# Analytical Reasoning Question I

I would appreciate it if someone could please help me understand what is being asked here and how to approach questions like the one below.

At the college entrance exam, a candidate is admitted according to whether he has passed or failed the test. Of the candidates who are really capable, 80 % pass the test and of the incapable, 25 % pass the test. Given that 40 % of the candidates are really capable, then the proportion of the really capable students who can pass the test to the total students who can pass is?

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Are we to assume that every student is either "really capable" or "incapable", or is there also a group of "just adequately capable" students? –  Henning Makholm Sep 6 '11 at 23:31
Out of all capable students, $80%$ pass, so since capable students are $40%$ of the total, they constitute $(.4)(.8)%$ of those who passed; similarly, out of the $60%$ that is incapable, $25%$ passed, so they form $15%$ of the total who passed. You can do this problem assuming there are 100 students and the solution will generalize to any number. –  gary Sep 6 '11 at 23:39
I don't know how to describe the general category, but maybe weighted averages fits reasonably-well. –  gary Sep 6 '11 at 23:40
This is a CAT question. This was just what was given –  user10695 Sep 7 '11 at 15:12
but I think your solution there makes a lot of sense. The final answer was 68% though. –  user10695 Sep 7 '11 at 15:18

You don’t need any fancy notation to solve the problem.

You know that $40$% of the candidates are really capable and that $80$% of those $40$% pass the test; $0.8 \cdot 0.4 = 0.32$, so $32$% of all candidates both pass the test and are really capable. Of the remaining $60$% of the candidates, $25$% pass; $0.25 \cdot 0.6 = 0.15$, so $15$% of all candidates both pass the test and aren’t really capable. Altogether, then, $32+15=47$% of the candidates can pass the test, and the fraction of those who are really capable is $32/47$.

Alternatively, you can do as gary suggested and imagine that you’re working with a specific number of candidates. Choose the number so that all of the percentages work out to whole numbers of people; in this case $100$ works. Then you have $40$ who are really capable, of whom $32$ pass, and $60$ who aren’t capable, of whom $15$ pass anyway. Thus, $47$ pass, of whom $32$ are really capable, and the desired proportion is $32/47$. As you can see, this is just doing with specific numbers what I did with the percentages in the previous paragraph.

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Thanks Brian. I did get 32%, just didn't know how to interpret that number. Is there a TRICK to solving these questions? A branch of math, or something that helps one better master these things? –  user10695 Sep 7 '11 at 15:14
@user10695: Not really. I’m assuming that this is from some sort of admissions test. In my experience the analytical reasoning questions on such tests rarely require any technical tools beyond what in the U.S. is high school mathematics; the difficulty lies not in the mathematical tools themselves, but in seeing which tools to use and how to use them (and perhaps to some extent in reading carefully). If you’re studying for such an exam, probably the best thing that you can do is what you’re doing: go through as many sample problems as you can find. –  Brian M. Scott Sep 7 '11 at 20:24

We have $$p(P|C) = 0.8$$ $$p(P|C') = 0.25$$ and $$p(C) = 0.4$$ and $$p(C') = 0.6$$

Note that $P$ denotes the event of passing and $C$ denotes the event that a person is capable. Thus $$p(P \cap C) = p(P|C)p(C) = (0.25)(0.4)$$

Also $$p(P) = p(P \cap C)+ p(P \cap C')$$

$$= p(P|C)p(C)+ p(P|C')p(C')$$ $$= (0.8)(0.4)+(0.25)(0.6)$$

So compute $$\frac{p(P \cap C)}{p(P)}$$

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Explaining your symbols wouldn't hurt. –  Henning Makholm Sep 6 '11 at 23:32
I would also suggest using a different letter for Probably and the event of passing the exam. –  JavaMan Sep 6 '11 at 23:36
The % of capable is 40% and the percentage of those Capable who passed is 80%. So wouldn't P n C be (0.8 * 0.4)? –  user10695 Sep 7 '11 at 15:16