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Is a < b? (1) |a – b| = b – a (2) a/b < 1

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mahdwen
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Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   04 Jun 2017, 14:20
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Is a < b?

(1) |a – b| = b – a
(2) a/b < 1
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C
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mahdwen
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   04 Jun 2017, 14:27
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Is a < b?
I. |a – b| = b – a
II. a/b < 1

I. |a-b| = b-a means b-a ≥ 0, or b ≥ a.
So a<b or a=b are both satisfactory answers => Insufficient

II. a/b<1, implying that b different than 0. We have two cases: if b > 0 then a < b, and if b < 0, then a > b => Insufficient

Both Statement together are sufficient to get to the conclusion a < b, answer C
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   04 Jun 2017, 23:16
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mahdwen wrote:

Detailed Answer



Is a < b?
I. |a – b| = b – a
II. a/b < 1

I. |a-b| = b-a means b-a ≥ 0, or b ≥ a.
So a<b or a=b are both satisfactory answers => Insufficient

II. a/b<1, implying that b different than 0. We have two cases: if b > 0 then a < b, and if b < 0, then a > b => Insufficient

Both Statement together are sufficient to get to the conclusion a < b, answer C


Thst's correct.

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Hard set on Abolute Values: https://gmatclub.com/forum/inequality-an ... 86939.html
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   05 Jun 2017, 04:24
Can someone please explain why the answer is not Option A.

I know why statement 2 is insufficient as the case changes for positive and negative numbers.
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   05 Jun 2017, 07:03
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sonikavadhera wrote:
Can someone please explain why the answer is not Option A.

I know why statement 2 is insufficient as the case changes for positive and negative numbers.


(1) |a – b| = b – a. So, |a – b| = -(a - b), which means that \(a \leq b\). So, a could be equal to b (answer NO) as well as a could be less than b (answer YES). Not sufficient.

Hope it's clear.
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   05 Jun 2017, 11:37
Bunuel wrote:
sonikavadhera wrote:
Can someone please explain why the answer is not Option A.

I know why statement 2 is insufficient as the case changes for positive and negative numbers.


(1) |a – b| = b – a. So, |a – b| = -(a - b), which means that \(a \leq b\). So, a could be equal to b (answer NO) as well as a could be less than b (answer YES). Not sufficient.

Hope it's clear.


Oh I forgot the equal possibility. Thanks.
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   05 Jun 2017, 19:51
Hi,
Why B is not sufficient.. given a/b<1 => a<b.. which is what we are trying to prove right.. or am I missing something more fundamental in inequalities?
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   05 Jun 2017, 22:53
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sasidharrs wrote:
Hi,
Why B is not sufficient.. given a/b<1 => a<b.. which is what we are trying to prove right.. or am I missing something more fundamental in inequalities?


We cannot multiply b on both sides as we do not know the sign on 'b'.
Never multiply/divide on both sides unless you are sure of the sign. When positive its the way you put it, if negative the inequality sign changes.
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   05 Jun 2017, 23:09
sasidharrs wrote:
Hi,
Why B is not sufficient.. given a/b<1 => a<b.. which is what we are trying to prove right.. or am I missing something more fundamental in inequalities?


Hi

In case of inequalities, you cannot multiply or divide both sides by a variable until and unless you know the sign of that variable (positive or negative)

So if you are given: a/b < 1 you will have to take both the cases:

Case 1. b is positive. In this case, we will multiply both sides by b, and since b is positive, the sign of inequality will Not change. So
a/b * b < 1*b or a < b

Case 2. b is negative. In this case, we will multiply both sides by b, and since b is negative, the sign of inequality Will change. So
a/b * b > 1*b or a > b

Hope this helps
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink] New post   01 Nov 2017, 02:19
(1) |a – b| = b – a => |a – b| = -a + b = - (a-b) => a-b ≤ 0 => insufficient
(2) a/b = 1 => (a-b)/ b < 0. We have 2 cases: b<0 and a<b or b>0 and a>b => insufficient

(1) + (2) => a-b <0 => a <b => sufficient. Hence, the answer is C
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Re: Is a < b? (1) |a – b| = b – a (2) a/b < 1 [#permalink]
01 Nov 2017, 02:19
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