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If a and b are integers, and a > b, is a b < a
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11 Sep 2009, 01:57
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If a and b are integers, and a > b, is a · b < a – b? (1) a < 0 (2) ab >= 0
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If a and b are integers, and a > b, is a b < a
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11 Sep 2009, 02:46
If a and b are integers, and a > b, is a · b < a – b? Given: \(a>b\) Question: is \(a*b<ab\)? (1) \(a<0\). If \(a=3\) and \(b=0\), then \(a*b=0>ab=3\) and the answer is NO but if \(a=3\) and \(b=1\), then \(a*b=3<ab=2\) and the answer is YES. Two different answers. Not sufficient. (2) \(ab\geq{0}\) Above example works here as well: \(a=3\) and \(b=0\) > \(a*b=0>ab=3\) > answer NO; \(a=3\) and \(b=1\) > \(a*b=3<ab=2\) > answer YES. Two different answers. Not sufficient. (1)+(2) Again the same example satisfies the stem and both statements and gives two different answers to the question whether \(a*b<ab\). Hence not sufficient. Answer E.
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Re: Absolute values
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10 Oct 2009, 11:54
If a and b are integers, and a > b, is a·b < a – b? 1. a < 0 2. ab >= 0 The key to solve this problem is to determine the relationship between the signs of ab and a·b, with the given condition that a>b. When determined the rest of problem will go smoothly. So, NOTE that when a>b: ab>0 IF and ONLY a>0, no matter what possible values will take b (possible means not violating the given condition a>b) ab<0 IF and ONLY a<0, no matter what possible values will take b (possible means not violating the given condition a>b) Above conclusion means that: when a>0 > a·b>0 and a – b>0 AND when a<0 > a·b<0 and a – b<0Also NOTE that even knowing the signs of LHS and RHS of our inequality its not possible to determine LHS<RHS or not. Generally speaking even not considering the statements, we can conclude, that if they are giving us ONLY the info about the signs of a and b, it won't help us to answer the Q. (in our case we can even not consider them separately or together, we know answer would be E, as the statements are only about the signs of the variables) But still let's look at the statement: (1) a<0 > ab<0 (a negative b positive) and we already determined that when a<0 ab<0 > so both are negative but we can not determine is a · b < a – b or not. Not Sufficient (2) ab >= 0 a>0 b=>0 (a can not be zero as a>b) or a<0, b<=0 a>0 b=>0 > ab>0 and a – b>0 > both are positive but we can not determine is a · b < a – b or not. a<0, b<=0 > ab<0 and ab<0 > both are negative but we can not determine is a · b < a – b or not. Not Sufficient (1)+(2) > a<0, b<0 same thing > ab<0 and ab<0 we can not determine is a · b < a – b or not. Answer E. Hope this helps.
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Re: If a and b are integers, and a > b, is a b < a
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24 Aug 2012, 13:14
I find it much easier to pick numbers to figure out the answer than do it algebraically.
WHAT WE KNOW: a & b are both integers a>b
Question Type: YES OR NO Is a*b<ab?
1) a < 0 meaning a is negative. But b could be anything as long as its absolute value is elss than a. Let's say a=3 B coudl equal 2, 0, or 2.
3*2<3(2)? 6<1? .......YES 3*0<30 0<3?..........NO 3*2<32 6<5...........YES >INSUFFICIENT
2) ab>=0 which means a b be either have the same sign, or one of them is 0. We know that a cannot be 0 because a>b. Therefore only b can be 0.
3*2<3(2).... YES 3*0<30...........NO 3*0<30............NO 3*2<32..............NO >INSUFFICIENT
1&2) Combined, if a is negative, and ab>=0, this means that b is either 0 or also negative. 3*2<3(2).... YES 3*0<30............NO >INSUFFICIENT
Answer is E.



