Maths-
General
Easy

Question

In the above question, the number of families which buy none of A comma B and C is

  1. (a) 4000  
  2. 3300  
  3. 4200  
  4. 5000  

The correct answer is: (a) 4000


    We have,
    Required number of families
    equals n left parenthesis A ´ intersection B ´ intersection C ´ right parenthesis
    equals n left parenthesis A union B union C right parenthesis ´
    equals N minus n left parenthesis A union B union C right parenthesis
    equals 10000 minus left curly bracket n open parentheses A close parentheses plus n open parentheses B close parentheses plus n open parentheses C close parentheses minus n left parenthesis A intersection B right parenthesis right curly bracket
    negative n open parentheses B intersection C close parentheses minus n open parentheses A intersection C close parentheses plus n left parenthesis A intersection B intersection C right parenthesis right curly bracket
    equals 10000 minus 4000 minus 2000 minus 1000 plus 500 plus 300 plus 400 minus 200
    equals 4000

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