Given,

• Out of 100 students, 80 passed in Science, 71 in Mathematics.
• 10 failed in both subjects.
• 7 did not appear in an examination.

Let

• \( S \) represent the set of students who passed in Science.
• \( M \) represent the set of students who passed in Mathematics.

Required Values:

• \( n(S) = 80 \)
• \( n(M) = 71 \)
• \( n\overline{(S \cup M)} = 10 \)
• \( n(U) = 100 \)

The total number of students is the sum of students who passed, students who failed, and students who did not appear in the examination.

\( \rm n(U) = n( S \cup M) + n \overline{n(S \cup M)}+ 7 \)

\( \rm or, n(U) = n(S) + n(M) - n(S \cap M) + n \overline{n(S \cup M)} + 7 \)

\( \rm or, 100 = 80 + 71 - n (S \cap M) + 10 + 7\)

\( \rm or, n(S \cap M) = 168 - 100 \)

\( \rm \therefore n(S \cap M) = 68 \)

The number of students who passed in either Science or Math are those students who did not fail in both subjects.  The number of such students (say x) is equal to the difference in total students who attended the examination and students who failed both subjects.

\( \rm x = n(U) - 7 - n \overline{n(S \cup M)}  \)

\( \rm x = 100 - 7 - 10 \)

\( \rm x = 83 \)

Hence,

\( n(S) = 80 \) and \( n(M) = 71 \).

Number of students who passed in either Science or Maths is 83.

Number of students who passed in both subjects: \( n(S \cap M) = 68 \).

Representing the above information in a Venn diagram