- Ch 1Q1. (a) Express 126 and 132 as products of prime factors.
$126 = $ 检查
$132 = $ 检查
(b) Find the HCF and LCM of 126 and 132.
HCF = 检查
LCM = 检查💡 思路
$126 = 2 \times 3^2 \times 7$；$132 = 2^2 \times 3 \times 11$。
HCF: 取最小幂次 = $2 \times 3 = 6$。
LCM: 取最大幂次 = $2^2 \times 3^2 \times 7 \times 11 = 4 \times 9 \times 77 = 2772$. - Ch 2Q2. Evaluate each of the following.
(a) $(-16) + (-3) \times [17 - (-5)^2] = $ 检查
(b) $\left(-2\tfrac{1}{3} - 4\tfrac{3}{4}\right) \div \left[(-3)^3 + 7\tfrac{1}{6}\right] = $ 检查💡 思路
(a) $[17 - 25] = -8$；$(-3)(-8) = 24$；$-16 + 24 = 8$。
(b) 分子：$-\tfrac{7}{3} - \tfrac{19}{4} = -\tfrac{28+57}{12} = -\tfrac{85}{12}$；分母：$-27 + \tfrac{43}{6} = -\tfrac{162-43}{6} = -\tfrac{119}{6}$；
商：$-\tfrac{85}{12} \div (-\tfrac{119}{6}) = \tfrac{85}{12} \cdot \tfrac{6}{119} = \tfrac{510}{1428} = \tfrac{5}{14}$（约分 GCD=102；或 85/119 = 5/7, ×6/12 = 1/2 → 5/14）。 - Ch 4Q3. The temperature of a room is $m$ °C and the temperature of a freezer is $-n$ °C, where $m$ and $n$ are positive numbers. The temperature of an oven is 150 °C higher than three times the temperature of the room. Express, in terms of $m$ and $n$,
(a) the difference between the temperature of the room and that of the freezer → °C 检查
(b) the average of the two temperatures of the room and the freezer → °C 检查
(c) the temperature of the oven → °C 检查💡 思路
(a) $m - (-n) = m + n$。
(b) $\dfrac{m + (-n)}{2} = \dfrac{m-n}{2}$。
(c) $3m + 150$。 - Ch 3Q4. (a) (i) Express $\tfrac{4}{7}$ as a recurring decimal. → 检查
(a)(ii) Express $\tfrac{4}{7}$ as a decimal correct to 3 significant figures. → 检查
(b) Using a calculator, evaluate $\dfrac{\sqrt[3]{17} - 1.236}{1.85^2}$, giving your answer correct to 4 significant figures. → 检查💡 思路
(a)(i) 4/7 = 0.571428571428… → recurring block 「571428」, write as $0.\dot{5}7142\dot{8}$.
(a)(ii) 3 s.f. → 0.571 (because next digit is 4 → round down).
(b) $\sqrt[3]{17} \approx 2.5713$；分子 $\approx 1.3353$；分母 $= 3.4225$；商 $\approx 0.3901$ (4 s.f.)。 - Ch 3Q5. The price of an apartment is $408 995.
(a) Find the price of the apartment, giving your answer correct to the nearest hundred dollars. → $ 检查
(b) Find the price of the apartment, giving your answer correct to 2 significant figures. → $ 检查💡 思路
(a) 看十位 (9 → 进位)：408,995 → 409,000.
(b) 2 s.f.：第二位是 0，第三位是 8 → 进位 → 4.1 × 10⁵ = 410,000. - Ch 4Q6. (a) Express the product of $x$ and the sum of $2y$ and $3z$ in terms of $x$, $y$ and $z$. → 检查
(b) If $x = 5$, $y = -4$ and $z = 2$, find the value of the product. → 检查💡 思路
(a) Product = $x \times (2y + 3z) = x(2y + 3z) = 2xy + 3xz$.
(b) $5(2 \cdot (-4) + 3 \cdot 2) = 5(-8 + 6) = 5 \cdot (-2) = -10$。 - Ch 3Q7. In a supermarket, a bottle of shampoo costs $7.95, a can of pineapple costs $3.20 and a box of frozen chicken wings costs $13.40.
(a) Estimate the total price of the three items by rounding off each price to the nearest dollar. → $ 检查
(b) If Mrs Foo has $24.50, based on the estimated total price in (a), is it possible to conclude if she has enough money to buy the three items? Explain your answer.
