SSC MTS Model Paper 7

(d) Part of the cistern filled in 1 hourwhen pipes P and S are open$= \frac{1}{4} - \frac{1}{10} = \frac{5-2}{20} = \frac{3}{20}$Hence, the cistern w ill be filled in$\frac{20}{3}$ hours $\approx 6.6$ hoursPart of the cistern filled in 1 hourwhen pipes P, R and S are open$= \frac{1}{4} + \frac{1}{12} - \frac{1}{10}$$= \frac{15+5-6}{60} = \frac{14}{60} = \frac{7}{30}$Hence, the cistern w ill be filled in$\frac{30}{7}$ hours ≈4.3 hoursPart of the cistern filled in 1 hourwhen pipes P, g and S are open$= \frac{1}{4} + \frac{1}{8} - \frac{1}{10}$$= \frac{10+5-4}{40} = \frac{11}{40}$Hence, the cistern will be filled in40—hours ≈ 3.6 hours