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If a and b are integers, and |a| > |b|, is a |b| < a [#permalink]
 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
...Please share your work......
Last edited by anindyat on 26 Dec 2006, 20:03, edited 1 time in total.
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Re: DS - Abs a and b [#permalink]
26 Dec 2006, 20:12
anindyat wrote: If a and b are integers, and |a| > |b|, is a · |b| < a – 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 · |b| < a – 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, 20:16
If a and b are integers, and |a| > |b|, is a · |b| < a – 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, 20:20
Swagatalakshmi wrote: If a and b are integers, and |a| > |b|, is a · |b| < a – 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, 20:32
Swagatalakshmi wrote: If a and b are integers, and |a| > |b|, is a · |b| < a – 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 · |b| < a – b
Now, lets pick a=-2 and b=0,
In this case, -2(0) is > than -2-(0) implying that a · |b| > a – b
My choice is E.
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Re: DS - Abs a and b [#permalink]
26 Dec 2006, 20:35
ncprasad wrote: Swagatalakshmi wrote: If a and b are integers, and |a| > |b|, is a · |b| < a – 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 · |b| < a – b Now, lets pick a=-2 and b=0, In this case, -2(0) is > than -2-(0) implying that a · |b| > a – 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 · |b| < a – b?
(1) a < 0
(2) ab >= 0
from (1) a<0
and |a| > |b|
a=-1
b= 0
a= -3,b = -2
a·|b| = -3.|-2| = -6
a-b = -3 +2=-1
a · |b| < a – b
a = -2,b = 1
a · |b| = -2*|1| = -2
a-b = -2 -1=-3 <-2
a · |b| > a – b
(1) not suff
for (2)
|a| > |b|
ab>=0
a=-2,b=0
a · |b| = 0
a-b = -2-0=-2
so a · |b| > a – b
a=3,b=2
a.|b| = 6
a-b = 3-2=1
a ·|b| > a – 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 · |b| < a – b?
(1) a < 0
(2) ab >= 0
from (1) a<0
and |a| > |b|
a=-1
b= 0
a= -3,b = -2
a·|b| = -3.|-2| = -6
a-b = -3 +2=-1
a · |b| < a – b
a = -2,b = 1
a · |b| = -2*|1| = -2
a-b = -2 -1=-3 <-2
a · |b| > a – b
(1) not suff
for (2)
|a| > |b|
ab>=0
a=-2,b=0
a · |b| = 0
a-b = -2-0=-2
so a · |b| > a – b
a=3,b=2
a.|b| = 6
a-b = 3-2=1
a ·|b| > a – 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, 06:40
anindyat wrote: If a and b are integers, and |a| > |b|, is a · |b| < a – 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, 12:53
anindyat wrote: If a and b are integers, and |a| > |b|, is a · |b| < a – 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 · |b| < a – 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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