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M15-37

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M15-37  [#permalink]

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New post 15 Sep 2014, 23:57
1
11
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

44% (01:08) correct 56% (01:05) wrong based on 140 sessions

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Re M15-37  [#permalink]

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New post 15 Sep 2014, 23:58
2
1
Official Solution:


Statement (1) by itself is insufficient. From S1 we know that \(a^2 - b^2 = 0\). From here we can only conclude that \(|a| = |b|\).

Statement (2) by itself is insufficient. From S2 we know that \(b\) is non-negative. But whether \(a\) is non-negative remains a question.

Statements (1) and (2) combined are insufficient. Consider \(a = b = c = 1\) (the answer to the question is "yes") and \(a = -1\), \(b = c = 1\) (the answer to the question is "no").


Answer: E
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Re: M15-37  [#permalink]

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New post 16 Sep 2014, 06:40
Hi

Can the original st a = |b| translated to a=sqrt(b^2) or a^2= b^2 ?
In which case St A will suffice:
b = |c|
b^2=c^2

a^2 - b^2 = b^2 - c^2
or, a^2 - c^2 = b^2 - c^2
or, a^2=b^2
Sufficient
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Re: M15-37  [#permalink]

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New post 16 Sep 2014, 12:49
sidinsin wrote:
Hi

Can the original st a = |b| translated to a=sqrt(b^2) or a^2= b^2 ?
In which case St A will suffice:
b = |c|
b^2=c^2

a^2 - b^2 = b^2 - c^2
or, a^2 - c^2 = b^2 - c^2
or, a^2=b^2
Sufficient


No. It's possible a^2 to be equal to b^2 and a not be equal to |b|. For example, a=-1 and b=1.
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Re: M15-37  [#permalink]

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New post 04 Jul 2015, 09:33
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. From S1 we know that \(a^2 - b^2 = 0\). From here we can only conclude that \(|a| = |b|\).

Statement (2) by itself is insufficient. From S2 we know that \(b\) is non-negative. But whether \(a\) is non-negative remains a question.

Statements (1) and (2) combined are insufficient. Consider \(a = b = c = 1\) (the answer to the question is "yes") and \(a = -1\), \(b = c = 1\) (the answer to the question is "no").


Answer: E


My doubt may sound silly but this is confusing me a lot.
what is the difference between

1. \(|a| = |b|\).
2. \(a = |b|\).
3. \(|a| = b\).

please explain the concept.
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Re: M15-37  [#permalink]

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New post 05 Jul 2015, 07:39
1
1
Mechmeera wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. From S1 we know that \(a^2 - b^2 = 0\). From here we can only conclude that \(|a| = |b|\).

Statement (2) by itself is insufficient. From S2 we know that \(b\) is non-negative. But whether \(a\) is non-negative remains a question.

Statements (1) and (2) combined are insufficient. Consider \(a = b = c = 1\) (the answer to the question is "yes") and \(a = -1\), \(b = c = 1\) (the answer to the question is "no").


Answer: E


My doubt may sound silly but this is confusing me a lot.
what is the difference between

1. \(|a| = |b|\).
2. \(a = |b|\).
3. \(|a| = b\).

please explain the concept.


1. \(|a| = |b|\) means that the distance from a to 0 is the same as the distance from b to 0. Or that the magnitudes of a and b are the same.

2. \(a = |b|\) means that the distance from b to 0 is a.

3. \(|a| = b\) means that the distance from a to 0 is b.
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Re: M15-37  [#permalink]

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New post 23 Nov 2015, 20:55
Hi

Can the solution be bit more elaborative ,Step by step. I'm finding it difficult to understand. :|
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Re: M15-37  [#permalink]

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New post 23 Jun 2016, 00:43
Hi,
The question asked is "Is a = mod b"
This can be written as is a=+b or a=-b

1) From 1, b = mod c

so, b = +c or b=-c

hence, b2=c2
hence the given equation, "a2-b2=b2-c2" or (a2+c2)/2=b2
becomes a2=b2
this would mean that a=b or a=-b

Similarly using 2nd also we can prove.

Why is my way wrong?
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Re: M15-37  [#permalink]

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New post 23 Jun 2016, 00:55
gauravprashar17 wrote:
Hi,
The question asked is "Is a = mod b"
This can be written as is a=+b or a=-b

1) From 1, b = mod c

so, b = +c or b=-c

hence, b2=c2
hence the given equation, "a2-b2=b2-c2" or (a2+c2)/2=b2
becomes a2=b2
this would mean that a=b or a=-b

Similarly using 2nd also we can prove.

