Question

Assume that 155 students are surveyed and every student takes at least one of the following languages. The results of the survey are as follows:
90 take French.

83 take German.

42 take French and German.

41 take German and Russian.

22 take French as their only foreign language.

22 take French, Russian, and German.

(1) How many take Russian?

(2) How many take French and Russian but not German?

Answer 1

Answer:

91 people take Russian

26 people take French and Russian but not German

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that take French.

-The set B represents the students that take German

-The set C represents the students that take Russian.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a is the number of students that take only Franch, A \cap B is the number of students that take both French and German , A \cap C is the number of students that take both French and Russian and A \cap B \cap C is the number of students that take French, German and Russian.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following subsets:

$a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

There are 155 people in my school. This means that:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155$

The problem states that:

90 take Franch, so:

$A = 90$

83 take German, so:

$B = 83$

22 take French, Russian, and German, so:

$A \cap B \cap C = 22$

42 take French and German, so:

$A \cap B = 42 - (A \cap B \cap C) = 42 - 22 = 20$

41 take German and Russian, so:

$B \cap C = 41 - (A \cap B \cap C) = 41 - 22 = 19$

22 take French as their only foreign language, so:

$a = 22$

Solution:

(1) How many take Russian?

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$C = c + (A \cap C) + 19 + 22$

$C = c + (A \cap C) + 41$

First we need to find $A \cap C$, that is the number of students that take French and Russian but not German. For this, we have to go to the following equation:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$90 = 22 + 20 + (A \cap C) + 22$

$(A \cap C) + 64 = 90$.

$(A \cap C) = 26$

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The number of students that take Russian is:

$C = c + 26 + 41$

$C = c + 67$

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Now we have to find c, that we can find in the equation that sums all the subsets:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155$

$22 + b + c + 20 + 26 + 19 + 22 = 155$

$b + c + 109= 155$

$b + c = 46$

For this, we have to find b, that is the number of students that take only German. Then we go to this eqaution:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$B = b + 19 + 20 + 22$

$B = b + 61$

$b + 61 = 83$

$b = 22$

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$b + c = 46$

$c = 46 - b$

$c = 24$

The number of people that take Russian is:

$C = c + 67$

$C = 24 + 67$

$C = 91$

91 people take Russian

(2) How many take French and Russian but not German?

$(A \cap C) = 26$

26 people take French and Russian but not German