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Inequality question

by Stockmoose16 » Mon Oct 27, 2008 3:38 pm
Does the following statement result in a finite set of numbers for X?

X^2-4X+3<0

My answer is "NO," but the actual answer is "yes."

Here's why I say no:

X^2-4X+3<0
= (X-3)(X-1)<0

For this statement to be true, one of the terms has to be negative, the other has to be positive.

So, either of the following should be correct:

X-3>0 & X-1<0

OR

X-3<0 & X-1>0

(i)By statement #1, X>3 AND X>1

(ii) By statement 2, X<3 AND X>1

(i) is not finite... so the answer should be that a finite result does not occur. I realize when you plug in numbers, (i) does not work out correctly, but can someone please explain the logic?
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Re: Inequality question

by Ian Stewart » Mon Oct 27, 2008 4:48 pm
Stockmoose16 wrote:Does the following statement result in a finite set of numbers for X?

X^2-4X+3<0

My answer is "NO," but the actual answer is "yes."

Here's why I say no:

X^2-4X+3<0
= (X-3)(X-1)<0

For this statement to be true, one of the terms has to be negative, the other has to be positive.

So, either of the following should be correct:

X-3>0 & X-1<0

OR

X-3<0 & X-1>0

(i)By statement #1, X>3 AND X>1

(ii) By statement 2, X<3 AND X>1

(i) is not finite... so the answer should be that a finite result does not occur. I realize when you plug in numbers, (i) does not work out correctly, but can someone please explain the logic?
One of your inequality signs got reversed during your simplifications - highlighted in red above - and that's the error in your solution. You correctly found that

(X-3)(X-1)<0

and analyzed this correctly: one factor must be positive, the other negative. At this stage, though, it is useful to make the following observation: x-3 must *always* be less than x-1. So, if exactly one of these two factors is negative, it must be true that x-3 is negative, and x-1 is positive. So you rule out one case quite quickly that way.

After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
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by cramya » Mon Oct 27, 2008 5:24 pm
After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
1<x<2 leaves us with just 2 correct. Am I misundertanding this?
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by cramya » Mon Oct 27, 2008 5:25 pm
After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?
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by logitech » Mon Oct 27, 2008 6:16 pm
cramya wrote:
After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?
Only if X is defined as INTEGER
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by Ian Stewart » Mon Oct 27, 2008 6:45 pm
cramya wrote:
After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?
It is what is called in mathematics a 'bounded' set, but it doesn't give a finite set of values for x; there is still an infinite number of possible values of x. For example, x could be any of the following:

3/2, 4/3, 5/4, 6/5, 7/6, 8/7, ...

which is an infinite set of different values in the required range. So there's something wrong with the wording of the question; a real GMAT question would never be worded that way. As logitech points out, if we knew x was an integer, then we would only have a finite number of solutions (in fact, we would know x=2).
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Re: Inequality question

by Stockmoose16 » Mon Oct 27, 2008 7:53 pm
Ian Stewart wrote:
Stockmoose16 wrote:Does the following statement result in a finite set of numbers for X?

X^2-4X+3<0

My answer is "NO," but the actual answer is "yes."

Here's why I say no:

X^2-4X+3<0
= (X-3)(X-1)<0

For this statement to be true, one of the terms has to be negative, the other has to be positive.

So, either of the following should be correct:

X-3>0 & X-1<0

OR

X-3<0 & X-1>0

(i)By statement #1, X>3 AND X>1

(ii) By statement 2, X<3 AND X>1

(i) is not finite... so the answer should be that a finite result does not occur. I realize when you plug in numbers, (i) does not work out correctly, but can someone please explain the logic?
One of your inequality signs got reversed during your simplifications - highlighted in red above - and that's the error in your solution. You correctly found that

(X-3)(X-1)<0

and analyzed this correctly: one factor must be positive, the other negative. At this stage, though, it is useful to make the following observation: x-3 must *always* be less than x-1. So, if exactly one of these two factors is negative, it must be true that x-3 is negative, and x-1 is positive. So you rule out one case quite quickly that way.

After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
Ian, am I missing something? Your answer is finite on the number line because it falls between 1 & 3, correct?
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by Stockmoose16 » Mon Oct 27, 2008 7:55 pm
Ian Stewart wrote:
cramya wrote:
After all of that, I must admit I don't understand the question. Perhaps it was worded differently in its original version, or perhaps it's just a badly framed question. We do find, if we solve the question completely, that 1 < x < 3, but that's still an infinite set of numbers. My answer is the same as Stock's, at least as the question is phrased: No, the solution set is not finite.
1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?
It is what is called in mathematics a 'bounded' set, but it doesn't give a finite set of values for x; there is still an infinite number of possible values of x. For example, x could be any of the following:

3/2, 4/3, 5/4, 6/5, 7/6, 8/7, ...

which is an infinite set of different values in the required range. So there's something wrong with the wording of the question; a real GMAT question would never be worded that way. As logitech points out, if we knew x was an integer, then we would only have a finite number of solutions (in fact, we would know x=2).
The original question asked, "which of the following inequalities have a finite range of values of "x" satisfying them? The equation I picked was the answer to the question. (Question from 4gmat)
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by Ian Stewart » Tue Oct 28, 2008 5:15 am
Stockmoose16 wrote:
The original question asked, "which of the following inequalities have a finite range of values of "x" satisfying them? The equation I picked was the answer to the question. (Question from 4gmat)
That wording makes more sense. Yes, if 1 < x < 3, the range of values for x is finite. This is a technicality that won't be important on anyone's GMAT, but the original wording of the question asked whether there was a finite set of solutions for x, and if 1 < x < 3, there is not: there is still an infinite number of possible values for x in that range, if x does not need to be an integer.
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Inequality question

by kumarkams » Thu Oct 30, 2008 2:23 pm
I'll pick up with (X-3)(X-1)<0

For this statement to be true, one of the terms has to be negative, the other has to be positive.

So, either of the following should be correct:

X-3>0 & X-1<0

OR

X-3<0 & X-1>0

(i)By statement 1, X>3 AND X<1. Clearly, there is no possible value of X, which can satisfy these two conditions. In other words, X doesn't have a solution and hence number of values for X is "0", which is finite.

(ii) By statement 2, X<3 AND X>1 e.g. 1<X<3. This will give us infinite number of solutions.

In brief, number of values for X can be finite or infinite and no single solution is possible.
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