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# M06-34

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Math Expert
Joined: 02 Sep 2009
Posts: 41698

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16 Sep 2014, 00:33
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Difficulty:

75% (hard)

Question Stats:

53% (01:33) correct 47% (01:42) wrong based on 129 sessions

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If $$a$$, $$b$$, and $$c$$ are positive and $$a^2 + c^2 = 202$$, what is the value of $$b - a - c$$?

(1) $$b^2 + c^2 = 225$$

(2) $$a^2 + b^2 = 265$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 41698

Kudos [?]: 124653 [0], given: 12079

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16 Sep 2014, 00:33
Expert's post
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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.

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Joined: 06 Aug 2014
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Concentration: Entrepreneurship, Marketing
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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^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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Math Expert
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23 Sep 2014, 02:07
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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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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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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Math Expert
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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.

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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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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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04 Aug 2016, 10:26
akshayshetty2720 wrote:
(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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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 high-quality question and I agree with explanation.

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Re M06-34   [#permalink] 03 Aug 2017, 23:08
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# M06-34

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