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1. Pipe A can fill a tank in 4 hours and pipe B can fill it 6 hours. If they are opened on alternate hours and if pipe A is opened first then in how many hours, the tank shall be full ?

A. $4 \frac{1}{2}$ hours

B. $4 \frac{2}{3}$ hours

C. $3 \frac{1}{2}$ hours

D. $3 \frac{1}{4}$ hours

Solution:

A → 4 hours
B → 6 hours
Pipes and Cistern mcq solution image

According to question,
⇒ For the first hour tap A is opened and B for second hour
⇒ Work done by both in 2 hours

\begin{aligned} \to (3 lit/h + 2 lit/h) \times 2 = 10 units \\ {| \times 2}_{4 hours}^{2 hours} {| \times 2}_{10 litres}^{5 liters} \end{aligned}

⇒ Remaining part
= 12 - 10 = 2 liters
⇒ Again 5th hour A will be opened Tap A will fill the 2 liters water with its efficiency =

\frac{2}{3}

hours
⇒ Therefore tank will be filled in
=

(4 + \frac{2}{3})

hours
=

4 \frac{2}{3}

hours

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