nickgailloux wrote: » Using this range you can eliminate all remaining answers but two: 136.28.90.34 and 136.28.86.63. Well we know that the latter of those two is already in use by Server1, so, the only available address not in use, thats on the same subnet, is 136.28.90.34.
nickgailloux wrote: » Since we know these have to be on the same subnet, one could make the assumption that the subnet mask will use the greatest amount of bits possible using these two addresses: 136.28.86.63 = xxxxxxxx.xxxxxxxx.01010110.xxxxxxxx 136.28.90.34 = xxxxxxxx.xxxxxxxx.01011010.xxxxxxxx Subnet Mask = = xxxxxxxx.xxxxxxxx.11110000.xxxxxxxx