3 women and a few men participated in a chess tournament. Each player played two matches with each of the other players. If the number of matches that men played among themselves is 78 more than those they played with the women, how many more men than women participated in the tournament?
Pick one.
A) 14  B) 13  C) 11  D) 10  E) 8
let the number of men=m..   so 3 women and m men will play= 2*3= 
<gadget id="calculator">2*3</gadget>
<output>6</output>
6m games   m will play each other in mC2 way so number of games= 2*mC2= m(m-1)   Difference in two = m(m-1)-6m=78   m^2-7m-78=0   m^2-13m+6m-78=0   (m-13)(m+6)=0   so m = 13 or -6   cannot be -ive so m=13   difference in number of women and men= 13-3=10.. Imp that we do not choose 13 as the answer   ANS 10
<result>D</result>
3 women and a few men participated in a chess tournament. Each player played two matches with each of the other players. If the number of matches that men played among themselves is 78 more than those they played with the women, how many more men than women participated in the tournament?