Table Tournament Results: Find Draws and Losses for Team D, Match B vs C

  • Thread starter Thread starter chawki
  • Start date Start date
  • Tags Tags
    Table
AI Thread Summary
Team D has no draws and three losses in the tournament. The results indicate that Team B won all its matches, including a match against Team A, which lost to B by a score of 0-1. To determine the outcomes of other matches, a process of elimination is recommended, starting with Team A's known record. The discussion emphasizes the importance of analyzing goals-for and goals-against to deduce match results. Overall, the conversation focuses on solving the tournament results through logical reasoning and elimination.
chawki
Messages
504
Reaction score
0

Homework Statement


Each team has played once against every other team in a tournament. The results were collected in the partly incomplete table below. (see attachments)

Homework Equations


a) Find the number of draws and the number of losses for team D.
b) What was the result of the match B against C?

The Attempt at a Solution


a)
drawn 0, lost 3 ?
 

Attachments

  • table.JPEG
    table.JPEG
    15.8 KB · Views: 509
Physics news on Phys.org
chawki said:
a)
drawn 0, lost 3 ?
Yes. Now what about part b?
 
this is really hard thinking...i need a good method to deal with these kind of questions :mad:
 
Nah, not really hard thinking, just process of elimination. :wink: Start with team A. It has a record of 1-1-1, and goals-for/goals against (GF-GA) of 1-1. Since B won all three games, A must have lost to B. Since A's GF-GA is only 1-1, it must have lost to B by a score of 0-1. Continuing from here, you should be able to figure out who beat whom, and by what scores.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top