SOLUTION: From a group of 440 students who took the exam, 300 passed English, 280 passed Spanish, 220 passed Math, 180 passed English and Spanish, 160 Spanish and Mathematics, 150 English an

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Question 1163106: From a group of 440 students who took the exam, 300 passed English, 280 passed Spanish, 220 passed Math, 180 passed English and Spanish, 160 Spanish and Mathematics, 150 English and Mathematics and 100 passed English, Spanish and Mathematics. How many altogether passed at least one of these three subjects?
Found 3 solutions by ikleyn, Edwin McCravy, greenestamps:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

You are given the sets 

    U - universal set of all 440 students
    E - set of students passed English (300)
    S - set of students passed Spanish (280)
    M - set of students passed Math    (220)

the in-pair intersections

    ES - English and Spanish           (180)
    SM - Spanish and Math              (160)
    EM - English and Math              (150)

and the triple intersection

    ESM - English, Spanish and Math    (100)


For the union E U S U M, there is a REMARKABLE formula from elementary set theory 


    n(E U S U M) = n(E) + n(S) + n(M) - N(ES) - n(SM) - n(EM) + n(ESM).    (1)


It says the number of elements in the union of three subsets is the alternate sum of the shown components.


By substituting the given values, you get the answer to the problem's question


    n(E U S U M) = 300 + 280 + 220 - 180 - 160 - 150 + 100 = 410.


ANSWER.  In all, 410 students passed at least one exam.

Solved.

Memorize the formula and the method of the solution (!)

=================

The last step is to prove the formula (1).

    It is totally clear to you why I add the first three addends in the formula (1).

    But when I add them, I count twice every term in each in-pair intersection.

    Therefore, I subtract the numbers of terms in each in-pair intersection.

    Next, when I add three first addends, I count thrice each term in the triple intersection;

    and when I subtract in-pair intersections, I cancel these terms thrice.

    Therefore, I must add the number of terms in the triple intersection one more time to restore the balance.

Thus the formula  (1)  is proved   //   and the solution is fully completed  (! !)


/\/\/\/\/\/\/\/

To see many other similar solved problems, look into the lessons
    - Counting elements in sub-sets of a given finite set
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in this site.

Look also into the links

https://www.algebra.com/algebra/homework/sets-and-operations/sets-and-operations.faq.question.1149313.html
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1132870.html
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1129554.html
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1129554.html
https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Rate-of-work-word-problems.faq.question.1126097.html
https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1126099.html

to similar solved problems in the archive of this forum.



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To tutor @greenestamps:

            Thanks for noticing my typos --- I just fixed them (!)



Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!


In Venn diagram problems, the clues are nearly always given in the reverse
order from the order that you need them, so we begin by rearranging them in
reverse order: 

100 passed English, Spanish and Mathematics, that's e
   So we put 100 for e
150 English and Mathematics, that's d+e
   So we put 150-100=50 for d
160 Spanish and Mathematics, that's, e+f
   So we put 160-100=60 for f 
180 passed English and Spanish, that's b+e
   So we put 180-100=80 for b 
220 passed Math, that's d+e+f+g
   So we put 220-50-100-60=10 for g 
280 passed Spanish, that's b+c+e+f 
   So we put 280-80-100-60=40 for c
300 passed English, that's a+b+d+e
   So we put 300-80-100-50=70 for a
440 students took the exam, that's a+b+c+d+e+f+g+h
   So we put 440-70-80-40-50-100-60-10=30 for h



The question is:

How many altogether passed at least one of these three subjects?

That's all but the h=30 who didn't pass any. That's 440-30=410

Edwin

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


For tutor @ikleyn....

There is a typo in your general formula that will cause confusion to the student:

" ... = n(ESM)" should be "... + n(ESM)"

Also -- less problematic -- there is a "U" missing on the left side of that equation.