Set 10B Q6
10101010 (twos complement) in decimal is...
=-128 + 32 + 8 + 2
= -128 + 42
= -86 which is answer A
Set 10C Q7
Multiplying by 4 in binary is achieved by....
Try an example, say 8 * 4 =32 in binary
8 = 00001000 in 8 bit binary
32 = 00100000 in 8 bit binary
To get from 00001000 to 00100000 we've shifted all the bits 2 places to the left, so D is the answer.
Set 10C Q8
Dividing by 8 in binary is achieved....
This is the opposite of Q7 except 8 is 2^3 so we shift 3 places to the right. (Answer B)
Set 10C Q9
1/5 is approximately 1/8 + 1/16 + 1/128 so the binary division
1100100 / 101 could be performed....
Finding 1/5 of something is equivalent to dividing by 5 and 101 is 5 in binary.
So an approximation for dividing by 5 is to divide the original thing by 8 (shift 3 places right in binary), divide the original thing by 16 (shift 4 to the right), divide the original thing by 128 (shift 7 to the right). And then add up these three results. Answer A.
Try it in decimal say, 20/5 (which equals 4)
20/8 + 20/16 + 20/128 = 2.5 + 1.25 + 0.15625 = 3.90625 which is about 4. The more fractions you use the better the answer becomes.
Set 10C Q10
-100 + -100 in 8 bit twos complement, ignoring overflows...
-100 = -128 + 16 + 8 + 4 which in 8 bit twos comp binary is 10011100
10011100 +
10011100
----------------
100111000 but we lose the leading one as we've only got 8 bits, so the result is 00111000 which in decimal is 32+16+8 = 56. So the answer is C.
We get the answer 56 because the system has "overflowed". It's gone round the loop, so to speak. The answer in decimal is -200, what's happened is that we've gone past -128 and back into the positives, through 127, 126, etc... in fact we've gone past 72 positives and ended up at 56. (NB. -128 - 72 = -200).
HTH
Sam
