Answer & Solution
Answer: Option D
Solution:
$$\eqalign{
& {\text{Part}}\,{\text{filled}}\,{\text{in}}\,{\text{4}}\,{\text{minutes}} \cr
& = 4\left( {\frac{1}{{15}} + \frac{1}{{20}}} \right) = \frac{7}{{15}} \cr
& {\text{Remaining}}\,{\text{part}} = {1 - \frac{7}{{15}}} = \frac{8}{{15}} \cr
& {\text{Part}}\,{\text{filled}}\,{\text{by}}\,B\,{\text{in}}\,{\text{1}}\,{\text{minute}} = \frac{1}{{20}} \cr
& \therefore \frac{1}{{20}}:\frac{8}{{15}}::1:x \cr
& x = {\frac{8}{{15}} \times 1 \times 20} \cr
& \,\,\,\,\,\, = 10\frac{2}{3}\,\min \cr
& \,\,\,\,\,\, = 10\min .\,40\,\sec . \cr
& \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\, \cr
& = {4\min . + 10\min . +\, 40\sec .} \cr
& = 14\min .\,40\sec . \cr} $$
