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Two water taps together can fill a tank in $8\frac{8}{3}$ hours. The tap of larger diameter takes $4$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
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Sol. Let tap of smaller diameter takes $x$ hours to fill the tank.
$\therefore$ tap of larger diameter takes $x - 4$ hours to fill the tank.
$\frac{1}{x} + \frac{1}{x-4} = \frac{9}{80}$ (2 Marks)
$\Rightarrow 9x^2-196x + 320 = 0$ (1 Mark)
$\Rightarrow (x-20)(9x – 16) = 0$ (1 Mark)
$\Rightarrow x = 20, \frac{16}{9}$
$x = \frac{16}{9}$ (rejected) (1/2 Mark)
$\therefore x = 20$
Hence, tap of smaller diameter and larger diameter takes $20$ hours & $16$ hours respectively, to fill the tank. (1/2 Mark)
$\therefore$ tap of larger diameter takes $x - 4$ hours to fill the tank.
$\frac{1}{x} + \frac{1}{x-4} = \frac{9}{80}$ (2 Marks)
$\Rightarrow 9x^2-196x + 320 = 0$ (1 Mark)
$\Rightarrow (x-20)(9x – 16) = 0$ (1 Mark)
$\Rightarrow x = 20, \frac{16}{9}$
$x = \frac{16}{9}$ (rejected) (1/2 Mark)
$\therefore x = 20$
Hence, tap of smaller diameter and larger diameter takes $20$ hours & $16$ hours respectively, to fill the tank. (1/2 Mark)