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GAC004 Mathematics I: Fundamentals

GAC Mathematics · Topic 1

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Worked Practice

Here is the map for this practice pack. Four questions, each tied to a sheet you already know: Q1 and Q2 belong to sheet one point one, Q3 to sheet one point…

English narration · English + 中文 subtitles burned in

1.1

What this module is, and how it is marked

A price can fall by 20% and then rise by 20% without returning to its starting value. Mathematics I helps you explain such results with a method, units and a clear interpretation. The six units connect arithmetic, algebra, graphs, geometry, trigonometry and exponential models.

Your centre's current assessment brief is the authority for tasks, weights, permitted tools and deadlines. An in-class test 课堂测验 is completed under stated classroom conditions. Projects 项目 can involve collecting data, calculating and reporting; an examination 考试 and coursework 平时作业 may have different instructions. Do not infer an official assessment pattern from these practice sheets.

A terminology logbook 术语记录本 can record the English term, a meaning in your own words and a small example. Follow your centre's instructions if it is submitted or assessed.

Read the command word 指令词: solve asks for values satisfying a condition, simplify for an equivalent expression, and justify for a reason. Show a valid method so a reader can check how the result follows. The mark scheme for each task determines its marks.

The six accompanying practice sheets are original GAC004-aligned material, not official papers. Their solutions award marks for stated mathematical steps rather than for the length of a written report.

Vocabulary Train
English
in-class test/ɪn klæs test/
projects/ˈprɒdʒekts/
examination/eɡˌzæmɪˈneɪʃn/
coursework/ˈkɔːsjuːɜːk/
terminology logbook/ˌtɜːmɪˈnɒlədʒi ˈlɒɡbʊk/
command word/kəˈmænd wɜːd/
1.1

Terminology and arithmetic review

Syllabus

Unit 1 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

Module purpose: On completion of this module, students should be able to demonstrate an understanding of the basic concepts of mathematics, and the language used, in preparation for tertiary study within an English-speaking environment.

The module outcomes this unit works towards:

Learning Objective GAC004.1: Solve elementary math problems using basic arithmetic operations.

Source: Cambridge International syllabus

The vocabulary of number is assumed by every later unit.

An integer 整数 is a whole number, including negative whole numbers and zero. A rational number 有理数 can be written as a fraction of two integers with a nonzero denominator; an irrational number 无理数 cannot, and $\pi$ and $\sqrt{2}$ are the standard examples. A prime number 质数 is a positive integer greater than 1 with exactly two positive factors, itself and 1.

The order of operations 运算顺序 fixes what a written expression means: brackets, then indices, then multiplication and division, then addition and subtraction, working left to right within a level.

$$3 + 4 \times 2^2 = 3 + 4 \times 4 = 3 + 16 = 19$$
  • A factor 因数 divides a number exactly; a multiple 倍数 is what you get by multiplying it. 6 is a factor of 24, and 24 is a multiple of 6.
  • The highest common factor (HCF) 最大公因数 and lowest common multiple (LCM) 最小公倍数 come from the prime factors: $24 = 2^3 \times 3$ and $36 = 2^2 \times 3^2$, so the HCF is $2^2 \times 3 = 12$ and the LCM is $2^3 \times 3^2 = 72$.
  • A percentage 百分比 is a fraction with denominator 100. An increase of 15% multiplies by $1.15$; a decrease of 15% multiplies by $0.85$. Reversing a percentage change means dividing, never subtracting the same percentage back.
  • Significant figures 有效数字 are the digits retained to express a value at a stated precision. They do not alone establish the accuracy of a measurement. $0.004\,072$ to three significant figures is $0.004\,07$.
  • Standard form 科学记数法 writes a number as $a \times 10^n$ with $1 \le |a| < 10$ for a nonzero number, with integer n. It is how a calculator shows a very large or very small answer.

Worked example. a decrease and increase use different bases

The worked price moves from 80 to 64 to 76.8 because each percentage uses the price at that stage.

Known: an invented price of 80 units falls by 20%, then rises by 20%. Why use successive multipliers? Each change is calculated from the current price.

$$P_{new}=P_{old}(1-r/100)$$
$$P_1=80(1-20/100)=64$$
$$P_{new}=P_{old}(1+r/100)$$
$$P_2=64(1+20/100)=76.8$$

The price is not back at 80. The decrease was 16, while the later increase was 12.8, because the bases differ.

Continue with practice sheet 1.1. Solve before opening the solutions, and check both the method and the final units.

Vocabulary Train
English
integer/ˈɪntɪdʒə/
rational number/ˈræʃənl ˈnʌmbə/
irrational number/ɪˈræʃənl ˈnʌmbə/
prime number/praɪm ˈnʌmbə/
order of operations/ˈɔːdə ɒv ˌɒpəˈreɪʃnz/
factor/ˈfæktə/
multiple/ˈmʌltɪpl/
highest common factor (HCF)/ˈhaɪɪst ˈkɒmən ˈfæktə/
lowest common multiple (LCM)/ˈləʊɪst ˈkɒmən ˈmʌltɪpl/
percentage/pəˈsentɪdʒ/
Significant figures/sɪɡˈnɪfɪkənt ˈfɪɡəz/
Standard form/ˈstændəd fɔːm/
1.2

Algebra I: introductory algebra

Syllabus

Unit 2 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC004.2: Perform basic algebraic operations and solve equations and inequations using algebraic methods.

