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# If a and b are integers, and |a| > |b|, is a |b| < a

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If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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11 Sep 2009, 00:57
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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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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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11 Sep 2009, 01:46
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Bunuel wrote:
thailandvc wrote:
If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

anybody have a solution for this (under 2 minute). testing value takes way too long here.

E

Given: $$|a|>|b|$$
Question: is $$a*|b|<a-b$$?

(1) $$a<0$$. If $$a=-3$$ and $$b=0$$, then $$a*|b|=0>a-b=-3$$ and the answer is NO but if $$a=-3$$ and $$b=-1$$, then $$a*|b|=-3<a-b=-2$$ and the answer is YES. Two different answers. Not sufficient.

(2) $$ab\geq{0}$$

Above example works here as well:
$$a=-3$$ and $$b=0$$ --> $$a*|b|=0>a-b=-3$$ --> answer NO;
$$a=-3$$ and $$b=-1$$ --> $$a*|b|=-3<a-b=-2$$ --> answer YES.

(1)+(2) Again the same example satisfies the stem and both statements and gives two different answers to the question whether $$a*|b|<a-b$$. Hence not sufficient.

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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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24 Aug 2012, 12:14
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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|<a-b?

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|<-3-0
0<-3?..........NO
-3*|2|<-3-2
-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|<-3-0...........NO
-3*|0|<-3-0............NO
3*|2|<3-2..............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|<-3-0............NO
--->INSUFFICIENT

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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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05 Jul 2010, 18:36
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Hi Kandi,

Equate those four conditions as greater or equal to zero and less or equal to zero

1 - a<0
we have two conditions
- a<b<0 a=-5, b=-3
- a<0<b a=-5, b=3
So 1 alone is not sufficient

2 - ab>=0
We also have two conditions
- a>b>0 a=5, b=3
- a<b<0 a=-5, b=-3
So again 2 alone is not sufficient

Taking 1 and 2, we have:
- a<b<0 a=-5, b=-3
So taking 1 and 2, we can answer the questions.

Hope it is clearer now.

regards,
Jack
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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06 Jan 2014, 00:04
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gmatgambler wrote:
If a & b are integers ,and |a| > |b|, is a*|b|< a-b ?

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

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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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28 Jul 2010, 10:11
utfan2424 wrote:
Is this really a 700+ question?

I got the question from MHGMAT; maybe it is a 700+ category question
under stressed condition, especially given the test conditions
required to establish a precise position.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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28 Jul 2010, 13:19
utfan2424 wrote:
Is this really a 700+ question?

Doesn't seems like to be 700 +

Its a tricky question and lots of people will choose C.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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28 Jul 2010, 15:50
Bunuel wrote:
Bunuel wrote:
thailandvc wrote:
If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

anybody have a solution for this (under 2 minute). testing value takes way too long here.

E

Given: $$|a|>|b|$$
Question: is $$a*|b|<a-b$$?

(1) $$a<0$$. If $$a=-3$$ and $$b=0$$, then $$a*|b|=0>a-b=-3$$ and the answer is NO but if $$a=-3$$ and $$b=-1$$, then $$a*|b|=-3<a-b=-2$$ and the answer is YES. Two different answers. Not sufficient.

(2) $$ab\geq{0}$$

Above example works here as well:
$$a=-3$$ and $$b=0$$ --> $$a*|b|=0>a-b=-3$$ --> answer NO;
$$a=-3$$ and $$b=-1$$ --> $$a*|b|=-3<a-b=-2$$ --> answer YES.

(1)+(2) Again the same example satisfies the stem and both statements and gives two different answers to the question whether $$a*|b|<a-b$$. Hence not sufficient.

|a| > |b|, is a · |b| < a – b?

(1) a < 0
==> a · |b|<0 since nothing about b we can not conclude ==> not sufficient
(2) ab >= 0

==> a&b neg or a and b neg but no information abt a relatif to b

1 +2 a<0 ==> b <0 not suff

Nevetheless if ab was neg then the ans would be C is tit correct?
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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28 Jul 2010, 16:55
Quote:

|a| > |b|, is a · |b| < a – b?

