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Is a  b < ab ? [#permalink]
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Is a  b < ab ? Statement 1: \(b^a\)<0 Statement 2: b>a
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Last edited by chetan2u on 23 Dec 2017, 20:34, edited 1 time in total.
Corrected the OA



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Re: Is a  b < ab ? [#permalink]
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03 Aug 2017, 09:35
DH99 wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a Dear DH99, I'm happy to respond. This is an intriguing question. I don't agree with the answer: I wonder if you posted the correct OA. Statement #1 is particularly interesting. \(b^a\)<0 This means a few things. The number b must be negative and it doesn't have to be an integer. The number a cannot be zero, but it could be any positive or negative odd integer. Example #1: \(a = 3\) and \(b = 2\) This yields \(b^a\) = \((2)^3\) = \( 8\) < \(0\), so these choices are consistent with Statement #1 Then \(a  b = 3  2 = 3  2 = 1\) and \(a  b = 3  (2) = 5 = 5\) This choice gives a "yes" answer to the prompt question. Example #2: \(a = 3\) and \(b = 2\) This yields \(b^a\) = \((2)^{3}\) = \( \tfrac{1}{8}\) < \(0\), so these choices are consistent with Statement #1 Then \(a  b = 3  2 = 3  2 = 1\) and \(a  b = 3  (2) = 1 = 1\) This choice gives a "no" answer to the prompt question. Two different choices consistent with the statement yield two different answer to the prompt question, so statement #1, alone and by itself, is insufficient. We can do algebraic work with Statement #1 \(b>a\) Subtract b from both sides \(0 > a  b\) So this difference now is always negative. Of course, a single absolute value is always positive (or zero of a = b, but that's excluded by this statement). Thus, \(a  b < 0 < a  b\) We absolutely have a "yes" answer to this statement. Because we have arrived at a definitive answer, we know that statement #2, alone and by itself, is sufficient. I find the answer (B). Again, I am not sure whether something was copied incorrect, but my answer doesn't match the OA posted. Does all this make sense? Mike
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Re: Is a  b < ab ? [#permalink]
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03 Aug 2017, 10:01
mikemcgarry wrote: DH99 wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a Dear DH99, I'm happy to respond. This is an intriguing question. I don't agree with the answer: I wonder if you posted the correct OA. Statement #1 is particularly interesting. \(b^a\)<0 This means a few things. The number b must be negative and it doesn't have to be an integer. The number a cannot be zero, but it could be any positive or negative odd integer. Example #1: \(a = 3\) and \(b = 2\) This yields \(b^a\) = \((2)^3\) = \( 8\) < \(0\), so these choices are consistent with Statement #1 Then \(a  b = 3  2 = 3  2 = 1\) and \(a  b = 3  (2) = 5 = 5\) This choice gives a "yes" answer to the prompt question. Example #2: \(a = 3\) and \(b = 2\) This yields \(b^a\) = \((2)^{3}\) = \( \tfrac{1}{8}\) < \(0\), so these choices are consistent with Statement #1 Then \(a  b = 3  2 = 3  2 = 1\) and \(a  b = 3  (2) = 1 = 1\) This choice gives a "no" answer to the prompt question. Two different choices consistent with the statement yield two different answer to the prompt question, so statement #1, alone and by itself, is insufficient. We can do algebraic work with Statement #1 \(b>a\) Subtract b from both sides \(0 > a  b\) So this difference now is always negative. Of course, a single absolute value is always positive (or zero of a = b, but that's excluded by this statement). Thus, \(a  b < 0 < a  b\) We absolutely have a "yes" answer to this statement. Because we have arrived at a definitive answer, we know that statement #2, alone and by itself, is sufficient. I find the answer (B). Again, I am not sure whether something was copied incorrect, but my answer doesn't match the OA posted. Does all this make sense? Mike Hi, I just checked there is no typo with this question and the OA is also given as E.It is possible that the OA is wrong.