💡 思路
(a) $7.95 ≈ 8$, $3.20 ≈ 3$, $13.40 ≈ 13$. 合计 $= 8 + 3 + 13 = \mathbf{\$24}$。
(b) 不能下结论。估算 $24 < \$24.50$ 看似够，但实际总价 $= 7.95 + 3.20 + 13.40 = \$24.55 > \$24.50$，所以钱不够。估算的误差可能让看似 "够" 的情况实际不够——做精确价格判断时要算实数。 - Ch 4Q8. The population $P$ of bacteria in a colony at time $t$ hours is given by the formula $P = ab^t$, where $a$ and $b$ are constants.
(a) Find the value of $P$ when $a = 456$, $b = 2$ and $t = 3$. → 检查
(b) Round off the answer in (a) to the nearest hundred. → 检查💡 思路
(a) $P = 456 \cdot 2^3 = 456 \cdot 8 = 3648$.
(b) 看十位 4 → 不进位 → 3600. - Ch 1Q9. (a) Express 315 as the product of its prime factors. → 检查
(b) John has 315 pieces of square tiles, each of side 30 cm long.
(i) Find the minimum number of such tiles he should buy more of so that he can lay a large square with all the tiles. → 检查
(ii) Hence find the area of the large square (in cm²). → cm² 检查
(c) There are 315 apples and 168 oranges. Each type of fruit is packed equally into small bags. Find the greatest possible number of fruit in a bag. → 检查💡 思路
(a) $315 = 3 \times 105 = 3 \times 3 \times 35 = 3^2 \times 5 \times 7$.
(b)(i) 最近的平方数 ≥ 315：$17^2 = 289 < 315$，$18^2 = 324 \geq 315$ → 多买 $324 - 315 = 9$ 块。
(b)(ii) 边长 $= 18 \times 30 = 540$ cm；面积 $= 540^2 = 291\,600$ cm² (= 29.16 m²)。
(c) 每袋数 × 袋数 = 总数；袋数 = HCF(315, 168) = 21 袋。每袋 = $315/21 + 168/21 = 15 + 8 = 23$ 个水果。 - Ch 1Q10. There is a group of soldiers. When they are lined in rows of 6, 8 or 15, there are 5 soldiers left.
(a) Find the minimum number of soldiers in the group. → 检查
(b) If the group can be exactly divided into small teams of 13 soldiers each, find the minimum number of soldiers in the group. → 检查💡 思路
(a) 余 5 → $N = \text{LCM}(6, 8, 15) \cdot k + 5 = 120k + 5$；最小 $k = 1 \Rightarrow N = 125$。
(b) 还要 $N$ 能被 13 整除。$120k + 5 \equiv 0 \pmod{13}$ → $3k + 5 \equiv 0$ → $k \equiv -5 \cdot 9 \equiv -45 \equiv 7 \pmod{13}$（因 $3 \cdot 9 = 27 \equiv 1$）。最小 $k=7 \Rightarrow N = 840 + 5 = 845$。Check: $845 / 13 = 65$ ✓。 - Ch 2Q11. In September, the local time of New York and that of Singapore was GMT−4 and GMT+8 respectively.
(a) When it is 13:00 on 2 September in Singapore, what is the local time in New York? → 检查
(b) An aeroplane takes 19 hours to fly between New York and Singapore.
(i) If it departs from New York at 23:00 on 3 September, find its date and time of arrival in Singapore. → 检查
(ii) If it departs from Singapore at 10:55 on 27 September, find its date and time of arrival in New York. → 检查💡 思路
SG 比 NY 快 12 小时（GMT+8 − GMT−4 = 12 hrs）。
(a) SG 13:00 → NY = 13:00 − 12hr = 01:00 on 2 Sept。
(b)(i) NY 起飞 23:00 (3 Sep) → NY 时间到达 23:00 + 19 = 18:00 (4 Sep) → SG 时间 = NY + 12 = 06:00 on 5 Sept。
(b)(ii) SG 起飞 10:55 (27 Sep) → SG 时间到达 10:55 + 19 = 05:55 (28 Sep) → NY 时间 = SG − 12 = 17:55 on 27 Sept。 - Ch 4Q12. There was a queue of 100 cars at the Woodlands Checkpoint. Assume that each car was $x$ m long and that the distance between any two adjacent cars was $y$ m. Let $z$ m be the distance from the front of the first car to the rear of the last car.