(1) a < 0
==> a · |b|<0 since nothing about b we can not conclude ==> not sufficient
(2) ab >= 0

==> a&b neg or a and b neg but no information abt a relatif to b

1 +2 a<0 ==> b <0 not suff

Nevetheless if ab was neg then the ans would be C is tit correct?

I would say it's still not sufficient.
If a<0 and ab<0 then b>0
For a=-3 and b=1 a|b| > a - b
For a=-3 and b=2 a|b| < a - b
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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01 Aug 2010, 08:34
If a and b are integers, and |a| > |b|, is a · |b| < a – b?
(1) a < 0
(2) ab >= 0
(1) since |a| > |b| => -a < b < a
now a<0 : if b<0 then a.|b| < a
also a - b > a
so clearly a.|b|< a - b
but if b=0 then a.|b| =0 and a - b = a thus a.|b| > a - b (0>a) --- so here itself we see that for two diff ranges of b , a<0 is insufficient to determine the inequality completely. So 1 is insuff

(2) ab>=0 --- this means either both a and b are negative (as seen above) OR b = 0 since a cannot be 0 because it is stated that |a|>|b| so a must be non-zero OR both a and b are greater than 0. As we have seen in (1) we get two different values for a<0, b<0 AND a<0 b=0 combo. So 2 also is insuff

combining 1 and 2 changes nothing - still the same conditon remains. So insufficient.

It's a damned tough problem. I wish I had the brains of Bunuel to pick up relevant numbers quickly and solve the problem as he has shown in his solution.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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01 Aug 2010, 21:50
1. a<0

-3|2|<-3-2 yes.
-3|-2|<-3+2 no.
Not Suff

2. ab >0
- - = -3|-2|<-3+2 yes.
+ + = 3|2|<3-2 no.
Not Suff

But together;
-3 |-2| <-3+2 yes

IMO C.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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29 Sep 2010, 02:23
is a - |b| < a - b
canceling a from both sides , the eqn becomes |b| > b
so basically the eqn is asking if b is -ve or not . Therefore , we will check for that

1 . a < 0 , does not talk about b
2. ab >= 0 implies a>=0 and b>=0
or a<=0 and b<=0

therefore not sufficient

combining both we get b<=0 which is also not sufficient since b=0 does not satisfy the
main question . I hope doing this way you feel faster.
Also I think that the condition |a| > |b| is irrelevant to the question.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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08 Apr 2012, 21:45
I believe that this question has been discussed before too.
Please do a search before posting in the forum.

1. a<0 but b can be =0 hence Insufficient.

2 for a=2| -2 and b = 1|-1 answer is yes and no. insufficient.

1+2
a= -2 b= -1 LHS < RHS

Hence C it is.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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24 Aug 2012, 12:53
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 [#permalink]

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24 Aug 2012, 14:00
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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<-b
Actually, 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 0-a>b-0
a pos, b neg: a>b where a-0>0-b
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 a-b is positive. a|b| is negative.

On combining ii and 1)
Two cases are emerging.
a-------0-------(-b)
a-------(-b)------0

We 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 a|b| will be a bigger negative than a-b
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 [#permalink]

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20 Dec 2012, 09:55
Any nice algebraic solution to this problem.

I picked numbers and got to the solution but still took over 3 minutes to do so.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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24 Sep 2013, 05:47
Thank you so much for the explanation. I completely ruled out the possibility for a and b to be equivalent to zero. I will keep that in mind for my future questions. Helps a lot! thanks
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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05 Jan 2014, 20:21
If a & b are integers ,and |a| > |b|, is a*|b|< a-b ?

A)a<0

B)$$ab>=0$$

How to solve this algebraically ?
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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06 Jan 2014, 01:52
gmatgambler wrote:
If a & b are integers ,and |a| > |b|, is a*|b|< a-b ?

A)a<0

B)$$ab>=0$$

How to solve this algebraically ?

Merging similar topics. Please refer to the solutions above.
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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Re: If a and b are integers, and |a| > |b|, is a |b| < a   [#permalink] 13 Feb 2015, 06:38

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