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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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26 Dec 2006, 19:36
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If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-and-b-are-integers-and-a-b-is-a-b-a-83804.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Sep 2013, 06:35, edited 2 times in total.
Edited the question and added the OA

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Re: DS - Abs a and b [#permalink]

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26 Dec 2006, 20:12
anindyat wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0

(2) ab >= 0

Picking #s is the best way I believe.

Option 1 : There are 2 cases here

since a<0 and given condition is |a| >|b|

the values of a and b can be a=-3,b=-2 or a=-3 b=2 (since the absolute value of a should be > abs value of b)

substituting in a Â· |b| < a â€“ b

A is SUFF

Option 2

ab>= 0 which means a or b can be 0 or they should be both positive or both -ve

You got 4 cases minimum here

take a=-1 and b=0 you get 0<-1 which is false
If you take a=-3 and b=-2 you get -6<-1 which is true

So B is INSUFF

A stands

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Re: DS - Abs a and b [#permalink]

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26 Dec 2006, 20:16
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0
b can be positive integer, negetive integer or zero.

if a=-5, b=4 ==> -20 <-9 --- true
if a=-5, b=-4 ==> -20 < -1 ---- true
if a=-5, b=0 ==> 0<-5 -------- false
Thereofore, (1) is insufficient.

(2) ab >= 0

if ab >0 Then either both positive or both negetive but none is zero.

when a=5, b=4 ==> 20<1 ---false
Therefore (2) insufficient.

Combining (1) and (2) we get both a and b are negetive integers.
and we already saw when a and b are both negetive, the relationship is true.

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Re: DS - Abs a and b [#permalink]

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26 Dec 2006, 20:20
Swagatalakshmi wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0
b can be positive integer, negetive integer or zero.

if a=-5, b=4 ==> -20 <-9 --- true
if a=-5, b=-4 ==> -20 < -1 ---- true
if a=-5, b=0 ==> 0<-5 -------- false
Thereofore, (1) is insufficient.

(2) ab >= 0

if ab >0 Then either both positive or both negetive but none is zero.

when a=5, b=4 ==> 20<1 ---false
Therefore (2) insufficient.

Combining (1) and (2) we get both a and b are negetive integers.
and we already saw when a and b are both negetive, the relationship is true.

Yes I fogot the 0 case in my option 1

Thanks Swagat.

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Re: DS - Abs a and b [#permalink]

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26 Dec 2006, 20:32
Swagatalakshmi wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0
b can be positive integer, negetive integer or zero.

if a=-5, b=4 ==> -20 <-9 --- true
if a=-5, b=-4 ==> -20 < -1 ---- true
if a=-5, b=0 ==> 0<-5 -------- false
Thereofore, (1) is insufficient.

(2) ab >= 0

if ab >0 Then either both positive or both negetive but none is zero.

when a=5, b=4 ==> 20<1 ---false
Therefore (2) insufficient.

Combining (1) and (2) we get both a and b are negetive integers.
and we already saw when a and b are both negetive, the relationship is true.

I agree with the reasoning that a<0 and that b <0.

But based on (2) ab can also be equal to 0 implying that b can also be 0.

Lets pick a=-2 and b=-1. This satisfies the question stem.

-2.|-1| is < than -2-(-1) implying that a Â· |b| < a â€“ b

Now, lets pick a=-2 and b=0,

In this case, -2(0) is > than -2-(0) implying that a Â· |b| > a â€“ b

My choice is E.
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Re: DS - Abs a and b [#permalink]

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26 Dec 2006, 20:35
Swagatalakshmi wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0
b can be positive integer, negetive integer or zero.

if a=-5, b=4 ==> -20 <-9 --- true
if a=-5, b=-4 ==> -20 < -1 ---- true
if a=-5, b=0 ==> 0<-5 -------- false
Thereofore, (1) is insufficient.

(2) ab >= 0

if ab >0 Then either both positive or both negetive but none is zero.

when a=5, b=4 ==> 20<1 ---false
Therefore (2) insufficient.

Combining (1) and (2) we get both a and b are negetive integers.
and we already saw when a and b are both negetive, the relationship is true.

I agree with the reasoning that a<0 and that b <0.

But based on (2) ab can also be equal to 0 implying that b can also be 0.

Lets pick a=-2 and b=-1. This satisfies the question stem.

-2.|-1| is < than -2-(-1) implying that a Â· |b| < a â€“ b

Now, lets pick a=-2 and b=0,

In this case, -2(0) is > than -2-(0) implying that a Â· |b| > a â€“ b

My choice is E.

