Answer:
a) The exact starting address for Job Q is 43008 bytes
b) The size of the memory block is 46080 bytes
c) Resulting fragmentation is external
Explanation:
a) Job Q starting address
42 k ⇒ 42 × 1024 = 43008.
b) memory block has 3 k in fragmentation ∴ the size of the memory block is calculated as 3 k ⇒ 3 × 1024 = 3072.
The fact that it is a block means that we have end address = 3072 + 43008 = 46080.
c) External Fragmentation: It will come in a system that uses the relocatable dynamic partitions scheme.