
Ten persons numbered 1, 2, 3…………10 plays a chess tournament, each player playing against every other player in exactly one game. It is known that no game ends in a draw. If $w_1, w_2, ......., w_{10}$ are the number of games won by players 1, 2, 3…………10 respectively, and ${l_1},{l_2},{l_{3,}}...............{l_{10}}$ are the number of games lost by the players 1, 2, 3,………………… 10 respectively then
$
(a){\text{ }}\sum {{w_i} = \sum {{l_i} = 45} } \\
(b){\text{ }}{{\text{w}}_i} + {l_i} = 9 \\
(c){\text{ }}\sum {{w_i}^2 = 81 + \sum {{l_i}^2} } \\
(d){\text{ }}\sum {{w_i}^2 = \sum {{l_i}^2} } \\
$
Answer
123.3k+ views
Hint – In this question first compute the total number of games played by each of these 10 players and it will be (10-1=9). Now each player plays against one another and a match is held between two so find the total number of games. Since no match can end in a draw, the sum of winning and losing for a particular ${i^{th}}$ player will be equal to the total game that can be played by a single player. This concept will help getting the right option.
Complete step-by-step answer:
There are a total of 10 players.
So each player will play (10 – 1) = 9 games.
Now there are 10 players and a game of chess is played between two candidates and it is given that each player will play against every other player so the total number of games = ${}^{10}{C_2} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}} = 45$
Now $w_1, w_2, ......., w_{10}$ and $l_1, l_2, ....., l_{10}$ are the number of games won and lost by players 1, 2, ...., 10 respectively.
Now each player plays 9 games without any draws.
Hence,
${w_i} + {l_i} = 9$....................... (1) (i.e. it’s the sum of winning and losing a game for a particular ith player is equal to 9).
Now the total number of wins = total number of losses.
$ \Rightarrow \sum {{w_i} = \sum {{l_i}} } $....................... (2)
Now apply summation on both sides in equation (1) we have,
$ \Rightarrow \sum {\left( {{w_i} + {l_i}} \right)} = \sum 9 $
$ \Rightarrow \sum {{w_i} + \sum {{l_i} = 9\left( {10} \right) = 90} } $ , as there are total 10 players therefore, \[\sum 9 = 9\left( {10} \right) = 90\]
Now from equation (2) we have,
$ \Rightarrow \sum {{w_i} + \sum {{w_i} = 90} } $
$ \Rightarrow \sum {{w_i}} = \dfrac{{90}}{2} = 45$
Hence,
$\sum {{w_i} = \sum {{l_i} = 45} } $......................... (3)
Now from equation (1) we have,
$ \Rightarrow {w_i} = 9 - {l_i}$
Now squaring on both sides we have,
$ \Rightarrow {\left( {{w_i}} \right)^2} = {\left( {9 - {l_i}} \right)^2}$
Now expand the square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so we have,
$ \Rightarrow {\left( {{w_i}} \right)^2} = {9^2} + {\left( {{l_i}} \right)^2} - 18{l_i}$
Now take summation on both sides we have,
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {\left( {{9^2} + {{\left( {{l_i}} \right)}^2} - 18{l_i}} \right)} $
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {81} - 18\sum {\left( {{l_i}} \right)} + \sum {{{\left( {{l_i}} \right)}^2}} $
Now there are total 10 players
Therefore $\sum {81} = 81\left( {10} \right) = 810$
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 18\sum {\left( {{l_i}} \right)} + \sum {{{\left( {{l_i}} \right)}^2}} $
Now from equation (3) we have,
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 18\left( {45} \right) + \sum {{{\left( {{l_i}} \right)}^2}} $
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 810 + \sum {{{\left( {{l_i}} \right)}^2}} $
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {{{\left( {{l_i}} \right)}^2}} $
So this is the required answer.
Hence option (A), (B) and (D) are correct.
Note – Now as the matches proceed the total number of winning has to be equal to the total number of losing , as when a match is played between two players there can only be two outcomes either person one wins and person 2 loses or person 1 loses and person 2 wins. Since the drawn cases need not to be taken into consideration thus under this circumstance only this equation holds true, and it’s the trick point of this problem.
Complete step-by-step answer:
There are a total of 10 players.
So each player will play (10 – 1) = 9 games.
Now there are 10 players and a game of chess is played between two candidates and it is given that each player will play against every other player so the total number of games = ${}^{10}{C_2} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}} = 45$
Now $w_1, w_2, ......., w_{10}$ and $l_1, l_2, ....., l_{10}$ are the number of games won and lost by players 1, 2, ...., 10 respectively.
Now each player plays 9 games without any draws.
Hence,
${w_i} + {l_i} = 9$....................... (1) (i.e. it’s the sum of winning and losing a game for a particular ith player is equal to 9).
Now the total number of wins = total number of losses.
$ \Rightarrow \sum {{w_i} = \sum {{l_i}} } $....................... (2)
Now apply summation on both sides in equation (1) we have,
$ \Rightarrow \sum {\left( {{w_i} + {l_i}} \right)} = \sum 9 $
$ \Rightarrow \sum {{w_i} + \sum {{l_i} = 9\left( {10} \right) = 90} } $ , as there are total 10 players therefore, \[\sum 9 = 9\left( {10} \right) = 90\]
Now from equation (2) we have,
$ \Rightarrow \sum {{w_i} + \sum {{w_i} = 90} } $
$ \Rightarrow \sum {{w_i}} = \dfrac{{90}}{2} = 45$
Hence,
$\sum {{w_i} = \sum {{l_i} = 45} } $......................... (3)
Now from equation (1) we have,
$ \Rightarrow {w_i} = 9 - {l_i}$
Now squaring on both sides we have,
$ \Rightarrow {\left( {{w_i}} \right)^2} = {\left( {9 - {l_i}} \right)^2}$
Now expand the square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so we have,
$ \Rightarrow {\left( {{w_i}} \right)^2} = {9^2} + {\left( {{l_i}} \right)^2} - 18{l_i}$
Now take summation on both sides we have,
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {\left( {{9^2} + {{\left( {{l_i}} \right)}^2} - 18{l_i}} \right)} $
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {81} - 18\sum {\left( {{l_i}} \right)} + \sum {{{\left( {{l_i}} \right)}^2}} $
Now there are total 10 players
Therefore $\sum {81} = 81\left( {10} \right) = 810$
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 18\sum {\left( {{l_i}} \right)} + \sum {{{\left( {{l_i}} \right)}^2}} $
Now from equation (3) we have,
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 18\left( {45} \right) + \sum {{{\left( {{l_i}} \right)}^2}} $
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = 810 - 810 + \sum {{{\left( {{l_i}} \right)}^2}} $
$ \Rightarrow \sum {{{\left( {{w_i}} \right)}^2}} = \sum {{{\left( {{l_i}} \right)}^2}} $
So this is the required answer.
Hence option (A), (B) and (D) are correct.
Note – Now as the matches proceed the total number of winning has to be equal to the total number of losing , as when a match is played between two players there can only be two outcomes either person one wins and person 2 loses or person 1 loses and person 2 wins. Since the drawn cases need not to be taken into consideration thus under this circumstance only this equation holds true, and it’s the trick point of this problem.
Recently Updated Pages
The real roots of the equation x23 + x13 2 0 are A class 11 maths JEE_Main

Find the reminder when 798 is divided by 5 class 11 maths JEE_Main

Let A and B be two sets containing 2 elements and 4 class 11 maths JEE_Main

A ray of light moving parallel to the xaxis gets reflected class 11 maths JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

If there are 25 railway stations on a railway line class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Physics Average Value and RMS Value JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 13 Statistics
