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Re M0634
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16 Sep 2014, 00:33



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23 Sep 2014, 02:01
Hi Bunuel,
In this question, if I take only the first statement and subtract the two equations given we get:
b^2a^2=23. Thus possible values positive values will be b=12, a=11. Upon substitution in any of the two equations we get c=9. Thus can we not answer the question using only statement 1.
Please kindly point out the flaw in this reasoning. Is it because the question does not give us that only integer values are possible?



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23 Sep 2014, 02:07
madhavmarda wrote: Hi Bunuel,
In this question, if I take only the first statement and subtract the two equations given we get:
b^2a^2=23. Thus possible values positive values will be b=12, a=11. Upon substitution in any of the two equations we get c=9. Thus can we not answer the question using only statement 1.
Please kindly point out the flaw in this reasoning. Is it because the question does not give us that only integer values are possible? The problem with your solution is that you assume, with no ground for it, that variables represent integers only. From b^2  a^2 = 23 you cannot say that b = 12 and a = 11. For example b could be \(\sqrt{24}\) and a could be 1.
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Tricky question but a good one.  I also had assumed that B = 12 , A = 11 C = 9.



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11 Apr 2015, 05:19
From st1, we get b^2a^2=23=>(b+a)(ba)=23, now as 23 is prime so (b+a)(ba)=23*1, only solution b=12 & a=11, from this we can get unique value of c as c is positive. So statement 1 is sufficient.
Bunuel please let me know where i am wrong.



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11 Apr 2015, 09:10
good question..I also did the same mistake of assuming the numbers as integers.



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27 Nov 2015, 16:17
To solve 3 different variables, you need 3 different equations. hence it was simple to identify 'C' as the option. Is the approach correct ? I did this question in less than 30 seconds



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(ab)(a+b)=23 which is a prime. Two numbers can multiple into a prime only when one of the number is itself and the other being 1. how does the solution go around this concept?



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18 Apr 2017, 22:24
HK17 wrote: To solve 3 different variables, you need 3 different equations. hence it was simple to identify 'C' as the option. Is the approach correct ? I did this question in less than 30 seconds I did it the same way without calculating the individual values. Is there a possiblity that this assumption will not hold?



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03 Aug 2017, 23:08
I think this is a highquality question and I agree with explanation.



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21 Oct 2017, 19:54
In my onion A is the answer
1) a^2 + c^2 = 202 ( given ) 2) b^2 + c^2 = 225 ( statement 1 )
subtracting 1 from 2, we get b^2  a^2 = 23 ( prime number )
its given that a,b and c are positive. therefore the only solution is b+c = 23 and bc = 1 > b= 12, c = 11
we can substitute the value of b in equation 2 > 12^2 + c^2 = 225 > c = 9 ( since 9 can be negated since c has to be +ve )
also similarly for statement 2, b^2  c^2 = 63; but 63 is not prime > therefore we cant get a solution ( it can be 21 * 3 or 9*7 )
I think the answer is A. High quality question > but wrong answer marked.
Ashwin



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21 Oct 2017, 22:57
TheGMATcracker wrote: In my onion A is the answer
1) a^2 + c^2 = 202 ( given ) 2) b^2 + c^2 = 225 ( statement 1 )
subtracting 1 from 2, we get b^2  a^2 = 23 ( prime number )
its given that a,b and c are positive. therefore the only solution is b+c = 23 and bc = 1 > b= 12, c = 11
we can substitute the value of b in equation 2 > 12^2 + c^2 = 225 > c = 9 ( since 9 can be negated since c has to be +ve )
also similarly for statement 2, b^2  c^2 = 63; but 63 is not prime > therefore we cant get a solution ( it can be 21 * 3 or 9*7 )
I think the answer is A. High quality question > but wrong answer marked.
Ashwin The problem with your solution is that you assume that the variables are integers. We are not given that. b^2  a^2 = 23 has infidelity many solutions for b ans a.
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22 Oct 2017, 09:56
Bunuel wrote: Official Solution:
(1) \(b^2+c^2=225\). Not sufficient on its own. (2) \(a^2+b^2=265\). Not sufficient on its own. (1)+(2) Subtract \(a^2+c^2=202\) from \(b^2+c^2=225\): \(b^2a^2=23\). Now, sum this with \(a^2+b^2=265\): \(2b^2=288\). So, \(b^2=144\), giving \(b=12\) (since it is given that \(b\) is a positive number). Since \(b=12\), then from \(b^2a^2=23\) we get that \(a=11\) and from \(a^2+c^2=202\) we get that \(c=9\). Sufficient.
Answer: C If it would've been given that a,b,c are integers then can we mark D? Explanation: if a,b,c are integers a^2+C^2=202 => a=9/11, C=11/9 or Solving with b^2+C^2=225 => b^2a^2=23 => (ba)(b+a)=23, 23 is prime number so b and a must be consecutive numbers.=> b=12,a=11 for c=9(integer value) Similarly, we can do for a^2+b^2=265



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24 Apr 2018, 16:11
I think this is a poorquality question and the explanation isn't clear enough, please elaborate.



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24 Apr 2018, 21:05



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08 Jun 2018, 00:06
I don't agree with the explanation. Hi, 1. Given a^2+b^2=202, b^2+c^2=225; c^2a^2=23; (ca)(c+a)=23; c+a=23, ca=1; c=12, a=1;
Same goes with option B.



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