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Re: If a and b are integers, and a > b, is a b < a
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24 Aug 2012, 13:53
Nice explanation Chris. But I don't think that picking up numbers is the best strategy. Can you tell how to decide which method , mathematics or picking numbers , to be chosen after seeing the question. Moreover can you tell me my mistake in the post just before your post.
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Re: If a and b are integers, and a > b, is a b < a
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24 Aug 2012, 15:00
siddharthasingh wrote: Nice explanation Chris. But I don't think that picking up numbers is the best strategy. Can you tell how to decide which method , mathematics or picking numbers , to be chosen after seeing the question. Moreover can you tell me my mistake in the post just before your post. Hi siddharthasingh, I think choosing between solving a problem conceptually, algebraically, or by picking numbers is really up to what you are comfortable with. For myself, when it comes to these types of questions with lots of absolute values, equations, and , I know I figure it out quicker either conceptually or by picking numbers. It's basically the best strategy for myself. I basically came to this conclusion after doing lots of practice problems with the help of my error log. Now, keep in mind I probably would not have done ALL those calculations in my post. I'm not aiming to solve the question, I just want to see if I have enough information to solve the question. Once I came up with a "yes" AND a "no" for each statement, I would have stopped. I just wrote it all out to explain. a>b => either i)a>b ii)a<bActually, there is a multitude of possibilities. What this statement means is that a is farther away from 0 than b is. Both positive: a>b both negative: a<b a neg, b pos: a<b where 0a>b0 a pos, b neg: a>b where a0>0b So I personally think it would be easier to view the statement conceptually, knowing that a>b means the distance of a is greater than the distance of b. Now, please keep in mind that the above scenarios are POSSIBILITIES. Only 1 is true. We want to figure out if they're both positive or not. 1) a<0 On combining i and 1, b<a<0. Therefore ab is positive. ab is negative. On combining ii and 1) Two cases are emerging. a0(b) a(b)0We don't know if b<a. we only know that a>b. knowing that a is a NEGATIVE gets rid of a few possiblities above. A<0, and be can be EITHER neg, pos, or both. i) b=pos while a>b (neg)pos<(neg)(pos)....neg<pos... YES ii)b=neg while a>b (neg)neg<(neg)(neg)....neg<neg....YES Keep in mind that in this case ab will be a bigger negative than ab iii)b=0 (neg)0<(neg)0....0<neg.... NO. 1 is insufficient. Not sure if this helped clear anything up....



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Re: If a and b are integers, and a > b, is a b < a
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17 Jan 2013, 00:18
yangsta8 wrote: If a and b are integers, and a > b, is a·b < a – b? 1. a < 0 2. ab >= 0
I noticed that these kinds of questions are important on the GMAT... I've seen many of these types that initially threw me off until I faced this type and practiced as much...My solution: 1. Test a is () and b is () : 21 < 2 + 1 ==> 2 < 1 YES! Test a is () and b is (+): 2 1 < 2  1 ==> 2 < 3 NO! We stop here since we know that the information is INSUFFICIENT.2. The info meant a and b must have the same sign or one of them is 0. Test a is () and b is (): From Statement 1 we know this is YES! Test a is (+) and b is (+): 21 < 2  1 ==> 2 < 1 NO! We stop here and we know that the information is INSUFFICIENT.Together: We know a is () and be is either () or (0) Test a is () and b is (): From Statement 1 we know this is YES! Test a is () and b is 0: 0 < 2 NO! Still information together is INSUFFICIENT!Answer: E



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Re: If a and b are integers, and a > b, is a b < a
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21 Jan 2013, 16:48
mbaiseasy wrote: yangsta8 wrote: If a and b are integers, and a > b, is a·b < a – b? 1. a < 0 2. ab >= 0
I noticed that these kinds of questions are important on the GMAT... I've seen many of these types that initially threw me off until I faced this type and practiced as much...My solution: 1. Test a is () and b is () : 21 < 2 + 1 ==> 2 < 1 YES! Test a is () and b is (+): 2 1 < 2  1 ==> 2 < 3 NO! We stop here since we know that the information is INSUFFICIENT.2. The info meant a and b must have the same sign or one of them is 0. Test a is () and b is (): From Statement 1 we know this is YES! Test a is (+) and b is (+): 21 < 2  1 ==> 2 < 1 NO! We stop here and we know that the information is INSUFFICIENT.Together: We know a is () and be is either () or (0) Test a is () and b is (): From Statement 1 we know this is YES! Test a is () and b is 0: 0 < 2 NO! Still information together is INSUFFICIENT!Answer: E This is the type of question that I have the most trouble with. Part of my problem is to think about the numbers that would prove the statement sufficient/insufficient. For example, in statement 1, if a=5 and b=2, then 5 l 2 l < 5  (2) Yes. So at this point, I think to myself what numbers can I plugin to prove that statement 1 insufficient. This is when I get in trouble because I spend too much time thinking about what numbers do I need to plug in next. So I just pick another random number: a=5 and b=2, then 5 l 2 l < 5  (2) Yes. By this time I'm probably close to the 2minute mark. How do you know what numbers to test efficiently? Can anybody suggest more of this type of questions or any materials that I can practice on? Any advice will be greatly appreciated. Thanks.



