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If a and b are integers, and a > b, is a b < a [#permalink]
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10 Oct 2009, 06:51
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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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Re: Absolute values [#permalink]
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10 Oct 2009, 07:45
yangsta8 wrote: If a and b are integers, and a > b, is a·b < a – b? 1. a < 0 2. ab >= 0
I managed to solve this by taking values for a and b that were both positive and negative. I managed to get the correct answer based on this approach but it took me almost 4 minutes, roughly 2 mins to solve for each of the clues. Is there any faster way to solve this?
I think this is a 700+ question. 1. a < 0 .... a is ve b is always +ve so a.b is always negative also a>b so even if b is +ve (ab) will always be ve. But a.b will always give a much smaller value than ab. So suff. 2. ab >= 0 . here a and b will always have the same sign unless and until one of them is 0. so if we have a and b both to be ve, our problem boils down to the one in case 1. i.e. the equation holds good. But consider a = 0, and b to be +ve and we see that it contradicts our previous result. So 2 is insuff. So A.
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10 Oct 2009, 07:56
jax91 wrote: 1. a < 0 .... a is ve b is always +ve so a.b is always negative also a>b so even if b is +ve (ab) will always be ve.
The bolded statement above is true. But of b is negative then (ab) can potentially be > 0 hence Statement 1 can be insuff.



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Re: Absolute values [#permalink]
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10 Oct 2009, 08:22
yangsta8 wrote: jax91 wrote: 1. a < 0 .... a is ve b is always +ve so a.b is always negative also a>b so even if b is +ve (ab) will always be ve.
The bolded statement above is true. But of b is negative then (ab) can potentially be > 0 hence Statement 1 can be insuff. a > b so if be is ve, like a 11 then b = 11 so a has to be > 11 say a = 12 if a is ve that makes a = 12 so (ab) = 12  (11) = 12 +11 = 1 and a.b = 12 . 11 = 12.(11) = 132 please correct me if i missed something.
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Re: Absolute values [#permalink]
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jax91 wrote: so (ab) = 12  (11) = 12 +11 = 1 and a.b = 12 . 11 = 12.(11) = 132 Your working for the above is correct. So in the above you prove that a.b < ab is true. However if you take a smaller number Let a = 2 let b = 1 a.b = 2 . 1 = 2.1 = 2 (ab) = 2  1 = 3 a.b < ab is false. Hence A is insufficient.



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Re: Absolute values [#permalink]
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10 Oct 2009, 08:35
yangsta8 wrote: jax91 wrote: so (ab) = 12  (11) = 12 +11 = 1 and a.b = 12 . 11 = 12.(11) = 132 Your working for the above is correct. So in the above you prove that a.b < ab is true. However if you take a smaller number Let a = 2 let b = 1 a.b = 2 . 1 = 2.1 = 2 (ab) = 2  1 = 3 a.b < ab is false. Hence A is insufficient. Missed that condition. Thanks for pointing it out.
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Re: Absolute values [#permalink]
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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: Absolute values [#permalink]
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10 Oct 2009, 23:19
yangsta8 wrote: If a and b are integers, and a > b, is a·b < a – b?
1. a < 0 2. ab >= 0
I managed to solve this by taking values for a and b that were both positive and negative. I managed to get the correct answer based on this approach but it took me almost 4 minutes, roughly 2 mins to solve for each of the clues. Is there any faster way to solve this?
I think this is a 700+ question. 1. If a < 0, b could be ve or 0 or +ve. If b = ve, (a·b) and (a – b) both are ve however the relationship between (a·b) and (a – b) cannot be established. If b = 0, (a·b) = 0 but (a – b) is still ve. so (a·b) < (a – b) is not true. If b = +ve, (a·b) and (a – b) bot hare ve however the relationship between (a·b) and (a – b) cannot be established. NSF... 2. If ab >= 0, a and b could be both ve or +ve or 0 or only one of either is 0 and the other is nonzero. The relationship between (a·b) and (a – b) cannot be established here too. NSF... Togather also the relationship between (a·b) and (a – b) cannot be established. NSF...... Thats E.
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Re: Absolute values [#permalink]
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11 Oct 2009, 08:38
i would go for c, pls post OA



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Re: Absolute values [#permalink]
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11 Oct 2009, 09:20
E for me
Condtion 2 does not bind anything, if B equals 0 or less than 1 and if B equals a large number with A being a bigger abs valued negative number it does not prove sufficient.



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Re: If a and b are integers, and a > b, is a b < a [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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16 Apr 2013, 22:56
VeritasPrepKarishma wrote: You got the question wrong. It is Is a * b < a  b ? (there is a dot there, not a minus sign) aaaaand that's why I have a bag over my head. See you at Harvard next year!



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