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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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If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 10 Aug 2018, 02:21
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 10 Aug 2018, 08:29
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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
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 10 Aug 2018, 13:00
1
Bunuel wrote:
If a < b, is a > 0?


(1) a^2 < b^2

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


1) any square number hide its signal. Not sufficient
2) ab > 0, so a > 0. Sufficient
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 05 Mar 2019, 01:30
Bunuel wrote:
If a < b, is a > 0?


(1) a^2 < b^2

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


(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<b) => 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
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If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post Updated on: 21 Aug 2019, 02:51
Bunuel wrote:
If a < b, is a > 0?

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


\(a<b…a-b<0\)

(1) \(a^2 < b^2…|a|<|b|\):
if \(a<0\) then \(b>0\) (\(b\) ≠ negative because \(a<b\) and \(|a|<|b|\))
if \(a>0\) 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 \(a<b\) this is not possible
if \(a>0\) then \(a-b<0…or…a<b\): this is sufi.

Answer (B).

Originally posted by exc4libur on 20 Aug 2019, 03:59.
Last edited by exc4libur on 21 Aug 2019, 02:51, edited 1 time in total.
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 20 Aug 2019, 04:33
Bunuel wrote:
If a < b, is a > 0?


(1) a^2 < b^2

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


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
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 21 Aug 2019, 02:14
exc4libur wrote:
Bunuel wrote:
If a < b, is a > 0?

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


VeritasKarishma hey how are you, could you check this out?

\(a<b…a-b<0\)

(1) \(a^2 < b^2…|a|<|b|\):
if \(a<0\) then \(b<0…or…b>0\)
if \(a>0\) 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 \(a<b\) this is not possible
if \(a>0\) then \(a-b<0…or…a<b\): this is sufi.

Answer (B).


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

Rest all is good.
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 21 Aug 2019, 02:49
VeritasKarishma wrote:
exc4libur wrote:
Bunuel wrote:
If a < b, is a > 0?

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


VeritasKarishma hey how are you, could you check this out?

\(a<b…a-b<0\)

(1) \(a^2 < b^2…|a|<|b|\):
if \(a<0\) then \(b<0…or…b>0\)
if \(a>0\) 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 \(a<b\) this is not possible
if \(a>0\) then \(a-b<0…or…a<b\): this is sufi.

Answer (B).


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

Rest all is good.


ah yes, bc of the \(|a|<|b|\)
thank you! :)
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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New post 23 Aug 2019, 07:07
Bunuel wrote:
If a < b, is a > 0?


(1) a^2 < b^2

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


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
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Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2   [#permalink] 23 Aug 2019, 07:07

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

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