Let's have symbols for all nodes. From P, the first node to the right shall be #P_1#, the next to the right shall be #P_2#, and the 3rd to the right is already known as Q. From R, the first node to the right shall be #R_1#, the next to the right shall be #R_2#, and the 3rd to the right is already known as S.
Since there is no connection from Q toward the right, the path thru the resistor between #P_2# and Q can have no current thru it. The same can be said about the path thru the resistor between #R_2# and S. Therefore we can ignore the 2 resistors farthest to the right.
Let's evaluate smaller portions of the circuit and calculate their equivalent resistance. And we will work our way thru the circuit using that equivalent resistance to form another portion of the circuit to form another equivalent resistance incorporating more of the original circuit. Call the first equivalent resistance determined #R_"eq1"#, the next #R_"eq2"# and so on.
Now, notice the 3 resistors in the path #P_1#, #P_2#, #R_1#, and #R_2#. Those 3 resistors are in series, so the equivalent resistance, #R_"eq1"#, of those 3 is given by
#R_"eq1" = 10Omega+10Omega+10Omega = 30 Omega#
Those #30 Omega#s are in parallel with the #10 Omega# directly between #P_1# and #R_1#. The equivalent resistance, #R_"eq2"#, of that parallel combination is given by
#R_"eq2" = (30 Omega * 10 Omega)/(30 Omega + 10 Omega) = 7.5 Omega#
The resistor between #P and P_1#, the equivalent resistance, #R_"eq2"#; and the resistor between#R and R_1#, form a series combination. The equivalent resistance, #R_"eq final"#, of that series combination will be the final answer to the question. It is given by
#R_"eq final" = 10Omega+7.5Omega+10Omega = 27.5Omega#
I hope this helps,
Steve
