GATE2011 MN: GA-64

Question

Four archers P, Q, R, and S try to hit a bull’s eye during a tournament consisting of seven rounds. As illustrated in the figure below, a player receives $10$ points for hitting the bull's eye, $5$ points for hitting within the inner circle and $1$ point for hitting within the outer circle.

The final scores received by the players during the tournament are listed in the table below.

$$\begin{array}{|c|c|c|c|c|} \hline \textbf{Round} & \textbf{P} & \textbf{Q} & \textbf{R} & \textbf{S} \\  \hline \textbf{1} & 1 & 5 & 1 & 10 \\ \hline \textbf{2} & 5 & 10 & 10 & 1 \\ \hline  \textbf{3} & 1 & 1 & 1 & 5 \\ \hline   \textbf{4} &  10 & 10 & 1 & 1  \\ \hline \textbf{5} & 1 & 5 & 5 & 10 \\  \hline \textbf{6} &10 & 5 & 1 & 1 \\ \hline \textbf{7} & 5 & 10 & 1 & 1  \\ \hline \end{array}$$

The most accurate and the most consistent players during the tournament are respectively

• A.  P and S
• B. Q and R
• C. Q and Q
• D. R and Q

Comments

I think C) will be P) and Q)

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Yes, it appears to be 'P' and 'Q'

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Option $(C)$ is saying that $Q$ is most accurate player as well as most consistent player. There is nothing wrong in option $(C).$

https://www.iitr.ac.in/gate/papers/GATE2011/GATE2011_MN.pdf?download_file=papers/GATE2011/GATE2011_MN.pdf

Answer 1

$\rightarrow$ Here the most accurate player will be the one who makes the highest score.

$\rightarrow$ $Q$ has the maximum score of $46$ points among all other players. So $Q$ is the most accurate player.

$\rightarrow$ Consistency for a series of data should mean their Standard Deviation is minimum. Standard deviation is given by the square root of the sum of the squares of the individual deviations from mean divided by the number of items. Here, mean values for $P,Q,R$ and $S$ are $\frac{33}{7}=4.71,\frac{46}{7}=6.57,\frac{20}{7}=2.85$ and $\frac{29}{7}=4.14$ respectively. 

• For $P$ the standard deviation is $\sqrt{\frac{3 \times 3.71^2 + 2 \times 0.29^2 + 2\times 5.29^2}{7}} = 3.73$
• For $Q$ the standard deviation is $\sqrt{\frac{1 \times 5.57^2 + 3 \times 1.57^2 + 3\times 3.43^2}{7}}=3.24$
• For $R$ the standard deviation is $\sqrt{\frac{5 \times 1.85^2 + 1 \times 2.15^2 + 1\times 7.15^2}{7}}=3.22$
• For $S$ the standard deviation is $\sqrt{\frac{4 \times 3.14^2 + 1 \times 0.86^2 + 2\times 5.86^2}{7}}=3.93$

$\rightarrow$ The most consistent player will be the one who has the minimum standard deviation.

$\rightarrow$ $R$ has the minimum standard deviation and is the most consistent.

$\therefore$ Option $B$ is the right answer.

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NOTE:- For calculating Standard deviation https://www.mathsisfun.com/data/standard-deviation-formulas.html.

Comments

Can we reason as follows? -: Q has the most 10s, so is the most accurate player, and R has the largest number of same-point rounds (1 point), so R is the most consistent (consistently scoring 1 point).

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Sachin has the maximum number of hundreds in tests but Bradman is the best batsman 😊

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So true… 😊

Answer 2

From the given table, we can formulate the below table.

Player	No of 1’s scored	No of 5’s scored	No of 10’s scored
P	3	2	2
Q	1	3	3
R	5	1	1
S	4	1	2

 

From the above table, it is clear that the player Q got the 10 and 5 points, the greatest number of times compared to others. Therefore, the player Q is most accurate.

The player R hit outside circle 5 times out of 7 rounds. Therefore, the plater R is most consistent.

Note: Consistency does not confirm the accuracy..