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Is 2a − 3b < a − b + a − 2b
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01 Aug 2012, 06:20
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Is 2a − 3b < a − b + a − 2b? (1) b = 3 (2) a < b Source: Go Gmat
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Re: Is 2a − 3b < a − b + a − 2b
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Updated on: 01 Aug 2012, 08:25
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 nonzero real numbers x and y, \(x+y\leqx+y\). 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
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Originally posted by EvaJager on 01 Aug 2012, 08:04.
Last edited by EvaJager on 01 Aug 2012, 08:25, edited 1 time in total.




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Re: Is 2a − 3b < a − b + a − 2b
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01 Aug 2012, 08:11
thank you very much, Eva! its clear now im not strong in modules



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Re: Is 2a − 3b < a − b + a − 2b
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01 Aug 2012, 08:44
Galiya wrote: thank you very much, Eva! its clear now im not strong in modules Welcome! I am supposed to know a few things about absolute value...I am a mathematician. I have sent you a PM, did you see it?
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Re: Is 2a − 3b < a − b + a − 2b
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27 Aug 2012, 07:41
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 nonzero real numbers x and y, \(x+y\leqx+y\). 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.



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Re: Is 2a − 3b < a − b + a − 2b
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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 nonzero real numbers x and y, \(x+y\leqx+y\). 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.
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Re: Is 2a − 3b < a − b + a − 2b
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16 Nov 2012, 00:14
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 = 43=1 and a  6 = 46 = 2 i.e. 1 and 2 which is +ve Consider case 1: a=2 a  2 = 42= 2 and a  6 = 26 = 4 i.e. 2 and 4 which is ve Hence the correct answer is E and not C..
jhe



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Re: Is 2a − 3b < a − b + a − 2b
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20 Nov 2012, 19:12
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 nonzero real numbers x and y, \(x+y\leqx+y\). 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 ab <0 and similarly a2b <0 . Therefore both X and Y are negative. In that case the inequality meets the condition. So shouldnt the answer be b



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Re: Is 2a − 3b < a − b + a − 2b
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26 Nov 2012, 13:04
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 nonzero real numbers x and y, \(x+y\leqx+y\). 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 ab <0 and similarly a2b <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.
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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 08:47
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 nonzero real numbers x and y, \(x+y\leqx+y\). 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 inequality 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, but does this help in saying E Or C ?? Im confused



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Re: Is 2a − 3b < a − b + a − 2b
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Updated on: 03 Jul 2013, 13:15
Is 2a − 3b < a − b + a − 2b?
(1) b = 3
2a − 3(3) < a − (3) + a − 2(3) 2a9 < a3 + a6
The checkpoints here are 4.5, 3, 6
The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6
a<3: (2a9) < (a3) + (a6) 2a+9 < a+3 + a+6 0 < 0 INVALID
3<a<4.5: (2a9) < (a3) + (a6) 2a+9 < a3 + a+6 2a < 6 a>3 VALID (a may fall within the range of 3<a<4.5)
4.5<a<6: (2a9) < (a3) + (a6) 2a9 < a3 + a+6 2a < 12 a < 6 VALID 9 (a may fall within the range of 4.5<a<6)
a>6: (2a9) < (a3) + (a6) 2a9 < a3 + a6 0 < 0 INVALID
Some solutions are sufficient, some are not. INSUFFICIENT
a<b
2a − 3b < a − b + a − 2b
If a<b then:
2(2)  3(3) < 23 + (2)2(3) 5 < 1 + 6 5<7 VALID 2(2)  3(3) < 23 + 2 2(3) 13 < 5 + 8 13<13 INVALID INSUFFICIENT
1+2) b=3 and a<b therefore a<3
Using the cases we found in #1, where a<3, the only solution where a<3 is invalid. SUFFICIENT
(C)
(Is that correct reasoning I am using?)



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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 12:03
WholeLottaLove wrote: Is 2a − 3b < a − b + a − 2b?
(1) b = 3
2a − 3(3) < a − (3) + a − 2(3) 2a9 < a3 + a6
The checkpoints here are 4.5, 3, 6
The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6
a<3: (2a9) < (a3) + (a6) 2a+9 < a+3 + a+6 0 < 0 INVALID
3<a<4.5: (2a9) < (a3) + (a6) 2a+9 < a3 + 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: (2a9) < (a3) + (a6) 2a9 < a3 + 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)
x>6: (2a9) < (a3) + (a6) 2a9 < a3 + a6 0 < 0 INVALID
All solutions are invalid SUFFICIENT
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 \(79<3.53+3.56\) or \(2<3\) => valid
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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 12:56
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? Zarrolou wrote: WholeLottaLove wrote: Is 2a − 3b < a − b + a − 2b?
(1) b = 3
2a − 3(3) < a − (3) + a − 2(3) 2a9 < a3 + a6
The checkpoints here are 4.5, 3, 6
The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6
a<3: (2a9) < (a3) + (a6) 2a+9 < a+3 + a+6 0 < 0 INVALID
3<a<4.5: (2a9) < (a3) + (a6) 2a+9 < a3 + 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: (2a9) < (a3) + (a6) 2a9 < a3 + 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)
x>6: (2a9) < (a3) + (a6) 2a9 < a3 + a6 0 < 0 INVALID
All solutions are invalid SUFFICIENT
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 \(79<3.53+3.56\) or \(2<3\) => valid



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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 13:07
Yes of course we need 2 a<b. With it the situation changes to: 3<a<4.5: (2a9) < (a3) + (a6) 2a+9 < a3 + 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: (2a9) < (a3) + (a6) 2a9 < a3 + 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
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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 13:19
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: (2a9) < (a3) + (a6) 2a+9 < a3 + 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: (2a9) < (a3) + (a6) 2a9 < a3 + 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



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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 13:22
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: (2a9) < (a3) + (a6) 2a+9 < a3 + 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: (2a9) < (a3) + (a6) 2a9 < a3 + 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 \(a<3\)
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Re: Is 2a − 3b < a − b + a − 2b
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03 Jul 2013, 13:36
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: (2a9) < (a3) + (a6) 2a+9 < a3 + 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: (2a9) < (a3) + (a6) 2a9 < a3 + 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 \(a<3\)



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Re: Is 2a − 3b < a − b + a − 2b
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26 Jul 2013, 10:38
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 2a9 < a3 + a6 (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?
Thanks!



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Re: Is 2a − 3b < a − b + a − 2b
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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 nonzero real numbers x and y, x+y\leqx+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\leqx+y to hold true, a  b and a  2b should of opposite signs. So: ab>0 and a2b<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



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Re: Is 2a − 3b < a − b + a − 2b
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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 nonzero real numbers x and y, \(x+y\leqx+y\). 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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