Is |2a − 3b| < |a − b| + |a − 2b| : GMAT Data Sufficiency (DS)
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# Is |2a − 3b| < |a − b| + |a − 2b|

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Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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01 Aug 2012, 06:20
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Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat
[Reveal] Spoiler: OA
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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01 Aug 2012, 08:04
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Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

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Last edited by EvaJager on 01 Aug 2012, 08:25, edited 1 time in total.
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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01 Aug 2012, 08:11
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thank you very much, Eva!
its clear now
im not strong in modules
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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01 Aug 2012, 08:44
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Galiya wrote:
thank you very much, Eva!
its clear now
im not strong in modules

Welcome!

I am supposed to know a few things about absolute value...I am a mathematician.

I have sent you a PM, did you see it?
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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27 Aug 2012, 07:41
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EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

That was a great explanation.
Mods is not a strong point for me.
It would be helpful if you can share some more tips and tricks on mods.

also please explain: If x and y have opposite signs, the inequality is strict.
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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27 Aug 2012, 09:00
manulath wrote:
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

That was a great explanation.
Mods is not a strong point for me.
It would be helpful if you can share some more tips and tricks on mods.

also please explain: If x and y have opposite signs, the inequality is strict.

Strict inequality means < , equality cannot hold.
For example |2+(-3)| < |2| + |-3| as 1 < 2 + 3.
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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16 Nov 2012, 00:14
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I am confused at this point...For (1) and (2) together: x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Now b= 3 and a<b.
Consider case 1: a=4 a - 3 = 4-3=-1 and a - 6 = 4-6 = -2 i.e. -1 and -2 which is +ve
Consider case 1: a=2 a - 2 = 4-2= 2 and a - 6 = 2-6 = -4 i.e. 2 and -4 which is -ve
Hence the correct answer is E and not C..

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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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20 Nov 2012, 19:12
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EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

Hi Eva

I am not sure on the solution mentioned above.

For statement 2 we have a< b, ie a-b <0 and similarly a-2b <0 . Therefore both X and Y are negative. In that case the inequality meets the condition. So shouldnt the answer be b
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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26 Nov 2012, 13:04
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azzhhuu wrote:
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

Hi Eva

I am not sure on the solution mentioned above.

For statement 2 we have a< b, ie a-b <0 and similarly a-2b <0 . Therefore both X and Y are negative. In that case the inequality meets the condition. So shouldnt the answer be b

a < b doesn't necessarily imply that a < 2b.
For example a = -4 < -3 = b, but a = -4 > 2(-3) = -6 = 2b.
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 08:47
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

hi mate,

here is my approach please correct me if im wrong,

the in-equality only holds good if they are opposite sign,
s1: b = 3 or one variable in const. but we can't predict anything with this so we need another values too so S1: NS

s2: a< b, even here we have many possibilities so s2 NS
now s1 and s2 combined :

we still don't know about a, as we can only determine B and a is still unknown , we can't drive any particular values .. we say NO,

but does this help in saying E Or C ??

Im confused
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 11:53
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3

|2a − 3(3)| < |a − (3)| + |a − 2(3)|
|2a-9| < |a-3| + |a-6|

The checkpoints here are 4.5, 3, 6

The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6

a<3: -(2a-9) < -(a-3) + -(a-6) -2a+9 < -a+3 + -a+6 0 < 0 INVALID

3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 VALID (a may fall within the range of 3<a<4.5)

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 VALID 9 (a may fall within the range of 4.5<a<6)

a>6: (2a-9) < (a-3) + (a-6) 2a-9 < a-3 + a-6 0 < 0 INVALID

Some solutions are sufficient, some are not.
INSUFFICIENT

a<b

|2a − 3b| < |a − b| + |a − 2b|

If a<b then:

|2(2) - 3(3)| < |2-3| + |(2)-2(3)| |-5| < |-1| + |-6| 5<7 VALID
|2(-2)| - 3(3)| < |-2-3| + |-2 -2(3)| |-13| < |-5| + |-8| 13<13 INVALID
INSUFFICIENT

1+2) b=3 and a<b therefore a<3

Using the cases we found in #1, where a<3, the only solution where a<3 is invalid.
SUFFICIENT

(C)

(Is that correct reasoning I am using?)

Last edited by WholeLottaLove on 03 Jul 2013, 13:15, edited 1 time in total.
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 12:03
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WholeLottaLove wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3

|2a − 3(3)| < |a − (3)| + |a − 2(3)|
|2a-9| < |a-3| + |a-6|

The checkpoints here are 4.5, 3, 6

The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6

a<3: -(2a-9) < -(a-3) + -(a-6) -2a+9 < -a+3 + -a+6 0 < 0 INVALID

3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)

x>6: (2a-9) < (a-3) + (a-6) 2a-9 < a-3 + a-6 0 < 0 INVALID

All solutions are invalid
SUFFICIENT

I know the above solution is incorrect but I cannot seem to figure out why. Can someone please explain?

Thanks!

Those two parts are not correct. Example for 3<a<4.5 you get a>3, so this could be a valid solution.

Just because $$a$$ COULD fall in the range, this makes that given range a possible valid solution (for a=3.5 for example).
$$|2a - 3b| < |a - b| + |a - 2b|$$ for a=3.5 and b=3 you get
$$|7-9|<|3.5-3|+|3.5-6|$$ or $$2<3$$ => valid
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 12:56
Interesting. I think I've had trouble with a few other problems with this very concept. Thanks for the pointer. Just to be clear, these would only be invalid if a (or whatever variable) fell entirely outside of the given range? (i.e. 2<a<3 and a>10 Would 2<a<3 and a<10 be valid?)

I take it that's why we need II. in addition to I. to solve this problem?