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Re: Is a  b < ab ? [#permalink]
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03 Aug 2017, 11:00
mikemcgarry wrote: DH99 wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a Dear DH99, I'm happy to respond. This is an intriguing question. I don't agree with the answer: I wonder if you posted the correct OA. Statement #1 is particularly interesting. \(b^a\)<0 This means a few things. The number b must be negative and it doesn't have to be an integer. The number a cannot be zero, but it could be any positive or negative odd integer. Example #1: \(a = 3\) and \(b = 2\) This yields \(b^a\) = \((2)^3\) = \( 8\) < \(0\), so these choices are consistent with Statement #1 Then \(a  b = 3  2 = 3  2 = 1\) and \(a  b = 3  (2) = 5 = 5\) This choice gives a "yes" answer to the prompt question. Example #2: \(a = 3\) and \(b = 2\) This yields \(b^a\) = \((2)^{3}\) = \( \tfrac{1}{8}\) < \(0\), so these choices are consistent with Statement #1 Then \(a  b = 3  2 = 3  2 = 1\) and \(a  b = 3  (2) = 1 = 1\) This choice gives a "no" answer to the prompt question. Two different choices consistent with the statement yield two different answer to the prompt question, so statement #1, alone and by itself, is insufficient. We can do algebraic work with Statement #1 \(b>a\) Subtract b from both sides \(0 > a  b\) So this difference now is always negative. Of course, a single absolute value is always positive (or zero of a = b, but that's excluded by this statement). Thus, \(a  b < 0 < a  b\) We absolutely have a "yes" answer to this statement. Because we have arrived at a definitive answer, we know that statement #2, alone and by itself, is sufficient. I find the answer (B). Again, I am not sure whether something was copied incorrect, but my answer doesn't match the OA posted. Does all this make sense? Mike Hi mikemcgarryWhile I do understand from your method why statement 2 is sufficient, could you please tell me what I am doing wrong. For a  b < ab to hold true , a and b should have opposite signs. Now in statement 2 b>a 3 > 2 { b = 3 and a =2 } 3 > 2 {b =3 and a =2 } since a and b can have the same sign as well as the opposite sign so by this logic, statement 2 is insufficient.



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Re: Is a  b < ab ? [#permalink]
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03 Aug 2017, 13:30
darn wrote: Hi mikemcgarry While I do understand from your method why statement 2 is sufficient, could you please tell me what I am doing wrong.
For a  b < ab to hold true , a and b should have opposite signs.
Now in statement 2
b>a
3 > 2 { b = 3 and a =2 } 3 > 2 {b =3 and a =2 }
since a and b can have the same sign as well as the opposite sign
so by this logic, statement 2 is insufficient. Dear darn, I'm happy to respond. The other problem similar to this could be deciphered according to whether a & b had the same or opposite signs, but that's not the case here. Here, the relative size of the absolute values of a & b is pertinent. It's true both {a = 2, b = 3} and {a = 2, b = 3}. For both of these, the left side of the prompt inequality, a  b, will equal 1, and the right side will equal something positive. Something positive is always greater than 1. Thus, both pairs produce a "yes" answer to the prompt. In this problem, regardless of the signs of a & b, if statement #2 is satisfied, then the answer to the prompt question is "yes." Does all this make sense? Mike
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Re: Is a  b < ab ? [#permalink]
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13 Dec 2017, 12:30
\(ab < ab\) ?
squaring on both sides, (safer to square both sides, since LHS > 0 and RHS > 0) => \(a^2 + b^2 2ab < a^2 + b^2  2ab\) ? => \(2ab < 2ab ?\) (as \(ab = ab\)) => \(ab > ab\) ? => \(ab < 0\) ? => a and b are of opposite sign?
Statement 1: \(b^a\) < 0 InSufficient, if b = 5, a = 3 => \(b^a < 0\) (a and b are of opposite sign) if b = 5, a = 3 => \(b^a < 0\) (but a and b are of same sign)
Statement 2: b > a => Clearly insufficient, no idea of signs of a and b
Statement 1 + 2: Examples taken for statement 1, also satisfy statement 2 so still insufficient to determine signs of a and b.
Answer (E)



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Re: Is a  b < ab ? [#permalink]
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13 Dec 2017, 22:12
DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a I too agree with Mikemcgarry here. Statement 2 is sufficient to answer the question. If b > a, then a  b < 0. So LHS is negative. But RHS can never be negative, as its ab, and modulus of a number can never be negative. We can check with 4 set of values for statement 2. Case 1. a=3, b=4. Here LHS = 1, RHS = 1 Case 2. a=3, b=4. Here LHS = 1, RHS = 7 Case 3. a=3, b=4. Here LHS = 1, RHS = 7 Case 4. a=3, b=4. Here LHS = 1, RHS = 1 So irrespective of the signs of a and b, as long as b < a, a  b < ab. Answer should be B.



