
There are two examination halls A and B. If 12 pupils are sent from A to B, the number of pupils in each room becomes the same. If 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B. Find the number of pupils in each room.
Answer
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Hint: We solve this question by first assuming the number of pupils in room A and room B as $x$ and $y$. Then from the given information we can find the number of pupils in each room in terms of $x$ and $y$ after the changes have been made from one room to another and we can make two equations with $x$ and $y$. Then by solving those two equations we can find the values of them, that is the number of pupils in each room.
Complete step by step answer:
We are given that there are two examination halls A and B
Let us assume that there are x pupils in room A and y pupils in room B.
We are given that if 12 pupils are sent from A to B, then the number of pupils in each room becomes the same.
So, if 12 pupils are sent from the room A to room B, then the number of pupils in each room becomes,
Number of pupils in room A = $x-12$
Number of pupils in room B = $y+12$
As, we are given that the number of pupils in both rooms are equal in this case, let us equate them. Then we get,
$\begin{align}
& \Rightarrow x-12=y+12 \\
& \Rightarrow x=y+24..........\left( 1 \right) \\
\end{align}$
We are also given that if 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B.
So, if 11 pupils are sent from room B to room A, then the number of pupils in each room becomes,
Number of pupils in room B = $y-11$
Number of pupils in room A = $x+11$
As, we are given that number of pupils in room A is double the number in room B, we can write it as,
$\begin{align}
& \Rightarrow x+11=2\left( y-11 \right) \\
& \Rightarrow x+11=2y-22 \\
& \Rightarrow x=2y-33...........\left( 2 \right) \\
\end{align}$
So, from equations (1) and (2), let us equate the values of $x$. Then we get,
$\begin{align}
& \Rightarrow y+24=2y-33 \\
& \Rightarrow 2y-y=24+33 \\
& \Rightarrow y=57 \\
\end{align}$
So, we get that the number of pupils in room B are 57.
Now, let us substitute this value in the equation (1). Then we get,
$\begin{align}
& \Rightarrow x=57+24 \\
& \Rightarrow x=81 \\
\end{align}$
So, the number of pupils in room A are 81.
Hence the answer is 81 and 57.
Note: The common mistake that one makes in this question is one might wrongly interpret the question and write the equation (2) above wrongly. They might interpret the case, if 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B, after the first case, that is one might write the equation (2) using the values,
Number of pupils in room B = $\left( y+12 \right)-11=y+1$
Number of pupils in room A = $\left( x-12 \right)+11=x-1$
But that is not the case. It is that, if 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B, in general case when the number of pupils in room A and B are $x$ and $y$.
Complete step by step answer:
We are given that there are two examination halls A and B
Let us assume that there are x pupils in room A and y pupils in room B.
We are given that if 12 pupils are sent from A to B, then the number of pupils in each room becomes the same.
So, if 12 pupils are sent from the room A to room B, then the number of pupils in each room becomes,
Number of pupils in room A = $x-12$
Number of pupils in room B = $y+12$
As, we are given that the number of pupils in both rooms are equal in this case, let us equate them. Then we get,
$\begin{align}
& \Rightarrow x-12=y+12 \\
& \Rightarrow x=y+24..........\left( 1 \right) \\
\end{align}$
We are also given that if 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B.
So, if 11 pupils are sent from room B to room A, then the number of pupils in each room becomes,
Number of pupils in room B = $y-11$
Number of pupils in room A = $x+11$
As, we are given that number of pupils in room A is double the number in room B, we can write it as,
$\begin{align}
& \Rightarrow x+11=2\left( y-11 \right) \\
& \Rightarrow x+11=2y-22 \\
& \Rightarrow x=2y-33...........\left( 2 \right) \\
\end{align}$
So, from equations (1) and (2), let us equate the values of $x$. Then we get,
$\begin{align}
& \Rightarrow y+24=2y-33 \\
& \Rightarrow 2y-y=24+33 \\
& \Rightarrow y=57 \\
\end{align}$
So, we get that the number of pupils in room B are 57.
Now, let us substitute this value in the equation (1). Then we get,
$\begin{align}
& \Rightarrow x=57+24 \\
& \Rightarrow x=81 \\
\end{align}$
So, the number of pupils in room A are 81.
Hence the answer is 81 and 57.
Note: The common mistake that one makes in this question is one might wrongly interpret the question and write the equation (2) above wrongly. They might interpret the case, if 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B, after the first case, that is one might write the equation (2) using the values,
Number of pupils in room B = $\left( y+12 \right)-11=y+1$
Number of pupils in room A = $\left( x-12 \right)+11=x-1$
But that is not the case. It is that, if 11 pupils are sent from room B to room A, then the number of pupils in room A is double their number in B, in general case when the number of pupils in room A and B are $x$ and $y$.
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