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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
01 Aug 2012, 08:04
26
This post received KUDOS
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Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?
(1) b = 3 (2) a < b
Source: Go Gmat
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Last edited by EvaJager on 01 Aug 2012, 08:25, edited 1 time in total.
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
27 Aug 2012, 07:41
1
This post was BOOKMARKED
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?
(1) b = 3 (2) a < b
Source: Go Gmat
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C
That was a great explanation. Mods is not a strong point for me. It would be helpful if you can share some more tips and tricks on mods.
also please explain: If x and y have opposite signs, the inequality is strict.
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
27 Aug 2012, 09:00
manulath wrote:
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?
(1) b = 3 (2) a < b
Source: Go Gmat
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C
That was a great explanation. Mods is not a strong point for me. It would be helpful if you can share some more tips and tricks on mods.
also please explain: If x and y have opposite signs, the inequality is strict.
Strict inequality means < , equality cannot hold. For example |2+(-3)| < |2| + |-3| as 1 < 2 + 3. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
16 Nov 2012, 00:14
1
This post received KUDOS
I am confused at this point...For (1) and (2) together: x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0, so the given inequality doesn't hold (definite answer is NO), therefore sufficient. Now b= 3 and a<b. Consider case 1: a=4 a - 3 = 4-3=-1 and a - 6 = 4-6 = -2 i.e. -1 and -2 which is +ve Consider case 1: a=2 a - 2 = 4-2= 2 and a - 6 = 2-6 = -4 i.e. 2 and -4 which is -ve Hence the correct answer is E and not C..
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
20 Nov 2012, 19:12
1
This post received KUDOS
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?
(1) b = 3 (2) a < b
Source: Go Gmat
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C
Hi Eva
I am not sure on the solution mentioned above.
For statement 2 we have a< b, ie a-b <0 and similarly a-2b <0 . Therefore both X and Y are negative. In that case the inequality meets the condition. So shouldnt the answer be b
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
26 Nov 2012, 13:04
2
This post received KUDOS
azzhhuu wrote:
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?
(1) b = 3 (2) a < b
Source: Go Gmat
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C
Hi Eva
I am not sure on the solution mentioned above.
For statement 2 we have a< b, ie a-b <0 and similarly a-2b <0 . Therefore both X and Y are negative. In that case the inequality meets the condition. So shouldnt the answer be b
a < b doesn't necessarily imply that a < 2b. For example a = -4 < -3 = b, but a = -4 > 2(-3) = -6 = 2b. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C
hi mate,
here is my approach please correct me if im wrong,
the in-equality only holds good if they are opposite sign, s1: b = 3 or one variable in const. but we can't predict anything with this so we need another values too so S1: NS
s2: a< b, even here we have many possibilities so s2 NS now s1 and s2 combined :
we still don't know about a, as we can only determine B and a is still unknown , we can't drive any particular values .. we say NO,
3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.
4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)
I know the above solution is incorrect but I cannot seem to figure out why. Can someone please explain?
Thanks!
Those two parts are not correct. Example for 3<a<4.5 you get a>3, so this could be a valid solution.
Just because \(a\) COULD fall in the range, this makes that given range a possible valid solution (for a=3.5 for example). \(|2a - 3b| < |a - b| + |a - 2b|\) for a=3.5 and b=3 you get \(|7-9|<|3.5-3|+|3.5-6|\) or \(2<3\) => valid _________________
It is beyond a doubt that all our knowledge that begins with experience.
Interesting. I think I've had trouble with a few other problems with this very concept. Thanks for the pointer. Just to be clear, these would only be invalid if a (or whatever variable) fell entirely outside of the given range? (i.e. 2<a<3 and a>10 Would 2<a<3 and a<10 be valid?)
I take it that's why we need II. in addition to I. to solve this problem?
3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.
4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)
I know the above solution is incorrect but I cannot seem to figure out why. Can someone please explain?
Thanks!
Those two parts are not correct. Example for 3<a<4.5 you get a>3, so this could be a valid solution.
Just because \(a\) COULD fall in the range, this makes that given range a possible valid solution (for a=3.5 for example). \(|2a - 3b| < |a - b| + |a - 2b|\) for a=3.5 and b=3 you get \(|7-9|<|3.5-3|+|3.5-6|\) or \(2<3\) => valid
With it the situation changes to: 3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.
