Two water taps together can fill a tank in 6 hours. The tap of a larger diameter takes 9 hours less than the smaller one to fill the tank separately. Each tap can separately fill the tank in

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19 hours and 10 hours

18 hours and 9 hours

17 hours and 8 hours

16 hours and 7 hours

answer is C.

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Detailed Solution

Given that two water taps together can fill a tank in 6 hours and a tap of a larger diameter takes 9 hours less than the smaller one to fill the tank separately.
Let the time taken by the tap with a larger diameter be

t

hours then,
Time taken by the tap with a smaller diameter be

(t - 9)

hours.
According to question,

⟹ \frac{1}{t} + \frac{1}{t - 9} = \frac{1}{6}

⟹ \frac{t + t - 9}{t (t - 9)} = \frac{1}{6}

⟹ \frac{2 t - 9}{t^{2} - 9 t} = \frac{1}{6}

⟹ 6 (2 t - 9) = t^{2} - 9 t

⟹ 12 t - 54 = t^{2} - 9 t

⟹ t^{2} - 12 t - 9 t + 54 = 0

⟹ t^{2} - 21 t + 54 = 0

⟹ t^{2} - 18 t - 3 t + 54 - 0

⟹ t (t - 18) - 3 (t - 18) = 0

⟹ (t - 3) (t - 18) = 0

⟹ t - 3 = 0 or t - 18 = 0

∴ t = 3 or t = 18

Therefore,

t = 18

hours. So,

(t - 9) = 18 - 9 = 9

hours
Hence, the time required for each tank separately is 18 hours and 9 hours respectively.
Hence, option (3) is correct.

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