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In the problem below algebra of sets is being evaluated, Venn diagrams are allowed.

There's a Modern Languages reading exam where 200 students are being evaluated. The exam content is in French, Spanish and English and the results were as follows:

80 students are able to read English, 100 read French, 80 read Spanish

55 students are able to read Spanish and French,55 read English and can't read French.

60 read English and can't read Spanish, 15 read English and Spanish but can't read French.

  • ¿How many students are able to read the three languages?
  • ¿How many can read just French?
  • ¿How many can't read none of the three languages?
  • ¿How Many students can read Spanish but can't read French nor English?

So, in order to give a solution to the problem I tried to represent the number of students by using a Venn diagram which I'm not certain wheter it's correct or not.

Venn Based on the numbers above, I could say the answers for question 1 and question 2 are: 15 students and 25 students, which seems logical but indeed any algebra has been applied which certainly would back up the answer (just in case). Would anyone be so kind of giving me a hand on this matter? Any help would be welcomed.

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  • $\begingroup$ 80 read English, but your E-circle contains $55+25+15=95$. The F-circle is also wrong. Oh, wait --- "I" is Ingles, "E" is Espanol? Confusing to have most of the problem in one language, the diagram in another. $\endgroup$ – Gerry Myerson Sep 16 '14 at 7:18
  • $\begingroup$ I do apologize for attaching a confusing diagram, you're right the labels are wrong. My mother tongue is Spanish but I study in English, so it was just a misunderstanding I still got confused now and then. $\endgroup$ – rickHdz Sep 16 '14 at 15:56
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The Venn diagram is more useful for formulating the problem than solving it.

enter image description here

Start with three sets for (E)nglish, (F)rench and (S)panish speaking and let E be English only, ES be English and Spanish, EFS English French and Spanish etc.

Now list what you are given:

  1. E + EF + ES + EFS = 80
  2. F + EF + SF + EFS = 100
  3. S + ES + SF + EFS = 80
  4. SF + EFS = 55 (Spanish and French)
  5. E + ES = 55 (English, no French)
  6. E + EF = 60 (English no Spanish)
  7. S + ES = 15 (Spanish no French)

Seven equations in seven unknowns - solve away.

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    $\begingroup$ ... and the number speaking none of the languages is 200 - (E + F + S + ...) $\endgroup$ – Tom Collinge Sep 16 '14 at 7:35

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