Question

Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes?
a.2 minutes
b.4 minutes
c.6 minutes
d.8 minutes

Answer 1

Hint: Let the capacity of the tank be 96 units (It is LCM of 24 and 32). You can assume any other number which is a multiple of these two numbers. Only B is closed before time, so A runs for full 18mins.

Complete step-by-step answer:

Let the total capacity of the tank be 96 units.
The rate of filling of water by pipe A $=\dfrac{96}{24}=4$ units per minute.
The rate of filling of water by pipe B =$=\dfrac{96}{32}=3$ units per minute.
Since pipe A is open for 18 minute water filled by it $=18\times 4=72$ units
Water filled by pipe B = 96 -72 = 24 units.
So the time for which it was open $=\dfrac{24}{3}=8$minutes.
Hence the pipe B must be closed after 8 minutes for the tank to be filled in 18 minutes.
Therefore, the correct option is (d).

Note: Alternatively, the question is solved as follows. The rate of A and B to fill tank alone are ${{\left( \dfrac{1}{24} \right)}^{th}}$ part and ${{\left( \dfrac{1}{32} \right)}^{th}}$ part of tank per minute respectively. Both are opened together and the tank must be full in 18min. Only B is closed before time, so A runs for full 18mins.
Let's assume B runs for x minutes.
$18\times \dfrac{1}{24}+x\times \dfrac{1}{32}=1$
$\dfrac{18\times 4+x\times 3}{8\times 3\times 4}=1$
$x=\dfrac{(8\times 3\times 4)-(18\times 4)}{3}=8$
Hence the pipe B must be closed after 8 minutes.