📚 Essential Maths 7 Core: Question Types and Answers Explained | Essential Maths 7 Core 答案题型解析
The "Essential Maths 7 Core" book is widely used in KS3 classrooms to build foundational skills. In this guide, we break down the typical question types found in its exercises, explaining step‑by‑step how to arrive at the correct answers. Whether you are a student checking your work or a parent helping at home, understanding these core problem‑solving methods will strengthen your mathematical confidence.
《Essential Maths 7 Core》是 KS3 阶段广泛使用的基础练习册。本文将书中常见题型进行分类解析，逐步展示如何得出正确答案。无论你是自查作业的学生，还是在家辅导的家长，掌握这些核心题型的解题方法都将助你提升数学信心。
1. Whole Number Calculations | 整数计算
Example problem: A library had 2,365 books last year. This year it received 1,847 new books and gave away 532 old books. How many books does the library have now?
例题：一个图书馆去年有 2,365 本书。今年它新购入了 1,847 本书，并淘汰了 532 本旧书。现在图书馆有多少本书？
= 11, write 1, carry 1; 3 + 8 + 1 = 12, write 2, carry 1; 2 + 1 + 1 = 4. The sum is 4,212.
步骤 1：将新书数量与去年存量相加。列竖式对齐数位，个位加起。5 + 7 = 12，个位写 2，向十位进 1。继续：6 + 4 + 1（进位）= 11，十位写 1，向百位进 1；3 + 8 + 1 = 12，百位写 2，向千位进 1；2 + 1 + 1 = 4。总和为 4,212。
rom thousands (making hundreds 11). 11 − 5 = 6. Thousands: 3 − 0 = 3. The answer is 3,680.
步骤 2：从 4,212 中减去淘汰的 532 本。列竖式，个位：2 − 2 = 0。十位：1 − 3 不够减，从百位借 1（十位变为 11），11 − 3 = 8。百位：被借后剩 1 − 5 不够，从千位借 1（百位变为 11），11 − 5 = 6。千位：3 − 0 = 3。答案为 3,680。
Common mistake: forgetting to carry or borrow correctly always leads to an error in the final result. Double‑check each column.
常见错误：进位或借位时疏忽会导致最终结果错误。务必逐列复查。
2. Fractions and Mixed Numbers | 分数与带分数
Example: Work out 3 ⅓ + 2 ¼. Give your answer as a mixed number in its simplest form.
例题：计算 3 ⅓ + 2 ¼。用最简带分数表示答案。
t both mixed numbers to improper fractions. 3 ⅓ = (3×3+1)/3 = 10/3. 2 ¼ = (2×4+1)/4 = 9/4.
步骤 1：将两个带分数转为假分数。3 ⅓ = (3×3+1)/3 = 10/3。2 ¼ = (2×4+1)/4 = 9/4。
for 3 and 4, which is 12. Convert: 10/3 = (10×4)/(3×4) = 40/12. 9/4 = (9×3)/(4×3) = 27/12.
步骤 2：找出 3 和 4 的公分母为 12。转换：10/3 = (10×4)/(3×4) = 40/12，9/4 = (9×3)/(4×3) = 27/12。
Step 3: Add the numerators: 40/12 + 27/12 = 67/12.
步骤 3：分子相加：40/12 + 27/12 = 67/12。
ve times (5×12=60), remainder 7. So 5 ⁷⁄₁₂. The fraction 7/12 is already in simplest form.
步骤 4：转回带分数。67 ÷ 12 = 5 余 7，所以为 5 ⁷⁄₁₂。分数 7/12 已是最简。
Remember to always reduce the final fraction if possible. In this case, 7 and 12 share no common factors.
切记最终分数需化简。本例中 7 和 12 无公因数。
3. Decimal Operations | 小数运算
Example: One chocolate bar costs £0.85. How much do 7 bars cost? If you pay with a £10 note, how much change do you receive?
例题：一块巧克力售价 £0.85。买 7 块要多少钱？若付 £10 钞票，应找回多少？
85 × 7 = 595. Since 0.85 has two decimal places, the product must have two decimal places: £5.95.
步骤 1：计算 0.85 × 7。先忽略小数点，视作 85 × 7 = 595。因 0.85 有两位小数，积也应有两位小数：£5.95。
Step 2: Find the change from £10. Set up subtraction: £10.00 − £ 10 − 0.05 = 0.05 in hundredths, 0.9 − 0.9 = 0 in tenths, 9 − 5 = 4 in units. The change is £4.05.
步骤 2：计算找回的零钱。£10.00 − £5.95，对齐小数点。借位：百分位 0 − 5 不够，向十分位借（十分位为 0，再向个位借），最终 10.00 变为 9 个 1、9 个 0.1 和 10 个 0.01。百分位：10 − 5 = 5；十分位：9 − 9 = 0；个位：9 − 5 = 4。找回 £4.05。
Alternatively, think of counting up from £5.95 to £10: add 5p to £6.00, then add £4.00 to reach £10, total £4.05.
另一种思路：从 £5.95 加到 £10。先加 5p 到 £6.00，再加 £4.00 到 £10，共 £4.05。
4. Percentages and Equivalent Values | 百分比与等值转换
Example: There are 650 students in a school. 24% of them walk to school. How many students walk? Also express 24% as a fraction in its simplest form.
例题：一校有 650 名学生，其中 24% 步行上学。步行人数是多少？同时将 24% 表为最简分数。
Step 1: To find 24% of 650, convert the percentage to a decimal: 24% = 650 × 20 + 650 × 4 = 13,000 + 2,600 = 15,600. Divide by 100: 156. So 156 students walk.
