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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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23 Dec 2013, 16:57
Hi, You can also use the number line to solve the problem The issue if to determine whether the distance between 2A and 3B < the 3*distance between AB (or the distance between A and B + the distance between a and 2B which leans 3AB) There are two main issues here: determine the exceptions (A=B, and A>B) and apply on the number line. Choice A: Not sufficient. A could be equal to B Choice B: If b=1 and A=0 than we have 3<3 Choice C permits to say that the equation is not true because we always find a number that contradicts with the equation. If a=2 and B=3 then we have 5<3... Therefore C is right because it enables us to say that the statement is false. Hope it helps
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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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25 Jan 2014, 08:58
From first statement: LHS=Mod(2a9)=mod(a3+a6) RHS=mod(a3)+mod(a6) We can put diff values of 'a' to get diff LHSRHS relationships.Insufficient.
From statement 2:aLHS=Mod(ab+a2b) RHS=mod(ab)+mod(a2b) Putting values:a=1;b=2 LHS=4 & RHS=4 a=1;b=4 LHS=10 & RHS=10 a=1 & b=1 LHS=4 & RHS=5 Again nothing can be said conclusively about the relationship.Insufficient.
Combining the 2 Statements: b=3 and aWe always get LHS=RHS.
Ans.C
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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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01 Aug 2014, 18:04
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 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?) in your range test, you missed a = 3, a =4.5, and a = 6. However, it does not change the result. Thanks for sharing your exp
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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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19 Aug 2014, 23:28
Hello all,
My approach to this question:
we are asked if 2a − 3b < a − b + a − 2b ? i.e. (ab) + (a2b) < a − b + a − 2b i.e. x + y < x + y, where x = ab and y = a2b
this is possible only in 2 cases: (i) x is +ve and y is ve ; example check with (x,y) = (3,4) or (4,3) (ii) x is ve and y is +ve ; example check with (x,y) = (3,4) or (4,3)
now for case (1) when x is +ve and y is ve i.e. ab > 0 and a2b <0 i.e. a > b and a < 2b i.e. b < a < 2b clearly this is only possible when a and b both are positive (for b<2b to hold true) and a lies within the positive range {b,2b}
now for case (ii) when x is ve and y is +ve i.e. ab < 0 and a 2b > 0 i.e. a < b and a > 2b i.e. 2b < a < b clearly this is only possible when both a and b are negaitve (for 2b<b to hold true) and a lies within the negative range {2b,b}
Now lets assess the choices: (1) b = 3 this does not give us any idea about if a < b or a > b. Hence insufficient
(2) a < b this does not provide us information about if a and b are positive or negative real numbers
(1) & (2) together b = 3 (positive real number) and a < b From our derivation we know that, when a < b, both a and b should be negative (case ii mentioned above). However here, we have b as a positive number. hence both (1) and (2) taken together disproves the inequality, hence sufficient.
PS: if this question had provided us the 2 choices as (1) a> b and (2) b = 3, than the choice (c) of finding sufficiency using both the options would still be insufficient. because even if a>b and b is positive = 3, we still need an 'a' which lies within b (i.e. 3) and 2b (i.e. 6). And this information is absent.
Hope it helps..



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Is 2a − 3b < a − b + a − 2b [#permalink]
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16 Oct 2014, 15:23
Quote: a < b doesn't necessarily imply that a < 2b. For example a = 4 < 3 = b, but a = 4 > 2(3) = 6 = 2b. unsure of this explanation. with b = 3 and a < b, how can ab be negative but a2b be positive? seems to me like both are negative and that would mean the inequality does not fail



