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Year 10 OCR Maths: Unit Test Mock Paper Solutions | 十年级 OCR 数学:单元测试模拟卷解析

📚 Year 10 OCR Maths: Unit Test Mock Paper Solutions | 十年级 OCR 数学:单元测试模拟卷解析

This mock paper walks you through key topics covered in the Year 10 OCR Mathematics course, including number operations, algebra, geometry, statistics and probability. Each section provides a worked example to help you revise effectively for your unit test.

本模拟卷带你梳理十年级 OCR 数学课程的重点,涵盖数与运算、代数、几何、统计与概率。每个部分都配有详细例题解析,帮助你高效备考单元测试。


1. Number Skills and BIDMAS | 数字技能与运算顺序

BIDMAS (Brackets, Indices, Division/Multiplication, Addition/Subtraction) sets the order for calculations. Always start with brackets, then powers or roots, then division or multiplication from left to right, and finally addition or subtraction.

BIDMAS(括号、指数、乘除、加减)规定了计算的顺序。先算括号,再算幂或方根,然后从左到右进行乘除,最后进行加减。

Question: Work out 3² + 8 ÷ 2 × (6 − 4)³.

问题:计算 3² + 8 ÷ 2 × (6 − 4)³。

Step 1: Evaluate the bracket first. (6 − 4) = 2.

第一步:先计算括号,(6 − 4) = 2。

Step 2: Apply the indices. 3² = 9 and 2³ = 8.

第二步:计算指数。3² = 9,2³ = 8。

Step 3: Perform division and multiplication in the order they appear. 8 ÷ 2 = 4, then 4 × 8 = 32.

第三步:按从左到右的顺序进行乘除。8 ÷ 2 = 4,然后 4 × 8 = 32。

Step 4: Finally, carry out the addition. 9 + 32 = 41.

第四步:最后加法。9 + 32 = 41。

The correct answer is 41.

正确答案是 41。


2. Algebraic Simplification | 代数化简

Simplifying an expression means collecting like terms — terms that have exactly the same variable and power. Constants are like terms with each other.

化简表达式就是合并同类项——变量和幂次完全相同的项。常数项之间也是同类项。

Question: Simplify 5x + 3y − 2x + 7y − 4.

问题:化简 5x + 3y − 2x + 7y − 4。

Identify the x terms: 5x and −2x. Combine them: 5x − 2x = 3x.

找出 x 的项:5x 和 −2x。合并:5x − 2x = 3x。

Now the y terms: 3y and +7y give 10y.

然后 y 的项:3y 和 +7y 得到 10y。

The constant term is −4 and it stays as it is.

常数项 −4 保持不变。

Hence the simplified expression is 3x + 10y − 4.

因此化简后的表达式是 3x + 10y − 4。


3. Solving Linear Equations | 解一元一次方程

When an unknown appears on both sides of an equation, first expand any brackets, then collect like terms, and isolate the variable.

当未知数出现在方程两侧时,先展开括号,然后移项合并,最后解出未知数。

Question: Solve 4(x − 3) = 2x + 8.

问题:解方程 4(x − 3) = 2x + 8。

4(x − 3) = 2x + 8

Expand the left-hand side: 4x − 12 = 2x + 8.

展开左边:4x − 12 = 2x + 8。

Subtract 2x from both sides: 2x − 12 = 8.

两边减去 2x:2x − 12 = 8。

Add 12 to both sides: 2x = 20.

两边加 12:2x = 20。

Divide by 2: x = 10.

除以 2:x = 10。

Check: substitute x = 10 into the original equation: 4(10−3)=4×7=28 and 2×10+8=28. It works.

检验:把 x = 10 代回原方程:4(10−3)=4×7=28,2×10+8=28,成立。


4. Angles in Parallel Lines | 平行线中的角

When a transversal crosses parallel lines, corresponding angles are equal, alternate interior angles are equal, and co-interior angles sum to 180°.

当一条横截线与两条平行线相交时,同位角相等,内错角相等,同旁内角互补(和为180°)。

Question: In the diagram, line AB is parallel to CD, and an angle formed by the transversal measures 70°. Find the corresponding angle x and the alternate angle y.

问题:在图中,AB 平行于 CD,横截线构成的一个角为 70°。求同位角 x 和内错角 y。

Angle x is the corresponding angle to the 70° angle, so x = 70°.

角 x 是 70° 的同位角,因此 x = 70°。

Angle y is the alternate interior angle to the 70° angle, so y = 70° as well.

角 y 是 70° 的内错角,因此 y 也是 70°。

If the question asks for the co-interior angle, it would be 180° − 70° = 110°.

如果题目要求同旁内角,则为 180° − 70° = 110°。


5. Area and Perimeter of Compound Shapes | 复合图形的面积与周长

A compound shape can be split into simpler rectangles, or considered as a larger rectangle with a cut-out. Calculate area by adding or subtracting areas of basic shapes, and perimeter by tracing the outer boundary.

复合图形可以分割成简单矩形,或看作大矩形减去一个缺口。面积通过加减基础图形面积计算,周长通过描摹外侧边缘求得。

Question: A shape is formed by removing a 4 cm × 3 cm rectangle from a corner of a 10 cm × 6 cm rectangle. Find its area and perimeter.

问题:从一个 10 cm × 6 cm 的矩形一角剪掉一个 4 cm × 3 cm 的小矩形,求剩余图形的面积和周长。

Area of large rectangle = 10 × 6 = 60 cm²

Area of cut-out = 4 × 3 = 12 cm².

小矩形面积 = 4 × 3 = 12 cm²。

Area of compound shape = 60 − 12 = 48 cm².

复合图形面积 = 60 − 12 = 48 cm²。

For perimeter, walk around the outside: starting from top-left, go right 10 cm, down 6 cm, left (10−4)=6 cm, up 3 cm, left 4 cm, and up (6−3)=3 cm back to start. Total perimeter = 10 + 6 + 6 + 3 + 4 + 3 = 32 cm.

周长沿外边界走一圈:从左上角开始,向右 10 cm,向下 6 cm,向左 (10−4)=6 cm,向上 3 cm,向左 4 cm,再向上 (6−3)=3 cm 回到起点。总周长 = 10 + 6 + 6 + 3 + 4 + 3 = 32 cm。


6. Fractions, Decimals and Percentages | 分数、小数与百分数

Converting between fractions, decimals and percentages is an essential skill. To change a fraction to a decimal, divide the numerator by the denominator. Multiply by 100 to get the percentage.

分数、小数和百分数之间的转换是必备技能。分数化小数,用分子除以分母;再乘以 100 得到百分数。

Question: Convert 3/8 to a decimal and a percentage. Also, which is larger: 0.35 or 2/5?

问题:将 3/8 转换为小数和百分数。另外,比较 0.35 和 2/5 哪个更大?

3/8 as a decimal: 3 ÷ 8 = 0.375.

3/8 化为小数:3 ÷ 8 = 0.375。

As a percentage: 0.375 × 100 = 37.5%.

化为百分数:0.375 × 100 = 37.5%。

Now compare 0.35 and 2/5. 2/5 = 2 ÷ 5 = 0.4. Since 0.4 > 0.35, therefore 2/5 is larger.

比较 0.35 和 2/5。2/5 = 2 ÷ 5 = 0.4。因为 0.4 > 0.35,所以 2/5 更大。


7. Ratio and Proportion | 比与比例

When a quantity is divided in a given ratio, think of the total number of parts. Each part gets an equal share of the total amount.

当一个量按给定比例分配时,考虑总份数,每一份占据总量的同等份额。

Question: Orange squash and water are mixed in the ratio 2:3. How

Published by TutorHao | Year 10 Mathematics Revision Series | aleveler.com

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