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How to STOP searching for values to prove a statement [#permalink]
19 Feb 2012, 05:45
When a statement can't hold good for all set of values, it is easy to find such values (at least in some time). But, when a statement holds good for all values, how do we prove it? It will take whole life finding values to disprove the statement, unless there is a clear pattern.
So, does GMAT include such kind of questions? _________________
You don't plug in numbers blindly here (or anywhere for that matter). If you try 4 numbers and it works for all 4, how do you know it will also work for the other infinite numbers? So you can do it in one of two ways: 1. Use logic. (As shown in my reply) 2. You find the transition points and then try values on either side. e.g. you see that 1/x = 2x when\(x = 1/\sqrt{2}\). So \(x = 1/\sqrt{2}\) is a transition point. The relation would change around it. On one side of \(1/\sqrt{2}\), 1/x will be greater. On the other side 2x will be greater.
Try a value less than \(1/\sqrt{2}\) and try one more than \(1/\sqrt{2}\) but less than 1 (because 1 is another transition point).
\(1/\sqrt{2} = \sqrt{2}/2 = 1.414/2 = .7\) apprx. You must try 1/2 and 8/10 (or something similar). _________________
Re: How to STOP searching for values to prove a statement [#permalink]
16 Feb 2013, 20:35
1
This post received KUDOS
Expert's post
Sachin9 wrote:
So , if we have a inequality, change it with '=' , find the transition point and test using values that are on either side of transition point .. But what analysis can we make out of the change by pluggin in the nos frm either side of the transition point?
You get the relation between the two expressions in various ranges of values. To give an example:
To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\). When \(x < 2, x^2 < 2x\) When \(x > 2, x^2 > 2x\) We have two ranges here x < 2 and x > 2. We know the relation between the two expressions in the two ranges.
Similarly \(1/x = x^2\) when \(x = 1\) When \(x < 1, 1/x > x^2\). When \(x > 1, 1/x > x^2\)
Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\) When \(x < 1/\sqrt{2}, 1/x > 2x\) When \(x > 1/\sqrt{2}, 1/x < 2x\)
So now you have three ranges in which the relation between the three expressions will be different: \(x < 1/\sqrt{2}\) \(1/\sqrt{2} < x < 1\) \(x > 1\)
If \(x < 1/\sqrt{2}\), \(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\) So \(x^2 < 2x < 1/x\) is possible.
If \(1/\sqrt{2} < x < 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible.
If \(x > 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. (Same as above) _________________
Re: How to STOP searching for values to prove a statement [#permalink]
07 Mar 2013, 15:30
1
This post received KUDOS
Chembeti,
I will make some general comments about your original question:
Yes, GMAT routinely includes problems, especially in data sufficiency, where the sufficiency can be established without having to resort to using numerical examples. On an official GMAT question if a particular statement is sufficient then 100% of the time there is a clean and logical way to demonstrate that without having to use numbers and examples. It is a different story that you may or may not see that method.
So what do we do in that case. The only option is to try several different numbers(example: negative, positives, very large numbers, numbers between -1 and 1, etc.), if you find the outcome to be a consistent Yes or No then your best bet is to select the statement to be insufficient. You could still be wrong because you did not consider a specific example that could establish insufficiency.
The entire purpose of these types of problems is to differentiate between students who can quickly and cleanly do the problem in a short time versus the ones who would waste a lot of time trying different examples and getting bogged down in computations. This is what makes one problem more difficult than another one.
The example you linked to falls in the category of the hardest GMAT problems. The reason is because students who are trying numbers have to be lucky to be testing numbers between 0.7 and 1 to get this problem right. This is what makes this problem a statistically difficult problem. The students who are comfortable with algebraic manipulation of the inequality would get to the answer quickly.
Re: How to STOP searching for values to prove a statement [#permalink]
20 Feb 2012, 00:36
Expert's post
Chembeti wrote:
When a statement can't hold good for all set of values, it is easy to find such values (at least in some time). But, when a statement holds good for all values, how do we prove it? It will take whole life finding values to disprove the statement, unless there is a clear pattern.
So, does GMAT include such kind of questions?
