Question

Two water taps are together can fill a tank in 9 3/8 hours.The tap of a larger diameter takes 10 hrs less than the smaller one to fill the tank separately find the time in which each top can separately fill the tank​

Answer 1

Answer:

smaller tap =25 Hours

Larger tap = 15 hours

Step-by-step explanation:

$the \: smaller \: tap \: b e\: x \: hours \\ larger \: tap \: be \: (x - 10) \: hours \\ both \: together \: = \frac{75}{8} hours \\ small \: tap \:for \: 1 \: hour = \frac{1}{x} \\ \\ larger \: tap \: for \: 1 \: hour \: = \frac{1}{x - 10} \\ part \: filled \: by \: both \: taps \: in \: 1 \: hour = \frac{8}{75} \\ \frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75} \\ \frac{x - 10 + x}{x(x - 10)} = \frac{8}{75} \\ \frac{2x - 10}{ {x}^{2} - 10x} = \frac{8}{75} \\ 150x - 750 = {8x}^{2} - 80x \\ 4 {x}^{2} - 115x + 375 = 0 \\ 4 {x}^{2} - 100x - 15x + 375 = 0 \\ 4x(x - 25) - 15(x - 25) = 0 \\( 4x - 15)(x - 25) = 0 \\ x = \frac{15}{4} \: \: \: x = 25 \\ smaller \: tap \: = x = 25 \: hours \\ larger \: tap \: = x - 10 = 15$