C seems the obvious answer, bt if that were the were teh case, then this question would be kinda stupid, right?
What happens when x is negative, such as x=-2?
[ sqrt((-2)^2) ] / -2 = sqrt (4) /-2 = 2/-2 = -1
Thus, C is not the answer - for negative xs, the numerator will be positive (anything squared is nonnegative, and a square root is always positive), but the denominator is negative.
Get used to plugging in on such questions to avoid mistakes. Here's the correct approach when you first see the question for accurate solution:
variables in answer choices mean "plug in and eliminate".
plug in a regular number such as x=2. You get
[ sqrt((2)^2) ] / 2 = sqrt (4) /2 = 2/2 = 1
so if x=2, the answer should be 1. Plug in x=2 into answer choices, and eliminate everything that doesn't equal 1 for this plug in. BE SURE TO CHECK ALL 5: both C and E will give you a result of 1 for x=2.
Since you still have more than one possible answer choice, plug in a second number to eliminate one or the other. Now's the time to plug in a negative number: For x=-2, the required goal is -1. C still equals 1, but E equals |-2|/-2 = 2/-2 = -1 - and tus C is eliminated and E remains the only non-eliminated answer choice, which must be the right one.
Inequality
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Geva@EconomistGMAT
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Thanks Geva, but a question for you - in such problems, do we not try and resolve the inequality first ? If we just go with the base equation and resolve it, we get x ^ 0 which equals 1.
Am I reading it correct?
Am I reading it correct?
Geva@MasterGMAT wrote:C seems the obvious answer, bt if that were the were teh case, then this question would be kinda stupid, right?
What happens when x is negative, such as x=-2?
[ sqrt((-2)^2) ] / -2 = sqrt (4) /-2 = 2/-2 = -1
Thus, C is not the answer - for negative xs, the numerator will be positive (anything squared is nonnegative, and a square root is always positive), but the denominator is negative.
Get used to plugging in on such questions to avoid mistakes. Here's the correct approach when you first see the question for accurate solution:
variables in answer choices mean "plug in and eliminate".
plug in a regular number such as x=2. You get
[ sqrt((2)^2) ] / 2 = sqrt (4) /2 = 2/2 = 1
so if x=2, the answer should be 1. Plug in x=2 into answer choices, and eliminate everything that doesn't equal 1 for this plug in. BE SURE TO CHECK ALL 5: both C and E will give you a result of 1 for x=2.
Since you still have more than one possible answer choice, plug in a second number to eliminate one or the other. Now's the time to plug in a negative number: For x=-2, the required goal is -1. C still equals 1, but E equals |-2|/-2 = 2/-2 = -1 - and tus C is eliminated and E remains the only non-eliminated answer choice, which must be the right one.
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Geva@EconomistGMAT
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Your approach is exactly what the test writers wanted you to do: Think that sqrt (x^2) is always equal to x, and miss out on The fact that x^2 is always positive, regardless of whether the x itself is positve or negative. The algebraic concept tested here is that sqrt(x^2) is equal x only when x is positive; when x is negative, then sqrt (x^2) is actually equal to -x.
This is not a concept I remembered immediately when I saw the question: I figured out that that was the issue AFTER I had solved by plugging in. You cannot reliably remember all the different algebraic concepts tested by the GMAT AND count on recognizing the required one for any particular question. We're not all math freaks like Rahul. The rest of us mortals need a differnt method than algebra to make sure we don't fall into the trap.
I have grown accustomed to resorting to the plugging in approach first WHENVER there are variables in the answer choices. Not because it faster (though it can be), but because it is SAFER. If you had plugged in and didn't stop until you eliminated four answer choices and were left only with one, there's no way you would've chosen C here: you would need to plug in different numbers until either C or E are eliminated, and that would've at least given you an indication that the question isn't a straightforward C. Only if plugging in proves too difficult, (too many variables, or difficulty choosing the right numbers) do I break out the algebra toolbox.
This is not a concept I remembered immediately when I saw the question: I figured out that that was the issue AFTER I had solved by plugging in. You cannot reliably remember all the different algebraic concepts tested by the GMAT AND count on recognizing the required one for any particular question. We're not all math freaks like Rahul. The rest of us mortals need a differnt method than algebra to make sure we don't fall into the trap.
I have grown accustomed to resorting to the plugging in approach first WHENVER there are variables in the answer choices. Not because it faster (though it can be), but because it is SAFER. If you had plugged in and didn't stop until you eliminated four answer choices and were left only with one, there's no way you would've chosen C here: you would need to plug in different numbers until either C or E are eliminated, and that would've at least given you an indication that the question isn't a straightforward C. Only if plugging in proves too difficult, (too many variables, or difficulty choosing the right numbers) do I break out the algebra toolbox.
pratyoosh wrote:Thanks Geva, but a question for you - in such problems, do we not try and resolve the inequality first ? If we just go with the base equation and resolve it, we get x ^ 0 which equals 1.
Am I reading it correct?
Geva@MasterGMAT wrote:C seems the obvious answer, bt if that were the were teh case, then this question would be kinda stupid, right?
What happens when x is negative, such as x=-2?
[ sqrt((-2)^2) ] / -2 = sqrt (4) /-2 = 2/-2 = -1
Thus, C is not the answer - for negative xs, the numerator will be positive (anything squared is nonnegative, and a square root is always positive), but the denominator is negative.
Get used to plugging in on such questions to avoid mistakes. Here's the correct approach when you first see the question for accurate solution:
variables in answer choices mean "plug in and eliminate".
plug in a regular number such as x=2. You get
[ sqrt((2)^2) ] / 2 = sqrt (4) /2 = 2/2 = 1
so if x=2, the answer should be 1. Plug in x=2 into answer choices, and eliminate everything that doesn't equal 1 for this plug in. BE SURE TO CHECK ALL 5: both C and E will give you a result of 1 for x=2.
Since you still have more than one possible answer choice, plug in a second number to eliminate one or the other. Now's the time to plug in a negative number: For x=-2, the required goal is -1. C still equals 1, but E equals |-2|/-2 = 2/-2 = -1 - and tus C is eliminated and E remains the only non-eliminated answer choice, which must be the right one.
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gmatusa2010
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Whats the source of this problem?
1) Can't there be two answer? E and the Zero?
2) E is just a simplification of the expression
3) Am I reading this wrong? X factorial = zero. X has to be zero. Or does the question say X cannot be equal to zero. On my computer, it says X!.
Edit: Just looked it up. 0! =1. In fact the definition of factorial is NON-ZERO POSITIVE. So what information can be extracted from X!=0????
1) Can't there be two answer? E and the Zero?
2) E is just a simplification of the expression
3) Am I reading this wrong? X factorial = zero. X has to be zero. Or does the question say X cannot be equal to zero. On my computer, it says X!.
Edit: Just looked it up. 0! =1. In fact the definition of factorial is NON-ZERO POSITIVE. So what information can be extracted from X!=0????
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KrazyKarl
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gmatusa I was confused on that too and I see the same question. I know that 0! is actually 1 (something about a null set in stats), so I don't think you can have a factorial equal to zero. Given what Geva said I don't think that x! = 0 is part of the problem...maybe it's just supposed to say tthat x does not equal 0 - which makes sense because then the problem would allow for division by 0 and that doesn't work.
I think if we read it as x doesn't equal 0 than the problem works with Geva's explanation.
I think if we read it as x doesn't equal 0 than the problem works with Geva's explanation.
