21
Number Bases
Paper 1 1.
3.
Numbers in base eight = 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, …
2368 = (2 × 8 2) + (3 × 8 1) + (6 × 8 0) = 15810 5 158 Remainder
Answer: D
2.
Digits used for the numbers of base base ve ve = 0, 1, 2, 3, 4
5
31 ........3
5
6 ........1
5
1 ........1 0 ... ...... .....1 ..1
Answer: B
Hence, 2368 = 11135 3.
110112 = (1 × 2 4) + (1 × 2 3) + (0 × 2 2) + (1 × 2 1) + (1 × 2 0)
Answer: C
Answer: A
4.
4.
4138 = (4 × 8 2) + (1 × 8 1) + (3 × 8 0) Answer: C
110012 − 112 –––––––––– 101102 –––––––––– Answer: B
5.
14235 = (1 × 5 3) + (4 × 5 2) + (2 × 5 1) + (3 × 5 0) 5.
Answer: D
53 + 4 = (1 × 5 3) + (0 × 5 2) + (0 × 5 1) + (4 × 5 0) = 10045 Answer: B
6.
Paper 1 1.
11 010 0112 = 3
2
38
111112 + 1102 ––––––––– 1001012 ––––––––– Answer: C
Answer: D
7. 2.
1101002 – 100112 –––––––––– 1000012 ––––––––––
110102 − 11012 ––––––––– 11012 ––––––––– Answer: A
Answer: D
8.
Value of digit 6 = 6 × 8 2 = 384 Answer: C
1
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Penerbitan Pelangi Sdn. Bhd.
Mathematics
9.
SPM
Chapter 21 3.
1100112 − 11012 –––––––––– 1001102 ––––––––––
Value of the rst digit 1 = 1 × 2 7 = 128 Answer: B
Answer: B
4. 10.
Value of digit 3 = 3 × 5 2 = 75
Value of digit 2 = 2 × 8 2 = 128 Answer: C
Answer: B
5. 11.
5(53 + 5) = 5 4 + 52 = (1 × 5 4) + (0 × 5 3) + (1 × 5 2) + (0 × 5 1) + (0 × 5 0) = 101005
Answer: D
6.
Answer: A
12.
1012 + 11012 –––––––– 100102 ––––––––
101112 = (1 × 2 4) + (0 × 2 3) + (1 × 2 2) + (1 × 2 1) + (1 × 2 0) = 16 + 4 + 2 + 1 = 2310 Answer: B
7.
Answer: C
13.
Value of digit 3 = 3 × 5 3 = 375
Value of digit 4 = 4 × 5 2 = 100
2145 = (2 × 5 2) + (1 × 5 1) + (4 × 5 0) = 50 + 5 + 4 = 59 10 Answer: A
Answer: B
8. 14.
110112 − 1102 ––––––––– 101012 ––––––––– Answer: C
10568 = (1 × 8 3) + (0 × 8 2) + (5 × 8 1) + (6 × 8 0) = 512 + 40 + 6 = 55810 Answer: C
9.
3245 = (3 × 5 2) + (2 × 5 1) + (4 × 5 0) = 75 + 10 + 4 = 89 10 Answer: B
Paper 1 1.
Value of digit 3 = 3 × 5 = 15
10. 1
Answer: C
5 164
Remainder
5
32 ........4
5
6 ........2
5
1 ........1 0 ... ...... .....1 ..1
2.
Value of digit 4 = 4 × 8 2 = 256
Hence, 16410 = 11245 Answer: B
Answer: B
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2
Mathematics
11.
2 182
16.
Remainder
x8
Chapter 21
= 1 111 1012
2
91 ........0
2
45 ........1
Hence, x = 175
2
22 ........1
Answer: C
2
11 ........0
2
5 ........1
2
2 ........1
2
1 ........0
= 1
17.
SPM
7
58
P10 =
6758 = (6 × 8 2) + (7 × 8 1) + (5 × 8 0) = 44510
Hence, P = 445
0 ... ...... .....1 ..1
Answer: B
Hence, 18210 = 101101102 18.
Answer: A
12.
5 500
Remainder
Hence, n = 9
5 100 ........0 5
20 ........0
5
4 ........0
Answer: D
19.
0 ... ...... .....4 ..4
20. Remainder
2
56 ........1
2
28 ........0
2
14 ........0
2
7 ........0
2
3 ........1
2
1 ........1
2
18
Answer: C
Answer: D
2 113
1
= 1 010 0012
Hence, 50010 = 40005
13.
1n610 = 3048 = (3 × 8 2) + (0 × 8 1) + (4 × 8 0) = 19610
778 + 18 = 1008 Answer: C
21.
Value of digit 2 = 2 × 8 2 Hence, m = 2 Answer: B
22.
0 ... ...... .....1 ..1
268 x8 328 …, 268, 27 8, 30 8, 318, 32 8, … Hence, x = 27, 30, 31
Hence, 11310 = 11100012
Answer: B
Answer: B
14.
25 + 22 + 1 = (1 × 25) + (0 + (1 × 2 0) = 1001012
23. ×
24) + (0 × 2 3) + (1 × 2 2) + (0 × 2 1)
2435 = (2 × 5 2) + (4 × 5 1) + (3 × 5 0) = 50 + 20 + 3 = 7310 8
Answer: A
15.
x
2 + 2 = 1000102 = (1 × 2 5) + (0 × 2 4) + (0 + (0 × 2 0) = 2 5 + 2
×
73
Remainder
8
9 ........1
8
1 ........1 0 ... ...... .....1 ..1
23) + (0 × 2 2) + (1 × 2 1) Hence, 2435 = 1118 Answer: A
Hence, x = 5 Answer: A
3
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Penerbitan Pelangi Sdn. Bhd.
Mathematics
24.
SPM
Chapter 21
1005 = (1 × 5 2) + (0 × 5 1) + (0 × 5 0) = 25 10
27.
Remainder
2
25
2
12 ........1
2
6 ........0
2
3 ........0
2
1 ........1
Answer: C
28.
0 ... ...... .....1 ..1 Hence,
x =
10102 + 11112 ––––––––– 110012 –––––––––
11001
1110112 1001012 − ––––––––––– 101102 ––––––––––– Answer: B
Answer: B
25.
4215 = (4 × 5 2) + (2 × 5 1) + (1 × 5 0) = 100 + 10 + 1 = 11110 8 111
29.
13 ........7
8
1 ........5
−
a2 =
110012 1101112 − 110012 = 111102
a2 =
Answer: C
Remainder
8
1101112
30.
0 ... ...... .....1 ..1 4215 = 1578
11112 +
m2
= 1010112 m2 = 1010112 = 111002
Answer: C
Hence, m = 5 Answer: C
26.
1101112 + 1110112 ––––––––––– 11100102 ––––––––––– Answer: B
32.
1110012 − 1101112 ––––––––––– 102 –––––––––––
1110102 = (1 × 2 5) + (1 × 2 4) + (1 × 2 3) + (0 × 2 2) + (1 × 2 1) + (0 × 2 0) = 32 + 16 + 8 + 2 = 5810 Remainder
5
58
5
11 ........3
5
2 ........1 0 ... ...... .....2 ..2
Answer: A
1110102 = 2135 Hence, x = 213 Answer: D
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31.
Penerbitan Pelangi Sdn. Bhd.
4
−
11112