(a) Express $z$ in terms of $x$ and $y$. → 检查
(b) If $x = 4.5$ and $y = 0.6$, find the value of $z$. → $z = $ m 检查💡 思路
(a) 100 辆车 + 99 个间隔 (相邻车之间) → $z = 100x + 99y$。
(b) $100(4.5) + 99(0.6) = 450 + 59.4 = 509.4$ m。 - Ch 1Q13. Let $m = 792$ and $n = 2 \times 3^4 \times 7^2 \times 11$.
(a) Express $m$ as the product of its prime factors. → 检查
(b) Find the HCF and LCM of $m$ and $n$.
HCF = 检查
LCM = 检查
(c) Given that $a$, $b$ and $c$ are prime numbers and $\left(mn \times \dfrac{bc}{a}\right)$ is a perfect cube, determine whether $(mn \times a^2 bc)$ is a perfect cube. Explain your answer.
💡 思路
(a) $792 = 2^3 \cdot 99 = 2^3 \cdot 3^2 \cdot 11$.
(b) $m = 2^3 \cdot 3^2 \cdot 11$, $n = 2 \cdot 3^4 \cdot 7^2 \cdot 11$。HCF = min 幂 = $2 \cdot 3^2 \cdot 11 = 198$。LCM = max 幂 = $2^3 \cdot 3^4 \cdot 7^2 \cdot 11 = 8 \cdot 81 \cdot 49 \cdot 11 = 349\,272$。
(c) 这是一道开放思考题。$mn = 2^4 \cdot 3^6 \cdot 7^2 \cdot 11^2$。$mn \cdot bc/a$ 是完全立方 → 每个质因数的幂都是 3 的倍数。当 a, b, c 都是质数时，需具体分析它们的取值（a 出现在分母，b, c 在分子），看每个质因数的幂能否凑成 3 的倍数。 Then $mn \cdot a^2 \cdot bc = (mn \cdot bc/a) \cdot a^3$，乘以 $a^3$ 不改变 "是不是立方" 的性质 → 仍然是完全立方。 - Ch 3Q14. (a) Let $x = 0.\dot{4}\dot{9}$ (i.e. 0.4949…).
(i) Find the value of $100x$. → 检查
(ii) Convert $0.\dot{4}\dot{9}$ to a fraction. → 检查
(b) Evaluate $\sqrt[3]{\dfrac{3375}{-8}} + 7.\dot{4}\dot{9}$ without using a calculator. Express your answer as a fraction in its simplest form. → 检查💡 思路
(a)(i) $100x = 49.4949\ldots = 49 + x$。
(a)(ii) $100x - x = 49 \Rightarrow 99x = 49 \Rightarrow x = \dfrac{49}{99}$。
(b) $3375 = 15^3$, $-8 = (-2)^3$，所以 $\sqrt[3]{\dfrac{3375}{-8}} = \dfrac{15}{-2} = -\dfrac{15}{2}$。
$7.\dot{4}\dot{9} = 7 + \dfrac{49}{99} = \dfrac{693 + 49}{99} = \dfrac{742}{99}$。
Total $= -\dfrac{15}{2} + \dfrac{742}{99} = \dfrac{-15 \cdot 99 + 742 \cdot 2}{2 \cdot 99} = \dfrac{-1485 + 1484}{198} = \dfrac{-1}{198} = -\dfrac{1}{198}$。 - Ch 4Q15. Gymnasium A charges a monthly fee of $20 and a charge of $5 per visit. Gymnasium B does not charge a monthly fee but the charge per visit is $8. Let $C_1$ and $C_2$ be the total monthly expenditure if you visit gymnasiums A and B respectively. Given that you go to a gymnasium $n$ times in a month,
(a) express $C_1$ in terms of $n$. → $C_1 = $ $ 检查
(b) express $C_2$ in terms of $n$. → $C_2 = $ $ 检查
(c) find the values of $C_1$ and $C_2$ when $n = 6$.
$C_1 = $ $ 检查; $C_2 = $ $ 检查
(d) find the possible (positive integer) values of $n$ such that $C_2 < C_1$.
$n \in $ 检查💡 思路
$C_2 < C_1 \Leftrightarrow 8n < 20 + 5n \Leftrightarrow 3n < 20 \Leftrightarrow n < \tfrac{20}{3} \approx 6.67$。
正整数 $n \in \{1, 2, 3, 4, 5, 6\}$。
即去 1–6 次时 B 更便宜；7 次起 A 更划算（因月费摊得开）。
🎉 完成 15 题恭喜！ 你已经走完了第 1–4 章的复习。下一站：第 5 章 Simple Equations in One Variable（一元简单方程）——把今天学的代数表达式真正变成方程来解。