Two water taps together can fill a tank in 9 (3/8) hours. The tap of larger diameter takes 10 hrs less than the smaller one to fil

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1. Answer:

   Let the smaller tap fill the tank in x hours.

   Then, the larger tap fills it in (x – 10) hours.

   Time taken by both together to fill the tank = 75/8 hours

   Part filled by the smaller tap in 1 hr = 1/x

   Part filled by the larger tap in 1 hr = 1/(x – 10)

   Part filled by both the taps in 1 hr = 8/75

   [tex] \sf \therefore \frac{1}{x} + \frac{1}{(x – 10)} = \frac{8}{75} \\ [/tex]

   [tex] \sf \longrightarrow \frac{(x – 10) + x}{x(x – 10)} = \frac{8}{75} \\ [/tex]

   [tex] \sf \longrightarrow \frac{(2x – 10)}{x(x – 10)} = \frac{8}{75} \\ [/tex]

   By cross multiplication

   [tex] \sf \longrightarrow 75(2x – 10) = 8x(x – 10)[/tex]

   [tex] \sf \longrightarrow 150x – 750 = 8 {x}^{2} – 80x[/tex]

   [tex] \sf \longrightarrow 8 {x}^{2} – 230x + 750 = 0[/tex]

   [tex] \sf \longrightarrow 4 {x}^{2} – 115x + 375 = 0[/tex]

   [tex] \sf \longrightarrow 4 {x}^{2} – 100x – 15x + 375 = 0[/tex]

   [tex] \sf \longrightarrow 4x(x – 25) – 15(x – 25) = 0[/tex]

   [tex] \sf \longrightarrow (x – 25)(4x – 15) = 0[/tex]

   [tex] \sf \longrightarrow x – 25 = 0 \: \: or \: \: 4x – 15 = 0[/tex]

   [tex] \sf \longrightarrow x = 25 \: \: or \: \: x = \frac{15}{4} \\ [/tex]

   [tex] \sf \longrightarrow x = 25[/tex]

   [tex] \because \sf \: x = \frac{15}{4} \\\\ \implies \sf(x – 10) < 0[/tex]

   Hence, the time taken by the smaller tap to fill the tank = 25 hours

   And, the time taken by the larger tap to fill the tank = (25 – 10) = 15 hours.

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