In a village ,50% of the students play football, 60% of the student play basketball,40% of the student play cricket. 20% play both football and basketball ,20% play both football and cricket,25% play both cricket and baseketball.5% of the student play all the three games and 100 students play none of the games.
1)How many students are there in the college?
2)How many students play exactly two games?
3)How many students play neither cricket nor football?
4)How many students play exactly one game?

• we all know n(aUbUc)=n(a)+n(b)+n(c)-n(aUb)-n(bUc)-n(cUa)+n(a intersection b intersection c)=5/10+6/10+4/10-2/10-2/10-2.5/10+.5/10=9/10
  
  the remaining 1/10 portion of people dont play anythng.
  if 1/10=100,then 1=1000.
  
  so there are total 1000 students.
  
  5
• 13 years agoHelpfull: Yes(15) No(3)
• the student play all three games=5%
  the student play only both football n bsketball=(20-5)%=15%
  the student play only both football n cricket=(20-5)%=15%
  the student play only both basketball n cricket=(25-5)=20%
  the student play only football=(50-15-15-5)%=15%
  the student play only basketball=(60-15-20-5)%=20%
  the student play only cricket=(40-15-20-5)%=0%
  so sdtudent play games=(0+20+15+15+15+20+5)%=90%
  so student does not play games=(100-90)%=10%
  as 10%=100 student den 100%=1000 student
  1)there are 1000 student in the college
  
  the student play exactly two games=(15+15+20)%=50%
  50%of1000=500
  2)500 student play exact two games
  
  no of students play neither football nor cricket mens no of student play only bsketbll
  so 20%of1000=200
  3)200student play neither cricket nor footbll
  
  students ply only one game=(15+20+0)%=35%
  35%of1000=350
  4)350students ply only one game
• 13 years agoHelpfull: Yes(7) No(1)
• Let total no of student in college = x
  n(f)=x/2
  n(b)=3x/5
  n(c)=2x/5
  n(f U b)= x/5
  n(f U c)= x/5
  n(b U c)= x/4
  n(f / b / c)= x/20
  100 student who plays nothing
  x-100=x/2+3x/5+2x/5-x/5-x/5-x/4+x/20
  x=1000
  (i) So 1000 students are in college
  n(f)=500
  n(b)=600
  n(c)=400
  n(f U b)= 200
  n(f U c)= 200
  n(b U c)= 500
  n(f / b / c)= 50
  now make vendiagram
  (ii) total no of student who plays exactly two games = 200+200+500-50
  =850
  (iii) student who plays neither cricket nor football = player who plays basketball and nothibng
  = 600 + 100 -500-200+50
  = 50
  (iV) student who plays only one game = 600+400+500-200-200-500+50= 650
  
• 12 years agoHelpfull: Yes(4) No(1)
• Answer :-
  1) 1000
  2) 500
  3) 200
  4) 350
  
  more detail mail me patel.paresh894@gmail.com
• 13 years agoHelpfull: Yes(3) No(1)
• 1)1000
  2)500
  3)300
  4)350
  F=football, B= basketball, C=cricket
  play all = 5
  play B&C=25%, play only b&c=25-5=20%
  similarly play only f&c=15%, only f&b=15%
  on the basis of same logic play only f=15%, only b=20%, only c=0%
  so total = 90%
  who dont play nything = 100 or 10%
  find the solutions of all.... :)
  regards
  Deepanshu Kalra
• 13 years agoHelpfull: Yes(2) No(1)
• F=football, B= basketball, C=cricket
  play all = 5
  play B&C=25%, play only b&c=25-5=20%
  similarly play only f&c=15%, only f&b=15%
  on the basis of same logic play only f=15%, only b=20%, only c=0%
  so total = 90%
  as given in question who dont play nothing = 100 or 10%
  so the stdnts who played 90%=900
  
  1) so the total number of students=1000
  
  2) b&c+f&c+f&b= 20%+15%+15% =50% =500
  
  3) no of students play neither football nor cricket mens no of student play only bsketbll
  so 20%of1000=200
  200 student play neither cricket nor footbll
  
  4)students ply only one game=(15+20+0)%=35%
  35%of1000=350
  
• 12 years agoHelpfull: Yes(2) No(0)
• ans is 1000....-total students
  ans of 2 is 500
  ans of 3 is 300
  ans of 4 is 350
  
• 12 years agoHelpfull: Yes(1) No(0)
• total student=100+90=190
  student who play two games=50
  neither cricket nor football=15
  only one game=35
• 13 years agoHelpfull: Yes(0) No(1)
• 1) 1000 students are there in college
  2) 600
  3) 450
  4) 400
• 13 years agoHelpfull: Yes(0) No(0)
• 1) n(aUbUc)= n(a)+n(b)+n(c)-n(aUb)-n(aUc)-n(bUc)+n(aIbIc)
  n(aUbUc)= 50+60+40-20-25-20+5= 90
  percentage of 100 who play none of the games= 10%
  total number of students= 100*100/10= 1000
  2) students playing exactly 2 games= n(aUb)+ n(aUc)+ n(bUc)- n(aIbIc)
  =200+200+250-50=600
• 13 years agoHelpfull: Yes(0) No(1)
• Lets say there are X students in the college.
  Thus number of students who play
  i) all three games - 0.05X
  ii) only cricket and basketball - 0.2X
  iii) only football and basket ball - 0.15X
  iv) only football and cricket - 0.15X
  v) only cricket - 0
  vi) only football - 0.15X
  vii) only basketball - 0.2X
  viii) neither of them - 100
  
  0.05X + 0.15X + 0.15X + 0.2X + 0.15X + 0.2X + 100 = X
  0.9X + 100 = X
  X = 1000
  
  Ans
  1) 1000 students in the college
  2) 500 students play exactly two games
  3) 300 students play neither cricket nor football
  4) 350 students play exactly one game
• 13 years agoHelpfull: Yes(0) No(0)
• 1. 1000 total students
  2. 200 play two games
  3. 300 students
  4.
• 13 years agoHelpfull: Yes(0) No(1)
• 1. 1000
  2. 500
  3. 300
  4. 350
• 13 years agoHelpfull: Yes(0) No(1)