Instructors plz help...Inqeuality DS 32

ern5231 wrote:32) Is (a^2) b + a (b^2) > 0 ?

(1) a <0

(2) a+b<0

The answer is E

How did you arrive at E? I feel the answer is B

I too think it is B

Can anyone explain how is it B?

IMO : E

Question :
is (a^2)b + a(b^2) > 0
==>is ab(a+b) > 0

for this to be true we have 2 cases :

1)ab>0 and (a+b) >0
2)ab<0 and (a+b) <0

Statement 1
Tells us a<0 but gives no information on b,hence insufficient

statement 2
Tells us (a+b) < 0 ,
This tells us nothig about a and b individually so we cant say if ab >0 or ab < 0,hence insufficient
Pick numbers to verify :
a=-1,b=-5 ==> -1-5 <0
a=-7,b=2 ==> =7+2 <0

Together again,

We know a<0 and we know a+b <0 but we dont know the sign of b and hence both statements togetjer are also insufficient.

Is (a^2) b + a (b^2) > 0 ?

(1) a <0

(2) a+b<0

Can we follow this approach -

(a^2)b>-a(b^2)
Hence, Is a>-b?

(1) does not make it suff
(2) a<-b looks suff

Hence B. Am I going wrong anywhere?

ern5231 wrote:Is (a^2) b + a (b^2) > 0 ?

(1) a <0

(2) a+b<0

Can we follow this approach -

(a^2)b>-a(b^2)
Hence, Is a>-b?

(1) does not make it suff
(2) a<-b looks suff

Hence B. Am I going wrong anywhere?

IMO E.

We can write (a^2) b + a (b^2) > 0 as ab(a+b) > 0.

a+b<0. it means,

a b
- -
- +
+ -

There are three possibilities.

If one is negative and the other is positive, the answer might be negative.

But if both are negative, the answer will be positive.

Since we're getting 2 kind of answers B is not sufficient.

Even if you take both the statements, you will get 2 answers. So, both the statements together are not sufficient.

Thanks Sanjana and Umaa. Your explanations look good but can you point out the mistake with my approach?

ern5231 wrote:Thanks Sanjana and Umaa. Your explanations look good but can you point out the mistake with my approach?

In the rephrase. Since you dont know whether a or be is negative or positive, you need to consider different possibilities.

IMO Rephrase is ab(a+b)>0

1. We dont the the sign of b
2. Given (a+b)<0. We dont know the sign of ab

1&2

From stmt 1, we know a<0. But we still dont know sign of b. Hence E.