Question

In a school where there was a compulsion to learn at least one foreign language from the choices given, namely German, French and Spanish: 28 students took French, 30 took German, and 32 took Spanish. 6 students learnt French and German, 8 students learnt German and Spanish, and 10 students learnt French and Spanish. 54 students learnt only one foreign language, while 20 students learnt only German. Find the number of students in the school.

• A. 60
• B. 62
• C. 70
• D. None of the above

Hint:

Split into only-one, exactly-two, and all-three regions, and use the given sums to solve for each unknown.

Answer 1

The Correct Option is C

Step 1: Recall the inclusion-exclusion formula for three sets.
$|G \cup F \cup S| = |G| + |F| + |S| - |G \cap F| - |G \cap S| - |F \cap S| + |G \cap F \cap S|$, where G, F, S are the sets of students learning German, French, and Spanish. Since every student learns at least one language, this union is the whole school.

Step 2: Plug in the totals and pairwise overlaps we are given.
$|G|=30$, $|F|=28$, $|S|=32$, $|G \cap F| = 6$, $|G \cap S| = 8$, $|F \cap S| = 10$. We still need the triple overlap $|G \cap F \cap S|$, call it Z.

Step 3: Find Z using the "only German" and "only one language" clues.
The German circle splits into: only German (20), German and French only ($6-Z$), German and Spanish only ($8-Z$), and all three (Z). Adding these must give the full German total: $20 + (6-Z) + (8-Z) + Z = 30$. This simplifies to $34 - Z = 30$, so $Z = 4$.

Step 4: Substitute Z back into the inclusion-exclusion formula.
Total = $30 + 28 + 32 - 6 - 8 - 10 + 4 = 90 - 24 + 4 = 70$.

Final Answer:
The school has 70 students. \[ \boxed{70} \]