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Re: If ab ≠ 0, is ab > a+b?
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07 Jan 2017, 09:07
ab>a+b will be true only when a and b have different signs, as when we are doing > ab we are actually adding the values of a and b and > a+b we are actually subtracting the values of a and b. STAT1 ab < 0 => a and b have different signs SUFFICIENT STAT2 a > b a and b can have the same sign or can have different signs too example: 1. a = 7 b = 2 => a> b but ab<a+b (5 < 9) 2. a = 7 b = 2 => a> b ab>a+b (9 > 5) So, INSUFFICIENT so, answer will be A Hope it helps! nayanparikh wrote: If ab ≠ 0, is ab>a+b?
1) ab < 0 2) a > b
Can any one please provide explanation on how to solve these type of questions ? @Bunnel ?
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Re: If ab ≠ 0, is ab > a+b?
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13 Mar 2017, 10:21
nayanparikh wrote: If ab ≠ 0, is ab>a+b?
1) ab < 0 2) a > b
Can any one please provide explanation on how to solve these type of questions ? @Bunnel ? Bunnel has already explained that squaring both the sides of this inequality is the easiest. What I did was that for the following to be true: ab>a+b Quick substitution of values tells us that the above inequality will be true only when a and b are of opposite signs. We can try a=1, b=2 OR a=2, b=1 So, the question basically is if a and b are of opposite signs. (1) says ab < 0. This clearly means that b are of opposite signs. So sufficient. (2) says a > b. This does not tell us that b are of opposite signs. So not sufficient.



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If ab ≠ 0, is ab > a+b?
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15 Apr 2017, 11:12
Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Hi Bunuel, to confirm, the sign of the sides matters because if we multiply by a ve, then the inequality sign would need to flip, correct?



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Re: If ab ≠ 0, is ab > a+b?
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15 Apr 2017, 11:17
Cez005 wrote: Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Hi Bunuel, to confirm, the sign of the sides matters because if we multiply by a ve, then the inequality sign would need to flip, correct? We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative.Check the link below for more: Inequality tips
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Re: If ab ≠ 0, is ab > a+b?
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17 Apr 2017, 23:54
Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Hai, I am new here. I have a doubt. Where are the options mentioned, I don't see any A,B,C,D and E beneath the questions.



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Re: If ab ≠ 0, is ab > a+b?
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18 Apr 2017, 02:11
subhash29 wrote: Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Hai, I am new here. I have a doubt. Where are the options mentioned, I don't see any A,B,C,D and E beneath the questions. This is a data sufficiency question. Options for DS questions are always the same. The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether— A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked. C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. D. EACH statement ALONE is sufficient to answer the question asked. E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed. I suggest you to go through the following post ALL YOU NEED FOR QUANT. Hope this helps.
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Re: If ab ≠ 0, is ab > a+b?
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18 Apr 2017, 09:15
Bunuel wrote: subhash29 wrote: Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Hai, I am new here. I have a doubt. Where are the options mentioned, I don't see any A,B,C,D and E beneath the questions. This is a data sufficiency question. Options for DS questions are always the same. The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether— A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked. C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. D. EACH statement ALONE is sufficient to answer the question asked. E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed. I suggest you to go through the following post ALL YOU NEED FOR QUANT. Hope this helps. Thnx bunuel... Sent from my A0001 using GMAT Club Forum mobile app



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Re: If ab ≠ 0, is ab > a+b?
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03 Aug 2017, 11:29



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Re: If ab ≠ 0, is ab > a+b?
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24 Jul 2018, 22:41
Took the following approach...
(1) ab < 0 > either a is neg or b is neg but not both...what happens to our equation in either case? > case 1 (a<0 and b>0) LHS gets bigger neg  pos and RHS gets smaller neg + pos > LHS > RHS > case 2 (a>0 and b<0) LHS gets bigger pos  neg and RHS gets smaller pos + neg > LHS > RHS > same outcome both cases; therefore, SUFFICIENT
(2) a > b > a and b could both be neg or a could be pos and b could be neg or a could be pos and b could be pos...what happens to our equation in each case? > case 1 (a<0 and b<0) LHS gets smaller neg  neg and RHS gets bigger neg + neg > LHS < RHS > case 2 (reference above logic in statement 1) > LHS > RHS (could stop here b/c two different outcomes) > case 3 (a>0 and b>0) LHS gets smaller pos  pos and RHS gets bigger pos + pos > LHS < RHS > different outcomes possible; therefore, NOT SUFFICIENT
Answer = A
If there are any holes in this logic, please let me know!



