ITPEC IP April 2026 Question 4
ITPEC IP April 2026 — Question 4 of 100
10011011 — decompose 155 into powers of two from 128 down to 1.
155 = 128 + 16 + 8 + 2 + 1. Therefore the bits for 128, 16, 8, 2, 1 are 1, while the bits for 64, 32, 4 are 0:
- •Weights:
128 64 32 16 8 4 2 1 - •Bits:
1 0 0 1 1 0 1 1
This gives 10011011.
Answer (a)
Why not others:
- (b) 10110011 = 128 + 32 + 16 + 2 + 1 = 179
- (c) 11001101 = 128 + 64 + 8 + 4 + 1 = 205
- (d) 11011001 = 128 + 64 + 16 + 8 + 1 = 217
Key rule: Label binary positions with powers of two and add the weights whose bits are 1; this both constructs and verifies the representation.
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