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Is 2a − 3b < a − b + a − 2b
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01 Aug 2012, 07:20
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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, 09:25
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, 09:04.
Last edited by EvaJager on 01 Aug 2012, 09:25, edited 1 time in total.




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Re: Is 2a − 3b < a − b + a − 2b
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26 Nov 2012, 14:04
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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01 Aug 2012, 09:44
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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01 Aug 2012, 09:11
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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27 Aug 2012, 08:41
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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16 Nov 2012, 01:14
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, 20:12
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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Updated on: 03 Jul 2013, 14:15
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, 13:03
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, 14:07
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, 14:19
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, 14:22
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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Is 2a − 3b < a − b + a − 2b?
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11 Mar 2015, 23:53
Answer is C. This DS question expects an answer in Yes or No. Consider the modulus expression x  a. As we know, Mod of any number gives out a positive value. But what is inside the Mod ("x  a" in this case) may have any sign or even have value zero. Ex: "x  a" will be positive for x > a, negative for x < a and equal to 0 for x = a. Two important things to note here. First, x = a is the point where sign of the expression gets reversed. Second, a negative value comes out of the Mod as positive, so when "x  a" is negative x  a = (x  a), just as  3  = (3) = 3. Now, coming to provided statements: Statement 1: Given b = 3. So the given question becomes, Is 2a  9 < a  3 + a  6 ?. Now as mentioned above about the modulus expression, each of the three expression inside Mod will change sign based on value of a (in this case). The expression in question a has 3 sign reversal points: 9/2 , 3 and 6 respectively for "2a  9", "a  3" and "a  6". On number line: < 3  9/2  6 > This is quite understandable that for each expression`s sign change, the complete equation will change. For a < 3 (First case of Statement 1), All the above "2a  9", "a  3" and "a  6" will be negative: so, 2a  9 =  (2a  9), a  3 = (a  3) and a  6 = (a  6). And the complete expression will be: L.H.S = 9  2a R.H.s = 3  a + 6  a = 9  2a. Here we can see that for any a < 3, L.H.S = R.H.S. So according to First case of Statement 1, answer to asked question 2a  9 < a  3 + a  6 ? is NO. Again, for 3 < a < 9/2 (Second case of Statement 1), Putting the values according to change signs (please try seeing which of the three have need to change sign), L.H.S = 9  2a R.H.S = 3. Just put any value between 3 and 9/2 for a. L.H.S is always less than R.H.S. So according to Second case of Statement 1, answer to asked question 2a  9 < a  3 + a  6 ? is YES. We can stop here, since 2 different conditions for statement 1 gives contradicting results. If this was not the case, we would have to check for all the ranges of values for a. For given situation statement 1 is clearly INSUFFICIENT. Statement 2: This also insufficient because we cannot decide for the signs of expressions inside Mod. INSUFFICIENT. Statement 1 + 2: b = 3 and a < b => a < 3. From our previous analysis we can see that for a < 3, we have one undoubted answer (No) for question Is 2a  9 < a  3 + a  6 ? So, Statement 1 + 2 is sufficient to answer the question. Hence C is correct. **It looks lengthy process, but just because it is explained. With practice you can reduce time taken to solve below 2 mins. Hope this helped.!!!!
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Re: Is 2a − 3b < a − b + a − 2b
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27 Jun 2015, 05:42
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 Can someone pls correct me if I am wrong in understanding the above statements in RED.  if the signs are same then statement only correct for < and is wrong for =  if the signs are opposite then the statement is only correct for = and not for < Based on this from 1 and 2 above since both signs are same we can definitely say the equation will be correct only for < sign(which is how it was given in the question). therefore C



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Re: Is 2a − 3b < a − b + a − 2b
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05 Sep 2015, 01:22
Q: Is 2a3b<ab+a2b? As rightly pointed out this is related to triangular inequality.
x=ab; y=a2b; x+y=2a3b
question is asking whether x and y have opposite signs or not.
=> whether: ab<0 & a2b>0 OR ab>0 & a2b<0
=> a<b & a>2b (Only possible when both a and b are negative) OR a>b & a<2b => b<a<2b Statement 1: b=3 b is positive. So, a<b and a>2b is not possible since we do not know whether b<a<2b we cannot answer the question. Not sufficient
Statement 2: a<b we don't know whether a and b both are positive or negative. We also don't know whether a<2b. Not sufficient
Statement 1 + 2: b=3 and a<b => neither a<b & a>2b (Only possible when both a and b are negative) NOR b<a<2b. Hence we can definitely say x+y is not less than x+y.



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Re: Is 2a − 3b < a − b + a − 2b
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27 Aug 2012, 10:00
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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03 Jul 2013, 09:47
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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03 Jul 2013, 13:56
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, 14:36
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 \(a<3\)




Re: Is 2a − 3b < a − b + a − 2b
[#permalink]
03 Jul 2013, 14:36



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