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M15-37

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M15-37  [#permalink]

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New post 16 Sep 2014, 00:57
1
13
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

40% (01:59) correct 60% (01:39) wrong based on 98 sessions

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Re M15-37  [#permalink]

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New post 16 Sep 2014, 00:58
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, 07:40
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, 13:49
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Re: M15-37  [#permalink]

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New post 04 Jul 2015, 10:33
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, 08:39
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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 22 Aug 2016, 04:40
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 19 Mar 2017, 22:09
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 20 Mar 2017, 00:50
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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 21 Mar 2017, 06:41
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
Amaresh
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Re: M15-37  [#permalink]

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New post 21 Mar 2017, 06:51
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 15 Apr 2018, 04:49
\(a^2\) -\(b^2\) =\(b^2-c^2\)

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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Re M15-37  [#permalink]

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New post 17 Jun 2018, 12:32
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, 08:03
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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Re: M15-37  [#permalink]

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New post 22 Jan 2019, 16:23
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Bunuel wrote:
\(a^2 - b^2 = b^2 - c^2\). Is \(a = |b|\)?


(1) \(b = |c|\)

(2) \(b = |a|\)


Target question: Is a = |b|?

Given: a² - b² = b² - c²

Statement 1: b = |c|
This tells us a few things, but with regard to this question, it tells us that b and c have the same magnitude
This also means that b² = c²
With this information, let's test some values that satisfy both statement 1 and the given information:
Case a: a = 0, b = 0 and c = 0. In this case, the answer to the target question is YES, a does EQUAL |b|
Case b: a = -1, b = 1 and c = -1. In this case, the answer to the target question is NO, a does NOT equal |b|
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: b = |a|
Let's test values again.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: a = 0, b = 0 and c = 0. In this case, the answer to the target question is YES, a does EQUAL |b|
Case b: a = -1, b = 1 and c = -1. In this case, the answer to the target question is NO, a does NOT equal |b|
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient.
So, the same counter-examples will satisfy the two statements COMBINED.

In other words,
Case a: a = 0, b = 0 and c = 0. In this case, the answer to the target question is YES, a does EQUAL |b|
Case b: a = -1, b = 1 and c = -1. In this case, the answer to the target question is NO, a does NOT equal |b|
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Re: M15-37  [#permalink]

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New post 31 Aug 2019, 13:32
Bunuel wrote:
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.



Hi Bunuel AjiteshArun chetan2u
can you please elaborate a bit.

1 - I get that b is non- negative but why is it not sufficient? Is it coz there is no new information about 'a' & 'c' (therefore we can't say whether 'a' is negative).
2 How did you combine both the statements? Is putting values the only way to do it OR is there some explanation behind it too.

Thanks in Advance.
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Re: M15-37  [#permalink]

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New post 01 Sep 2019, 03:20
bratbg wrote:
Bunuel wrote:
amargad0391 wrote:
[quote="Bunuel"


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.



Hi Bunuel AjiteshArun chetan2u
can you please elaborate a bit.

1 - I get that b is non- negative but why is it not sufficient? Is it coz there is no new information about 'a' & 'c' (therefore we can't say whether 'a' is negative).
2 How did you combine both the statements? Is putting values the only way to do it OR is there some explanation behind it too.

Thanks in Advance.
Bharat


Have you checked the discussion above? I think your doubts should be addressed there.
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Re: M15-37   [#permalink] 01 Sep 2019, 03:20
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