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