Answer
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Hint: In order to find the number of persons in \[II\] group, we must be considering the variable values for both the groups. Then calculating the mean of both the groups as the total persons of both the groups is \[160\]. We must be equating this mean equation to \[Rs.430\]. And then on solving, we obtain the required answer.
Complete step-by-step solution:
Now let us learn about meaning. The mean of data is nothing but the average of the observations which means as the central value of the data. It is generally calculated by the sum of observations divided by the number of observations. Mean is a part of measures of central tendency. There are three types of mean. They are: Arithmetic mean, geometric mean and harmonic mean.
Now let us find the number of persons in group \[II\].
Let us consider the number of persons in group \[I\] is \[x\]
Let us consider the number of persons in group \[II\] as \[y\].
We are given that \[x+y=160\] \[\to \left( 1 \right)\]
So now let us calculate the income of both the groups.
Total income of group \[I\]is \[400x\]
Total income of group \[II\] is \[480y\].
We are given that mean income of both the groups is \[430\]
We can express this as
\[\Rightarrow \dfrac{400x+480y}{x+y}=430\]
Upon solving this, we get
\[\begin{align}
& \Rightarrow 400x+480y=430x+430y \\
& \Rightarrow 50y=30x \\
& \Rightarrow 50y-30x=0 \\
\end{align}\]
\[50y-30x=0\to \left( 2 \right)\]
Here we multiply the equation 1 with 30 and add to equation 2, so we get $\{30x+30y=4800\}+\{50y-30x=0\}$
$\Rightarrow 50y+30y=4800$
$\Rightarrow y=\dfrac{4800}{80}=60$
Now we substitute x=60 in equation 1, so we get
$60+y=160$ $\Rightarrow y=100$
Thus Upon solving the first and second equations, we get
\[\begin{align}
& x=100 \\
& y=60 \\
\end{align}\]
\[\therefore \] Number of persons in group \[II\] is \[60\]
Hence, option C is the correct answer.
Note: We must always calculate the values accurately for obtaining the accurate answers. In the problem above, we have considered variables for the unknown values and we must note that this method is applicable to all such problems. We can use mean in finding our marks, or calculating the total goods sold per day and many more.
Complete step-by-step solution:
Now let us learn about meaning. The mean of data is nothing but the average of the observations which means as the central value of the data. It is generally calculated by the sum of observations divided by the number of observations. Mean is a part of measures of central tendency. There are three types of mean. They are: Arithmetic mean, geometric mean and harmonic mean.
Now let us find the number of persons in group \[II\].
Let us consider the number of persons in group \[I\] is \[x\]
Let us consider the number of persons in group \[II\] as \[y\].
We are given that \[x+y=160\] \[\to \left( 1 \right)\]
So now let us calculate the income of both the groups.
Total income of group \[I\]is \[400x\]
Total income of group \[II\] is \[480y\].
We are given that mean income of both the groups is \[430\]
We can express this as
\[\Rightarrow \dfrac{400x+480y}{x+y}=430\]
Upon solving this, we get
\[\begin{align}
& \Rightarrow 400x+480y=430x+430y \\
& \Rightarrow 50y=30x \\
& \Rightarrow 50y-30x=0 \\
\end{align}\]
\[50y-30x=0\to \left( 2 \right)\]
Here we multiply the equation 1 with 30 and add to equation 2, so we get $\{30x+30y=4800\}+\{50y-30x=0\}$
$\Rightarrow 50y+30y=4800$
$\Rightarrow y=\dfrac{4800}{80}=60$
Now we substitute x=60 in equation 1, so we get
$60+y=160$ $\Rightarrow y=100$
Thus Upon solving the first and second equations, we get
\[\begin{align}
& x=100 \\
& y=60 \\
\end{align}\]
\[\therefore \] Number of persons in group \[II\] is \[60\]
Hence, option C is the correct answer.
Note: We must always calculate the values accurately for obtaining the accurate answers. In the problem above, we have considered variables for the unknown values and we must note that this method is applicable to all such problems. We can use mean in finding our marks, or calculating the total goods sold per day and many more.
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