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Re: Is a  b < ab ? [#permalink]
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13 Dec 2017, 22:14
hellosanthosh2k2 wrote: \(ab < ab\) ?
squaring on both sides, (safer to square both sides, since LHS > 0 and RHS > 0) => \(a^2 + b^2 2ab < a^2 + b^2  2ab\) ? => \(2ab < 2ab ?\) (as \(ab = ab\)) => \(ab > ab\) ? => \(ab < 0\) ? => a and b are of opposite sign?
Statement 1: \(b^a\) < 0 InSufficient, if b = 5, a = 3 => \(b^a < 0\) (a and b are of opposite sign) if b = 5, a = 3 => \(b^a < 0\) (but a and b are of same sign)
Statement 2: b > a => Clearly insufficient, no idea of signs of a and b
Statement 1 + 2: Examples taken for statement 1, also satisfy statement 2 so still insufficient to determine signs of a and b.
Answer (E) Hi hellosanthosh2k2 We cannot square both sides. While we know that RHS (ab) is clearly non negative, we cannot say the same about LHS (which is a  b). If b > a, then LHS will be negative.



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Re: Is a  b < ab ? [#permalink]
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20 Dec 2017, 13:27
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mikemcgarry wrote: darn wrote: Hi mikemcgarry While I do understand from your method why statement 2 is sufficient, could you please tell me what I am doing wrong.
For a  b < ab to hold true , a and b should have opposite signs.
Now in statement 2
b>a
3 > 2 { b = 3 and a =2 } 3 > 2 {b =3 and a =2 }
since a and b can have the same sign as well as the opposite sign
so by this logic, statement 2 is insufficient. I'm happy to respond. similar to this could be deciphered according to whether a & b had the same or opposite signs, but that's not the case here. Here, the relative size of the absolute values of a & b is pertinent. It's true both {a = 2, b = 3} and {a = 2, b = 3}. For both of these, the left side of the prompt inequality, a  b, will equal 1, and the right side will equal something positive. Something positive is always greater than 1. Thus, both pairs produce a "yes" answer to the prompt. In this problem, regardless of the signs of a & b, if statement #2 is satisfied, then the answer to the prompt question is "yes." Does all this make sense? Mike Hey Mike, In second condition, if a and b both are equal then ab<ab is No. For rest of the cases it is "Yes". Hence The condition 2 will be insufficient. Experts please comment. Thanks, Piyush Please give Kudos if it helps



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Re: Is a  b < ab ? [#permalink]
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20 Dec 2017, 17:47
Piyush0489 wrote: Hey Mike,
In second condition, if a and b both are equal then ab<ab is No. For rest of the cases it is "Yes". Hence The condition 2 will be insufficient.
Experts please comment.
Thanks, Piyush Dear Piyush0489, I'm happy to respond. The second statement is: Statement 2: b>aGiven that statement, there's no way that a could equal b. Does this make sense? Mike
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Re: Is a  b < ab ? [#permalink]
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22 Dec 2017, 06:15
The answer should be B right ??



