Lesson More complicated problems on finding number of elements in finite subsets

More complicated problems on finding number of elements in finite subsets

Problem 1

Let x students study only History; 

    y students study only Geography;

    z students study both subjects.


Then we have these three equations

     x + y + z = 30      (1)

     x + z = 2*(y+z)     (2)

     y = 8               (3)


Substitute y= 8 from equation (3) into equations (1) and (2)

    x + 8 + z = 30       (1')

    x + z = 2*(8+z)      (2')


Simplify

    x + z = 22           (1'')

    x - z = 16           (2'')


From equations (1'')  and (2'') you get by adding       2x = 22+16 = 38,  x = 38/2 = 19.

From equations (1'')  and (2'') you get by subtracting  2z = 22-16 =  6,  z = 6/2 = 3.


Thus 19 students study only History;  8 students study only Geography, and 3 students study both subjects.


ANSWER.  3 students study both subjects.

Problem 2

From the condition, 30-16 = 14 failed the first quiz, and

                    30-20 = 10 failed the second quiz,

             while           4 failed both quizzes.


Hence,  the number of those who failed at least one quiz is  14 + 10 - 4 = 20.


It means that 30-20 = 10 students passed both quizzes.    ANSWER

Problem 3

Since  20%  have neither blue eyes nor dark hair, we conclude that 

100 -20% = 80% are those who either has blue eyes OR dark hair.


So, now we have  60%  of students have blue eyes n(B) = 0.6, 
55% have dark hair  n(D) = 0.55,  and their union  (B U D)  is  80%:  n(B U D) = 0.8.


Having it, we write

      0.6 + 0.55 - n(B and D) = 0.8,

      0.6 + 0.55 - 0.8 = n(B and D),

      n(B and D) = 0.35.


Therefore, the probability that a randomly selected student will have blue eyes and dark hair is

    P =  = 0.35 = 35%.    ANSWER