Question

(1) Whose armour was Joan prepared to wear?
Ans.
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Answer 1

$\tt Let \:the \:speed \:of \:water\: current \\ \tt be\: x \:km/hr. \\ \\ \tt Speed \:of \:boat\: is\: 12\: km/hr. \\ \\ \tt In \: upstream, \: speed \: of \: the \: water \\ \tt current \: decreases \: the \: speed \: of \: the \\ \tt boat \: and \: it \: is \: the \: opposite \: in \: downstream.\\ \\ \tt \therefore Speed \:of \:boat\:in\: upstream = (12-x) \: km\hr \\ \tt Speed \:of \:boat\:in\:downstream = (12+x) \: km\hr \\ \\ \tt Now,\:Time = \frac{Distance}{speed} \\ \\ \tt Time\: required \:to \:cover\: 36\: km \: upstream = \frac{36}{12−x}\:hrs \\ \tt Time\: required\: to \:cover\: 36 \:km \: downstream = {36}{x+12} \\ \\ \tt According\:to \:the\: given\: condition, \\ \\ \tt \frac{36}{12-x} + \frac{36}{12+x} = 8 \\ \\ \tt 36( \frac{1}{12-x} + \frac{1}{12+x} ) = 8 \\ \\ \frac{1}{12-x} + \frac{1}{12+x} = \frac{8}{36} \\ \\ \tt \frac{12+x+12-x}{(12+x)(12-x)} = \frac{2}{9} \\ \\ \tt \frac{24}{144-{x}^{2}} = \frac{2}{9} \\ \\ \tt 144 - {x}^{2} = 12 × 9 \\ \\ \tt 144 - {x}^{2} = 108 \\ \\ \tt {x}^{2} = 144 - 108 \\ \\ \tt {x}^{2} = 36 \\ \\ \tt x = 6 \\ \\ \boxed{\bold{\underline{\red{\tt Speed\: of\: water \: current \: is \: 6 \: km/hr}}}}$