# Ex.3.1 Q5 Data Handling - NCERT Maths Class 7

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## Question

Following table shows the points each player scored in four games: -

 Player Game 1 Game 2 Game 3 Game 4 $$A$$ $$14$$ $$16$$ $$10$$ $$10$$ $$B$$ $$10$$ $$8$$ $$6$$ $$4$$ C $$8$$ $$11$$ Didn’t play $$13$$

Now answer the following questions: -

(i) Find the mean to detetextine $$A’s$$ average number of points scored per game.

(ii) To find the mean number of points per game for $$C,$$ would you divide the points by $$3$$ or $$4$$? Why?

(iii) $$B$$ played in all the four games. How would you find the mean?

(iv) Who is the best performer?

Video Solution
Data Handling
Ex 3.1 | Question 5

## Text Solution

Reasoning:

What is known?

Point of each player scored in four games.

What is unknown?

Average score, Mean

Reasoning:\begin{align}\text{Mean}=\frac{\text{Sum of score }}{\text{Number of games}}\end{align}

Steps:

Total number of games played by $$A = 4$$

Scores obtained by $$A = 14, 16, 10, 10$$

Mean of player $$A$$

\begin{align}&= \frac{{14 + 16 + 10 + 10}}{4}\\ &= \frac{{50}}{4}\\ &= 12.5\end{align}

ii) We should divide total points by $$3$$ because player $$C$$ played only three games.

iii) Total number of games played by $$B = 4$$

Scores obtained by $$B = 0, 8, 6, 4$$

Mean of player $$B$$

\begin{align}&= \frac{{{\text{Sum of scores by B}}}}{{{\text{No}}{\text{. of games played by B}}}}\\ &= \frac{{0 + 8 + 6 + 4}}{4}\\ &= \frac{{18}}{4}\\ &= 4.5\end{align}

iv) To find the best performer,

We should find the mean of all players.

Mean of player $$A = 12.5$$

Mean of player $$B = 4.5$$

Mean of player $$C$$

\begin{align} &= \frac{{{\text{sum of scores by C}}}}{{{\text{No}}{\text{. of games played by C}}}}\\ &=\frac{{8 + 11 + 13}}{3}\\ &= \frac{{32}}{3}\\ &= 10.67\end{align}

Therefore, on comparing means of all players, $$A$$ is the best performer.

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