A different way

I did it by looking at the units digit in the numerator, compared the choices, and got (b). This should work right?

Hey @professionalbronco, that's a really interesting method you have there! Unfortunately, I don't think this trick works. There is no correlation between the sum of \(w, x, y\) and the units digit of the numerator. The only way to confirm your answer would be to just solve the problem. If you're unsure on how to do that, it's basically just a repeated process of taking out the largest integer possible from the fraction and taking the reciprocal of what's left.
Here's what it looks like:
Going back and looking at the problem, we can see that \(w=5,x=9,y=2\), so the sum of the three numbers is \(5+9+2=\boxed{16}\). 
@quacker88 I meant the mini question for the second part. Sorry if I didn't make the post clear.

@professionalbronco Ohh, I see! Very clever to think about units digits, since with bigger numbers, it can save you a lot of time. Still, I think that the best way to do the problem is to notice that all of \(\frac{18}{396},\frac{15}{396},\frac{9}{396}\) aren't simplified.
BUT, let's talk about the units digit method that you used, since it's a very smart method to bring up. I'm not sure if you just compared the units digit or if you actually found the units digit of the whole fraction, because there's actually a difference.
You can't simply just look at the units digit of the numerators on both fractions. The fact that it worked out that way is a coincidence. However, if we go a bit deeper, we can see why it actually turned out to work.
Let's look at the first answer choice: \(\frac{7937}{396}=20+\frac{18}{396}\)
To compare the units digits, I'm going to turn \(20\) into a fraction with denominator \(396\). \(20=\frac{7920}{396}\), so we have \(\frac{7937}{396}=\frac{7920}{396}+\frac{18}{396}\). Clearly this doesn't work.
But wait it's really slow to multiply out \(396\cdot20\). This is where we can use the units digit as a shortcut! I don't care about the rest of the digits of \(396\cdot20\), all that matters is that it has units digit \(0\). So all we needed is the fact that
\(\frac{7937}{396}=\frac{***0}{396}+\frac{18}{396}\)
A number with units digit \(0\) plus a number with units digit \(8\) definitely can't add to a number with units digit \(7\). So, we can rule out the first option.We can repeat this process with the other three options and it gives us another quick method to find the solution!
for the second option:
\(\frac{8053}{396}=20+\frac{133}{396}\)
again, \(20\cdot396\) has units digit 0, so
\(\frac{8053}{396}=\frac{***0}{396}+\frac{133}{396}\)
\(0+3\) is indeed \(3\). However, since we didn't find exactly what \(20\cdot396\) is, we can't be 100% sure that this is right. All we know is that it's a possibility.for the third:
\(\frac{9521}{396}=24+\frac{15}{396}\)
for \(24\cdot396\), it will have the same units digit as \(4\cdot6\), which is \(4\), so
\(\frac{9521}{396}=\frac{***4}{396}+\frac{15}{396}\)
This can't work because \(4+5=9\).last but not least:
\(\frac{5953}{396}=15+\frac{9}{396}\)
for \(15\cdot396\), it will have the same units digit as \(5\cdot6\), which is \(0\), so
\(\frac{5953}{396}=\frac{***0}{396}+\frac{9}{396}\)
This can't work because \(0+9=9\).So, since the question doesn't have a "none of the above option", we can be safe with (b) since the other three are clearly wrong. Otherwise, just double check (b) since that's the only possibility and you'll have your answer!
(by the way, do you see how you can't just compare only the numerators of the two fractions? a few extra steps does the trick though :))It might look long, but once you get the hang of it, it's a really clever shortcut! And in general, it's very smart to look out for units digit to simplify things. Great idea, well done!