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Re: If a and b are integers, and a > b, is a b < a
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21 Jan 2013, 21:34
Samwong wrote: This is the type of question that I have the most trouble with. Part of my problem is to think about the numbers that would prove the statement sufficient/insufficient. For example, in statement 1, if a=5 and b=2, then 5 l 2 l < 5  (2) Yes. So at this point, I think to myself what numbers can I plugin to prove that statement 1 insufficient. This is when I get in trouble because I spend too much time thinking about what numbers do I need to plug in next. So I just pick another random number: a=5 and b=2, then 5 l 2 l < 5  (2) Yes. By this time I'm probably close to the 2minute mark.
How do you know what numbers to test efficiently? Can anybody suggest more of this type of questions or any materials that I can practice on? Any advice will be greatly appreciated. Thanks. Such questions are best solved using part logic and part number plugging. When you plug numbers, keep in mind that you have to try to get the answer but keep life as simple as possible for yourself. Let me explain what I mean: If a and b are integers, and a > b, is a·b < a – b? 1. a < 0 2. ab >= 0 Read the question stem: "If a and b are integers"  very good .. no fractions to worry about "and a > b"  absolute value of a is greater than absolute value of b.. we don't know anything about their signs "is a·b < a – b" There is no obvious relation between the left hand side (LHS) and the right hand side(RHS) so I need to move on and look at the statements. 1. a < 0 a is negative means LHS is negative or 0 (if b is 0). RHS could be negative or positive depending on value of b. Assume b = 0 (makes life simple) We get 0 < a. Inequality does not hold since a is negative. Assume b is a large negative number say, 10 and a is a number a little smaller than b. LHS is a large negative number and RHS is a small negative number. e.g. a = 11, b = 10 110 < 1 Inequality holds. Insufficient. 2. ab >= 0 This means EITHER a is negative, b is negative (or one or both are 0) OR a is positive, b is positive (or one or both are 0) Now notice that even with both statements together, you cannot say whether the inequality holds. With statement 1, we saw two cases  b large negative, a is a little smaller than b (inequality holds) a negative, b is 0 (inequality does not hold) According to this statement as well, both cases are possible. Hence this statement alone is not sufficient and both together are also not sufficient. Hence answer must be (E) Check out a related post that discusses how to choose the numbers you should plug: http://www.veritasprep.com/blog/2012/12 ... gameplan/(Edited)
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Re: If a and b are integers, and a > b, is a b < a
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01 Feb 2013, 16:25
VeritasPrepKarishma wrote: Samwong wrote: This is the type of question that I have the most trouble with. Part of my problem is to think about the numbers that would prove the statement sufficient/insufficient. For example, in statement 1, if a=5 and b=2, then 5 l 2 l < 5  (2) Yes. So at this point, I think to myself what numbers can I plugin to prove that statement 1 insufficient. This is when I get in trouble because I spend too much time thinking about what numbers do I need to plug in next. So I just pick another random number: a=5 and b=2, then 5 l 2 l < 5  (2) Yes. By this time I'm probably close to the 2minute mark.
How do you know what numbers to test efficiently? Can anybody suggest more of this type of questions or any materials that I can practice on? Any advice will be greatly appreciated. Thanks. Such questions are best solved using part logic and part number plugging. When you plug numbers, keep in mind that you have to try to get the answer but keep life as simple as possible for yourself. Let me explain what I mean: If a and b are integers, and a > b, is a·b < a – b? 1. a < 0 2. ab >= 0 Read the question stem: "If a and b are integers"  very good .. no fractions to worry about "and a > b"  absolute value of a is greater than absolute value of b.. we don't know anything about their signs "is a·b < a – b" There is no obvious relation between the left hand side (LHS) and the right hand side(RHS) so I need to move on and look at the statements. 1. a < 0 a is negative means LHS is negative or 0 (if b is 0). RHS could be negative or positive depending on value of b. Assume b = 0 (makes life simple) We get 0 < a. Inequality does not hold since a is negative. Assume b is a large negative number say, 10 and a is a small negative number. LHS is negative and RHS is positive. Inequality holds.Insufficient. 2. ab >= 0 This means EITHER a is negative, b is negative (or one or both are 0) OR a is positive, b is positive (or one or both are 0) Now notice that even with both statements together, you cannot say whether the inequality holds. With statement 1, we saw two cases  a small negative, b large negative (inequality holds) a negative, b is 0 (inequality does not hold) According to this statement as well, both cases are possible. Hence this statement alone is not sufficient and both together are also not sufficient. Hence answer must be (E) Check out a related post that discusses how to choose the numbers you should plug: http://www.veritasprep.com/blog/2012/12 ... gameplan/Hi Karishma, Thank you for answering my question. However, in statement 1, I don't see how the RHS can be positive when "a" is negative with the given condition a > b. If b = 10 then "a" has to be less than 10 (ie 11, 12, 13...) Thus, 11  ( 10) = 1 RHS = negative. Please check whether I miss something. Thanks.



