Understanding the Problem: Calculating Exam Failure Percentage

The problem asks us to find the percentage of candidates who failed in at least one subject out of a total of 500 candidates. We are given data about candidates failing in English, Hindi, and Mathematics, specifically detailing failures in only one subject, in two subjects, and in all three subjects.

Breaking Down the Given Data

We have the following information:

• Total number of candidates = 500
• Candidates who failed in English only = 30
• Candidates who failed in Hindi only = 75
• Candidates who failed in Mathematics only = 50
• Candidates who failed in both English and Hindi = 15
• Candidates who failed in both Hindi and Mathematics = 17
• Candidates who failed in both Mathematics and English = 17
• Candidates who failed in all three tests = 5

Let E represent the set of candidates who failed in English, H the set who failed in Hindi, and M the set who failed in Mathematics.

Based on the standard interpretation of such problems, the data can be represented using set notation:

• $|E \text{ only}| = 30$
• $|H \text{ only}| = 75$
• $|M \text{ only}| = 50$
• $|E \cap H| = 15$
• $|H \cap M| = 17$
• $|M \cap E| = 17$
• $|E \cap H \cap M| = 5$

Identifying Candidates Failed in at Least One Subject

The candidates who failed in at least one subject are those belonging to the union of the three sets, i.e., $|E \cup H \cup M|$. We can find the size of this union by summing the number of candidates in each disjoint region of the Venn diagram.

The disjoint regions are:

• Failed in English only ($|E \text{ only}|$)
• Failed in Hindi only ($|H \text{ only}|$)
• Failed in Mathematics only ($|M \text{ only}|$)
• Failed in English and Hindi only ($|(E \cap H) \text{ only}|$)
• Failed in Hindi and Mathematics only ($|(H \cap M) \text{ only}|$)
• Failed in Mathematics and English only ($|(M \cap E) \text{ only}|$)
• Failed in English, Hindi, and Mathematics ($|E \cap H \cap M|$)

Calculating the Sizes of Disjoint Regions

We are given the 'only' values for single subjects and the total intersection values for pairs and all three. We need to calculate the 'only' values for the pairwise intersections:

• Failed in English and Hindi only: These are candidates in $E \cap H$ but not in M. This is $|E \cap H| - |E \cap H \cap M| = 15 - 5 = 10$.
• Failed in Hindi and Mathematics only: These are candidates in $H \cap M$ but not in E. This is $|H \cap M| - |E \cap H \cap M| = 17 - 5 = 12$.
• Failed in Mathematics and English only: These are candidates in $M \cap E$ but not in H. This is $|M \cap E| - |E \cap H \cap M| = 17 - 5 = 12$.

Now we have the numbers for all the disjoint regions:

Calculating Total 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 the disjoint regions:

$|E \cup H \cup M| = |E \text{ only}| + |H \text{ only}| + |M \text{ only}| + |(E \cap H) \text{ only}| + |(H \cap M) \text{ only}| + |(M \cap E) \text{ only}| + |E \cap H \cap M|$

Total failed $= 30 + 75 + 50 + 10 + 12 + 12 + 5$

Total failed $= 105 + 50 + 10 + 12 + 12 + 5$

Total failed $= 155 + 10 + 12 + 12 + 5$

Total failed $= 165 + 12 + 12 + 5$

Total failed $= 177 + 12 + 5$

Total failed $= 189 + 5$

Total failed $= 194$

So, 194 candidates failed in at least one subject.

Calculating the Percentage of Failed Candidates

To find the percentage of candidates who failed in at least one subject, we divide the total number of failed candidates by the total number of candidates and multiply by 100.

Percentage of failed candidates $= \left( \frac{\text{Total failed}}{\text{Total candidates}} \right) \times 100$

Percentage of failed candidates $= \left( \frac{194}{500} \right) \times 100$

Percentage of failed candidates $= \left( \frac{194}{5} \right)$

Percentage of failed candidates $= 38.8\%$

Conclusion

The percentage of candidates who failed in at least one subject is 38.8%.

Revision Table: Key Concepts for Exam Problems

Additional Information: Principle of Inclusion-Exclusion

Another way to calculate the size of the union of three sets is using the Principle of Inclusion-Exclusion:

$|E \cup H \cup M| = |E| + |H| + |M| - (|E \cap H| + |E \cap M| + |H \cap M|) + |E \cap H \cap M|$

To use this formula, we first need to find the total number of candidates who failed in each individual subject. We can derive these from the given data:

• Candidates failed in English ($|E|$): This includes those who failed in E only, E and H only, E and M only, and E, H, and M.
  $|E| = |E \text{ only}| + |(E \cap H) \text{ only}| + |(E \cap M) \text{ only}| + |E \cap H \cap M|$
  $|E| = 30 + (15-5) + (17-5) + 5 = 30 + 10 + 12 + 5 = 57$
• Candidates failed in Hindi ($|H|$):
  $|H| = |H \text{ only}| + |(E \cap H) \text{ only}| + |(H \cap M) \text{ only}| + |E \cap H \cap M|$
  $|H| = 75 + (15-5) + (17-5) + 5 = 75 + 10 + 12 + 5 = 102$
• Candidates failed in Mathematics ($|M|$):
  $|M| = |M \text{ only}| + |(M \cap E) \text{ only}| + |(H \cap M) \text{ only}| + |E \cap H \cap M|$
  $|M| = 50 + (17-5) + (17-5) + 5 = 50 + 12 + 12 + 5 = 79$

Now, plug these values into the Inclusion-Exclusion Principle formula:

$|E \cup H \cup M| = 57 + 102 + 79 - (15 + 17 + 17) + 5$

$|E \cup H \cup M| = 238 - 49 + 5$

$|E \cup H \cup M| = 189 + 5$

$|E \cup H \cup M| = 194$

This confirms the number of candidates who failed in at least one subject is 194, which matches the result obtained by summing the disjoint regions. Both methods lead to the same percentage calculation: $\left( \frac{194}{500} \right) \times 100 = 38.8\%$.