步骤 1：求 650 的 24%，将百分比转为小数：24% = 0.24。相乘：650 × 0.24。可先算 650 × 24 ÷ 100。650 × 24 = 650×20 + 650×4 = 13,000 + 2,600 = 15,600。除以 100 得 156。因此 156 名学生步行。
Step 2: Express 24% as a fraction: 24% dividing numerator and denominator by 4: 24÷4 = 6, 100÷4 = 25. The simplest form is 6/25.
步骤 2：将 24% 表为分数：24/100。用 4 约分：24÷4=6，100÷4=25，最简分数为 6/25。
It is useful to memorise common equivalents. Here is a quick reference table.
熟记常用等值关系会很有帮助，以下为速查表。
| Fraction | Decimal | Percentage |
|---|---|---|
| 1/2 | 0.5 | 50% |
| 1/4 | 0.25 | 25% |
| 3/4 | 0.75 | 75% |
| 1/5 | 0.2 | 20% |
| 1/10 | 0.1 | 10% |
5. Introducing Algebra | 代数入门表达式
Example: 5p + 3q − 2p + 7q. Then find the value of the simplified expression when p = 4 and q = 2.
例题：化简表达式：5p + 3q − 2p + 7q，并求当 p = 4、q = 2 时代数式的值。
Step 1: Group like terms. Terms with p: 5p and −2p; terms with q: 3q and +7q.
步骤 1：合并同类项。含 p 的项：5p 与 −2p；含 q 的项：3q 与 +7q。
Step 2: Combine them: 5p − 2p = 3p; 3q + 7q = 10q. The simplified expression is 3p + 10q.
步骤 2：合并计算：5p − 2p = 3p；3q + 7q = 10q。化简后为 3p + 10q。
Step 3: Substitute p = 4 and q = 2 into 3p + 10q.
3p + 10q = 3(4) + 10(2) = 12 + 20 = 32
步骤 3：将 p = 4 和 q = 2 代入 3p + 10q。
3p + 10q = 3(4) + 10(2) = 12 + 20 = 32
mon error: forgetting to multiply the number with the variable, e.g. 3p is 3×p, not 3 + p.
务必直接用系数与变量值相乘。常见错误：忘记乘法关系，例如 3p 表示 3×p，而非 3 + p。
6. Solving Simple Equations | 解简单方程
Example: Solve the equation 4x − 3 = 17. Show your working and check the answer.
例题：解方程 4x − 3 = 17。写出解题步骤并验算。
Step 1: Isolate the term with x by undoing the subtraction. Add 3 to both sides:
4x − 3 + 3 = 17 + 3 → 4x = 20
步骤 1：通过逆运算消去常数项，两边同时加 3：
4x − 3 + 3 = 17 + 3 → 4x = 20
Step 2: Undo the multiplication by 4. Divide both sides by 4:
4x ÷ 4 = 20 ÷ 4 → x = 5
步骤 2：用除法消去系数 4。两边同时除以 4：
4x ÷ 4 = 20 ÷ 4 → x = 5
l equation: 4(5) − 3 = 20 − 3 = 17, which matches the right side. The solution is correct.
步骤 3：将 x = 5 代回原方程验算：4(5) − 3 = 20 − 3 = 17，与等号右边一致。解答正确。
For two‑step equations, always undo addition/subtraction first, then multiplication/division. This method works for all similar equations in Essential Maths 7 Core.
解两步方程时，总是先处理加减法，再处理乘除法。这个方法适用于《Essential Maths 7 Core》中的所有同类方程。
7. Geometry: Angles and Symmetry | 几何：角与对称
Example: In a triangle, two angles are 35° and 82°. Find the third angle x. Explain why a triangle cannot have two right angles.
例题：一个三角形中两个角分别为 35° 和 82°，求第三个角 x。并解释为何一个三角形不可能有两个直角。
Step 1: Use the fact that angles in a triangle sum to 180°. x = 180° − (35° + 82°).
步骤 1：利用三角形内角和 180°。x = 180° − (35° + 82°)。
ep 2: Add the known angles: 35 + 82 = 117. Subtract from 180: 180 − 117 = 63°. So x = 63°.
步骤 2：加上已知角：35 + 82 = 117。180 − 117 = 63°，所以 x = 63°。
Step 3: If a triangle had two right angles (90° each), their sum would already be 180°, leaving 0° for the third angle. That is impossible, so a triangle cannot contain two right angles.
步骤 3：若一个三角形有两个直角（各 90°），其和已达 180°，第三个角只能为 0°，这是不可能的。因此三角形不能有两个直角。
Similarly, angles on a straight line add to 180°. For example, if one angle is 120°, the other must be 60°.
类似地，平角直线上的角之和为 180°。例如，若一角为 120°，则另一角必为 60°。
8. Data Handling: Averages and Charts | 数据处理：平均数与图表
Example: The test scores of 10 students are: 7, 9, 6, 8, 9, 4, 7, 10, 8, 7. Find the mean, median, mode and range.
例题：10 名学生的测验分数为：7, 9, 6, 8, 9, 4, 7, 10, 8, 7。求平均数、中位数、众数和极差。
Step 1: Mean – add all scores: 7+9+6+8+9+4+7+10+8+7 = 75. Divide by 10: 75 ÷ 10 = 7.5.
步骤 1：平均数 – 所有分数相加：7+9+6+8+9+4+7+10+8+7 = 75。除以 10 得 7.5。
Step 2: Median – lues (even number), the median is the average of the 5th and 6th values. 5th = 7, 6th = 8,
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