If ax + b = 0, is x > 0?

a. a+b > 0
b. a-b > 0

Oks, Ans is E

In your solution, make sure you consider the case a=2(positive) and b=-1 (negative)when evaluating a-b>0

Crazykrans, I am not sure I follow.

If a=2(positive) and b=-1, then a is > 0, so the answer should be B?

oks wrote:If ax + b = 0, is x > 0?

a. a+b > 0
b. a-b > 0

ax + b = 0
ax = -b

x > 0 ?? or is x = positive

Statement I

a+b > 0
a - ax > 0
a( 1-x) > 0

either a & (1-x) both positive or both negative.

x could be positive or negative

Insufficient.

Statement II
a-b > 0
a+ax > 0
a (1+x) > 0

either a & (1+x) both positive or both negative.

x could be either negative or positive

Insufficient.

Combining I & II

We only know that x could be positive or negative.

Hence E.

Hope this helps.

Got E..
procedure is similar to parallel_chase. Let me add a little more detail.

from the stem
ax = -b OR b = -ax

stmt 1:
a+b > 0
=> a -ax > 0
=> a(1 -x) > 0

=> either a is +ve OR (1 -x) is +ve
=> ie 1 -x > 0 => x < 1. X could be +ve (values 0 and 1. maybe decimals) OR -ve (x can take any value < 1...)
So Not sufficient.


stmt2:
a -b >0
a - (ax) > 0
a(1 +x) > 0
a +ve OR 1 +x is +ve
1 +x > 0 => x > -1
Again X could be +ve or -ve.

Combining the two.
Here we have 2 possibilities.
1) a is +ve. Then we dont know anything about x. so not sufficient.

2) x > -1 and x < 1. There could be an infinite decimal numbers between these two values; some are +ve and some are -ve.
So insufficient as well.

From both cases we cant determine x +ve or -ve.
so E


HT Helps

PC / Vittal,
Good explanations!

Regards,
Cramya

from stem x = -b/a

so one needs to determine whether a n b are of opposite sign, as that is the only way to make x positive...

from 1. a+b>0 a n b can be choosen not to satisfy req i.e a = 3 b =2 x-ve
a = 3 b = -1 x +ve

from 2. a-b >0 => a>b so still a = 3 b = 2 holds true x -ve
a=3 b=-1 x +ve
combine also doesnt help as for a = 3 b=2 x -ve
a = 3 b = -1 x +ve