Question

At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take only? Bulgarian, and 20 take only Chinese. In? addition, 14 are taking both Arabic and? Bulgarian, some of whom also take Chinese. How many students are taking all three? languages? None of these three? languages?

Answer 1

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$

Thus, the number of students who take all the three languages is 4 and the number of students who take none of the languages is 100-68 = 32.