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Is a  b < ab ? [#permalink]
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23 Dec 2017, 09:55
DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a 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. a  b < ab ⇔ a < ab + b ⇔ ab + b < ab + b ⇔ ab + b^2 < (ab + b)^2 since both sides are nonnegative in the above inequality. ⇔ (ab + b)^2 < (ab + b)^2 ⇔ (ab)^2 + 2b(ab) + b^2 < ab^2 + 2abb + b^2 ⇔ (ab)^2 + 2b(ab) + b^2 < (ab)^2 + 2abb + b^2 ⇔ 2b(ab) < 2(ab)b ⇔ b(ab) < (ab)b ⇔ b(ab) < 0 Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. Conditions 1) & 2) From the first condition 1), we have b < 0. From the the inequality b > a, we have b = b > a since b < 0. If a ≥ 0, a ≥ 0 > b since b < 0 If a < 0, a = a < b from the above inequality and we have a > b or a < b.. Thus b(ab) < 0 since b < 0 and ab < 0 Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) a = 1, b = 1 : Yes a = 1, b = 1 : No The condition 1) is not sufficient. Condition 2) If b ≥ 0, b = b > a ≥ a from the b>a. If b < 0 and b = b > a ≥ a or b > a. Thus a  b > 0 The condition 2) is sufficient. Therefore, B is the answer. Normally, in problems which require 2 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 a  b < ab ? [#permalink]
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23 Dec 2017, 10:16
MathRevolution wrote: DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a 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. a  b < ab ⇔ a < ab + b ⇔ ab + b < ab + b ⇔ ab + b^2 < (ab + b)^2 since both sides are nonnegative in the above inequality. ⇔ (ab + b)^2 < (ab + b)^2 ⇔ (ab)^2 + 2b(ab) + b^2 < ab^2 + 2abb + b^2 ⇔ (ab)^2 + 2b(ab) + b^2 < (ab)^2 + 2abb + b^2 ⇔ 2b(ab) < 2(ab)b ⇔ b(ab) < (ab)b ⇔ b(ab) < 0 Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. b = 2, a = 1 : Yes b = 4, a = 3 : No Thus, both conditions 1) & 2) are not sufficient. Therefore, E is the answer. Normally, in problems which require 2 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. This calculation seems to be convincing. But can you prove insufficiency of statement (2) with any values of a and b, without the derivation. From a logical point of view, LHS is always negative with statement (2) and RHS is always positive, no matter what.



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Re: Is a  b < ab ? [#permalink]
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23 Dec 2017, 19:30
MathRevolution wrote: DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a 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. a  b < ab ⇔ a < ab + b ⇔ ab + b < ab + b ⇔ ab + b^2 < (ab + b)^2 since both sides are nonnegative in the above inequality. ⇔ (ab + b)^2 < (ab + b)^2 ⇔ (ab)^2 + 2b(ab) + b^2 < ab^2 + 2abb + b^2 ⇔ (ab)^2 + 2b(ab) + b^2 < (ab)^2 + 2abb + b^2 ⇔ 2b(ab) < 2(ab)b ⇔ b(ab) < (ab)b ⇔ b(ab) < 0 Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. b = 2, a = 1 : Yes b = 4, a = 3 : NoThus, both conditions 1) & 2) are not sufficient. Therefore, E is the answer. Normally, in problems which require 2 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. Hi MathRevolutionThere seems to be a discripency in the highlighted portion if b=4 & a=3, then b>a AND a  b < ab => 34<3(4) i.e 1<1, which is true. so as per your logic also answer should be B. Essentially if b>a then ab<0, this should hold true for any condition. Hi Bunuel, the OA seems to be incorrect. Can you provide a more cogent explanation



