Okay... let's redo q25 a i and ii again step by step:

The packets are:

**0101100101 0110100111**
Rules: the 1st digit and the last digit are start and end bits.

the 2nd digit is the direction X and Y.

In X direction: 1 is right, 0 is left

In Y direction: 1 is up, 0 is down

i. Spliting the digits away they become:

0 (start bit) 1 (right) 0110010 (mm) 1 (end bit)

0110010 = 50 base 10

therefore x direction: right 50 mm

similarly with the 2nd packet:

0 (start bit) 1 (up) 1010011 (mm) 1 (end bit)

1010011 = 83 base 10

therefore y direction: up 83 mm

ii. as for the checksum for the data above, the checksum bits are the sum of the two packets... In the exam i didn't include the direction into the calculation. I did 83 + 50 = 133 / 13 and get a reminder of 3 (sorry when i said i got 13 as the remainder).

If you included the directions into the addition, i.e:

10110010 + 11010011 then you should get 0 as the remainder.

The question is now... were u suppose to include them or not?