A MAC address is how many bits long?

The correct answer is A, Attack Surface. The attack surface of a system is the total collection of every point at which an unauthorised person could try to enter it or pull data out of it: open ports, running services, user accounts, login pages, removable drives and even the employees who can be tricked. Network administrators reduce risk by shrinking this surface, that is by closing unused ports and removing software that is not needed. Option B is wrong because an attack signature is the recognisable pattern of bytes or behaviour by which antivirus and intrusion detection software identify a known attack. Option C is wrong because an attack pattern is a general description of how a class of attacks is carried out, used while designing defences. Option D is wrong because an attack path, also called an attack vector chain, is the particular route a single intruder actually followed from entry to target, not the whole set of possible entry points.

The correct answer is A, Modulator. A modulator changes the form of a signal: it loads the digital bits produced by a computer onto an analog carrier wave so that they can travel over a telephone line, a cable or a radio link. The reverse device, a demodulator, recovers the bits at the far end, and the two together make the modem, whose very name is short for modulator-demodulator. Option B is wrong because a mouse is a pointing input device; it sends position and click data to the computer but does not convert one form of signal into another. Option C is wrong because a plotter is an output device that draws large line diagrams, maps and engineering drawings with pens on paper. Option D is wrong because a keyboard is an input device that turns keystrokes into character codes for the processor, which is coding rather than signal conversion between digital and analog forms.

The correct answer is A, General Packet Radio Service. GPRS was added to the 2G cellular system so that the same radio channel could carry data in packets instead of holding a circuit open the way a voice call does. Because the user is charged for the volume of data rather than for the minutes of connection, it was called an always-on service, and it made mobile browsing, WAP pages and picture messaging possible; the faster EDGE and then 3G grew out of it. Option B is wrong because the R stands for radio, the medium used, not for route. Option C is wrong because the unit of data in this system is a packet, and package is not a networking term here. Option D is wrong because radiation has nothing to do with the expansion; it only resembles the word radio and is placed there as a distractor.

The correct answer is C, Star topology. In a star network every computer has its own cable running to a central hub or switch, so the shape resembles the spokes of a wheel. Adding or removing one machine disturbs nobody else and a broken cable affects only that one node, which is why almost every office and laboratory LAN is wired this way; the weakness is that failure of the central device stops the whole network. Option A is wrong because a bus topology has all nodes tapping one common backbone cable with terminators at its two ends. Option B is wrong because in a ring the machines form a closed loop, each joined to two neighbours, and data circulates in one direction using a token. Option D is wrong because a mesh links every node directly to every other node, needing n(n-1)/2 cables for n nodes and having no central device at all.

The correct answer is B, 15. In a full mesh every node is joined directly to every other node, so the number of links equals the number of pairs that can be formed from the nodes, given by the formula n(n-1)/2. Putting n equal to 6 gives 6 multiplied by 5, that is 30, divided by 2, which equals 15 cables; each machine also needs 5 network ports, so the port count is n(n-1) in total. Option A is wrong because 12 would be twice the number of nodes and has no basis in the formula. Option C is wrong because 18 is three times the number of nodes and again does not follow from counting pairs. Option D is wrong because 30 is the value of n(n-1), which counts each link twice, once from either end, and must be halved since one cable serves both directions.