Question

The average score of $10$ football players is $3$ in ${1^{{\text{st}}}}$ off, $4$ in ${2^{{\text{nd}}}}$ off and $5$ in ${3^{{\text{rd}}}}$ off. Find the average score of these $6$ football players in $5$ matches.
$\left( {\text{A}} \right)\;{\text{3}}$
$\left( {\text{B}} \right)\;{\text{2}}$
$\left( {\text{C}} \right)\;{\text{4}}$
$\left( {\text{D}} \right)\;{\text{1}}$

Answer 1

Hint: Here, we have to find the average score of the $6$ football players in $5$ matches.
First, we need to find the actual score of $10$ football players.
Then, we have to find the actual and average score of $6$ football players in $5$ matches
Finally, we will get the required answer.

Formula used: ${\text{Average score}}{\text{ = }}\dfrac{{{\text{Actual score}}}}{{{\text{Number of players}}}}$
${\text{Actual score}}{\text{ = Average score }} \times {\text{ Number of players}}$

Complete step-by-step solution:
It is given that the average score of $10$ football players is $3$ in off, $4$ in ${2^{{\text{nd}}}}$ off and $5$ in ${3^{{\text{rd}}}}$ off.
\[{\text{The average score of 10 football player in }}{{\text{1}}^{{\text{st}}}}{\text{off = }}\dfrac{{{\text{The actual score of 10 football players in }}{{\text{1}}^{{\text{st}}}}{\text{off}}}}{{{\text{10}}}}\] Now, putting the values and we get,
\[3 = \dfrac{{{\text{The actual score of 10 football players in }}{{\text{1}}^{{\text{st}}}}{\text{off}}}}{{{\text{10}}}}\]
Taking cross multiplication we get,
The actual score of $10$ football players in ${1^{{\text{st}}}}$ off = \[10 \times 3 = 30\]
Similarly, we have to find out the average score of \[{\text{10}}\] football player in \[{{\text{2}}^{{\text{nd}}}}\]off
\[{\text{The average score of 10 football player in }}{{\text{2}}^{{\text{nd}}}}{\text{off = }}\dfrac{{{\text{The actual score of 10 football players in }}{{\text{2}}^{{\text{nd}}}}{\text{off}}}}{{{\text{10}}}}\]Putting the values and we get,
\[4 = \dfrac{{{\text{The actual score of 10 football players in }}{{\text{2}}^{nd}}{\text{off}}}}{{{\text{10}}}}\]
Taking cross multiplication we get,
The actual score of $10$ football players in ${2^{{\text{nd}}}}$ off =\[10 \times 4 = 40\]
Also, we have to find out the average score of \[{\text{10}}\] football player in \[{3^{rd}}\] off
\[{\text{The average score of 10 football player in }}{{\text{3}}^{{\text{rd}}}}{\text{off = }}\dfrac{{{\text{The actual score of 10 football players in }}{{\text{3}}^{{\text{rd}}}}{\text{off}}}}{{{\text{10}}}}\]
Putting the values and we get,
\[5 = \dfrac{{{\text{The actual score of 10 football players in }}{{\text{3}}^{rd}}{\text{off}}}}{{{\text{10}}}}\]
Taking cross multiplication we get,
The actual score of $10$ football players in ${3^{{\text{rd}}}}$ off =\[10 \times 5 = 50\]
From the question it is understandable that the football match consists of 3 off, then the total score is the sum of the actual score of all the $3$ off.
\[{\text{Total score}} = 30 + 40 + 50\]
On adding we get,
\[{\text{Total score = 120}}\]
Now, actual score of $10$ football players = $120$
 Here we have to consider the total score is $120$ for $6$ players.
The average score of 6 football players =\[\dfrac{{120}}{6}\]
On dividing the terms and we get
\[ \Rightarrow 20\]
Now, according to the question, we need to find the average score of $6$ football players in $5$ matches.
As $20$ is the actual score of $6$ football players,
Now we use the same formula to find that:
\[{\text{The average score of 6 football player in 5 matches}} = \dfrac{{{\text{The actual score of 6 football players}}}}{{{\text{No of matches}}}}\]
The average score of \[{\text{6}}\] football in \[{\text{5}}\] matches \[ = \dfrac{{20}}{5}\]
On dividing we get,
$ \Rightarrow 4$
Hence the average score of \[{\text{6}}\] football in \[{\text{5}}\] matches is \[4\]

$\therefore $The correct option is \[\left( {\text{C}} \right)\]

Note: In this question, students may go wrong in the last step i.e., finding the average score of \[6\] football players in matches. It is because at first we used a number of players to find the average and taking that in mind you may use the same method.
First is calculated on the basis of the number of players as there is just one game and average is found in and among that one particular game. But the latter is different; if scores of \[5\] games are given we should find the average on the basis of the number of matches.