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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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17 Oct 2014, 01:25
bsmith37 wrote: Quote: a < b doesn't necessarily imply that a < 2b. For example a = 4 < 3 = b, but a = 4 > 2(3) = 6 = 2b. unsure of this explanation. with b = 3 and a < b, how can ab be negative but a2b be positive? seems to me like both are negative and that would mean the inequality does not fail We have that b = 3 and a < b, so a < 3. So, the question becomes is 2a − 9 < a − 3 + a − 6? Now, since a < 3, then 2a − 9 < 0, a − 3 < 0 and a − 6 <0. Now the question becomes, is (2a − 9) < (a − 3)  (a − 6)? > is 9 < 9? The answer to this question is NO.
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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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18 Oct 2014, 11:14
This DS is testing your knowledge of wellknown formula ǀx+yǀ ≤ǀxǀ +ǀyǀ. In the given case we don’t have “=” sign. Hence, 2a − 3b < a − b + a − 2b is NOT true Only When, 2a − 3b = a − b + a − 2b is true, For this we have to remember another key side of the formula which states: ǀx+yǀ =ǀxǀ +ǀyǀ when xy ≥0; So, 2a − 3b = a − b + a − 2b is true when (ab)(a2b)≥0 is true; We can’t go further here as we have two unknown variables. Thus we have to use (1) b = 3 now. Now when we solve (a3)(a6)≥0, and we find out that a has to be any number among (∞; 3][6: ∞). Still we don’t have enough info to prove 2a − 3b = a − b + a − 2b Now we have to use 2) a < b , which gives us a<3 (remember (1) b = 3) in other worlds aЄ(∞; 3) . Hence, using (1) b = 3 and (2) a < b together is SUFFICIENT to answer 2a − 3b = a − b + a − 2b, which can answer NO to the original question “Is 2a − 3b < a − b + a − 2b?” Thus C is the answer.
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Is 2a − 3b < a − b + a − 2b? [#permalink]
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11 Mar 2015, 23:53
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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 [#permalink]
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27 Jun 2015, 05:42
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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 [#permalink]
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05 Sep 2015, 01:22
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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 [#permalink]
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21 Dec 2017, 11:38
Galiya wrote: Is 2a − 3b < a − b + a − 2b?
(1) b = 3 (2) a < b
Source: Go Gmat Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question. We can modify the original condition and question using the following property. \(x+y < x + y ⇔ xy < 0\) The question asks if \((ab)(a2b) < 0\) since \(2a3b = (ab)+(a2b) < ab + a2b\) by replacing \(x = ab\) and \(y = a2b\). Since we have 2 variables (\(x\) and \(y\)) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. Condition 1) & 2): We have \(a < b < 2b\), since \(b = 3\). \(a  b < 0\) and \(a  2b < 0\). Thus \((ab)(a2b) > 0\). By CMT(Common Mistake Type) 1, since "No" is also an answer, both conditions 1) & 2) together are sufficient. Since this question is an absolute valeu question (one of the key question areas), CMT 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) \(a = 1, b = 3\) : \((ab)(a2b) > 0\) \(a = 4, b = 3\) : \((ab)(a2b) < 0\) The condition 1) is not sufficient. Condition 2) \(a = 1, b = 3\) : \((ab)(a2b) > 0\) \(a = 4, b = 3\) : \((ab)(a2b) < 0\) The condition 2) is not sufficient. Therefore, C is the answer. Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is 2a − 3b < a − b + a − 2b [#permalink]
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23 Dec 2017, 07:30
\(2a3b < ab + a2b\) ?
Squaring both sides, as it is safe to square on both side, since modulus positive or 0 always
\(4a^2 + 9b^2  12ab < a^2 + b^2 2ab + a^2 + 4b^2  4ab\) ? \(2a^2 + 4b^2 < 6ab\) ? \(a^2 + 2b^2 < 3ab\) ?
Statement 1: b = 3
question becomes : \(a^2 + 18 < 9a\) => \(a^2  9a + 18 < 0\) => \((a  6)(a  3) < 0\) => or \(3 < a < 6\) ? we don't know about a => insufficient
Statement 2: \(a < b\), there could be values for which \(a^2 + 2b^2 < 3ab\) may hold or may not hold true => insuff
(1) + (2), since b = 3, a < 3, answer to question 3 < a < 6 ? is NO => sufficient => (C)




Re: Is 2a − 3b < a − b + a − 2b
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