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Answers to Student’s Book
Exam-style questions and sample answers have been written by the authors. References to
assessment and/or assessment preparation are the publisher’s interpretation of the syllabus
requirements and may not fully reflect the approach of Cambridge Assessment International
Education. Cambridge International recommends that teachers consider using a range of teaching
and learning resources in preparing learners for assessment, based on their own professional
judgement of their students’ needs.
Cambridge Assessment International Education bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.
The questions, example answers, marks awarded and/or comments that appear in this digital
material were written by the author(s). In examination, the way marks would be awarded to
answers like these may be different.
1 Data representation
Answers to activities
1.1
a 51
b 127
c 153
d 116
e 255
f 15
g 143
h 179
i 112
j 238
k 487
l 1364
m 3855
n 1992
o 2047
p 31984
q 16141
r 49983
s 34952
t 32767
1.2
a 00101001
b 01000011
c 01010110
d 01100100
e 01101111
f 01111111
g 10010000
h 10111101
i 11001000
j 11111111
k 1000000011101000
l 0000001101111000
m 0000111111111111
n 0100000000010000
o 1111001101100011
1.3
a C3
b F7
c 27F
d 4EE
e 1E1
f 89E
g (0)4FE
h (0)E9C
i FF7D
j 67AE
1.4
a 0110 1100
b 0101 1001
c 1010 1010
d 1010 0000 0000
e 0100 0000 1110
f 1011 1010 0110
g 1001 1100 1100
h 0100 0000 1010 1010
i 1101 1010 0100 0111
j 0001 1010 1011 0000
1.5
a 107
b 156
c 74
d 255
e 511
f 2561
g 2996
h 3240
i 4782
j 44425
1.6
a 62
b E3
c 1EA
d 1FF
e 33A
f 3E8
g A4A
h E9F
i FA7
j 1388
1.7
1 Student investigation
2 a 35, 37, 8B
b C9, 7A, CC
c C, 6F, 51
1.8
a 10000011
b 01100110
c 01111011
d 10110110
e 01101111
f 10110111
g 01111110
h 01110000
i 11111110
j 11011101
1.9
a 98 = 01100010 and 15 = 00001111; sum = 01110001 (= 113)
b 29 = 00011101 and 88 = 01011000; sum = 01110101 (= 117)
c 49 = 00110001 and 100 = 01100100; sum = 10010101 (= 149)
d 51 = 00110011 and 171 = 10101011; sum = 11011110 (= 222)
e 82 = 01010010 and 69 = 01000101; sum = 10010111 (= 151)
f 100 = 01100100 and 140 = 10001100; sum = 11110000 (= 240)
g 19 = 00010011 and 139 = 10001011; sum = 10011110 (= 158)
h 203 = 11001011 and 30 = 00011110; sum = 11101001 (= 233)
i 66 = 01000010 and 166 = 10100110; sum = 11101000 (= 232)
j 211 = 11010011 and 35 = 00100011; sum = 11110110 (= 246)
1.10
1 a 89 = 01011001 and 175 = 10101111; sum = 1 00001000 (= 8)
overflow error has occurred (9 bits)
b 168 = 10101000 and 99 = 01100011; sum = 1 00001011 (= 11)
overflow error has occurred (9 bits)
c 88 = 01011000 and 215 = 11010111; sum = 1 00101111 (= 47)
overflow error has occurred (9 bits)
2 a 1101 1111 0010 1010 – result is correct
b 1 1110 1011 1110 0100 – overflow error has occurred
1.11
1 a denary value = 64 + 32 + 8 = 104
0 0 0 0 1 1 0 1
b denary value is 13 (104 ÷ 8) therefore shift three places right is the same as division
by 23 (i.e. 8)
c denary value = 15
1 1 1 1 0 0 0 0
d denary value is 240 (15 × 16) therefore shift four places left is the same as multiplication
by 24 (i.e. 16)
2 a 29 = 00011101 and 51 = 00110011
b sum = 01010000
c 00001010
d 75 = 01001011
e 01010101
f 10101010
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