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(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.

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?

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.
_________________

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.

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.

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

(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?

(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?

This question is answered above.
_________________

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?