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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 \(B=\sqrt{24}\).