169 Easy inequality

[ern5231]

A<B. Is A<0?
(1) 2A>AB
(2) B<0

The OA given for this problem is D but I feel it is B. What is you opinion?

[mp2437]

I don't think Statement 1 is sufficient.

For statement 1, using a positive fraction and a negative number as 2 examples will contradict, so it isn't sufficient on its own.

[ern5231]

for the stmt 1 can't we jus divide by A on both sides and prove that it is insufficient(since it will be reduced to 2>B which just cannot prove A<0

[euwe]

it's obvious that on the test they (or you) omitted to state that both A and B are integers with which the answer should be D.

[Katrusya]

A<B
Let's check first option:

2A>AB
A>(AB/2)
Consider these numbers:
A=1, B=1.2
1>(1*1.2)/2=0.6 - true
OR
A=-5, B=10
-5>(-5*10/2)=-25 - true
So, first option is not suffisient, because A can be <0 or >0
I think the answer is B

[samsachd]

1)2A>AB
=>2A-AB>0
=>A(2-B)>0
=>2-B>0 and A<0 or B<2 and A<0. Sufficent

2)B<0 and given A<B
=>A<0.Sufficent

Hence D.

[Katrusya]

samsachd,
you made a mistake:
1)2A>AB
=>2A-AB>0
=>A(2-B)>0
=>2-B>0 and A<0 - doesn't work! A must be >0 also, in order the product of these to be positive!
So, it's either
A>0 and 2-B>0, B<2
OR
A<0 and 2-B<0, B>2
Even though we are given that A<B it could satisfy both of aforementioned options, hence, A could be both <0 or >0

[NikolayZ]

samsachd,
you made a mistake:
1)2A>AB
=>2A-AB>0
=>A(2-B)>0
=>2-B>0 and A<0 - doesn't work! A must be >0 also, in order the product of these to be positive!
So, it's either
A>0 and 2-B>0, B<2
OR
A<0 and 2-B<0, B>2
Even though we are given that A<B it could satisfy both of aforementioned options, hence, A could be both <0 or >0

Agree with you.
Is OA really "D" ?

[samsachd]

I am refering to one of the explanations from Stuart Kovinsky here:

https://www.beatthegmat.com/inequalities ... tml#193824

If this explanation can be used here(and i am saying this considering both A and B as Intergers) then i think D is correct answer.Correct me if i am wrong.

[mehravikas]

Pick numbers for statement 1 -

A = 1, B = 1.5

2A > AB - true and A > 0

A = -1, B = 6

2A > AB - true and A < 0

Statement 1 is insufficient

[Stuart@KaplanGMAT]

A<B. Is A<0?
(1) 2A>AB
(2) B<0

The OA given for this problem is D but I feel it is B. What is you opinion?

Let's look at (1) algebraically.

(1) 2A>AB

With inequalities, we have to be very careful about dividing through by variables; to be safe, we should look at both the positive and negative cases (where applicable).

If A>0, we can divide both sides by A to get:

2 > B

Now, b also has to be greater than a (according to the question stem, so we get the inequality:

2 > B > A > 0

Are there values for B and A that fit this inequality? Sure, nothing says we have to pick integer values. Therefore, A could be > 0 to give us a "no" answer.

If A<0, we can divide by sides by A to get:

2 < B

(we have to swap the direction of the inequality since we're dividing by a negative number).

So now we get the inequality:

B > 2 > 0 > A

are there values for A and B that fit this inequality? Sure - so we can get a "yes" answer as well.

Therefore, (1) is insufficient.

If we knew that A and b were integers, we could ignore the positive case (since there aren't two distinct integers between 2 and 0), and (1) would be sufficient, but we don't have that information.

[Talkativetree]

Stuart wrote what I was going to say, so I'm just going to throw my opinion behind his.