At the college entrance examination, each candidate is admitted or rejected according to whether he has passed or failed the tests. Of the candidates who are really capable, 80% pass the tests and of the incapable, 25% pass the test. Given that 40% of the candidates are really capable, the proportion of capable college students is about

### Concept & Logic This problem is a practical application of **Conditional Probability / Weighted Averages**. The phrase "proportion of capable college students" specifically refers to the students who actually made it *into* the college (i.e., those who passed). We need to create a hypothetical pool of $100$ candidates, find out exactly how many passed from both the "capable" and "incapable" groups, combine them to get the total number of admitted students, and then see what fraction of that final admitted group came from the capable pool. ### Step-by-Step Solution Let the total number of candidates appearing for the exam be $100$. First, split the candidates by capability based on the $40\%$ given: * **Capable Candidates:** $40$ * **Incapable Candidates:** $100 - 40 = 60$ Next, determine how many from each group actually pass (and thus become "college students"): * **Capable students who pass:** $80\%$ of $40 = 0.8 \times 40 = 32$ * **Incapable students who pass:** $25\%$ of $60 = 0.25 \times 60 = 15$ Now, find the total number of students admitted to the college: * **Total students passed (admitted):** $32 + 15 = 47$ The question asks for the proportion of *capable* students out of the admitted students: $$ \text{Proportion} = \left( \frac{\text{Capable and Passed}}{\text{Total Passed}} \right) \times 100 $$ $$ \text{Proportion} = \left( \frac{32}{47} \right) \times 100 \approx 68.08\% $$ Rounding to the nearest whole number given in the options, we get $68\%$. ### Exam Strategy & Shortcut Use ratio methods. Capable to Incapable ratio is $40:60$ or $2:3$. Pass rate is $80\%$ for the $2$-part group and $25\%$ for the $3$-part group. Passed capable $= 2 \times 0.8 = 1.6$ ratio units. Passed incapable $= 3 \times 0.25 = 0.75$ ratio units. Total passed units $= 1.6 + 0.75 = 2.35$. Required percentage $= \frac{1.6}{2.35} \times 100 = \frac{160}{235} \times 100$. Quick estimation: $160$ is slightly more than two-thirds ($66.6\%$) of $235$. The only option slightly above $66\%$ is $68\%$. This saves you from doing long division on $3200 / 47$. ### Common Pitfall The most common mistake is calculating $80\%$ of $40\%$ ($32\%$) and stopping there, thinking it's the final answer. However, $32\%$ is the proportion of capable passing students relative to the *entire applicant pool*. The question asks for the proportion relative to the *admitted college students* pool. You must divide by the total passing rate ($47\%$). ### Final Answer Therefore, the correct answer is **68%**.