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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

OA:C OE: Stmt 1 & 2 are insufficient. Both together are sufficient. Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

as you stated - 2a^2 = 242 => a^2 = 121 so a can be +11 or -11

since they mention positive integers, we take a=11 and hence get answer....

Highlighted part is not mentioned in the question. Its not mentioned that a is an integer. So a can be \sqrt{121},& this will give a^2=121 _________________

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

OA:C OE: Stmt 1 & 2 are insufficient. Both together are sufficient. Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.

You are thinking too much about this problem as there is no reason for confusion: \sqrt{121}=11 as 11^2=121 (the same way as \sqrt{4}=2).

Given: a>0 and a^2=121 --> so either a=-\sqrt{121}=-11, but this solution is out as a>0 OR a=\sqrt{121}=11. So a=11.

Isn't it possible to: (-A^2 - B^2 = -202) + (B^2 + C^2 = 225) = B^2 - A^2 = 23 which gives us (B - A)*(B+ A)=23 which implies A=11 B=12 no? So wouldn't 1 be enough to solve? Please explain, thank you.

Bunuel wrote:

Hussain15 wrote:

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

OA:C OE: Stmt 1 & 2 are insufficient. Both together are sufficient. Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.

You are thinking too much about this problem as there is no reason for confusion: \sqrt{121}=11 as 11^2=121 (the same way as \sqrt{4}=2).

Given: a>0 and a^2=121 --> so either a=-\sqrt{121}=-11, but this solution is out as a>0 OR a=\sqrt{121}=11. So a=11.

Isn't it possible to: (-A^2 - B^2 = -202) + (B^2 + C^2 = 225) = B^2 - A^2 = 23 which gives us (B - A)*(B+ A)=23 which implies A=11 B=12 no? So wouldn't 1 be enough to solve? Please explain, thank you.

Bunuel wrote:

Hussain15 wrote:

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

OA:C OE: Stmt 1 & 2 are insufficient. Both together are sufficient. Subtract S1 from a^2 + c^2 = 202 to get a^2-b^2= -23 Now add this equation to Stmt 2. It wil give a^2=121 & thus a=11 , with the help of a, b & c can also be found.

How can a=11, as its not given that a is an integer. a can be \sqrt{121}

Please help.

You are thinking too much about this problem as there is no reason for confusion: \sqrt{121}=11 as 11^2=121 (the same way as \sqrt{4}=2).

Given: a>0 and a^2=121 --> so either a=-\sqrt{121}=-11, but this solution is out as a>0 OR a=\sqrt{121}=11. So a=11.

Hope it's clear.

Notice that we are not told that A and B are positive integers, so from B^2-A^2=23 it's possible for instance that A=1 and B=\sqrt{24}.