Is lxl< 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0
Besides plugging values, can someone tell me another method how to solve absolute values questions.
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gmattester
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There is one way if you want. But it's a bit odd.
ST 1. In order to avoid model square(sorry can't express) the both sides
(x+1)^2=(2(x-1))^2
x^2+2x+1=4x^2-8x-4
3x^2-10x-5=0
appr. x1=-1/3 x2=3.5 INSUFF
ST 2. Same way
(x-3)^2 ≠0
x ≠3 INSUFF
BOTH INSUFF
I go for E
ST 1. In order to avoid model square(sorry can't express) the both sides
(x+1)^2=(2(x-1))^2
x^2+2x+1=4x^2-8x-4
3x^2-10x-5=0
appr. x1=-1/3 x2=3.5 INSUFF
ST 2. Same way
(x-3)^2 ≠0
x ≠3 INSUFF
BOTH INSUFF
I go for E
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gmattester
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No this is not odd....
This is an excellent method but I am not sure I can apply this method for every absolute value question.
I mean to say something generic which I can apply to all absolute questions
This is an excellent method but I am not sure I can apply this method for every absolute value question.
I mean to say something generic which I can apply to all absolute questions
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Stuart@KaplanGMAT
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Math is a bit off:bomond wrote:There is one way if you want. But it's a bit odd.
ST 1. In order to avoid model square(sorry can't express) the both sides
(x+1)^2=(2(x-1))^2
x^2+2x+1=4x^2-8x-4
(x+1)^2=(2(x-1))^2
x^2 + 2x + 1 = (2x - 2)^2
x^2 + 2x + 1 = 4(x^2) - 8x + 4
0 = 3(x^2) - 10x + 3
0 = (3x - 1) (x - 3)
3x = 1 or x = 3
x = 1/3 or x = 3
x may or may not be between -1 and +1.. insufficient.
However, when we combine with:
(2) x ≠ 3
we know that the only possible value of x is 1/3, so choose (c): together the statements are sufficient.
In fact, (c) is a great strategic guess on this question, long before doing all the math.
Statement (2) seems pretty random, so either we're being given a totally useless piece of information or x ≠ 3 will eliminate a possibility from (1). As soon as we see that (1) can be turned into a quadratic, we should anticipate that (1) will give us two values for x (so will be insufficient) and (2) will narrow it down to just one possibility.
The better you know the GMAT, the less work you'll need to do to confidently pick up points. Knowing how and when to guess strategically is a huge advantage on test day.
To go back to the original poster's question, yes, squaring both sides of an absolute value equation is often a great approach to these questions.
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Ian Stewart
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There is another way, which I use on pretty much every absolute value GMAT question, but you'd want to practice it a bit before beginning to use it on a test. Remember that |x| is the distance between x and zero on the number line, and |x-y| is the distance between x and y on the number line. If you understand these two facts, you can change many problems about absolute value into problems about distances on the number line, and you can bypass the algebra altogether. Doing that here:gmattester wrote:Is lxl< 1?
(1) lx + 1l = 2lx - 1l
(2) lx - 3l ≠ 0
Besides plugging values, can someone tell me another method how to solve absolute values questions.
Is |x| < 1? --> Is x less than 1 away from zero?
1) lx + 1l = 2lx - 1l ---> lx - (-1)l = 2lx - 1l
In words this says "the distance between x and -1 is twice the distance between x and 1", or phrased more simply, "x is twice as far from -1 as x is from 1". If you draw a number line, you'll see there are two places x could be: between -1 and 1 (but closer to 1), or to the right of 1. It's not difficult to see that x = 3 is one of the two solutions. Insufficient.
2) tells us x is not 3. Insufficient.
Together, there is only one possible solution, the point between -1 and 1, so the answer is yes, |x| < 1, and the two statements are sufficient together. C.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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I'm sorry gmattester. I made a calculation mistake. Thanks to Stuart, I got it.
Ian Stewart! Do you think your method is always helpful
How do you approach to this problem: |x-1|=|2x+3| (instead lx + 1l = 2lx - 1l)
Ian Stewart! Do you think your method is always helpful
How do you approach to this problem: |x-1|=|2x+3| (instead lx + 1l = 2lx - 1l)
