Q:

Odd BJT circuit

This circuit came in my today's exam. It asked to find every base, emitter and source current.Base emitter voltage .8  and \$\beta\$ for both BJT is 49.
first I tried normally where \$I_B\$ found that it is coming out from n type BJT and go into p type BJT which is not possible. 

simulate this circuit – Schematic created using CircuitLab
I also try for \$Q_p\$ saturation \$Q_n\$ active ; \$Q_n\$ saturation \$Q_p\$ active and both saturation. Every time I've found some impossible values which cancel that option. 
same thing happen to my other friends.
Is there any other condition I didn't consider?
Is there any other situation or special situation happen?

A:

Note that Qp is flipped collector to emitter from normal operation.  However, it is still a P-N-P sandwich, so will still function as a PNP transistor with collector and emitter flipped, although its gain will be lower.  It's C-B junction is forward biased, so you can assume the usual junction drop.
I'd start to analyze this assuming the gain of both transistors is infinite with something like 700 mV B-E drop and 200 mV in saturation.  For this purpose, simply pretend the E and C of Qp are swapped.  As I said above, it will still function as a transistor that way, just with lower gain than if used normally.
After you get a good idea what this circuit is doing with the above simplifying assumptions, you can go back and use some plausible finite values for the gain, if you think the answer is expected to have that level of detail.
Real transistor circuits usually need to be designed to work with transistor gain from some minimum value to infinity.  Good circuits won't change their operating points much over that range.  A really good answer to this question would be showing what the range of operating points are as the gain of Qp varies from 20 to inifinity and Qn from 50 to infinity, unless of course you were given more specific parameters you haven't told us about.

A:

The collector junction of Qp and emitter junction of Qn - forward biased. 
The collector junction of Qn and emitter junction of Qp - reverse biased. 
now the base current of Qn is, $$I_{Bn} = \frac{V_1 - V_{BEn} - V_{CBp}}{R1+R5+(\beta+1)R4}$$
then,
$$I_{Cn} = \mathrm{min}(\beta I_{Bn}, I_{Csat})$$
where $$I_{Csat} = \frac{V_1-V_{CEn}}{R_4+R_3}$$
The voltages you can calculate as. 
$$V_{En} = R4\times(I_{Bn}+I_{Cn})$$
$$V_{Cn} = V_1 - R3\times(I_{Cn})$$
$$\text{so on ...}$$

