BJT differential pair
An approach you might take is to simply make an assumption either way and then test it out. If it succeeds, then look further. If it fails, drop the idea and try out another one. A lot of questions like this one can be quickly figured out that way. The problem asks a simple "this or that" question. So assume one and test it out. Doesn't matter which you choose. See how it pans out.

For example: The base of \$Q_1\$ is \$+500\:\textrm{mV}\$. If it is ON (and given the specified assumption about \$\vert V_{BE}\vert=700\:\textrm{mV}\$), then it must be the case that \$Q_1\$'s emitter is at \$V_{B_1}-V_{BE}=500\:\textrm{mV}-\left(-700\:\textrm{mV}\right)=1.2\:\textrm{V}\$, right? But if so, then that would place all that voltage across \$Q_2\$'s base-emitter junction and wouldn't that mean that if \$Q_1\$ is ON then \$Q_2\$ must be so much more-so ON? Keep in mind that for every \$60\:\textrm{mV}\$ that the magnitude of \$V_{BE}\$ increases, the collector current increases by a factor of \$10\times\$.

Since with this assumption \$\vert V_{{BE_2}}\vert=\vert V_{{BE_1}}\vert+500\:\textrm{mV}\$, that's more than \$ 10^8\times\$ the collector current in \$Q_2\$ than in \$Q_1\$. Or, put otherwise, \$Q_2\$ gets all the current and \$Q_1\$ gets practically none of it. And since the resistors are all the same value, the collector of \$Q_2\$ will be about as much lower than its base as its emitter is above its base. So \$V_E\approx +700\:\textrm{mV}\$ and \$V_{C_2}\approx -700\:\textrm{mV}\$. \$Q_1\$'s collector resistor has no current at all, so that collector voltage will be at the negative rail, or \$V_{C_1}\approx -5\:\textrm{V}\$.



Part of learning electronics is about applying logic. Part is about applying quantitative theory. Part still more is about imagination and not taking it personally when it fails, but instead just marching on with the next idea.

There is nothing to be ashamed of, in being wrong and having your errors handed to you on a silver platter for all to see. Happens to all of us. Just pick yourself up and move forward again. But there is something to be ashamed of in not exposing your thinking and imagining it is instead better to protect yourself from criticism by not letting others see into your mind much.

As an aside, you might take a look at this very much related answer that I wrote, about something similar. It isn't at all a complete description even of something as simple as your circuit, but it will provide a taste of where you are headed.