With \(b>a\), \(a>3\) => \(b>a>3\) so \(b>3\). But \(b =3\), so it's NOT more than 3=> Invalid
4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)
With \(b>a\), \(a>4.5\) so this means that \(b>4.5\) as well. But \(b =3\), so it's NOT more than 4.5 => Invalid _________________
It is beyond a doubt that all our knowledge that begins with experience.
In my revised solution (edited from the one above) I used a<b and b=3 so doesn't that mean a<3?
Zarrolou wrote:
Yes of course we need 2 a<b.
With it the situation changes to: 3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.
With \(b>a\), \(a>3\) => \(b>a>3\) so \(b>3\). But \(b =3\), so it's NOT more than 3=> Invalid
4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)
With \(b>a\), \(a>4.5\) so this means that \(b>4.5\) as well. But \(b =3\), so it's NOT more than 4.5 => Invalid
In my revised solution (edited from the one above) I used a<b and b=3 so doesn't that mean a<3?
Zarrolou wrote:
Yes of course we need 2 a<b.
With it the situation changes to: 3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.
With \(b>a\), \(a>3\) => \(b>a>3\) so \(b>3\). But \(b =3\), so it's NOT more than 3=> Invalid
4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)
With \(b>a\), \(a>4.5\) so this means that \(b>4.5\) as well. But \(b =3\), so it's NOT more than 4.5 => Invalid
Yes, good catch! I overlook that, so the analysis can be reduced to just the case \(a<3\) _________________
It is beyond a doubt that all our knowledge that begins with experience.
Haha! I got one right for once! Thank you very much for all of your help - I would be completely lost on these 700+ questions without it!
Zarrolou wrote:
WholeLottaLove wrote:
In my revised solution (edited from the one above) I used a<b and b=3 so doesn't that mean a<3?
Zarrolou wrote:
Yes of course we need 2 a<b.
With it the situation changes to: 3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.
With \(b>a\), \(a>3\) => \(b>a>3\) so \(b>3\). But \(b =3\), so it's NOT more than 3=> Invalid
4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)
With \(b>a\), \(a>4.5\) so this means that \(b>4.5\) as well. But \(b =3\), so it's NOT more than 4.5 => Invalid
Yes, good catch! I overlook that, so the analysis can be reduced to just the case \(a<3\)
Reviewing this question, I found a problem with my reasoning.
Looking above to my solution I found that there were valid ranges of a for 3<a<4.5 and 4.5<a<6. The problem is, every single value of a I plug into the statement |2a-9| < |a-3| + |a-6| (after plugging in for b) makes the inequality true. Normally I would say that #1 is sufficient but obviously it is not. Could someone explain this to me?
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
07 Dec 2013, 17:34
EvaJager's explanation seems correct and let me make it more simple algebraically:
EvaJager's explanation:
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, |x+y|\leq|x|+|y|. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by x = a - b, y = a - 2b, and the given inequality becomes |x+y|<|x|+|y|. So, the question is asking whether x and y are of opposite signs, or a - b and a - 2b are of opposite signs
~~~~~~~~~~`
Here I will add my two cents True that for |x+y|\leq|x|+|y| to hold true, a - b and a - 2b should of opposite signs. So: a-b>0 and a-2b<0 which means 2b<a<b Option B contradicts that above statement. It states: a<b: Thus Combining the two options, the inequality fails to stand true. Hence C
Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]
09 Dec 2013, 06:50
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?
(1) b = 3 (2) a < b
Source: Go Gmat
There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, \(|x+y|\leq|x|+|y|\). Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.
In our case, we can denote by \(x = a - b, y = a - 2b\), and the given inequality becomes \(|x+y|<|x|+|y|\). So, the question is asking whether x and y are of opposite signs, or \(a - b\) and \(a - 2b\) are of opposite signs.
Then, we can see that neither (1) nor (2) alone is sufficient. For (1) and (2) together: \(x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0\), so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Answer C
Excellent man! Keep sharing, keep helping. Got to learn something new today after a couple of weeks. _________________
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