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If ab ≠ 0, is ab > a+b?
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15 Feb 2019, 19:28
The best way to solve this is by using Bunuel's method. The next best way is to use smart numbers and plug them into the various equations. However, if you don't see that Bunuel's method is possible during the ~ 2min you have to solve the problem or if your smart numbers aren't working, then the following method is a pretty standard way to arrive at the right solution. Keep in mind though that this method does take longer and should be used as a kind of last resort. \(\lvert a  b\rvert > \lvert a + b\rvert\) \(\lvert a  b\rvert = a  b\), when \(a  b > 0\) \(\lvert a  b\rvert = (a  b)\), when \(a  b < 0\) \(\lvert a + b\rvert = a + b\), when \(a + b > 0\) \(\lvert a + b\rvert = (a + b)\), when \(a + b < 0\) Case 1:if \(a  b > 0, a + b > 0\), then \(a  b > a + b\) \(0 > 2b\) \(b < 0\) Also, if we add \(a + b > 0\) and \(a  b > 0\) we get: \(a > 0\) Case 2:if \(a  b > 0, a + b < 0\), then \(a  b > (a + b)\) \(a  b > a  b\) \(2a > 0\) \(a > 0\) Also, if we subtract \(a + b < 0\) from \(a  b > 0\) we get: \(b < 0\) Case 3:if \(a  b < 0, a + b > 0\), then \((a  b) > a + b\) \(a + b > a + b\) \(0 > 2a\) \(a < 0\) Also, if we subtract \(a + b > 0\) from \(a  b < 0\) we get: \(b > 0\) Case 4:if \(a  b < 0, a + b < 0\), then \((a  b) > (a + b)\) \(a + b > a  b\) \(2b > 0\) \(b > 0\) Also, if we add \(a + b < 0\) and \(a  b < 0\) we get: \(a < 0\) For all of our cases, if \(a < 0\), then \(b > 0\) or if \(a > 0\), then \(b < 0\). This is the same as \(ab < 0\). More on this method here and here.



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Re: If ab ≠ 0, is ab > a+b?
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15 Feb 2019, 20:47
Sallyzodiac wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
1) \(ab < 0\)
2) \(a > b\) See if you have time, you can actually open the mod and get the cases, such as b<0 , a> 0 , a<0 or b>0 Now from 1) we can easily say that all the 4 cases will be satisfied a can be +ive or ive b can be +ive or ive from 2) we cannot be sure on aspects such as a and b 3> 2 or 3> 2 These 2 cases will give us a Yes and a No, on the above expressions A
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Re: If ab ≠ 0, is ab > a+b?
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25 Feb 2019, 18:11
In the first statement there are only two possibilities either a= ve and b= +ve or b=ve and a = +ve no matter what the LHS is always going to add up and RHS is always going to subtract Hence Sufficient
But in second take two values a=1 b=2 a=1 b=2
Yes and No Hence, not sufficient.



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Re: If ab ≠ 0, is ab > a+b?
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15 May 2019, 12:58
Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Seems bold to do this bunuel..



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Re: If ab ≠ 0, is ab > a+b?
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24 Aug 2019, 00:44
Sallyzodiac wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
1) \(ab < 0\)
2) \(a > b\) Given: \(ab ≠ 0\) Asked: Is \(ab > a+b\)? 1) \(ab < 0\) Either a>0 and b<0 => ab>a+b Since b<0<a or b.....0.......a or a<0 and b>0 => ab > a+b since a<0<b or a......0.......b SUFFICIENT 2) \(a > b\) If a>0>b => ab>a+b since b.....0......a but if a>b>0 => ab<a+b since 0.....b......a NOT SUFFICIENT IMO A
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Re: If ab ≠ 0, is ab > a+b?
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24 Aug 2019, 01:50
Sallyzodiac wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
1) \(ab < 0\)
2) \(a > b\) Given: \(ab ≠ 0\) a & b are not 0 Asked: Is \(ab > a+b\)? This is TRUE only if a & b are of different signs 1) \(ab < 0\) a & b are of different signs SUFFICIENT 2) \(a > b\) If a>0>b => \(ab > a+b\) But if a>b>0 or 0>a>b => \(ab < a+b\) NOT SUFFICIENT IMO A
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Re: If ab ≠ 0, is ab > a+b?
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