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Re: Is a  b < ab ? [#permalink]
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23 Dec 2017, 20:33
DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a Lot of debate on OA, where clearly it is B, as also pointed by mikemcgarryIs a  b < ab ? II. b>a it says absolute value of b > that of a, so LHSa  b will be NEGATIVE also b>a....a  b < 0 Right Hand Side is ABSOLUTE value so always POSITIVE so LHS<RHS Sufficient OA being edited accordingly
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Re: Is a  b < ab ? [#permalink]
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24 Dec 2017, 10:56
MathRevolution wrote: DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a 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. a  b < ab ⇔ a < ab + b ⇔ ab + b < ab + b ⇔ ab + b^2 < (ab + b)^2 since both sides are nonnegative in the above inequality. ⇔ (ab + b)^2 < (ab + b)^2 ⇔ (ab)^2 + 2b(ab) + b^2 < ab^2 + 2abb + b^2 ⇔ (ab)^2 + 2b(ab) + b^2 < (ab)^2 + 2abb + b^2 ⇔ 2b(ab) < 2(ab)b ⇔ b(ab) < (ab)b ⇔ b(ab) < 0 Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. b = 2, a = 1 : Yes b = 4, a = 3 : No Thus, both conditions 1) & 2) are not sufficient. Therefore, E is the answer. Normally, in problems which require 2 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. Dear MathRevolution, My colleague, with all due respect, I am going to disagree on a few points. You first made three opening statements with which I disagree 1) Forget conventional ways of solving math questions. 2) For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. 3) Remember that equal numbers of variables and independent equations ensure a solution. In response to #1, I would assert that the heart of advanced problemsolving is elegance. In response to #2, I would say that the balance of your post seems a rebuttal to this statement. In response to #3, I would sayyes, this is a good rule, but the caveat should be mention that if two equations are dependent, we are not guaranteed a solution. I studied your algebra carefully. I don't find a single step that's off: your inequality may be equivalent to the prompt inequality. b(ab) < 0 Your claim is that this is equivalent to the prompt inequality. I cannot gainsay that claim. Then you picked two pairs of numbers: Pair #1: b = 2, a = 1 Pair #2: b = 4, a = 3 But here's the thing. Both of these pairs satisfy your inequality. In fact, I would submit that it's impossible to find a pair of numbers consistent with Statement #2 that doesn't satisfy your inequality. Both of these pairs definitely satisfy the prompt inequality. Prompt Inequality: a  b < ab Pair #1: a  b = 1  2 = 1 ab = 1  (2) = 3 = +3 Prompt inequality works. Pair #1: a  b = 3  4 = 3  4 = 1 ab = 3  (4) = 1 = +1 Prompt inequality works. If both pairs satisfy both your inequality and the prompt inequality, they do not constitute counterexamples. I make the claim that counterexamples consistent with Statement #2 do not exist. Please provide one if I am mistaken in this claim. Finally, I will say that the great irony is that this problem can be resolved an astonishingly elegant solution. Statement 1 can be dismissed with numerical counterexamples Statement 2: b>a Since these are not equal, we know a ≠ b, so a  b > 0 b>a is equivalent to the inequality 0 > a  b Combining, we get a  b < 0 < a  b, which directly implies a  b < a  b. With remarkable elegance, Statement #2 directly implies the prompt inequality and is sufficient. OA = (B). My colleague, I am open to anything you would like to say in reply. With sincere respect, Mike McGarry
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Re: Is a  b < ab ? [#permalink]
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25 Dec 2017, 06:15
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mikemcgarry wrote: MathRevolution wrote: DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a 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. a  b < ab ⇔ a < ab + b ⇔ ab + b < ab + b ⇔ ab + b^2 < (ab + b)^2 since both sides are nonnegative in the above inequality. ⇔ (ab + b)^2 < (ab + b)^2 ⇔ (ab)^2 + 2b(ab) + b^2 < ab^2 + 2abb + b^2 ⇔ (ab)^2 + 2b(ab) + b^2 < (ab)^2 + 2abb + b^2 ⇔ 2b(ab) < 2(ab)b ⇔ b(ab) < (ab)b ⇔ b(ab) < 0 Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. b = 2, a = 1 : Yes b = 4, a = 3 : No Thus, both conditions 1) & 2) are not sufficient. Therefore, E is the answer. Normally, in problems which require 2 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. Dear MathRevolution, My colleague, with all due respect, I am going to disagree on a few points. You first made three opening statements with which I disagree 1) Forget conventional ways of solving math questions. 2) For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. 3) Remember that equal numbers of variables and independent equations ensure a solution. In response to #1, I would assert that the heart of advanced problemsolving is elegance. In response to #2, I would say that the balance of your post seems a rebuttal to this statement. In response to #3, I would sayyes, this is a good rule, but the caveat should be mention that if two equations are dependent, we are not guaranteed a solution. I studied your algebra carefully. I don't find a single step that's off: your inequality may be equivalent to the prompt inequality. b(ab) < 0 Your claim is that this is equivalent to the prompt inequality. I cannot gainsay that claim. Then you picked two pairs of numbers: Pair #1: b = 2, a = 1 Pair #2: b = 4, a = 3 But here's the thing. Both of these pairs satisfy your inequality. In fact, I would submit that it's impossible to find a pair of numbers consistent with Statement #2 that doesn't satisfy your inequality. Both of these pairs definitely satisfy the prompt inequality. Prompt Inequality: a  b < ab Pair #1: a  b = 1  2 = 1 ab = 1  (2) = 3 = +3 Prompt inequality works. Pair #1: a  b = 3  4 = 3  4 = 1 ab = 3  (4) = 1 = +1 Prompt inequality works. If both pairs satisfy both your inequality and the prompt inequality, they do not constitute counterexamples. I make the claim that counterexamples consistent with Statement #2 do not exist. Please provide one if I am mistaken in this claim. Finally, I will say that the great irony is that this problem can be resolved an astonishingly elegant solution. Statement 1 can be dismissed with numerical counterexamples Statement 2: b>a Since these are not equal, we know a ≠ b, so a  b > 0 b>a is equivalent to the inequality 0 > a  b Combining, we get a  b < 0 < a  b, which directly implies a  b < a  b. With remarkable elegance, Statement #2 directly implies the prompt inequality and is sufficient. OA = (B). My colleague, I am open to anything you would like to say in reply. With sincere respect, Mike McGarry Yeah mike, I agree with your mentioned points. Thanks, Piyush Kudos +1



