total 100 members are writing exam. In the 48 members are writing first exam. 45 members are writing second exam. 38 members are writing third exam. 5 members are writing all the three exams.How many members are writing 2 exams?

• Ans) 21
  It's basic set theory problem using venn diagram.
  Draw three intersecting circles 1,2,3. and name all the parts a,b,c,d,e,f,g. 'e' being common in all circles.
  these will be the equation.
  a+b+c+d+e+f+g= 100 ........(1)
  a+b+d+e=48.......................(2) (members are writing first exam)
  b+e+f+c=45........................(3) (members are writing second exam)
  d+e+f+g=38........................(4) (members are writing third exam)
  e=5......................................(5) (members are writing all the three exams)
  we have to find the value of d+b+f=?
  add eq 2,3,4 and put the value of a+c+g from 1.
  you will get d+b+f=21.
  
  
• 9 years agoHelpfull: Yes(30) No(8)
• P(A U B U C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C)
  this gives 36 ( we calculate here P(A∩B) - P(B∩C) - P(C∩A)=)
  is this ryt.
• 9 years agoHelpfull: Yes(13) No(9)
• let P,Q,R writing a only first exm x,y,zis writing a only two exm and 5 write a three exam
  P+Q+R+x+y+z+5=100--------------->4
  p+x+y+5=48-------->1
  Q+y+z+5=45---------->2
  R+z+x+5=38------------->3
  1+2+3
  P+Q+R+x+y+z+15=131------------------>5
  4-5
  x+y+z=21
  so 21 is the ans
  
• 9 years agoHelpfull: Yes(8) No(1)
• Total number of exams written by 100 students = 48 + 45 + 38 = 131
  Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
  Therefore, x + 2y + 3z = 131 also x + y + z = 100.
  Given that z = 5. So x + 2y = 116 and x + y = 95.
  Solving we get y = 21.
  So 21 members are writing exactly 2 exams.
• 8 years agoHelpfull: Yes(7) No(0)
• let a=48,b=45,c=38
  since 5 are writing 3 exams
  so subtract 5 from each value,
  then,a1=43,b1=40,c1=33
  a1+b1+c1=116
  required value = 116-100 = 16
• 9 years agoHelpfull: Yes(4) No(1)
• 21 members are writing 2 exams
• 9 years agoHelpfull: Yes(3) No(4)
• a+b+d+5=48
  b+c+e+5=45
  d+e+f+5=38
  Solving above three equation we get
  b+d+e=21
• 9 years agoHelpfull: Yes(1) No(1)
• Answer is 26
  Total members =100
  Total members giving different exam =48+45+38=131
  Total members giving 2exams =((total members giving different exam-members giving all 3 exams)-total members in exam)
  =((131-5)-100)=26
• 8 years agoHelpfull: Yes(1) No(0)
• we can try 2 approach
  P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)
  =>100=43+40+33-x-y-z+5
  =>x+y+z=21(ans)
  2nd approach is venn diagram
  =>100=43-(x+z)+40-(x+y)+33-(y+z)+x+y+z+5
  =>x+y+z=21(ans)
• 7 years agoHelpfull: Yes(1) No(0)
• ans is 16. because 131-15=116. so 16 exam is extra and it is clear that 16 students gave two exam.
• 9 years agoHelpfull: Yes(0) No(1)
• venn diagram ? vikas ?
• 9 years agoHelpfull: Yes(0) No(1)
• Total number of exams written by 100 students = 48 + 45 + 38 = 131
  Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
  Therefore, x + 2y + 3z = 131 also x + y + z = 100.
  Given that z = 5. So x + 2y = 116 and x + y = 95.
  Solving we get y = 21.
  So 21 members are writing exactly 2 exams.
• 6 years agoHelpfull: Yes(0) No(0)