You are darn RIGHT. I read the second condition as ab>0 !!!
Stupid me .... Grrrrr ......
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26 Dec 2006, 20:42
Got (B)

If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0

(2) ab >= 0

from (1) a<0
and |a| > |b|

a=-1
b= 0

a= -3,b = -2
aÂ·|b| = -3.|-2| = -6
a-b = -3 +2=-1
a Â· |b| < a â€“ b

a = -2,b = 1
a Â· |b| = -2*|1| = -2
a-b = -2 -1=-3 <-2

a Â· |b| > a â€“ b

(1) not suff

for (2)
|a| > |b|
ab>=0

a=-2,b=0
a Â· |b| = 0
a-b = -2-0=-2
so a Â· |b| > a â€“ b

a=3,b=2

a.|b| = 6
a-b = 3-2=1
a Â·|b| > a â€“ b

(2) suff

ANS (B)

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26 Dec 2006, 20:49
mkl_in_2001 wrote:
Got (B)

If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0

(2) ab >= 0

from (1) a<0
and |a| > |b|

a=-1
b= 0

a= -3,b = -2
aÂ·|b| = -3.|-2| = -6
a-b = -3 +2=-1
a Â· |b| < a â€“ b

a = -2,b = 1
a Â· |b| = -2*|1| = -2
a-b = -2 -1=-3 <-2

a Â· |b| > a â€“ b

(1) not suff

for (2)
|a| > |b|
ab>=0

a=-2,b=0
a Â· |b| = 0
a-b = -2-0=-2
so a Â· |b| > a â€“ b

a=3,b=2

a.|b| = 6
a-b = 3-2=1
a Â·|b| > a â€“ b

(2) suff

ANS (B)

I was wrong
for (2)
if a =-3 ,b=-2
a.|b| =-6
a-b = -3 +2=-1
a.|b|<a-b
this makes (B) also insuff

S0 it should be E!!

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Re: DS - Abs a and b [#permalink]

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29 Dec 2006, 06:40
anindyat wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0

(2) ab >= 0

Its E

Stmt 1) Consider a=-6 and b=-5
so a. |b| < a-b is -30 < -1 OK
Consider a=6 and b=5
so a. |b| < a-b is 30 < 1 So insufficient

Stmt 2) Consider a=-6 and b=-5
so a. |b| < a-b is -30 < -1 OK
Consider a=6 and b=5
so a. |b| < a-b is 30 < 1 So insufficient

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Re: DS - Abs a and b [#permalink]

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29 Dec 2006, 12:53
anindyat wrote:
If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b?

(1) a < 0

(2) ab >= 0

-------------------------------------------------------
I am getting E

1) a=-2, b=-1 -> stem false; a=-2,b=0-> stem false, a=-3, b=-1-> stem true
TF, insuffi..

2) -> a=0 or b =0 or both +ve or both -ve
stem and 2) -> 2,0 -> stem true
-2,0-> stem false

insuffi..

both 1) and 2) together: take -2,0->false ;-2,-1-> true..insuffi.

TF, E

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31 Dec 2006, 11:19
(1) and (2) are insufficient alone. How do they fare together?

(1) a < 0 AND (2) ab >= 0 is equivalent to a<0 AND b <=0.

a Â· |b| < a â€“ b?

|b| = -b; then: -ab < a-b. Reordering the expression: -a/(a-1)>b. The term to the left is (-), and as b is (-) itself, we cannot ascertain that the expression always holds (it could be that -1/(a-1) = -3/4 and b = -1/2 or -1, yielding the inequality false or true, respectively).

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07 Jan 2007, 08:58
very nice rally folks, i say E too

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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, 06:30
for |a| > |b| with plugging in we get four possibilities

a = -8 , b = 3
a = -8 , b = -3
a = 8 , b = -3
a = 8 , b=3

A) If a < 0

a = -8 , b =3

-8 |3| < -8-3
-24<-11

a = -8 , b = -3

-8 |-3| < -8-(-3)
-24 < -5

A is sufficient

2) if ab>=0

a=-3, b=-3

-8 |-3| < -8-(-3)
-24 < -5

a=3, b=3

8|3|<8-3
24<5

Insufficient

Hence, the answer should be A. Isn't it?

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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, 06:36
2
KUDOS
Expert's post
aakrity wrote:
for |a| > |b| with plugging in we get four possibilities

a = -8 , b = 3
a = -8 , b = -3
a = 8 , b = -3
a = 8 , b=3

A) If a < 0

a = -8 , b =3

-8 |3| < -8-3
-24<-11

a = -8 , b = -3

-8 |-3| < -8-(-3)
-24 < -5

A is sufficient

2) if ab>=0

a=-3, b=-3

-8 |-3| < -8-(-3)
-24 < -5

a=3, b=3

8|3|<8-3
24<5

Insufficient

Hence, the answer should be A. Isn't it?

If a and b are integers, and |a| > |b|, 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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-and-b-are-integers-and-a-b-is-a-b-a-83804.html
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Re: If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]

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10 Aug 2015, 01:05
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Re: If a and b are integers, and |a| > |b|, is a |b| < a   [#permalink] 10 Aug 2015, 01:05
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