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M06-34

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M06-34 [#permalink]

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Re M06-34 [#permalink]

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(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 \(b^2+c^2=225\): \(b^2-a^2=23\).

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 \(b^2-a^2=23\) we get that \(a=11\) and from \(a^2+c^2=202\) we get that \(c=9\). Sufficient.


Answer: C
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Re: M06-34 [#permalink]

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New post 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^2-a^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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Re: M06-34 [#permalink]

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New post 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^2-a^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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M06-34 [#permalink]

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New post 29 Nov 2014, 22:25
Tricky question but a good one. -- I also had assumed that B = 12 , A = 11 C = 9.
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Re: M06-34 [#permalink]

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New post 11 Apr 2015, 05:19
From st1, we get b^2-a^2=23=>(b+a)(b-a)=23, now as 23 is prime so (b+a)(b-a)=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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Re: M06-34 [#permalink]

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New post 11 Apr 2015, 08:16
goGMAT2015 wrote:
From st1, we get b^2-a^2=23=>(b+a)(b-a)=23, now as 23 is prime so (b+a)(b-a)=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.


Please read the whole thread: m06-183731.html#p1419052
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Re: M06-34 [#permalink]

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New post 11 Apr 2015, 09:10
good question..I also did the same mistake of assuming the numbers as integers.
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Re: M06-34 [#permalink]

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New post 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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M06-34 [#permalink]

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New post 04 Aug 2016, 09:45
(a-b)(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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Re: M06-34 [#permalink]

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Re: M06-34 [#permalink]

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New post 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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Re M06-34 [#permalink]

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New post 03 Aug 2017, 23:08
I think this is a high-quality question and I agree with explanation.
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Re: M06-34 [#permalink]

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New post 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 b-c = 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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Re: M06-34 [#permalink]

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New post 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 b-c = 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M06-34 [#permalink]

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New post 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 \(b^2+c^2=225\): \(b^2-a^2=23\).

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 \(b^2-a^2=23\) 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^2-a^2=23 => (b-a)(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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Re M06-34 [#permalink]

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New post 24 Apr 2018, 16:11
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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New post 24 Apr 2018, 21:05
Re: M06-34   [#permalink] 24 Apr 2018, 21:05
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