📚 Essential Maths Book 7F: Compressed Question Types Explained | 核心数学练习册7F：核心题型全解析
The Essential Maths Book 7F for Key Stage 3 covers a broad range of foundation topics. This compressed review highlights the most common question types found in the book, from place value and arithmetic to algebra, geometry and data handling. Each section below shows the question style you will actually meet, the method expected by the mark scheme, and a worked example. Working through these examples will help Year 7 students master key skills and prepare confidently for tests.
针对 KS3 编写的 Essential Maths Book 7F 覆盖了广泛的基础主题。这篇浓缩复习梳理了书中最常见的题型，涉及位值、算术、代数、几何和数据处理。下文每一节都会展示你在练习中真正会遇到的题型、评分标准所要求的方法，以及一道完整的例题解析。通过练习这些精选例题，七年级学生能够掌握核心技能，并为考试做好充分准备。
1. Whole Numbers and Place Value | 整数与位值
Typical questions ask you to read and write large numbers, such as 'Write 560 082 in words' or 'What is the value of the digit 7 in 3 472 105?'. The answer to the second question is 70 000, because the 7 sits in the ten-thousands column. You must be able to name every place value column: millions, hundred-thousands, ten-thousands, thousands, hundreds, tens and units.
常见题型要求读、写大数，例如 '用文字写出 560 082' 或 '数字 3 472 105 中，数位 7 的值是多少？'。第二题的答案是 70 000，因为 7 位于万位。你必须能够说出每一个位值栏的名称：百万、十万、万、千、百、十、个。
The worked answer to the first question is 'five hundred and sixty thousand and eighty-two'. Notice that the word 'and' links the thousands group to the final group, and that the zero in the hundreds column is not read out as a separate word. Questions like this test both reading accuracy and spelling.
第一题的完整答案是 'five hundred and sixty thousand and eighty-two'（五十六万零八十二）。注意 'and' 用来连接千位组与最后一组，而百位上的 0 不需要单独读出。这类题目同时考查阅读的准确性和拼写。
| Millions 百万 | Hundred-thousands 十万 | Ten-thousands 万 | Thousands 千 | Hundreds 百 | Tens 十 | Units 个 |
| 3 | 4 | 7 | 2 | 1 | 0 | 5 |
Another common question type asks you to order a list of numbers from smallest to largest, or to insert the correct symbol (less than or greater than) between two numbers. Always line the digits up by place value before comparing, and remember that a number with more digits is not automatically larger when decimals are involved.
另一类常见题型要求把一组数从小到大排序，或在两个数之间填入正确的符号（小于或大于）。比较之前一定要按位值对齐数字，并记住：当涉及小数时，位数更多的数并不一定更大。
2. Multiplying and Dividing by 10, 100 and 1000 | 乘除 10、100、1000
ch is the same as adding two zeros on the end. For example, 34 × 100 = 3400. For decimals you move the decimal point instead: 5.6 × 100 = 560.
乘或除以 10、100、1000 是另一项核心技能。当你把整数乘以 100，每个数字向左移动两个位值，效果等同于在末尾加两个零。例如，34 × 100 = 3400。对于小数则移动小数点：5.6 × 100 = 560。
0.38. The most common mistake is losing count of the zeros, so say the direction out loud: 'times means left, divide means right'.
除法方向相反。除以 100 会让每个数字向右移动两位，所以 4.7 ÷ 100 = 0.047，380 ÷ 1000 = 0.38。最常见的错误是数错零的个数，因此做题时把方向讲出来：'乘向左，除向右'。
34 × 100 = 3400 5.6 × 100 = 560 4.7 ÷ 100 = 0.047
Exam questions often dress this up as a word problem, for instance 'A box holds 100 pens. How many pens are in 34 boxes?' The arithmetic is identical, but you must choose the correct operation and remember to include the unit in your answer.
考试题经常把它包装成应用题，例如 '一盒装 100 支笔，34 盒共有多少支笔？' 运算完全相同，但你需要选择正确的运算，并记得在答案中写上单位。
3. Column Addition and Subtraction | 竖式加法与减法
Basic exercises require the standard written methods: column addition, column subtraction, short and long multiplication, and short division. For addition, align the numbers by place value and carry whenever a column totals 10 or more. A question such as 'Calculate 345 + 678' is worked step by step as follows.
