Given,
Tap A can fill a tank in= 10 hours
Tap B can fill the tank in = 15 hours
Work done by Tap A in 1 hour \(\frac{1}{10}\)
Work done by tap B in 1 hour \(=\frac{1}{15}\)
∴ Work done by both tap in 1 hour \(=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\)part
Work done by both tap in 4 hours \(={4}\times\frac{1}{6}=\frac{2}{3}\)part
Remaining part \(={1}-\frac{2}{3}=\frac{1}{3}\)part
Hence, time taken by A to fill remaining part \(=\frac{\frac{1}{3}}{\frac{1}{10}}=\frac{10}{3}={3}\frac{1}{3}\)hours