UGC NET Paper 1 GA 2012 June Solved Paper

Upto 9 A.M. Ist pipe has worked for 3 hrs.
2nd pipe for 2 hrs. and 3rd pipe for 1 hr.
∴ Part filled up to 9 A.M. $=3\times \frac{1}{15}+2\times \frac{1}{20}+\frac{1}{30}$ $=\frac{1}{3}$
Unfilled part at 9 A.M $=1-\frac{1}{3}=\frac{2}{3}$
From 9 A.M. onwards all four pipes are functioning. Part filled (by all four pipes) in 1 hr. $=\frac{1}{30}+\frac{1}{15}+\frac{1}{20}+\frac{1}{60}$ $=\frac{1}{6}$
∴ $\frac{2}{3}$ of reservoir will be filled up in $6\times \frac{2}{3}=4\mathrm{hr}.$
Counting 4 hrs. from 9 A.M. reservoir will be full at 1.00 P.M.