Understanding the Problem: Exam Results Analysis

The question provides data about 500 candidates who took an examination with three tests: English, Hindi, and Mathematics. We are given the number of candidates who failed in various combinations of these subjects. We need to find the total number of candidates who passed in two or more subjects.

Information from the Question

Total candidates = 500

Number of candidates who failed in specific subjects:

• Failed in English only: 30
• Failed in Hindi only: 75
• Failed in Mathematics only: 50
• Failed in both English and Hindi: 15
• Failed in both Hindi and Mathematics: 17
• Failed in both Mathematics and English: 17
• Failed in all three tests: 5

Note: In Venn diagram problems, "failed in both English and Hindi" typically refers to the intersection of the failure sets for English and Hindi, which includes those who also failed Mathematics, unless specified as "only". We will use this standard interpretation.

Using Venn Diagrams for Failure Analysis

Let E, H, and M be the sets of candidates who failed in English, Hindi, and Mathematics, respectively. We can use a Venn diagram to represent the number of candidates in each disjoint region based on the given failure data.

Calculating Disjoint Regions of Failures

We are given the number of candidates who failed in each subject 'only' and the intersections (including the center).

• Failed in English only (\( |E \text{ only}| \)): 30
• Failed in Hindi only (\( |H \text{ only}| \)): 75
• Failed in Mathematics only (\( |M \text{ only}| \)): 50
• Failed in English and Hindi and Mathematics (\( |E \cap H \cap M| \)): 5

Now we calculate the number of candidates who failed in exactly two subjects:

• Failed in English and Hindi only (\( |E \cap H \text{ only}| \)) = |E $\cap$ H| - |E $\cap$ H $\cap$ M| = 15 - 5 = 10
• Failed in Hindi and Mathematics only (\( |H \cap M \text{ only}| \)) = |H $\cap$ M| - |E $\cap$ H $\cap$ M| = 17 - 5 = 12
• Failed in Mathematics and English only (\( |M \cap E \text{ only}| \)) = |M $\cap$ E| - |E $\cap$ H $\cap$ M| = 17 - 5 = 12

Total Candidates Failed in at Least One Subject

The total number of candidates who failed in at least one subject is the sum of the numbers in all disjoint regions of the failure Venn diagram:

\( \text{Total Failed} = |E \text{ only}| + |H \text{ only}| + |M \text{ only}| + |E \cap H \text{ only}| + |H \cap M \text{ only}| + |M \cap E \text{ only}| + |E \cap H \cap M| \)

\( \text{Total Failed} = 30 + 75 + 50 + 10 + 12 + 12 + 5 \)

\( \text{Total Failed} = 105 + 50 + 10 + 12 + 12 + 5 \)

\( \text{Total Failed} = 155 + 10 + 12 + 12 + 5 \)

\( \text{Total Failed} = 165 + 12 + 12 + 5 \)

\( \text{Total Failed} = 177 + 12 + 5 \)

\( \text{Total Failed} = 189 + 5 \)

\( \text{Total Failed} = 194 \)

So, 194 candidates failed in at least one subject.

Candidates Who Passed All Three Subjects

The candidates who passed in all three subjects are those who did not fail in any subject. This is the total number of candidates minus those who failed in at least one subject.

\( \text{Passed in All Three} = \text{Total Candidates} - \text{Total Failed in At Least One} \)

\( \text{Passed in All Three} = 500 - 194 \)

\( \text{Passed in All Three} = 306 \)

Candidates Who Passed Exactly Two Subjects

A candidate who passed in exactly two subjects failed in exactly one subject. For example, a candidate who passed English and Hindi but failed Mathematics passed exactly two subjects (English and Hindi).

The number of candidates who passed in exactly two subjects corresponds to the number of candidates who failed in exactly one subject:

• Passed English and Hindi, failed Mathematics = Failed in Mathematics only = 50
• Passed Hindi and Mathematics, failed English = Failed in English only = 30
• Passed Mathematics and English, failed Hindi = Failed in Hindi only = 75

\( \text{Passed in Exactly Two} = (\text{Failed E only}) + (\text{Failed H only}) + (\text{Failed M only}) \)

\( \text{Passed in Exactly Two} = 30 + 75 + 50 \)

\( \text{Passed in Exactly Two} = 155 \)

Calculating Candidates Who Passed Two or More Subjects

The question asks for the number of candidates who passed in two or more subjects. This includes candidates who passed in exactly two subjects and candidates who passed in exactly three subjects.

\( \text{Passed in Two or More} = (\text{Passed in Exactly Two}) + (\text{Passed in Exactly Three}) \)

\( \text{Passed in Two or More} = 155 + 306 \)

\( \text{Passed in Two or More} = 461 \)

Final Answer

The number of candidates who passed in two or more subjects is 461.

Revision Table: Key Concepts & Calculations

Additional Information: Understanding Exam Results Analysis

Problems like this are classic examples of using the Principle of Inclusion-Exclusion or simply mapping out the regions of a Venn diagram. The key is to carefully distinguish between "failing in a subject" and "failing in that subject only," and similarly for combinations like "failing in both E and H" versus "failing in both E and H only."

When data is given for categories like "failed in both E and H" (without the word "only"), it usually refers to the entire intersection of sets E and H, which includes the overlap with set M (failing in all three). The "only" descriptions specify the regions outside the triple intersection.

Calculating the number of candidates in each distinct region of the Venn diagram (based on failure status) allows us to account for all candidates and accurately determine various groups, such as those who passed specific combinations of subjects.

Passing in 'k' subjects is equivalent to failing in '(Total Subjects - k)' subjects. This relationship is crucial for translating information about failures into information about passes.

• Passed 3 subjects = Failed 0 subjects
• Passed 2 subjects = Failed 1 subject
• Passed 1 subject = Failed 2 subjects
• Passed 0 subjects = Failed 3 subjects

By calculating the number of candidates in each 'failed exactly k subjects' category, we can easily find the number in each 'passed exactly (Total Subjects - k) subjects' category.