I can't see any reason why the answer should be any of the possible values:
"type" is two bits, so assigning a value of 7 should leave type == 3
"count" is twelve bits, so assigning a value of 0x16 should leave count == 22
"reserved" is unassigned, so the value is random.
Bit breakdown:
Bit number
3 2 1 0
10987654321098765432109876543210
..............................xx type
..................xxxxxxxxxxxx.. count
xxxxxxxxxxxxxxxxxx.............. reserved
Puting the values in:
Bit number
3 2 1 0
10987654321098765432109876543210
..............................11 type == 0x03
..................000000010110.. count == 0x16
xxxxxxxxxxxxxxxxxx.............. reserved == unassigned
xxxxxxxxxxxxxxxxxx00000001011011 Total of "apple"
So Apple will end 0x05B, with some random crap at the top.
Even if for some weird reason the compiler packed the bits the other way
Bit number
3 2 1 0
10987654321098765432109876543210
11.............................. type == 0x03
..000000010110.................. count == 0x16
..............xxxxxxxxxxxxxxxxxx reserved == unassigned
11000000010110xxxxxxxxxxxxxxxxxx Total of "apple"
Then apple would start with 0xC05 then some random crap.
I think they got it wrong, and maybe the test was: Will he have the confidence to say "You are wrong"?