基础练习要求使用标准的笔算方法：竖式加法、竖式减法、短乘/长乘法、短除法。加法时，按位值对齐数字，当某一位之和为 10 或更多时需进位。如 '计算 345 + 678' 逐步运算如下。
5 + 8 = 13, write 3 carry 1 → 4 + 7 + 1 = 12, write 2 carry 1 → 3 + 6 + 1 = 10
So the total is 1023. Writing the carried digit small and above the next column keeps your working tidy and earns method marks even if the final answer slips.
因此总和为 1023。把进位数字写得小一些，放在下一栏的上方，可以保持卷面整洁，即使最终答案出错也能拿到方法分。
is 0, you must borrow from the hundreds first. The answer is 344. Questions that look like '702 − 358' are sometimes given as missing-number puzzles, such as '345 + □ = 678', where you simply subtract to find the missing value.
减法使用借位。计算 702 − 358 时，无法从 2 中减去 8，需要向十位借位；由于十位是 0，必须先向百位借位。答案是 344。像 '702 − 358' 这样的题目有时会以填空形式出现，例如 '345 + □ = 678'，此时只需做减法即可求出缺失的数。
8 = 702. This one-second habit catches the majority of careless errors in the examination.
做完减法一定要用加法验算：344 + 358 = 702。这个只要一秒钟的习惯，能在考试中抓出大部分粗心错误。
4. Short and Long Multiplication | 短乘法与长乘法
3 = 24, giving 2422. Carrying is the only real difficulty, so keep the carried digits consistent in size and position.
短乘法是用一个数字去乘一个数。计算 346 × 7 时从右往左算：6 × 7 = 42，写 2 进 4；4 × 7 + 4 = 32，写 2 进 3；3 × 7 + 3 = 24，得到 2422。唯一的难点是进位，所以要统一进位数字的大小和位置。
47, first work out 23 × 7 = 161, then 23 × 40 = 920, and finally 161 + 920 = 1081. The zero in the second row is not optional decoration: it shows that you are multiplying by 4 tens, not by 4 units.
两位数长乘法需要分别乘以十位和个位，再把两个部分积相加。计算 23 × 47 时，先算 23 × 7 = 161，再算 23 × 40 = 920，最后 161 + 920 = 1081。第二行末尾的 0 不是可有可无的装饰：它表明你是乘以 4 个十，而不是 4 个一。
23 × 7 = 161 23 × 40 = 920 161 + 920 = 1081
A frequent exam twist is to ask for an estimate before the exact answer, for example 'Estimate 23 × 47 by rounding to the nearest ten' firms that 1081 is sensible. Estimations are usually worth one mark and are easy to score.
考试中常见的变式是先要求估算、再求精确值，例如 '把 23 × 47 四舍五入到最近的十位来估算'。这里 20 × 50 = 1000，可以验证 1081 是合理的。估算通常值 1 分，是最容易拿的分。
5. Short Division, Remainders and Long Division | 短除法、余数与长除法
hen 48 ÷ 6 = 8. The quotient is 78. When there is a remainder, the question may ask for it explicitly, as in 47 ÷ 5 = 9 remainder 2.
除以一位数时使用短除法。计算 468 ÷ 6 时从左往右算：4 ÷ 6 不够除，把 4 带到下一位组成 46；46 ÷ 6 = 7 余 4；再算 48 ÷ 6 = 8。商为 78。当出现余数时，题目可能明确要求写出余数，例如 47 ÷ 5 = 9 余 2。
get 14, bring down the 7 to make 147, then 21 × 7 = 147 exactly. The answer is 47 with no remainder.
除数是两位数时需要长除法。计算 987 ÷ 21 时，先想 98 里有几个 21（4 个），在 8 上方写 4，计算 21 × 4 = 84 并相减得到 14，把 7 落下来组成 147，再算 21 × 7 = 147，正好整除。答案是 47，没有余数。
Word problems in this section often require interpretation of the remainder. If 47 students travel in cars that each hold 5, the answer is not 9 remainder 2 but 10 cars, because the remaining 2 students still need a car. Reading the question carefully decides whether you round up or simply state the remainder.
