This question seems to be very popular on other GRE forums, unless there is no source specified anywhere.

However, here is my try:

We know 9% of the total are >20 and susceptible (so 9 out of our hypothetical 100 people). We know these 9 are 90% of the susceptible (b/c we're told that 10% of the susceptible are <20). That means there must be 10 total susceptible.

How many of the remaining 90 are not susceptible? 40% according to the question, so 36.

Then, we can take this 36 and divide by the total number of people >20:

36/[9 + 36] = 0.8, or 80%.

Answer E!

We can divide the students into the following 4 groups (with the variables denoting the number in each):

1) ≥ 20 years of age and susceptible to tuberculosis = a

2) ≥ 20 years of age and not susceptible to tuberculosis = b

3) < 20 years of age and susceptible to tuberculosis = c

4) < 20 years of age and not susceptible to tuberculosis = d

We are given that 0.1(a + c) = c and 0.4(b + d) = b and 0.09(a + b + c + d) = a. We need to find b/(a + b).

Multiplying the first equation by 10, we have a + c = 10c → a = 9c → c = a/9.

Multiplying the second equation by 5, we have 2b + 2d = 5b → 2d = 3b → d = 3b/2.

Multiplying the third equation by 100, we have 9a + 9b + 9c + 9d = 100a. Substituting c = a/9 and d = 3b/2, we have:

9a + 9b + 9(a/9) + 9(3b/2) = 100a

9a + 9b + a + 27b/2 = 100a

18a + 18b + 2a + 27b = 200a

45b = 180a

b = 4a

So b/(a + b) = 4a/(a + 4a) = 4a/(5a) = 4/5 = 80%

Answer: E