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?
Inequality question
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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 thatStockmoose16 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?
(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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1<x<2 leaves us with just 2 correct. Am I misundertanding this?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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1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?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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Only if X is defined as INTEGERcramya wrote:1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?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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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:cramya wrote:1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?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.
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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Ian, am I missing something? Your answer is finite on the number line because it falls between 1 & 3, correct?Ian Stewart wrote: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 thatStockmoose16 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?
(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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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)Ian Stewart wrote: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:cramya wrote:1<x<3 leaves us with just 2 which is finite, correct. Am I misunderstanding this?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.
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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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.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)
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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.
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.