If a and b are integers, and a > b, is a * b < a – b?
(1) a < 0
(2) ab >= 0
OA: E
Anyone for an easy explination?
Another inequality challange?
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Givena>bcallmemo wrote:If a and b are integers, and a > b, is a · b < a – b?
(1) a < 0
(2) ab 0
OA: E
Anyone for an easy explination?
four possiblitiesa=3,b=2 a=3,b=2 a=3,b=2 a=3,b=2
stmt 1:a<0
two possiblities left a=3,b=2 a=3,b=2
stmt 2:ab=0
it means a.b=0
0<ab?
we ve four possiblits for a and bso insufficent
Combine 1 and 2,two possibilies again
Again we are unable to conclude the signs for a and b
Answer should be E...hope it helps
I. Not Sufficient
a < 0 => b can be either positive or negative
eg. a = 3, b = 2 or + 2
if b = 2; a * b < a – b => true
if b = 2; a * b < a – b => false
II. Sufficient
ab >= 0 => Given a > b; a or b can't be zero. so a,b must be positive.
a=1, b = 2; a * b < a – b => false
a=3, b = 2; a * b < a – b => false
But OA is E. What am I missing?
a < 0 => b can be either positive or negative
eg. a = 3, b = 2 or + 2
if b = 2; a * b < a – b => true
if b = 2; a * b < a – b => false
II. Sufficient
ab >= 0 => Given a > b; a or b can't be zero. so a,b must be positive.
a=1, b = 2; a * b < a – b => false
a=3, b = 2; a * b < a – b => false
But OA is E. What am I missing?
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 Brent@GMATPrepNow
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The missing part is your conclusion that a and b must be positive.ab >= 0 => Given a > b; a or b can't be zero. so a,b must be positive.
a=1, b = 2; a * b < a – b => false
a=3, b = 2; a * b < a – b => false
But OA is E. What am I missing?
We could have a= 3 and b=2 (as you suggested earlier in your solution)
Also, your first set of numbers to plug in (a=1 and b=2) breaks the condition that a > b
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Hi guys,
do you think this to be a difficult question, a so called "700 level" question?
thanks!
do you think this to be a difficult question, a so called "700 level" question?
thanks!
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I'd classify this question somewhere in the 600700 range.
My $0.02
My $0.02
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Here's my take on this q:
If a and b are integers, and a > b, is a * b < a – b?
(1) a < 0
a =6, b = 2
12 < 8 ? yes
a = 3, b = 0
0 < 3? no
Hence, (1) is INSUFFICIENT.
(2) ab >= 0
There can be two cases that satisfy this condition.
Case 1: a <=0 and b <= 0
Case 2: a >=0 and b >= 0
Take case 1: a <=0 and b<=0
a=3,b=1
3 < 2 ? Yes
a = 5, b = 0
0 < 5 ? No
No need to go any further. Hence, (2) is INSUFFICIENT as well.
(3) Taking both (1) and (2) together, we find that a < 0 and b <=0. From case 1 in 2) we know that this is INSUFFICIENT.
So, my answer is E).
If a and b are integers, and a > b, is a * b < a – b?
(1) a < 0
a =6, b = 2
12 < 8 ? yes
a = 3, b = 0
0 < 3? no
Hence, (1) is INSUFFICIENT.
(2) ab >= 0
There can be two cases that satisfy this condition.
Case 1: a <=0 and b <= 0
Case 2: a >=0 and b >= 0
Take case 1: a <=0 and b<=0
a=3,b=1
3 < 2 ? Yes
a = 5, b = 0
0 < 5 ? No
No need to go any further. Hence, (2) is INSUFFICIENT as well.
(3) Taking both (1) and (2) together, we find that a < 0 and b <=0. From case 1 in 2) we know that this is INSUFFICIENT.
So, my answer is E).