Let there are two pipes A and B and diameter of A is larger than B.Now suppose that Pipe A take \\\\(x\\\\) hours and Pipe B takes \\\\(y\\\\) hours to fill the pool separately .In 1 hour pipe A can fill the pool \\\\(=\\dfrac{1}{x}\\\\)In 1 hour pipe B can fill the pool\\\\(=\\dfrac{1}{y}\\\\)If both pipe are together then they take \\\\(12\\\\) hour to fill the pool.If the are togther then in 1 hour they fill the pool.\\\\(\\dfrac{1}{x} \\\\)+ \\\\(\\dfrac{1}{y}\\\\)=\\\\(\\dfrac{1}{12}----(1)\\\\)Now In \\\\(4\\\\) hour pipe A can fill the pool is\\\\(=\\dfrac{4}{x}\\\\)In \\\\(9\\\\) hour pipe B can fill the pool is\\\\(=\\dfrac{9}{y}\\\\) than they fill half of the pool,so\\\\(\\dfrac{4}{x}\\\\)+\\\\(\\dfrac{9}{y}\\\\) \\\\(=\\dfrac{1}{2}-----(2)\\\\) Now Let \\\\(\\dfrac{1}{x}=P\\\\) &\\\\(\\dfrac{1}{y}=Q\\\\) and puuting in equation 1 &2, than\\\\(P+Q\\\\)=\\\\(\\dfrac{1}{12}\\\\) \\\\(\\Rightarrow12P+12Q=1----(3)\\\\)And \\\\(4P+9Q\\\\) \\\\(=\\dfrac{1}{2}\\\\)\\\\(\\Rightarrow8P+18Q=1-----(4)\\\\)Now solving equation 3 & 4Equation (3) multiplying by 2 and (4) multiplying by 3 and substract.\\\\(\\Rightarrow -30Q=-1\\\\)\\\\(\\Rightarrow Q\\\\)\\\\(=\\dfrac{1}{30}\\\\)now put value of \\\\(Q\\\\) in equation \\\\(3\\\\)We get \\\\(P\\\\)\\\\(=\\dfrac{1}{20}\\\\)So we found\\\\(P\\\\)\\\\(=\\dfrac{1}{20}\\\\) & \\\\( Q\\\\)\\\\(=\\dfrac{1}{30}\\\\)So \\\\(x=20\\\\) and \\\\(y=30\\\\)Hence the A pipe would take 20 hours and the pipe B would take 30 hours separately .