📚 9 Core Answers for Prelims Paper 1 & 2 | 预考卷1与卷2九大核心题解答
This revision guide covers nine high-frequency question types in the Core prelims for Paper 1 and Paper 2. Each answer is worked through step by step so you can use the same method in the real exam. Work through all nine, then cover the solutions and repeat the questions under timed conditions.
本复习指南涵盖卷1与卷2预考中九类高频试题。每道核心题均给出分步骤详解，帮助你在正式考试中套用相同方法。建议先完成全部九题，再遮住答案，在限时条件下重做一遍。
1. Solving Linear Equations | 解线性方程
Solve the equation 3x – 7 = 2x + 5. Begin by collecting the x terms on one side and the constant terms on the other side.
解方程 3x – 7 = 2x + 5。先把含 x 的项集中到等式一边，常数项移到另一边。
Subtract 2x from both sides to get x – 7 = 5. Then add 7 to both sides to isolate x.
两边同时减去 2x，得到 x – 7 = 5；再两边同时加 7，即可解出 x。
x = 12
Always check your answer by substituting x = 12 back into the original equation. Both sides give 29, so the solution is correct.
务必把 x = 12 代回原方程检验。两边都得到 29，因此答案正确。
2. Linear Inequalities and Number Lines | 一元一次不等式与数轴表示
Solve the inequality 2x – 3 ≤ 7 and represent the solution on a number line. The same rules as linear equations apply, except that the direction changes if you multiply or divide by a negative.
解不等式 2x – 3 ≤ 7 并在数轴上表示解集。除乘以或除以负数需要改变不等号方向外，其余步骤与线性方程相同。
Add 3 to both sides to obtain 2x ≤ 10. Then divide both sides by 2 to get x ≤ 5.
两边加 3 得 2x ≤ 10；两边除以 2 得 x ≤ 5。
x ≤ 5
On a number line, place a closed circle at 5 and shade everything to the left. A closed circle means 5 is included in the solution set.
在数轴上用实心圆标出 5，并向左侧画阴影。实心圆表示 5 包含在解集内。
3. Factorising Quadratic Equations | 因式分解法解二次方程
Solve the quadratic equation x² – 5x + 6 = 0 by factorisation. Look for two numbers that multiply to the constant term 6 and add to the coefficient -5.
用因式分解法解二次方程 x² – 5x + 6 = 0。找出两个数，使它们的积为常数项 6，和为系数 -5。
The numbers are -2 and -3 because -2 × -3 = 6 and -2 + -3 = -5. Therefore the factorised form is (x – 2)(x – 3) = 0.
这两个数是 -2 和 -3，因为 -2 × -3 = 6，且 -2 + -3 = -5。因此因式分解为 (x – 2)(x – 3) = 0。
(x – 2)(x – 3) = 0 → x = 2 or x = 3
Set each factor equal to zero and solve separately. Both values must be substituted into the original equation to confirm they work.
分别令每个因式为零并解出 x。两个解都必须代回原方程，以确认它们成立。
4. Simultaneous Linear Equations | 二元一次方程组
Solve the simultaneous equations 2x + y = 10 and x – y = 1. Since the y-coefficients are +1 and -1, elimination is the quickest method.
解二元一次方程组 2x + y = 10 和 x – y = 1。由于 y 项系数分别为 +1 和 -1，使用消元法最快。
Add the two equations to eliminate y: (2x + y) + (x – y) = 10 + 1. This gives 3x = 11, so x = 11/3.
将两式相加消去 y：(2x + y) + (x – y) = 10 + 1，得到 3x = 11，因此 x = 11/3。
Substitute x = 11/3 into the second equation: 11/3 – y = 1. Rearranging gives y = 11/3 – 1 = 8/3.
把 x = 11/3 代入第二个方程：11/3 – y = 1，移项得 y = 11/3 – 1 = 8/3。
x = 11/3, y = 8/3
Elimination works best when one pair of coefficients is equal or opposite. Otherwise, multiply one or both equations first, or use substitution.
当某一未知数的系数相等或互为相反数时，消元法最有效。否则可先对方程进行倍增，或改用代入法。
5. Pythagoras Theorem in Right-Angled Triangles | 直角三角形中的勾股定理
A right-angled triangle has shorter sides of length 6 cm and 8 cm. Find the length of the hypotenuse. Pythagoras’ theorem states a² + b² = c², where c is the hypotenuse.
一个直角三角形的两条直角边长分别为 6 cm 和 8 cm。求斜边长度。勾股定理为 a² + b² = c²，其中 c 表示斜边。
Substitute the given values: c² = 6² + 8² = 36 + 64 = 100. Taking the positive square root gives c = 10.
