HCU MCA 2013 Paper

(d) 6 technicians work in 1 h = $\frac{1}{10}$11 am to 5 pm (6 h) work in 6 technicians = $\frac{1}{10} \times 6 = \frac{3}{5}$∴ Remaining work $= 1 - \frac{3}{5} = \frac{2}{5}$∵ 1 technician added after 5 pm in each hour∴ 1 technician in 1 h work $= \frac{1}{10} \times \frac{1}{6} = \frac{1}{60}$∵5 am to 6 am (7 technicians) 1 h work = $\frac{7}{60}$6 am to 7 am (8 technicians) 1 h work = $\frac{8}{60}$and 7 am to 8 am (9 technicians) 1 h = $\frac{9}{60}$ Form work = $\frac{7}{60} + \frac{8}{60} + \frac{9}{60} = \frac{24}{60} = \frac{2}{5}$∵Remaining work = form work∴Total time = 8 pm