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A large tanker can be filled by two pipes \[A\] and \[B\] in 60 minutes and 40 minutes, respectively. How many minutes will it take to fill the empty tanker if only \[B\] is used in the first-half of the time and \[A\] and \[B\] are both used in the second-half of the time?
A. 15
B. 20
C. \[27.5\]
D. 30

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Last updated date: 25th Jun 2024
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Answer
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Hint: Here we need to find the time taken to fill the empty tanker. For that, we will first find the part of the tanker filled by the first pipe in one minute and the part filled by the second pipe in one minute. Then we will find the part filled by both the pipe together in one minute. Then we will assume the total time taken to fill the tanker by both the pipes to be any variable. We will then form an equation according to the given information. After solving the equation, we will get the required answer.

Complete step by step solution:
It is given that pipe \[A\] takes 60 minutes to fill the tanker and pipe \[B\] takes 40 minutes to fill the tanker.
Therefore, part filled by pipe \[A\] in 1 minute \[ = \dfrac{1}{{60}}\]
Part filled by pipe \[B\] in 1 minute \[ = \dfrac{1}{{40}}\] …………… \[\left( 1 \right)\]
Part filled by both pipes \[A\] and \[B\] in 1 minute \[ = \dfrac{1}{{60}} + \dfrac{1}{{40}}\]
On adding these fractions, we get
Part filled by both pipes \[A\] and \[B\] in 1 minute \[ = \dfrac{1}{{60}} + \dfrac{1}{{40}} = \dfrac{1}{{24}}\] ……………. \[\left( 2 \right)\]
Let the total time taken by the pipes to fill the tanker be \[x\] minutes.
As it is given that the pipe \[B\] is used in the first-half of the time and \[A\] and \[B\] are both used in the second-half of the time.
Therefore, we can say that to fill the tanker from empty state, \[B\] is used for \[\dfrac{x}{2}\] minutes and \[A\] and \[B\] together is used for the rest \[\dfrac{x}{2}\] minutes.
Hence, the total part of the tanker can be written as \[\dfrac{x}{2} \times \dfrac{1}{{40}} + \dfrac{x}{2} \times \dfrac{1}{{24}}\]
Also we can write it as
\[ \Rightarrow \dfrac{x}{2} \times \dfrac{1}{{40}} + \dfrac{x}{2} \times \dfrac{1}{{24}} = 1\]
On multiplying the terms, we get
\[ \Rightarrow \dfrac{x}{{80}} + \dfrac{x}{{48}} = 1\]
On adding these fractions, we get
\[\begin{array}{l} \Rightarrow \dfrac{{3x + 5x}}{{240}} = 1\\ \Rightarrow \dfrac{{8x}}{{240}} = 1\end{array}\]
On cross multiplying the terms, we get
\[ \Rightarrow 8x = 240\]
Now, we will divide both sides by 5.
\[\begin{array}{l} \Rightarrow \dfrac{{8x}}{8} = \dfrac{{240}}{8}\\ \Rightarrow x = 30\end{array}\]
Hence, the total time taken to fill the tanker is equal to 30 minutes.

Hence, the correct option is option D.

Note:
Here we have used the cross multiplication to multiply the terms. Cross multiplication means the multiplication of the numerator of the first fraction with the denominator of the second fraction and the multiplication of the numerator of the second fraction with the denominator of the first fraction. We need to keep in mind that in the first half only pipe B is used and not both, so we will not multiply the part filled by A to the time. Also, in the second half, both the pipes are used so we will multiply the parts filled by them to half the time.