Question: Rajah 1 menunjukkan empat keping kad nombor.
Diagram 1 shows four number cards.
312_5 1043_5 34_5 203_5
Rajah 1 / Diagram 1
Cari beza, dalam asas 5, antara nombor terbesar dengan nombor terkecil.
Find the difference, in base 5, between the largest and the smallest number.
Jawapan/Answer:
[2 markah/marks]
3 Sebuah van sekolah perlu berulang-alik sebanyak tiga kali untuk menghantar murid ke sekolah. Rajah 2
merupakan tempoh masa, dalam minit, yang diambil bagi ketiga-tiga perjalanan itu.
A school van needs to make three trips to drop students at school. Diagram 2 shows the period of time, in minutes,
for all three trips.
401_5 1212_4 254_6
Rajah 2 / Diagram 2
Tentukan tempoh masa, dalam minit, yang paling pendek van tersebut menghantar murid ke sekolah,
dalam asas dua.
Determine the shortest period of time, in minutes, for the van to send to students at school, in base two.
[3 markah/marks]
Jawapan/Answer:
1. **Problem 1:** Find the difference in base 5 between the largest and smallest numbers from the cards: $312_5$, $1043_5$, $34_5$, $203_5$.
2. Convert each base 5 number to base 10 to compare:
- $312_5 = 3 \times 5^2 + 1 \times 5^1 + 2 \times 5^0 = 3 \times 25 + 1 \times 5 + 2 = 75 + 5 + 2 = 82$
- $1043_5 = 1 \times 5^3 + 0 \times 5^2 + 4 \times 5^1 + 3 \times 5^0 = 1 \times 125 + 0 + 20 + 3 = 148$
- $34_5 = 3 \times 5^1 + 4 \times 5^0 = 15 + 4 = 19$
- $203_5 = 2 \times 5^2 + 0 \times 5^1 + 3 \times 5^0 = 50 + 0 + 3 = 53$
3. Identify the largest and smallest numbers in base 10:
- Largest: $148$
- Smallest: $19$
4. Find the difference in base 10:
$$148 - 19 = 129$$
5. Convert $129$ back to base 5:
Divide $129$ by $5$:
- $129 \div 5 = 25$ remainder $4$
- $25 \div 5 = 5$ remainder $0$
- $5 \div 5 = 1$ remainder $0$
- $1 \div 5 = 0$ remainder $1$
So, $129_{10} = 1004_5$
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6. **Problem 2:** Find the shortest time in minutes for the van trips given times in different bases: $401_5$, $1212_4$, $254_6$, and express the shortest time in base 2.
7. Convert each time to base 10:
- $401_5 = 4 \times 5^2 + 0 \times 5^1 + 1 \times 5^0 = 4 \times 25 + 0 + 1 = 101$
- $1212_4 = 1 \times 4^3 + 2 \times 4^2 + 1 \times 4^1 + 2 \times 4^0 = 64 + 32 + 4 + 2 = 102$
- $254_6 = 2 \times 6^2 + 5 \times 6^1 + 4 \times 6^0 = 72 + 30 + 4 = 106$
8. Identify the shortest time in base 10:
- Shortest: $101$ minutes (from $401_5$)
9. Convert $101$ to base 2:
Divide $101$ by $2$:
- $101 \div 2 = 50$ remainder $1$
- $50 \div 2 = 25$ remainder $0$
- $25 \div 2 = 12$ remainder $1$
- $12 \div 2 = 6$ remainder $0$
- $6 \div 2 = 3$ remainder $0$
- $3 \div 2 = 1$ remainder $1$
- $1 \div 2 = 0$ remainder $1$
Reading remainders from bottom to top: $1100101_2$
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**Final answers:**
- Difference in base 5: $1004_5$
- Shortest time in base 2: $1100101_2$