So I tried to solve this problem in the 1955 AHSME:
In order to pass $B$ going $40$ mph on a two-lane highway, $A$, going $50$ mph, must gain $30$ feet. Meantime, $C, 210$ feet from $A$, is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, then, in order to pass safely, $A$ must increase his speed by:
$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$
I do not understand some parts of the problem clearly, but I did try to solve the problem. So I came up with answer B. But the correct answer is C. After some research, C only makes sense if the cars are mathematical points, but if they have a volume, then C shouldn't be correct.
So, why is C the correct answer? And why isn't B the correct answer?
Let $x$ be speed of $A$ at which it can pass safely. Then we can show that $$\frac{210}{50+x} = \frac{30}{x-40}=t,$$ where $t$ is time of passing. Solving this equation gives us $x = 55$, and if A was going $50$ mph, then answer is $55-50=5 \to C$