In a bank examination, 68% of candidates passed in GK and 28% failed in quantitative aptitude. If 18% failed in both sections and 1102 candidates passed in both sections, find the number of candidates who did not fail in quantitative aptitude.

To solve this problem, let's define the variables first:

- Let \( n \) be the total number of candidates.
- Let \( P(A) \) be the percentage of candidates who passed General Knowledge (GK).
- Let \( P(B) \) be the percentage of candidates who passed Quantitative Aptitude (QA).
- Let \( P(\text{Fail in GK}) = 1 - P(A) \).
- Let \( P(\text{Fail in QA}) = 1 - P(B) \).
- Let \( P(\text{Fail in both}) \) be the percentage of candidates who failed in both sections.
- Let \( P(\text{Pass in both}) \) be the percentage of candidates who passed in both sections.

We are given the following information:

- \( P(A) = 68\% \) (since 68% passed in GK, 100% - 68% = 32% failed in GK)
- \( P(\text{Fail in QA}) = 28\% \), so \( P(B) = 72\% \)
- \( P(\text{Fail in both}) = 18\% \)
- The number of candidates who passed in both sections is 1102.

### Step 1: Calculate the total number of candidates

The number of candidates who passed in both sections is given by:
\[
P(\text{Pass in both}) = P(A \cap B)
\]

We know that:
\[
P(A \cap B) = \frac{1102}{n}
\]

We also know:
\[
P(A \cap B) = P(A) + P(B) - P(A \cup B)
\]
But we can simplify this by using the formula for the intersection of two events:
\[
P(A \cap B) = P(A) + P(B) - P(A \cup B)
\]

Given that \( P(A) = 68\% \) and \( P(B) = 72\% \), and knowing that \( P(\text{Fail in both}) = 18\% \):
\[
P(A \cup B) = 1 - P(\text{Fail in both}) = 82\%
\]

Now:
\[
P(A \cap B) = P(A) + P(B) - P(A \cup B) = 68\% + 72\% - 82\% = 58\%
\]

Thus:
\[
\frac{1102}{n} = 0.58
\]

So, the total number of candidates \( n \) is:
\[
n = \frac{1102}{0.58} \approx 1900
\]

### Step 2: Calculate the number of candidates who did not fail in Quantitative Aptitude

The number of candidates who did not fail in Quantitative Aptitude is simply the total percentage that passed QA:
\[
P(B) \times n = 0.72 \times 1900 = 1368
\]

Thus, the number of candidates who did not fail in Quantitative Aptitude is **1368**.

Would you like a more detailed explanation, or do you have any further questions?

---

1. What if the percentage of those passing both sections changes?
2. How would the problem change if 15% failed in both sections?
3. Can we determine how many failed in only GK?
4. How many candidates passed only in GK?
5. What if 25% failed in both subjects?
6. How do you calculate the total number of candidates if only percentages are given?
7. Can the intersection formula be used for more than two subjects?
8. What happens if the pass percentage for GK increases?

**Tip:** Always double-check the steps when calculating percentages and totals, as small errors can lead to incorrect results.

Solve the bank examination problem involving percentages of candidates passing General Knowledge (GK) and Quantitative Aptitude (QA), along with those failing in both sections. Learn how to calculate total candidates and those not failing in QA with step-by-step solutions.

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