Hi,
In my view, when given only one specified key, on average, it will be half of the time. For example, given N blocks that we will check one by one (a linear search), if a key in the first block the time cost is 1, and if the key the last block it will be N. On average, the cost is N/2. But if there are multiple keys, on average we might check all the blocks, thus it is B(D+RC).
1.25 is from:
80% page occupancy => File size = 1.25 data size
Cheers,
Yu