2012/13 AHSMC Part 1 Problem 7

The fraction of problems solved only by Karla was $\dfrac{4}{5} - \dfrac{1}{2} = \dfrac{3}{10}$ so that the total number of problems is a multiple of $10$. The fraction of problems solved by Klaus was at most $1 - \dfrac{3}{10} = \dfrac{7}{10}$. Thus the total number of problems was at least $50$. If it were $50$, then $10$ problems were solved by Klaus alone, and as Karla solved $\dfrac{4}{5} \times 50 = 40$ problems, the number of problems solved by neither is $0$. The total number of problems could only be as large as $70$ since $35$ problems would be solved by both. In this case, the number of problems solved by neither was $\dfrac{1}{5} \times 70 = 14$. It follows that the total number of problems must be $60$, of which $30$ were solved by both, $5$ by Klaus alone, $\dfrac{3}{10} \times 60 = 18$ problems by Karla alone, and $60 - 30 - 5 - 18 = 7$ by neither of them. The answer is (D).