Zarrolou wrote:
WholeLottaLove wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3

|2a − 3(3)| < |a − (3)| + |a − 2(3)|
|2a-9| < |a-3| + |a-6|

The checkpoints here are 4.5, 3, 6

The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6

a<3: -(2a-9) < -(a-3) + -(a-6) -2a+9 < -a+3 + -a+6 0 < 0 INVALID

3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)

x>6: (2a-9) < (a-3) + (a-6) 2a-9 < a-3 + a-6 0 < 0 INVALID

All solutions are invalid
SUFFICIENT

I know the above solution is incorrect but I cannot seem to figure out why. Can someone please explain?

Thanks!

Those two parts are not correct. Example for 3<a<4.5 you get a>3, so this could be a valid solution.

Just because $$a$$ COULD fall in the range, this makes that given range a possible valid solution (for a=3.5 for example).
$$|2a - 3b| < |a - b| + |a - 2b|$$ for a=3.5 and b=3 you get
$$|7-9|<|3.5-3|+|3.5-6|$$ or $$2<3$$ => valid
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 13:07
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Yes of course we need 2 a<b.

With it the situation changes to:
3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.

With $$b>a$$, $$a>3$$ => $$b>a>3$$ so $$b>3$$. But $$b =3$$, so it's NOT more than 3=> Invalid

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)

With $$b>a$$, $$a>4.5$$ so this means that $$b>4.5$$ as well. But $$b =3$$, so it's NOT more than 4.5 => Invalid
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 13:19
1
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In my revised solution (edited from the one above) I used a<b and b=3 so doesn't that mean a<3?

Zarrolou wrote:
Yes of course we need 2 a<b.

With it the situation changes to:
3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.

With $$b>a$$, $$a>3$$ => $$b>a>3$$ so $$b>3$$. But $$b =3$$, so it's NOT more than 3=> Invalid

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)

With $$b>a$$, $$a>4.5$$ so this means that $$b>4.5$$ as well. But $$b =3$$, so it's NOT more than 4.5 => Invalid
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 13:22
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WholeLottaLove wrote:
In my revised solution (edited from the one above) I used a<b and b=3 so doesn't that mean a<3?

Zarrolou wrote:
Yes of course we need 2 a<b.

With it the situation changes to:
3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.

With $$b>a$$, $$a>3$$ => $$b>a>3$$ so $$b>3$$. But $$b =3$$, so it's NOT more than 3=> Invalid

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)

With $$b>a$$, $$a>4.5$$ so this means that $$b>4.5$$ as well. But $$b =3$$, so it's NOT more than 4.5 => Invalid

Yes, good catch! I overlook that, so the analysis can be reduced to just the case $$a<3$$
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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03 Jul 2013, 13:36
Haha! I got one right for once! Thank you very much for all of your help - I would be completely lost on these 700+ questions without it!

Zarrolou wrote:
WholeLottaLove wrote:
In my revised solution (edited from the one above) I used a<b and b=3 so doesn't that mean a<3?

Zarrolou wrote:
Yes of course we need 2 a<b.

With it the situation changes to:
3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 INVALID (a may fall within the range of 3<a<4.5 but it may be greater than it as well.

With $$b>a$$, $$a>3$$ => $$b>a>3$$ so $$b>3$$. But $$b =3$$, so it's NOT more than 3=> Invalid

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 INVALID 9 a may fall within the range of 4.5<a<6 but it may be less than it as well)

With $$b>a$$, $$a>4.5$$ so this means that $$b>4.5$$ as well. But $$b =3$$, so it's NOT more than 4.5 => Invalid

Yes, good catch! I overlook that, so the analysis can be reduced to just the case $$a<3$$
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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26 Jul 2013, 10:38
Reviewing this question, I found a problem with my reasoning.

Looking above to my solution I found that there were valid ranges of a for 3<a<4.5 and 4.5<a<6. The problem is, every single value of a I plug into the statement |2a-9| < |a-3| + |a-6| (after plugging in for b) makes the inequality true. Normally I would say that #1 is sufficient but obviously it is not. Could someone explain this to me?

Thanks!
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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07 Dec 2013, 17:34
EvaJager's explanation seems correct and let me make it more simple algebraically:

EvaJager's explanation:

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, |x+y|\leq|x|+|y|. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by x = a - b, y = a - 2b, and the given inequality becomes |x+y|<|x|+|y|. So, the question is asking whether x and y are of opposite signs, or a - b and a - 2b are of opposite signs

~~~~~~~~~~`

Here I will add my two cents
True that for |x+y|\leq|x|+|y| to hold true, a - b and a - 2b should of opposite signs. So: a-b>0 and a-2b<0 which means 2b<a<b
Option B contradicts that above statement. It states: a<b: Thus Combining the two options, the inequality fails to stand true. Hence C
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Re: Is |2a − 3b| < |a − b| + |a − 2b| [#permalink]

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09 Dec 2013, 06:50
EvaJager wrote:
Galiya wrote:
Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3
(2) a < b

Source: Go Gmat

There is a well known inequality, called the "triangle inequality", which states that for any non-zero real numbers x and y, $$|x+y|\leq|x|+|y|$$. Equality holds if and only if either x and y are both positive, or x and y are both negative. If x and y have opposite signs, the inequality is strict.

In our case, we can denote by $$x = a - b, y = a - 2b$$, and the given inequality becomes $$|x+y|<|x|+|y|$$. So, the question is asking whether x and y are of opposite signs, or $$a - b$$ and $$a - 2b$$ are of opposite signs.

Then, we can see that neither (1) nor (2) alone is sufficient.
For (1) and (2) together: $$x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0$$, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.

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Re: Is |2a − 3b| < |a − b| + |a − 2b|   [#permalink] 09 Dec 2013, 06:50

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