代入已知值：c² = 6² + 8² = 36 + 64 = 100。开平方取正值，得 c = 10。
c = √(6² + 8²) = √100 = 10 cm
The hypotenuse is always the side opposite the right angle and is the longest side of the triangle. Do not forget to write the correct unit in the final answer.
斜边正对直角，是三角形中最长的一条边。最后答案不要忘记写上正确单位。
6. Trigonometry in Right-Angled Triangles | 直角三角形中的三角比
In a right-angled triangle, the angle is 30° and the hypotenuse is 12 cm. Find the length of the side opposite the 30° angle. Use sin θ = opposite / hypotenuse.
在直角三角形中，一个角为 30°，斜边长为 12 cm。求 30° 角的对边长度。使用 sin θ = 对边 / 斜边。
Since sin 30° = 1/2, the equation becomes opposite / 12 = 1/2. Multiply both sides by 12 to get opposite = 6 cm.
因为 sin 30° = 1/2，方程化为 对边 / 12 = 1/2。两边同乘 12，得到对边 = 6 cm。
opposite = 12 × sin 30° = 12 × 1/2 = 6 cm
Identify the correct ratio by looking at the given side, the unknown side and the given angle. Avoid rounding trigonometric values until the final calculation step.
根据已知边、未知边和已知角选择合适的三角比。在最终计算前，不要先对三角函数值取近似。
7. Probability Tree Diagrams with Replacement | 有放回的概率树状图
A bag contains 3 red balls and 2 blue balls. Two balls are drawn one at a time with replacement. Find the probability of drawing at least one red ball.
袋中有 3 个红球和 2 个蓝球。每次抽出一个球后放回，共抽两次。求至少抽到一个红球的概率。
For each draw, P(red) = 3/5 and P(blue) = 2/5. The fastest method is to use the complement: P(at least one red) = 1 – P(two blue).
每次抽取，P(红球) = 3/5，P(蓝球) = 2/5。最快的方法是使用补集：P(至少一个红) = 1 – P(两个蓝)。
P(two blue) = 2/5 × 2/5 = 4/25. Therefore the required probability is 1 – 4/25 = 21/25.
P(两个蓝) = 2/5 × 2/5 = 4/25。因此所求概率为 1 – 4/25 = 21/25。
P(at least one red) = 21/25
For ‘at least one’ questions, the complement rule is usually faster than listing all favourable paths. Along each branch, multiply probabilities; add between different paths if needed.
对于“至少一个”类问题，补集法通常比罗列所有有利路径更快。沿同一路径相乘，不同路径之间相加。
8. Statistics: Mean, Median, Mode and Range | 统计：平均数、中位数、众数和极差
A data set is given as 4, 7, 8, 8, 10, 12, 15. Find the mean, median, mode and range. Put the values in ascending order first for median and range.
一组数据为 4, 7, 8, 8, 10, 12, 15。求平均数、中位数、众数和极差。求中位数与极差前，先把数据按升序排列。
Mean = (4 + 7 + 8 + 8 + 10 + 12 + 15) / 7 = 64/7 ≈ 9.14. The median is the middle value, which is the 4th value, giving 8.
平均数 = (4 + 7 + 8 + 8 + 10 + 12 + 15) / 7 = 64/7 ≈ 9.14。中位数是中间位置的数，即第 4 个数，为 8。
The mode is the most frequent value, which is 8. The range is the maximum minus the minimum: 15 – 4 = 11.
众数是出现次数最多的值，为 8。极差为最大值减最小值：15 – 4 = 11。
Mean = 64/7, Median = 8, Mode = 8, Range = 11
Always arrange the data set in order before locating the median and range. A data set can have no mode, one mode or more than one mode.
求中位数和极差前一定要先排序。一组数据可以没有众数、有一个众数或多个众数。
9. Area and Circumference of a Circle | 圆的面积与周长
A circle has radius 5 cm. Find its area and circumference, giving your answers in terms of π. Use the formulae A = πr² and C = 2πr.
一个圆的半径为 5 cm。求其面积和周长，答案用 π 表示。使用公式 A = πr² 和 C = 2πr。
Substitute r = 5 into the area formula: A = π × 5² = 25π cm². For the circumference, C = 2π × 5 = 10π cm.
将 r = 5 代入面积公式：A = π × 5² = 25π cm²。周长公式中，C = 2π × 5 = 10π cm。
Area = 25π cm², Circumference = 10π cm
If the question says ‘in terms of π’, do not convert to a decimal. Leaving the answer in exact form avoids rounding errors and saves time.
如果题目要求“用 π 表示”，不要转化为小数。保留精确形式可以避免舍入误差，并节省考试时间。
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