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Squares of squares [need expert help]

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by voodoo_child » Wed Sep 28, 2011 5:47 am
If a, b and C are positive; a^2 + c^2 = 202; what's b-a-c?

1) b^2 + c^2 = 225
2) a^2 + b^2 = 265

Here's what I did: I found all possible combinations for b,c to be positive and such that sum of their squares equal to 225.

Possible values = b|c = 12|9 or 9|12; however, if you add another equation (a^2 ...) you would get unique a, b and c

Similarly, we can do it for #2).

My answer was D). OA is C.

Can anyone please explain why?

Thanks
Voodoo
Last edited by voodoo_child on Wed Sep 28, 2011 6:54 am, edited 1 time in total.
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Source: — Problem Solving |

by shankar.ashwin » Wed Sep 28, 2011 5:56 am
You sure of the question?

The question says a^2 + b^2 = 202 and statement B is a^2 + b^2 = 265 ?
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by voodoo_child » Wed Sep 28, 2011 6:54 am
corrected :a^2 + c^2 = 202;
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by shankar.ashwin » Wed Sep 28, 2011 7:08 am
a,b and c are not mentioned to be integers.

So you wouldn't want to restrict values of b and c to just 12(or)9.

It could be 5 and sqrt200?

So, that leads to a different soln. Hence C IMO.

But if its mentioned as integers, what you say makes sense,
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by navami » Wed Sep 28, 2011 7:28 am
If a, b and C are positive; a^2 + c^2 = 202; what's b-a-c?

1) b^2 + c^2 = 225
2) a^2 + b^2 = 265

FROM a^2 + c^2 = 202
a = 9 and c = 11
c = 11 and a = 9

Now option 1.

1) b^2 + c^2 = 225
lets say c = 9 b = 12

again put c = 11 b = 10.2~

again from option 2)

2) a^2 + b^2 = 265


a = 9

b = 13.5

again a = 11

b = 12


only combining option 1 and option 2 we can findout b = 12 .a = 11 c = 9.. and value of A and C
This time no looking back!!!
Navami
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