If a 0? (1) a^2

Question

If a < b, is a > 0?

(1) a^2 < b^2

(2) a^2 < ab < b^2

Gemelo90 · Answer

If a < b, is a > 0?

(1) a^2 < b^2

(2) a^2 < ab < b^2

Choosing different values for (1): a=-2 < b=3 or a=2 < b=3 Hence, not sufficient

Choosing different values for (2): a=2 < 2*3 < 3 but a=-2 <-2*3 <3 violates the statement. Hence, B is sufficient

devlin · Answer

1) any square number hide its signal. Not sufficient
2) ab > 0, so a > 0. Sufficient

Skyline393 · Answer

(1) |a| < |b| a = 5, b = 7 =>|a| < |b| => a > 0 a = -5, b = 7 => |a| < |b| => a < 0 => Stm1 insufficient. (2) a^2 < ab < b^2 => ab > 0 => a & b are both negative or positive a = 5, b = 7 => a^2 < ab < b^2 => a > 0 a= -5, b = -3 ( a this violates a^2 < ab < b^2 (since 25 can't < 15 can't < 9) => This case can not be existed... => a & b are only positive => a > 0 => Sufficient Hence B

exc4libur · Answer

a0 ( b ≠ negative because a0 then b>0 both cases are possible, insufi. (2) a^2 < ab < b^2…a^2 < ab…a^2-ab<0…a(a-b)<0 : if a<0 then a-b>0…or…a>b : but given a0 then a-b<0…or…a

Kinshook · Answer

Given: a < b Asked: Is a > 0? (1) a^2 < b^2 |a|<|b| a can be +ve or -ve and b must be +ve NOT SUFFICIENT (2) a^2 < ab < b^2 ab > a^2 > 0 ab>0 a & b have same sign a^2 < ab < b^2 => |a|<|b| a can be +ve or -ve and b must be +ve But a & b have same sign b = +ve => a = +ve a>0 SUFFICIENT IMO B

KarishmaB · Answer

In stmnt 1, in the highlighted part, b cannot be negative. 
a can be positive or negative, as you mentioned.

Rest all is good.

exc4libur · Answer

ah yes, bc of the |a|<|b| thank you!

Archit3110 · Answer

test with values ;
a,b ; 2,3 , 1/4,1,3 , -2,-1

#1
a^2 < b^2
not sufficeint as a >0 as 1/4 and -ve as -2
#2a^2 < ab < b^2
can be re written as ; a<1<b
sufficient to say that a is -ve
IMO B

GMATGuruNY · Answer

Statement 1: Case 1: a=1 and b=2 In this case, a>0, so the answer to the question stem is YES. Case 2: a=0 and b=2 In this case, a=0, so the answer to the question stem is NO. INSUFFICIENT. Statement 2: Since a^20 . Thus: 0 a-b<0 Statement 2: a^2 a^2-b^2<0 --> (a+b)(a-b)<0 Since a-b<0 and (a+b)(a-b)<0 , we know that a+b>0 . Since a+b>0 and a and b have the same sign, they must both be POSITIVE. Thus, the answer to the question stem is YES. B

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If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2

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