Most GMAT questions can be solved by using logic. You don't need to try 10 different values. Or, one or two obvious values will help. Also, if you do need to use numbers numbers, you need to try out numbers smartly say, you know numbers behave differently in the following regions: < -1, -1 < x < 0, 0<x< 1 and x > 1, so you try out one number from each range. If you need to prove something, there will be a pattern. Obviously, GMAT will not give you a question where the only way is to plug in numbers and where you can never be sure. No standard test prep company will give such a question either. If you have a question in mind, let us know. There definitely will be a catch. _________________
Re: How to STOP searching for values to prove a statement [#permalink]
20 Feb 2012, 01:46
VeritasPrepKarishma wrote:
Chembeti wrote:
When a statement can't hold good for all set of values, it is easy to find such values (at least in some time). But, when a statement holds good for all values, how do we prove it? It will take whole life finding values to disprove the statement, unless there is a clear pattern.
So, does GMAT include such kind of questions?
Most GMAT questions can be solved by using logic. You don't need to try 10 different values. Or, one or two obvious values will help. Also, if you do need to use numbers numbers, you need to try out numbers smartly say, you know numbers behave differently in the following regions: < -1, -1 < x < 0, 0<x< 1 and x > 1, so you try out one number from each range. If you need to prove something, there will be a pattern. Obviously, GMAT will not give you a question where the only way is to plug in numbers and where you can never be sure. No standard test prep company will give such a question either. If you have a question in mind, let us know. There definitely will be a catch.
Thanks for the reply, Karishma. My problem is selecting smart numbers. I understand ranges: -1<x<0, 0<x<1 and x>1. But these ranges are really very vast. Problem is selecting particular number. For e.g., for a problem, for the range 0<x<1, I selected 0.1 and 0.2 but it did not work. But after seeing replies in the forum, 0.9 worked. So, there seems to be some logic like numbers close to one side of the range work and numbers close to other end do not work. Have we figured out a smart way for this? _________________
Re: How to STOP searching for values to prove a statement [#permalink]
20 Feb 2012, 03:50
Expert's post
Chembeti wrote:
VeritasPrepKarishma wrote:
Chembeti wrote:
When a statement can't hold good for all set of values, it is easy to find such values (at least in some time). But, when a statement holds good for all values, how do we prove it? It will take whole life finding values to disprove the statement, unless there is a clear pattern.
So, does GMAT include such kind of questions?
Most GMAT questions can be solved by using logic. You don't need to try 10 different values. Or, one or two obvious values will help. Also, if you do need to use numbers numbers, you need to try out numbers smartly say, you know numbers behave differently in the following regions: < -1, -1 < x < 0, 0<x< 1 and x > 1, so you try out one number from each range. If you need to prove something, there will be a pattern. Obviously, GMAT will not give you a question where the only way is to plug in numbers and where you can never be sure. No standard test prep company will give such a question either. If you have a question in mind, let us know. There definitely will be a catch.
Thanks for the reply, Karishma. My problem is selecting smart numbers. I understand ranges: -1<x<0, 0<x<1 and x>1. But these ranges are really very vast. Problem is selecting particular number. For e.g., for a problem, for the range 0<x<1, I selected 0.1 and 0.2 but it did not work. But after seeing replies in the forum, 0.9 worked. So, there seems to be some logic like numbers close to one side of the range work and numbers close to other end do not work. Have we figured out a smart way for this?
You need to give the exact question. There would be clues which should make you think of a particular section of a range. _________________
You don't plug in numbers blindly here (or anywhere for that matter). If you try 4 numbers and it works for all 4, how do you know it will also work for the other infinite numbers? So you can do it in one of two ways: 1. Use logic. (As shown in my reply) 2. You find the transition points and then try values on either side. e.g. you see that 1/x = 2x when\(x = 1/\sqrt{2}\). So \(x = 1/\sqrt{2}\) is a transition point. The relation would change around it. On one side of \(1/\sqrt{2}\), 1/x will be greater. On the other side 2x will be greater.
Try a value less than \(1/\sqrt{2}\) and try one more than \(1/\sqrt{2}\) but less than 1 (because 1 is another transition point).
\(1/\sqrt{2} = \sqrt{2}/2 = 1.414/2 = .7\) apprx. You must try 1/2 and 8/10 (or something similar).
So , if we have a inequality, change it with '=' , find the transition point and test using values that are on either side of transition point .. But what analysis can we make out of the change by pluggin in the nos frm either side of the transition point? _________________
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