本节的应用题常常需要解释余数的意义。如果 47 名学生乘坐每辆坐 5 人的汽车，答案不是 9 余 2，而是 10 辆车，因为剩下的 2 名学生仍然需要一辆车。仔细读题决定你是要向上取整，还是直接写出余数。
6. Negative Numbers and Order of Operations | 负数与运算顺序
line, so −3 + 7 = 4; subtracting a larger number moves you to the left, so 5 − 9 = −4. Multiplying or dividing two negatives gives a positive result: (−4) × (−3) = 12 and (−20) ÷ (−5) = 4.
负数题通常以温度或银行余额为背景。加上一个正数相当于在数轴上向右移动，所以 −3 + 7 = 4；减去一个更大的数则向左移动，所以 5 − 9 = −4。负数乘负数或负数除负数得正数：(−4) × (−3) = 12，(−20) ÷ (−5) = 4。
, not 35, while (3 + 4) × 5 = 35. Indices and roots are also examined: 2³ = 8, 5² = 25 and √49 = 7.
运算顺序依据 BIDMAS 规则考查：括号、乘方、乘除，最后加减。因此 3 + 4 × 5 = 23，而不是 35；而 (3 + 4) × 5 = 35。乘方与开方也在考查范围内：2³ = 8，5² = 25，√49 = 7。
3 + 4 × 5 = 23 (3 + 4) × 5 = 35 2³ = 8 √49 = 7
A common trap is a question such as 'insert brackets to make the statement true'. Work backwards from the target answer and test the two or three possible positions of the brackets rather than guessing at random.
常见的陷阱题是 '添加括号使等式成立'。这时应从目标答案倒推，逐一测试两三种可能的加括号位置，而不是随意乱猜。
7. Factors, Multiples, Primes and HCF/LCM | 因数、倍数、质数与最大公因数/最小公倍数
Factor questions ask which numbers divide exactly into a given number. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. Multiples are the results of the times table for that number: the first five multiples of 7 are 7, 14, 21, 28 and 35. A prime number has exactly two factors, itself and 1, so 2, 3, 5, 7, 11 and 13 are prime while 1 is not prime and 9 is not prime.
因数题问的是哪些数能整除给定的数。24 的因数是 1、2、3、4、6、8、12 和 24。倍数则是该数乘法口诀的结果：7 的前五个倍数是 7、14、21、28 和 35。质数恰好有两个因数，即它本身和 1，所以 2、3、5、7、11、13 是质数，而 1 不是质数，9 也不是质数。
The highest common factor (HCF) and lowest common multiple (LCM) are examined regularly. For 18 and 24 the HCF is 6 and the LCM is 72. A reliable method is to list the factors or multiples in order until you find the largest or smallest match, or to use prime factor decomposition.
最大公因数（HCF）与最小公倍数（LCM）是常考内容。18 和 24 的最大公因数是 6，最小公倍数是 72。可靠的做法是按顺序列出因数或倍数，直到找到最大或最小的匹配项，或者使用质因数分解。
HCF(18, 24) = 6 LCM(18, 24) = 72 360 = 2³ × 3² × 5
for the HCF or LCM of two larger numbers such as 360 and 84, so practise building the factor trees quickly and writing the final line with indices.
质因数分解要用指数形式书写，例如 360 = 2³ × 3² × 5。典型题目会要求求两个较大数（如 360 和 84）的最大公因数或最小公倍数，因此要练习快速画出因数树，并用指数写出最后一行。
8. Fractions, Decimals and Percentages | 分数、小数与百分数
Converting between fractions, decimals and percentages is one of the most heavily examined skills in Book 7F. To turn 3/8 into a decimal, divide 3 by 8 to get 0.375; multiply by 100 to obtain 37.5%. To turn 0.6 into a fraction, write 6/10 and cancel to 3/5.