Why is my way wrong?


Check here: m15-184069.html#p1416891
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Re M15-37  [#permalink]

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New post 22 Aug 2016, 03:40
I think this is a high-quality question and I agree with explanation.
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Re: M15-37  [#permalink]

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New post 18 Dec 2016, 10:59
Bunuel,
I would like to add one more point to the wonderful explanation provided by you.
Using S1, we can actually deduce |A| = |B| = |C|.
Please correct me if wrong.
Thanks !
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Re: M15-37  [#permalink]

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New post 19 Mar 2017, 21:09
Is there a link explaining the mod theory and related problems? I have been getting most of the questions wrong in this section.
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Re: M15-37  [#permalink]

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New post 19 Mar 2017, 23:50
1
1
gmat2k17 wrote:
Is there a link explaining the mod theory and related problems? I have been getting most of the questions wrong in this section.




Hope it helps.
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Re: M15-37  [#permalink]

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New post 20 Mar 2017, 05:39
Bunuel wrote:
gmat2k17 wrote:
Is there a link explaining the mod theory and related problems? I have been getting most of the questions wrong in this section.




Hope it helps.


Perfect, thanks Bunuel
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Re: M15-37  [#permalink]

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New post 21 Mar 2017, 05:41
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. From S1 we know that \(a^2 - b^2 = 0\). From here we can only conclude that \(|a| = |b|\).

Statement (2) by itself is insufficient. From S2 we know that \(b\) is non-negative. But whether \(a\) is non-negative remains a question.

Statements (1) and (2) combined are insufficient. Consider \(a = b = c = 1\) (the answer to the question is "yes") and \(a = -1\), \(b = c = 1\) (the answer to the question is "no").


Answer: E




Hi Bunuel,

In Statement 1, how are we arriving at a^2-b^2=0 ? And would you please elaborate explanation for statement 2 as well.

Thanks
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Re: M15-37  [#permalink]

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New post 21 Mar 2017, 05:51
amargad0391 wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is insufficient. From S1 we know that \(a^2 - b^2 = 0\). From here we can only conclude that \(|a| = |b|\).

Statement (2) by itself is insufficient. From S2 we know that \(b\) is non-negative. But whether \(a\) is non-negative remains a question.

Statements (1) and (2) combined are insufficient. Consider \(a = b = c = 1\) (the answer to the question is "yes") and \(a = -1\), \(b = c = 1\) (the answer to the question is "no").


Answer: E




Hi Bunuel,

In Statement 1, how are we arriving at a^2-b^2=0 ? And would you please elaborate explanation for statement 2 as well.

Thanks
Amaresh


From the stem we know that \(a^2 - b^2 = b^2 - c^2\). From (1) \(b = |c|\) means that \(b^2=c^2\). Substitute: \(a^2 - b^2 = b^2 - b^2=0\)

As for (2): \(b = |a|\). Given that b is equal to an absolute value of a number. An absolute value cannot be negative, theretofore \(b\) is non-negative.

Hope it's clear.
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Re: M15-37  [#permalink]

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New post 26 Jul 2017, 12:17
This is really the toughest DS question i solved
thank you Bunuel for this question
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Re: M15-37  [#permalink]

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New post 15 Apr 2018, 03:49
\(a^2\) -\(b^2\) =\(b^2-c^2\)

so, \((a-b)(a+b)=(b-c)(b+c)\)

its given that (1) \(b=|c|\)

so, \((a-b)(a+b) = 0\)
i.e, \(a=b\) or \(a = -b\), or \(a= |b|\),

So , isn't (1) sufficient ?
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New post 17 Jun 2018, 11:32
I think this is a high-quality question and I agree with explanation.
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M15-37  [#permalink]

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New post 10 Nov 2018, 07:03
samji wrote:
\(a^2\) -\(b^2\) =\(b^2-c^2\)

so, \((a-b)(a+b)=(b-c)(b+c)\)

its given that (1) \(b=|c|\)

so, \((a-b)(a+b) = 0\)
i.e, \(a=b\) or \(a = -b\), or \(a= |b|\),

So , isn't (1) sufficient ?


U need one more step and u will get the correct answer.
so u found a=b OR a=-b

We HAVE to consider both these above scenarios SEPARATELY while solving the sum.
But notice that just by your first statement a=b we can prove the above as insufficient.
so if a = b and a = b = -2. Then \(a=|b|\) is not possible.
but if a = b = 2. Then \(a=|b|\) is possible.

You only missed considering both the OR statements SEPARATELY.

Hope it helps :)
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