Two pipes running together can fill a tank in 1119 minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, the time in which each pipe would fill the tank separately are

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Two pipes running together can fill a tank in $11 \frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, the time in which each pipe would fill the tank separately are

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21 minutes and 26 minutes

22 minutes and 27 minutes

20 minutes and 25 minutes

23 minutes and 28 minutes

answer is A.

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Detailed Solution

Given that the time taken by both the pipes

= 11 \frac{1}{9} = \frac{100}{9}

.
Let's assume that the time taken by one pipe is

t_{1}

minutes and time taken by other pipe is

t_{2}

minutes. Then,
Time taken by one pipe in one minute

= \frac{1}{t_{1}}

.
Since, one pipe takes 5 minutes more than another pipe to fill the tank separately. Then,

⟹ t_{2} = t_{1} + 5

Time taken by another pipe in one minute will be

= \frac{1}{t_{1} + 5}

.
According to question,
The total time taken by both the pipes in one minute will be

⟹ \frac{1}{t_{1}} + \frac{1}{t_{1} + 5} = \frac{9}{100}

⟹ \frac{t_{1} + 5 + t_{1}}{t_{1} (t_{1} + 5)} = \frac{9}{100}

⟹ \frac{2 t_{1} + 5}{t_{1}^{2} + 5 t_{1}} = \frac{9}{100}

⟹ 100 (2 t_{1} + 5) = 9 (t_{1}^{2} + 5 t_{1})

⟹ 200 t_{1} + 500 = 9 t_{1}^{2} + 45 t_{1}

⟹ 9 t_{1}^{2} + 200 t_{1} - 45 t_{1} - 500 = 0

⟹ 9 t_{1}^{2} + 155 t_{1} - 500 = 0

⟹ 9 t_{1}^{2} - 180 t_{1} + 25 t_{1} - 500 = 0

⟹ 9 t_{1} (t_{1} - 20) + 25 (t_{1} - 20) = 0

⟹ (9 t_{1} + 25) (t_{1} - 20) = 0

⟹ 9 t_{1} + 25 = 0 or t_{1} - 20 = 0

∴ t_{1} = - \frac{25}{9} or t_{1} = 20

We will choose the value

t_{1} = 20

because time cannot be negative.
Therefore, one pipe will take 20 min to fill the tank and the other will take 25 min to fill the tank.
Hence, option (1) is correct.

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