combinatory task about languages studying

Question

Problem:

1. $95$% of student study foreign languages: English, German, and French
2. $75$% study English
3. $70$% French or German
4. $3$% of students study three languages at the same time
5. $36$% of students study English and German
6. $10$% percent of student study French and German
7. Nobody studies only German

How many people in percents study no less than two languages?

I have to solve this problem using set theory. but I am more about general logic, transaction to the set theory should not be difficult.

So what I was thinking of:

$36+10 = 46$% these people study at least two languages, as per problem's description we need two or more.

$36+10+3 = 49$ % - two or more languages.

$75 - 49 = 26$% only English $70 - 49 = 21$% either French or German

But I am not sure if my logic is correct at all, and also not sure how to deal with German here.

Answer 1

A Venn diagram may help you. Assume that there are $100$ students overall. Set $7$ variables:

• $a_1$ people study only English

• $a_2$ people study only German (given $a_2=0$)

• $a_3$ people study only French

• $a_4$ people study only English and German

• $a_5$ people study only German and French

• $a_6$ people study only French and English

• $a_7$ people study them all (given $a_7=3$)

The task asked us to find $a_4+a_5+a_6+a_7$ (that is two or more languages).

I will say something about your logic first. You said "at least two languages" is $36\%+10\%=46\%$, this is not correct because in the diagram above, this method is equivalent to $(a_4+a_7)+(a_5+a_7)$, as $a_7$ is counted twice and not $a_6$.

$36\%+10\%+3\%=49\%$ is even more wrong, because now $a_7$ is counted three times.

You also said "only English" is $75\%-49\%=26\%$, I would say that this is equivalent to $(a_1+a_4+a_6+a_7)-(a_4+a_7+a_5+a_7+a_7)\ne a_1$, this cannot be true either.

For $7$ things listed in the task respectively, we have

${\begin{cases}a_1+a_2+a_3+a_4+a_5+a_6+a_7=95\\a_1+a_4+a_6+a_7=75\\a_2+a_3+a_4+a_5+a_6+a_7=70\\a_7=3\\a_4+a_7=36\\a_5+a_7=10\\a_2=0\end{cases}}$

From first and third equations, we have $a_1=25$.

From fourth, fifth, sixth equations, we have $a_4=33$; $a_5=7$.

From $a_1=25$; $a_4=33$; second and fourth equations, we have $a_6=14$.

From $a_1=25;a_2=0;a_4=33;a_5=7;a_6=14;a_7=13$ and first and third equations, we have $a_3=3$.

The number people study in at least two subjects is $a_4+a_5+a_6+a_7=67$.