分数、小数与百分数之间的互化是 7F 中考查频率最高的技能之一。把 3/8 化成小数，用 3 除以 8 得 0.375；再乘以 100 得到 37.5%。把 0.6 化成分数，写成 6/10 并约分为 3/5。
Percentage of an amount questions are answered by converting the percentage to a decimal and multiplying, or by finding 10% and 1% first. For 45% of 260: 10% is 26, so 40% is 104, and 5% is 13; adding gives 117.
求一个量的百分之几，可以先把百分数化成小数再相乘，也可以先求 10% 和 1%。例如 260 的 45%：10% 是 26，所以 40% 是 104，5% 是 13，相加得到 117。
3/8 = 0.375 = 37.5% 45% of 260 = 117
Other regular question styles include ordering a mixed set of values from smallest to largest, finding the original amount after a percentage change, and simple percentage increase or decrease such as 'Increase £80 by 15%'. For a 15% increase, multiply by 1.15 to get £92.
其他常见题型包括把混合数值从小到大排序、已知变化后的量反求原量，以及简单的百分数增减，例如 '把 80 英镑增加 15%'。增加 15% 时乘以 1.15，得到 92 英镑。
e directly across the numerators and denominators: ½ × ⅔ = 2/6 = ⅓. Cancelling before multiplying saves time and reduces errors.
分数运算出现在同一节。计算 ½ + ⅓ 需要公分母 6，得到 3/6 + 2/6 = 5/6。分数乘法直接分子乘分子、分母乘分母：½ × ⅔ = 2/6 = ⅓。先约分再相乘可以节省时间并减少错误。
9. Ratio, Proportion and the Unitary Method | 比、比例与单位法
n problem. To simplify 18 : 24, divide both parts by their HCF of 6 to get 3 : 4. To share £240 in the ratio 3 : 5, add the parts (3 + 5 = 8), divide to find one part (£240 ÷ 8 = £30), then multiply: £90 and £150.
比例题通常要求化简比、按比分配一个总量，或解决比例问题。化简 18 : 24，把两部分同时除以最大公因数 6，得到 3 : 4。把 240 英镑按 3 : 5 分配，先求份数之和（3 + 5 = 8），再求一份的量（240 ÷ 8 = 30 英镑），最后相乘得到 90 英镑和 150 英镑。
18 : 24 = 3 : 4 £240 ÷ 8 = £30 £30 × 3 = £90 £30 × 5 = £150
method solves proportion questions by finding the value of one unit first. If 5 pens cost £3.75, then 1 pen costs £3.75 ÷ 5 = £0.75, and 7 pens cost £0.75 × 7 = £ es, such as scaling a recipe for 4 people up to 10 people, use exactly the same technique.
单位法通过先求一个单位的量来解决比例问题。如果 5 支笔售价 3.75 英镑，那么 1 支笔是 3.75 ÷ 5 = 0.75 英镑，7 支笔就是 0.75 × 7 = 5.25 英镑。按比例放大食谱（例如把 4 人份改为 10 人份）用的正是同一种方法。
Always check whether your answer is sensible. When a ratio has more parts on one side, that share must be larger, and the two shares must add back to the original total. This quick check catches reversed ratios before the marker sees them.
一定要检查答案是否合理。当比的一侧份数更多时，那一份必须更大，而且两份相加必须还原为原来的总量。这个快速检查能在阅卷老师看到之前就发现比写反的错误。
10. Basic Algebra: Simplifying and Solving | 基础代数：化简与解方程
ans 4 × x × 3, so only the coefficients (4 and 3) are multiplied, giving 12x and not 12x².
7F 中的代数题从合并同类项开始。5a + 3a = 8a，7b − 2b = 5b，4x × 3 = 12x。记住 4x × 3 表示 4 × x × 3，因此只有系数（4 和 3）相乘，结果是 12x，而不是 12x²。
ery common error is to write 10a − 3 instead of 10a − 15 by forgetting to multiply both terms.
展开括号要用分配律：3(x + 4) = 3x + 12，5(2a − 3) = 10a − 15。注意负号；非常常见的错误是忘了乘第二项，写成 10a − 3 而不是 10a − 15。
Solving linear equations is done by doing the same operation to both sides. For 4x +
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