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Re: If a and b are integers, and a > b, is a b < a
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03 Feb 2013, 23:09
Samwong wrote: Hi Karishma,
Thank you for answering my question. However, in statement 1, I don't see how the RHS can be positive when "a" is negative with the given condition a > b. If b = 10 then "a" has to be less than 10 (ie 11, 12, 13...) Thus, 11  ( 10) = 1 RHS = negative. Please check whether I miss something. Thanks.
Yes, you are right. a cannot be smaller than b so we should take numbers where a is a little less than b to get a small negative on RHS. The LHS will be negative with a greater absolute value. a = 11, b = 10 110 < 11  (10) 110 < 1 (inequality holds)
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Re: If a and b are integers, and a > b, is a b < a
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16 Apr 2013, 14:39
OK, my process for solving this was way simpler than all this number plugging and stuff I'm seeing, but maybe I'm doing something wrong and I just happened to get the right answer. Here's my thought process.
1) Manipulate the question stem
is a  b < a  b ?  subtract 'a' from both sides to get
is b < b ? the answer is 'yes' if 'b' is negative, and 'no' if 'b' is nonnegative, so in order to be sufficient the data needs to tell us definitively the sign of 'b'
1) a < 0 tells us nothing about the sign of 'b', INSUFFICIENT
2) ab >= 0 tells us that 'a' and 'b' have the same sign, or that 'a' or 'b' or both are zero. From the original question we know that a > b, so 'a' cannot be zero, which means this statement is telling us that either 'a' and 'b' have the same sign, or b=0. This does nothing to establish the sign of 'b', INSUFFICIENT
1 & 2 together: applying " a < 0 " to "ab >= 0 " tells us that 'b' must nonpositive, but can still equal zero or any negative number greater than 'a'. Since there are multiple possibilities for the sign of 'b', we cannot answer the original question of " is b < b ?", so the answer is E, statements 1 and 2 are INSUFFICIENT
Is this valid logic? I think it hinges on the question of whether or not you can manipulate the question stem by subtracting 'a' from both sides.



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Re: If a and b are integers, and a > b, is a b < a
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16 Apr 2013, 22:33
dustwun wrote: OK, my process for solving this was way simpler than all this number plugging and stuff I'm seeing, but maybe I'm doing something wrong and I just happened to get the right answer. Here's my thought process.
1) Manipulate the question stem
is a  b < a  b ?  subtract 'a' from both sides to get
is b < b ? the answer is 'yes' if 'b' is negative, and 'no' if 'b' is nonnegative, so in order to be sufficient the data needs to tell us definitively the sign of 'b'
1) a < 0 tells us nothing about the sign of 'b', INSUFFICIENT
2) ab >= 0 tells us that 'a' and 'b' have the same sign, or that 'a' or 'b' or both are zero. From the original question we know that a > b, so 'a' cannot be zero, which means this statement is telling us that either 'a' and 'b' have the same sign, or b=0. This does nothing to establish the sign of 'b', INSUFFICIENT
1 & 2 together: applying " a < 0 " to "ab >= 0 " tells us that 'b' must nonpositive, but can still equal zero or any negative number greater than 'a'. Since there are multiple possibilities for the sign of 'b', we cannot answer the original question of " is b < b ?", so the answer is E, statements 1 and 2 are INSUFFICIENT
Is this valid logic? I think it hinges on the question of whether or not you can manipulate the question stem by subtracting 'a' from both sides. You got the question wrong. It is Is a * b < a  b ? (there is a dot there, not a minus sign)
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Re: If a and b are integers, and a > b, is a b < a
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06 Jan 2014, 01:04
gmatgambler wrote: If a & b are integers ,and a > b, is a*b< ab ?
A)a<0
B)\(ab>=0\)
How to solve this algebraically ? Statement I is insufficient 'a' is a negative number. (Negative) (Positive) < Negative  Number Not always 2(3) < 2  3 AND 1(3) > 1  3 Statement II is insufficient A and B have the same signs. (Negative) (Positive) < Negative  Negative (Positive) (Positive) < Positive  Positive Not always the sign will hold as (+3)(9) > 3  9 AND (2)(+3) < 2 +3 Combining is insufficient a is negative and b is also negative (AB>0) Now it comes to (Negative) (Positive) < Negative  Negative Not always: (3)(2) < 3 + 2 AND (0.2) (0.1) > 0.2 + 0.1 Hence the answer is E.
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Re: If a and b are integers, and a > b, is a b < a
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09 May 2016, 23:25
I personally am always prefer the algebraic method to solve rather than mumber plugin. I came up with this way to solve. Please comment if it is acceptable or not.
Attachments
IMG20160510WA0001.jpg [ 115.68 KiB  Viewed 3067 times ]
File comment: I have not solved for the second statement and also combining both the statements because they are the subset as computed for the statement 1.
So the answer ie E
IMG20160510WA0003.jpg [ 139.11 KiB  Viewed 3081 times ]



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Re: If a and b are integers, and a > b, is a b < a
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