if x < -1, then 2|x-1| => -2(x-1)
Thanks.
Can somebody confirm what the answer is for this?
Given the cases, it seems that x cannot be greater than -1 and cannot be less than 1. Hence, C? That is NO?
Thanks in advance.
Given the cases, it seems that x cannot be greater than -1 and cannot be less than 1. Hence, C? That is NO?
Thanks in advance.
Morgoth wrote:Agreed! nice spot.nitin86 wrote: CASE IV is not valid. This situation will never arrive since the range of value of x can be in this case.
x > 1
-1 < x < 1
x < -1
so, for case IV it cannot be that |x+1| => - (x+1) and 2|x-1| => 2(x-1), because if x < -1, then 2|x-1| => -2(x-1)
Thanks.
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statement (1)
for EQUATIONS involving absolute value, like this one, the key realization is that the absolute value of a quantity can signify either the quantity itself or the opposite of the quantity. therefore, if you try each of the sign combinations (pos/neg) of the absolute values in the problem, you'll be guaranteed to find all of the solutions.
(note: in what follows, "+" means leaving the expression within the absolute value bars alone; "-" means reversing the sign of that expression)
in this equation, there are ostensibly four sign combinations, +/+, +/-, -/+, -/-, but it's only necessary to try two of them:
** first, either +/+ or -/-, in which both or neither of the absolute value expressions are flipped. may as well go with +/+ (i.e., leaving both of the absolute value expressions alone while removing the bars): x + 1 = 2(x - 1), or x = 3. plugging this back into the original equation shows that it works.
** second, either +/- or -/+, in which one of the absolute value expressions is flipped. let's go (at random) with flipping the first one: -x - 1 = 2(x - 1), or x = 1/3. plugging this into the original equation also shows that it works.
therefore, statement 1 means that x = 3 or x = 1/3.
--
statement (2)
two ways to interpret absolute value inequalities like this one:
** memorize the template of the solution (preferred for efficiency's sake): you should just know that |expression| > a means "either expression > a or expression < -a".
** conceptualize absolute value as distance: in this case, |x - 3| means the distance between x and 3. therefore, this statement means that the distance between x and 3 is greater than 0 (in either direction).
either of these interpretations means that x < 3 or x > 3, or, equivalently, x is not equal to 3.
statement 1 is insufficient, because 1/3 gives a "yes" and 3 gives a "no". statement 2 is also insufficient, because every number except 3 is possible. taken together, though, the two statements are sufficient because they yield a unique value, 1/3, for x.
notice that there's no reason even to figure out whether 1/3 gives "yes" or "no" at this point; it's one value, meaning that it is guaranteed to be sufficient no matter what the answer.
answer = c
for EQUATIONS involving absolute value, like this one, the key realization is that the absolute value of a quantity can signify either the quantity itself or the opposite of the quantity. therefore, if you try each of the sign combinations (pos/neg) of the absolute values in the problem, you'll be guaranteed to find all of the solutions.
(note: in what follows, "+" means leaving the expression within the absolute value bars alone; "-" means reversing the sign of that expression)
in this equation, there are ostensibly four sign combinations, +/+, +/-, -/+, -/-, but it's only necessary to try two of them:
** first, either +/+ or -/-, in which both or neither of the absolute value expressions are flipped. may as well go with +/+ (i.e., leaving both of the absolute value expressions alone while removing the bars): x + 1 = 2(x - 1), or x = 3. plugging this back into the original equation shows that it works.
** second, either +/- or -/+, in which one of the absolute value expressions is flipped. let's go (at random) with flipping the first one: -x - 1 = 2(x - 1), or x = 1/3. plugging this into the original equation also shows that it works.
therefore, statement 1 means that x = 3 or x = 1/3.
--
statement (2)
two ways to interpret absolute value inequalities like this one:
** memorize the template of the solution (preferred for efficiency's sake): you should just know that |expression| > a means "either expression > a or expression < -a".
** conceptualize absolute value as distance: in this case, |x - 3| means the distance between x and 3. therefore, this statement means that the distance between x and 3 is greater than 0 (in either direction).
either of these interpretations means that x < 3 or x > 3, or, equivalently, x is not equal to 3.
statement 1 is insufficient, because 1/3 gives a "yes" and 3 gives a "no". statement 2 is also insufficient, because every number except 3 is possible. taken together, though, the two statements are sufficient because they yield a unique value, 1/3, for x.
notice that there's no reason even to figure out whether 1/3 gives "yes" or "no" at this point; it's one value, meaning that it is guaranteed to be sufficient no matter what the answer.
answer = c
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron