📚 7H Homework Answers | 7H 数学作业答案
This answer key covers a typical 7H mathematics homework set. Each section shows a worked example followed by the correct answer, so you can check your method as well as your final result.
本答案覆盖一套典型 7H 数学作业。每节给出示例步骤和正确答案，帮助你同时检查方法与最终结果。
1. Fraction Arithmetic | 分数四则运算
west common denominator, which is 12. Then 2/3 = 8/12 and 1/4 = 3/12, so the sum is 11/12.
题目 1：计算 2/3 + 1/4。最小公分母为 12，因此 2/3 = 8/12，1/4 = 3/12，两者之和为 11/12。
Question 2: Write 17/5 as a mixed number. Divide 17 by 5 to get 3 remainder 2, so the answer is 3 2/5.
题目 2：将 17/5 写成带分数。17 除以 5 得 3 余 2，因此答案为 3 2/5。
: Multiply 3/4 × 2/5. Multiply numerators and denominators: (3 × 2)/(4 × 5) = 6/20 = 3/10.
题目 3：计算 3/4 × 2/5。分子、分母分别相乘：(3 × 2)/(4 × 5) = 6/20 = 3/10。
| Question | Answer |
|---|---|
| 2/3 + 1/4 | 11/12 |
| 17/5 as a mixed number | 3 2/5 |
| 3/4 × 2/5 | 3/10 |
2. Solving Linear Equations | 解一元一次方程
Question 1: Solve x + 3 = 7. Subtract 3 from both sides to isolate x, giving x = 4.
题目 1：解方程 x + 3 = 7。两边同时减去 3，得到 x = 4。
Question 2: Solve 5x = 20. Divide both sides by 5, so x = 4.
题目 2：解方程 5x = 20。两边同时除以 5，得到 x = 4。
on 3: Solve 2x + 1 = 9. First subtract 1 from both sides: 2x = 8. Then divide by 2: x = 4.
题目 3：解方程 2x + 1 = 9。先两边减去 1：2x = 8，再除以 2：x = 4。
| Equation | Solution |
|---|---|
| x + 3 = 7 | x = 4 |
| 5x = 20 | x = 4 |
| 2x + 1 = 9 | x = 4 |
3. Area and Perimeter | 面积与周长
7 cm and width 4 cm. Its perimeter is 2 × (7 + 4) = 22 cm, and its area is 7 × 4 = 28 cm².
题目 1：一个矩形的长为 7 cm，宽为 4 cm。其周长为 2 × (7 + 4) = 22 cm，面积为 7 × 4 = 28 cm²。
square has side length 6 cm. Its perimeter is 4 × 6 = 24 cm, and its area is 6² = 36 cm².
题目 2：一个正方形的边长为 6 cm。其周长为 4 × 6 = 24 cm，面积为 6² = 36 cm²。
Question 3: A triangle has base 10 cm and height 5 cm. Its area is (10 × 5)/2 = 25 cm².
题目 3：一个三角形的底为 10 cm，高为 5 cm。其面积为 (10 × 5)/2 = 25 cm²。
| Shape | Perimeter | Area |
|---|---|---|
| Rectangle 7 cm by 4 cm | 22 cm | 28 cm² |
| Square side 6 cm | 24 cm | 36 cm² |
| Triangle base 10 cm, height 5 cm | — | 25 cm² |
4. Negative Numbers | 负数运算
Question 1: Work out -3 + 8. Start at -3 and move 8 places to the right, giving 5.
题目 1：计算 -3 + 8。从 -3 开始向右移动 8 个单位，得到 5。
Question 2: Work out -5 - 4. Subtract 4 from -5 by moving further left, giving -9.
题目 2：计算 -5 - 4。从 -5 减去 4，继续向左移动，得到 -9。
stion 3: Work out -6 × -2. A negative times a negative is a positive, so the answer is 12.
题目 3：计算 -6 × -2。负数乘负数得正数，因此答案为 12。
| Operation | Answer |
|---|---|
| -3 + 8 | 5 |
| -5 - 4 | -9 |
| -6 × -2 | 12 |
5. Ratio and Proportion | 比与比例
Question 1: Share £60 in the ratio 2:3. The total number of parts is 2 + 3 = 5, so one part is £60 ÷ 5 = £12. The shares are £24 and £36.
题目 1：按 2:3 的比例分配 60 英镑。总份数为 2 + 3 = 5，每份为 60 ÷ 5 = 12 英镑，因此两份分别为 24 英镑和 36 英镑。
1. If 240 g of flour is used, then one part is 240 ÷ 3 = 80 g, so 80 g of sugar is needed.
题目 2：一份食谱中面粉与糖的比例为 3:1。如果用了 240 g 面粉，则一份为 240 ÷ 3 = 80 g，因此需要 80 g 糖。
is 5:4. If there are 20 boys, then one part is 20 ÷ 5 = 4, so there are 4 × 4 = 16 girls.
题目 3：班级中男生与女生的比例为 5:4。如果男生有 20 人，则一份为 20 ÷ 5 = 4，因此女生有 4 × 4 = 16 人。
| Problem | Answer |
|---|---|
| Share £60 in ratio 2:3 | £24 and £36 |
| Flour to sugar ratio 3:1, flour 240 g | Sugar = 80 g |
| Boys to girls ratio 5:4, boys 20 | Girls = 16 |
6. Simplifying Algebraic Expressions | 化简代数式
Both terms contain the same variable a, so combine the coefficients: 3 + 5 = 8, giving 8a.
题目 1：化简 3a + 5a。两项都含有相同字母 a，合并系数：3 + 5 = 8，得到 8a。
4x + 2y + 3x - y. Group like terms: 4x + 3x = 7x and 2y - y = y, so the result is 7x + y.
题目 2：化简 4x + 2y + 3x - y。合并同类项：4x + 3x = 7x，2y - y = y，结果为 7x + y。
Question 3: Expand 3(x + 4). Multiply 3 by each term inside the bracket: 3x + 12.
题目 3：展开 3(x + 4)。把 3 乘以括号内每一项：3x + 12。
| Expression | Simplified form |
|---|---|
| 3a + 5a | 8a |
| 4x + 2y + 3x - y | 7x + y |
| 3(x + 4) | 3x + 12 |
7. Angles in Triangles and Quadrilaterals | 三角形与四边形内角
le. The angles in a triangle add up to 180°, so the third angle is 180° - 45° - 55° = 80°.
题目 1：一个三角形的两个角分别为 45° 和 55°。求第三个角。三角形内角和为 180°，因此第三个角为 180° - 45° - 55° = 80°。
in a quadrilateral add up to 360°, so the missing angle is 360° - 90° - 80° - 110° = 80°.
题目 2：一个四边形的三个角分别为 90°、80° 和 110°。求第四个角。四边形内角和为 360°，因此缺失角为 360° - 90° - 80° - 110° = 80°。
osceles triangle has one angle of 40°. The two equal angles are each (180° - 40°)/2 = 70°.
题目 3：一个等腰三角形有一个角为 40°。两个相等角各为 (180° - 40°)/2 = 70°。
| Problem | Missing angle |
|---|---|
| Triangle with 45° and 55° | 80° |
| Quadrilateral with 90°, 80°, 110° | 80° |
| Isosceles triangle with one angle 40° | Each equal angle = 70° |
8. Averages and Range | 平均数与极差
n of 4, 7, 8, 10, 11. Add all values: 4 + 7 + 8 + 10 + 11 = 40, then divide by 5 to get 8.
题目 1：求 4、7、8、10、11 的平均数。所有数相加：4 + 7 + 8 + 10 + 11 = 40，再除以 5，得到 8。
Question 2: Find the median of 3, 9, 5, 12, 6. First order the data: 3, 5, 6, 9, 12. The middle value is 6, so the median is 6.
题目 2：求 3、9、5、12、6 的中位数。先将数据排序：3、5、6、9、12。中间的数为 6，因此中位数为 6。
22, 8, 19, 30. The largest value is 30 and the smallest is 8, so the range is 30 - 8 = 22.
题目 3：求 15、22、8、19、30 的极差。最大值为 30，最小值为 8，因此极差为 30 - 8 = 22。
| Data set | Mean | Median | Range |
|---|---|---|---|
| 4, 7, 8, 10, 11 | 8 | 8 | 7 |
| 3, 9, 5, 12, 6 | 7 | 6 | 9 |
| 15, 22, 8, 19, 30 | 18.8 | 19 | 22 |
9. Percentages | 百分数
Question 1: Find 20% of 80. Convert 20% to a decimal by dividing by 100, giving 0.20, then multiply: 0.20 × 80 = 16.
题目 1：求 80 的 20%。把 20% 除以 100 转化为小数 0.20，然后相乘：0.20 × 80 = 16。
Question 2: Increase 50 by 10%. First find 10% of 50, which is 5. Add this to the original amount: 50 + 5 = 55.
题目 2：把 50 增加 10%。先求 50 的 10%，即 5。再加到原数上：50 + 5 = 55。
Question 3: Decrease 90 by 30%. Find 30% of 90, which is 27. Subtract from the original: 90 - 27 = 63.
题目 3：把 90 减少 30%。求 90 的 30%，即 27。从原数中减去：90 - 27 = 63。
| Percentage problem | Answer |
|---|---|
| 20% of 80 | 16 |
| Increase 50 by 10% | 55 |
| Decrease 90 by 30% | 63 |
10. Coordinates and Linear Graphs | 坐标与线性图像
Question 1: Plot the point (3, 4). Start at the origin, move 3 units to the right along the x-axis, then 4 units up along the y-axis.
题目 1：描点 (3, 4)。从原点出发，沿 x 轴向右移动 3 个单位，再沿 y 轴向上移动 4 个单位。
+ 6)/2 = 4. Add the y-coordinates and divide by 2: (5 + 11)/2 = 8. The midpoint is (4, 8).
题目 2：求点 (2, 5) 与 (6, 11) 的中点。x 坐标相加除以 2：(2 + 6)/2 = 4；y 坐标相加除以 2：(5 + 11)/2 = 8。中点为 (4, 8)。
e graph of y = 2x to find y when x = 3. Substitute x = 3 into the equation: y = 2 × 3 = 6.
题目 3：利用 y = 2x 的图像，求当 x = 3 时 y 的值。将 x = 3 代入方程：y = 2 × 3 = 6。
| Coordinate problem | Answer |
|---|---|
| Midpoint of (2, 5) and (6, 11) | (4, 8) |
| y = 2x when x = 3 | y = 6 |
| Plot (3,
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