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\(a^2  b^2 = b^2  c^2\). Is \(a = b\)? (1) \(b = c\) (2) \(b = a\)
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16 Sep 2014, 00:58
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 nonnegative. But whether \(a\) is nonnegative 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: M1537
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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: M1537
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16 Sep 2014, 13:49
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: M1537
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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 nonnegative. But whether \(a\) is nonnegative 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: M1537
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05 Jul 2015, 08:39
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 nonnegative. But whether \(a\) is nonnegative 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: M1537
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23 Nov 2015, 21:55
Hi Can the solution be bit more elaborative ,Step by step. I'm finding it difficult to understand.



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Re: M1537
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23 Jun 2016, 01:43
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, "a2b2=b2c2" 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: M1537
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23 Jun 2016, 01:55
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, "a2b2=b2c2" 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: m15184069.html#p1416891
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Re M1537
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22 Aug 2016, 04:40
I think this is a highquality question and I agree with explanation.



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Re: M1537
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18 Dec 2016, 11:59
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: M1537
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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: M1537
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20 Mar 2017, 00:50
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: M1537
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20 Mar 2017, 06:39
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: M1537
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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 nonnegative. But whether \(a\) is nonnegative 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^2b^2=0 ? And would you please elaborate explanation for statement 2 as well. Thanks Amaresh



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Re: M1537
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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 nonnegative. But whether \(a\) is nonnegative 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^2b^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 nonnegative. Hope it's clear.
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Re: M1537
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26 Jul 2017, 13:17
This is really the toughest DS question i solved thank you Bunuel for this question



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Re: M1537
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15 Apr 2018, 04:49
\(a^2\) \(b^2\) =\(b^2c^2\)
so, \((ab)(a+b)=(bc)(b+c)\)
its given that (1) \(b=c\)
so, \((ab)(a+b) = 0\) i.e, \(a=b\) or \(a = b\), or \(a= b\),
So , isn't (1) sufficient ?



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17 Jun 2018, 12:32
I think this is a highquality question and I agree with explanation.



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samji wrote: \(a^2\) \(b^2\) =\(b^2c^2\)
so, \((ab)(a+b)=(bc)(b+c)\)
its given that (1) \(b=c\)
so, \((ab)(a+b) = 0\) 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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