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Re: Is a  b < ab ? [#permalink]
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18 Jan 2018, 14:17
hellosanthosh2k2 wrote: \(ab < ab\) ?
squaring on both sides, (safer to square both sides, since LHS > 0 and RHS > 0) => \(a^2 + b^2 2ab < a^2 + b^2  2ab\) ? => \(2ab < 2ab ?\) (as \(ab = ab\)) => \(ab > ab\) ? => \(ab < 0\) ? => a and b are of opposite sign?
Statement 1: \(b^a\) < 0 InSufficient, if b = 5, a = 3 => \(b^a < 0\) (a and b are of opposite sign) if b = 5, a = 3 => \(b^a < 0\) (but a and b are of same sign)
Statement 2: b > a => Clearly insufficient, no idea of signs of a and b
Statement 1 + 2: Examples taken for statement 1, also satisfy statement 2 so still insufficient to determine signs of a and b.
Answer (E) Answer must be clean B. I missed the point if b > a, then a  b must be less than zero, and ab must be greater than or equal to zero. in which case, sufficient to answer the question as a  b < ab ? as YES



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Re: Is a  b < ab ? [#permalink]
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22 Jan 2018, 11:15
DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a 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. a  b < ab ⇔ a < ab + b ⇔ ab + b < ab + b ⇔ ab + b^2 < (ab + b)^2 since both sides are nonnegative in the above inequality. ⇔ (ab + b)^2 < (ab + b)^2 ⇔ (ab)^2 + 2b(ab) + b^2 < ab^2 + 2abb + b^2 ⇔ (ab)^2 + 2b(ab) + b^2 < (ab)^2 + 2abb + b^2 ⇔ 2b(ab) < 2(ab)b ⇔ b(ab) < (ab)b ⇔ b(ab) < 0 Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first. Conditions 1) & 2) From the first condition 1), we have b < 0. From the the inequality b > a, we have b = b > a since b < 0. If a ≥ 0, a ≥ 0 > b since b < 0 If a < 0, a = a < b from the above inequality and we have a > b or a < b.. Thus b(ab) < 0 since b < 0 and ab < 0 Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) a = 1, b = 1 : Yes a = 1, b = 1 : No The condition 1) is not sufficient. Condition 2) If b = 0, we don't have a solution a with 0 > a If b > 0, we have b = b > a ≥ a or b > a from the b>a and so b(ab) < 0 If b < 0 and b = b > a ≥ a or b > a and so a  b > 0 and b(ab) < 0 The condition 2) is sufficient. Therefore, B is the answer. In the previous solution, there was a mistake. Normally, in problems which require 2 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 a  b < ab ? [#permalink]
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01 Feb 2018, 07:27
DHAR wrote: Is a  b < ab ?
Statement 1: \(b^a\)<0 Statement 2: b>a We can reduce the question to is magnitude of a grater than magnitude of bFrom A: b is negative and a is odd but a can be less or more than a, so in suffucuent From B: clearly magnitude of B grater than A Sufficient
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