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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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Joined: 02 Sep 2009
Posts: 64318
If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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10 Aug 2018, 02:21
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75% (hard)

Question Stats:

53% (01:55) correct 47% (02:06) wrong based on 145 sessions

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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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Manager
Joined: 06 May 2018
Posts: 56
Re: If a < b, is a > 0? (1) a^2 < b^2 (2) a^2 < ab < b^2  [#permalink]

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10 Aug 2018, 08:29
1
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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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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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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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.

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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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

(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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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.

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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21 Aug 2019, 02:49
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.

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