Exam Analysis: Calculating Total Candidates from Failure Data

This problem asks us to find the total number of candidates who appeared in an examination, given the percentages of candidates who failed in Mathematics, English, and both subjects, as well as the specific number of candidates who passed in both subjects.

Understanding the Problem with Percentages

We are given the following failure percentages:

• Failed in Mathematics only or Mathematics and English: 25%
• Failed in English only or Mathematics and English: 12%
• Failed in Both Mathematics and English: 10%

We need to find the percentage of candidates who failed in at least one subject to then determine the percentage who passed in both subjects.

Using the Principle of Inclusion-Exclusion

To find the percentage of candidates who failed in at least one subject (Mathematics or English or both), we can use the formula based on the principle of inclusion-exclusion:

Percentage Failed in at least one subject = Percentage Failed in Mathematics + Percentage Failed in English - Percentage Failed in Both Subjects

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

Percentage Failed in \(M \cup E\) = Percentage Failed in M + Percentage Failed in E - Percentage Failed in \(M \cap E\)

\(\text{Percentage Failed in at least one subject} = 25\% + 12\% - 10\%\)

\(\text{Percentage Failed in at least one subject} = 37\% - 10\%\)

\(\text{Percentage Failed in at least one subject} = 27\%\)

This means 27% of the candidates failed in Mathematics only, or English only, or both Mathematics and English.

Calculating Percentage Passed in Both Subjects

The candidates who passed in both subjects are those who did not fail in at least one subject. Therefore, the percentage of candidates who passed in both subjects is:

Percentage Passed in Both = Total Percentage - Percentage Failed in at least one subject

Percentage Passed in Both = \(100\% - 27\%\)

Percentage Passed in Both = \(73\%\)

Relating Percentage to the Number of Candidates

We are given that 292 candidates passed in both subjects. This number corresponds to the 73% we just calculated.

Let \(T\) be the total number of candidates appeared in the examination.

We can set up the equation:

\(73\% \text{ of } T = 292\)

Converting the percentage to a decimal or fraction:

\(\frac{73}{100} \times T = 292\)

Solving for the Total Number of Candidates

To find the total number of candidates \(T\), we can rearrange the equation:

\(T = \frac{292 \times 100}{73}\)

We can perform the division:

\(292 \div 73\)

Let's test some multiples of 73:

• \(73 \times 1 = 73\)
• \(73 \times 2 = 146\)
• \(73 \times 3 = 219\)
• \(73 \times 4 = 292\)

So, \(292 \div 73 = 4\).

Now substitute this back into the equation for \(T\):

\(T = 4 \times 100\)

\(T = 400\)

Conclusion

The total number of candidates who appeared in the examination is 400.

We know that 73% corresponds to 292 candidates.

Let Total Candidates = \(X\).

\(73\% \text{ of } X = 292\)

\(\frac{73}{100} \times X = 292\)

\(X = \frac{292 \times 100}{73}\)

\(X = 4 \times 100\)

\(X = 400\)

Revision Table: Key Percentages in Examination Results

Additional Information: Understanding Venn Diagrams for Percentages

This type of problem can often be visualized using a Venn diagram. We have two overlapping circles, one for Math failures and one for English failures. The overlapping section represents failures in both.

• The center overlap is 10% (Failed in Both).
• The part of the Math circle *only* (Failed in Math only) is \(25\% - 10\% = 15\%\).
• The part of the English circle *only* (Failed in English only) is \(12\% - 10\% = 2\%\).

The total percentage of those who failed in at least one subject is the sum of these three distinct regions:

\(15\% (\text{Math only}) + 2\% (\text{English only}) + 10\% (\text{Both}) = 27\%\)

The candidates outside these circles passed in both subjects. Their percentage is \(100\% - 27\% = 73\%\).

Knowing that 73% equals 292 candidates allows us to find the total number of candidates by setting up a proportion or equation, as demonstrated in the solution.