In a squash competition, each player played once with everyone else.

Question

In a squash competition, each player played once with everyone else.
Gordon played with 4 times as many boys as girls.
Jayanthi played with 5 times as many boys as girls.
How many more boys than girls were there?

Answer 1

The answer is 19. The explanation is

b_G = number of boys Gordon plays       b_J = number of boys Jayanthi plays

g_G = number of girls Gordon plays        g_J = number of girls Jayanthi plays

There are a total of _nC₂ games played. Each player plays n-1 games.

b_G+g_G = n-1

b_G = 4g_G

5g_G=n-1 and so g_G=(n-1)/5 and b_G = 4(n-1)/5

 

b_J+g_J = n-1

b_J=5g_J

6g_J = n-1 and so g_J=(n-1)/6 and b_J = 5(n-1)/6

 The number of boys minus the number of girls is

(b_G+1)-g_G = (4(n-1)/5+1)-(n-1)/5 = 3(n-1)/5 +1

b_J-(g_J+1) = 5(n-1)/6-((n-1)/6 +1)=4(n-1)/6 -1

That these last are equal gives that

3(n-1)/5 +1 = 4(n-1)/6 -1 or that (n-1)/15=2 or n=31

Now:3(n-1)/5 +1=90/5+1=19 or 4(n-1)/6 -1 = 120/6-1 = 19