Question

A cistern can be filled by two pipes in 30 and 40 min respectively. Both the pipes are opened at once, When the first pipe be turned off so that the tank may be just filled in 18 min?
A. 12.5 min
B. 13.5 min
C. 14.5 min
D. 16.5 min

Answer 1

Hint: To solve this problem, we should know the concept of volume filled vs minutes problems. A pipe can fill the tank in x minutes means that the pipe can fill $\dfrac{1}{x}$ portion of the tank in 1 minute. By using this analogy, we can get the portion of the tank filled by the two pipes in a minute as $\dfrac{1}{30},\dfrac{1}{40}$ respectively. By assuming that the first pipe A is turned on for x min and the second pipe is turned on for 18 min, the tank is filled completely. We can write that the sum of portions filled by the first pipe in x min and second pipe in 18 min is equal to 1. The equation is $\dfrac{x}{30}+\dfrac{18}{40}=1$. From this, we can get the value of x which is the required answer.

Complete step-by-step solution:
In the question, it is given that the first pipe fills the tank in 30 min and the second pipe fills the tank in 40 min.
We can write that a pipe filling the tank in x minutes means that the pipe can fill $\dfrac{1}{x}$ portion of the tank in 1 minute.
Using the above statement, we can find the portion of the tank filled by the pipes individually in 1 min.
Portion of the tank filled by the first pipe when x = 30 min is = $\dfrac{1}{30}$
Portion of the tank filled by the second pipe when x = 40 min is = $\dfrac{1}{40}$
Let us consider a pipe that fills $\dfrac{1}{x}$ portion of the tank in 1 minute is open for y minutes. Then the portion of the tank filled by it is given by $\dfrac{y}{x}$.
Let us consider that the first pipe is open for x minutes. The second pipe is open for the whole 18 minutes. The individual portions filled by the pipes, from the above relation, is given by
The portion of the tank filled by the first pipe in x minutes is $\dfrac{x}{30}$
The portion of the tank filled by the second pipe in 18 minutes is $\dfrac{18}{40}$
We can infer from the question that at the end of the 18 minutes, the tank is filled completely. From this, we can write that the sum of the portions filled by the two pipes will be equal to 1. Mathematically, it is written as
$\dfrac{x}{30}+\dfrac{18}{40}=1$
Subtracting $\dfrac{18}{40}$on both sides, we get
$\begin{align}
  & \dfrac{x}{30}=1-\dfrac{18}{40} \\
 & \dfrac{x}{30}=\dfrac{40-18}{40}=\dfrac{22}{40} \\
\end{align}$
Multiplying by 30 on both sides, we get
$x=\dfrac{22\times 30}{40}=\dfrac{22\times 3}{4}=5.5\times 3=16.5$
$\therefore $ The first pipe should be open for 16.5 min for the tank to be filled completely by the end of 18 min. The answer is option-D

Note: The alternate way to do the question is to directly subtract the portion of the tank filled by the second pipe from 1 to get the portion filled by the first pipe in x min. This gives us $\dfrac{x}{30}=1-\dfrac{18}{40}$. Some students forget the fact that the second pipe is open even after the close of the first pipe and end up getting the wrong answer. Understanding the question properly will nullify such mistakes.