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If a and b are integers, and |a| > |b|, is a |b| < a [#permalink ]
26 Dec 2006, 18:36

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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
Last edited by

Bunuel on 24 Sep 2013, 05:35, edited 2 times in total.

Edited the question and added the OA

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Re: DS - Abs a and b [#permalink ]
26 Dec 2006, 19:12

anindyat wrote:

If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b? (1) a < 0 (2) ab >= 0...Please share your work......

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 ]
26 Dec 2006, 19:16

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.

(C) is my answer.

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Re: DS - Abs a and b [#permalink ]
26 Dec 2006, 19:20

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. (C) is my answer.

Yes I fogot the 0 case in my option 1

Thanks Swagat.

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Re: DS - Abs a and b [#permalink ]
26 Dec 2006, 19:32

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. (C) is my answer.

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 ]
26 Dec 2006, 19:35

ncprasad wrote:

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. (C) is my answer.

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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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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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 ]
29 Dec 2006, 05:40

anindyat wrote:

If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b? (1) a < 0 (2) ab >= 0...Please share your work......

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 ]
29 Dec 2006, 11:53

anindyat wrote:

If a and b are integers, and |a| > |b|, is a Â· |b| < a â€“ b? (1) a < 0 (2) ab >= 0...Please share your work......

-------------------------------------------------------

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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(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).
The answer is E.

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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 ]
24 Sep 2013, 05:30

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 ]
24 Sep 2013, 05:36
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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.

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|<a-b . Hence not sufficient.

Answer E. 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
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24 Sep 2013, 05:36

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