combinatory task about languages studying Problem:


*

*$95$% of student study foreign languages: English, German, and French

*$75$% study English

*$70$% French or German

*$3$% of students study three languages at the same time

*$36$% of students study English and German

*$10$% percent of student study French and German

*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.
 A: 
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 $a_4+a_5+a_6+a_7=67$.