Source: Cambridge International syllabus

Algebra is arithmetic with letters standing for numbers, and it has its own vocabulary.

In $5x^2 - 3x + 7$, the whole thing is an expression 表达式; $5x^2$, $-3x$ and $7$ are its terms 项; 5 is the coefficient 系数 of $x^2$; and 7 is a constant 常数.

An equation 方程 says two expressions are equal and is solved for a value. An identity 恒等式 is true for every value in its domain. An inequality 不等式 compares two expressions with $<$, $\le$, $>$ or $\ge$.

To expand 展开 is to remove brackets; to factorise 因式分解 is to put them back.

$$(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$$
  • Solving a linear equation 一元一次方程 means doing the same operation to both sides until the letter stands alone.
  • A quadratic equation 一元二次方程 has the form $ax^2+bx+c=0$ with $a\ne0$. Factorise when suitable, or use $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
  • Simultaneous equations 联立方程 are two equations in two unknowns, solved by substitution or by elimination. For two linear equations, a unique solution is their intersection. Parallel distinct lines have no solution; identical lines have infinitely many.
  • ⚠ An inequality reverses 反向 when you multiply or divide both sides by a negative number: from $-2x > 6$ it follows that $x < -3$.

Worked example. reverse the inequality sign when dividing by a negative

A number line shows x less than 2, with an open circle at 2.

Known: solve $7-3x>1$. Subtract 7 from both sides, then divide by negative 3. The last operation reverses the inequality.

$$7-3x>1\quad\Longrightarrow\quad -3x>-6$$
$$x<\frac{-6}{-3}=2$$

The open circle excludes 2. Check $x=1$: $7-3(1)=4>1$; the boundary $x=2$ gives equality and is excluded.

Continue with practice sheet 1.2. Solve before opening the solutions, and check both the method and the final units.

Vocabulary Train
English
expression/ekˈspreʃn/
terms/tɜːmz/
coefficient/ˌkəʊɪˈfɪʃənt/
constant/ˈkɒnstənt/
equation/ɪˈkweɪʒn/
identity/aɪˈdentɪti/
inequality/ɪniːˈkwɒlɪti/
expand/ekˈspænd/
factorise/ˈfæktəraɪz/
linear equation/ˈlɪnɪə ɪˈkweɪʒn/
quadratic equation/kwɒˈdrætɪk ɪˈkweɪʒn/
Simultaneous equations/ˌsɪməlˈteɪnɪəs ɪˈkweɪʒnz/
reverses/rɪˈvɜːsɪz/
1.3

Algebra: graphs and coordinate geometry

Syllabus

Unit 3 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC004.3: Graph algebraic relations and use coordinate geometry to solve problems.

Source: Cambridge International syllabus

A graph turns an equation into a picture, and coordinate geometry measures that picture.

A nonvertical straight line can be written $y = mx + c$, where $m$ is the gradient 斜率 and $c$ the y-intercept y 轴截距. The gradient is the rise divided by the run.

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
  • Distinct nonvertical lines are parallel 平行 when their gradients are equal. Two nonvertical lines are perpendicular 垂直 when their gradients multiply to $-1$. Vertical lines have undefined gradient; a vertical and a horizontal line are perpendicular.
  • The midpoint 中点 of a segment is the average of the endpoints, and the distance 距离 between two points comes from Pythagoras: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
  • A quadratic graphs as a parabola 抛物线. Its vertex 顶点 is the turning point, and the roots 根 are the x-values where it meets the $x$-axis, the solutions of $y=0$. A repeated root touches the axis without crossing; some quadratics have no real roots.
  • To solve two equations graphically 用图像求解, draw both and read the coordinates of the intersection. The answer is only as accurate as the drawing, which is why an algebraic check matters.

Worked example. a rising line through two given points

The line passes through A at (1,2) and B at (4,8), with dashed coordinate-change guides.

Known: $A=(1,2)$ and $B=(4,8)$. Use gradient because the line is nonvertical.

$$m=\frac{y_B-y_A}{x_B-x_A}=\frac{8-2}{4-1}=2$$

The line equation is $y=mx+c$. Substitute A to find the intercept.

$$c=y_A-mx_A=2-2(1)=0$$

Thus $y=2x$. The diagram's horizontal change is 3 and vertical change is 6.

Continue with practice sheet 1.3. Solve before opening the solutions, and check both the method and the final units.

Vocabulary Train
English
gradient/ˈɡreɪdɪənt/
y-intercept/waɪ ˌɪntəˈsept/
parallel/ˈpærəlel/
perpendicular/ˌpɜːpənˈdɪkjʊlə/
midpoint/ˈmɪdpɔɪnt/
distance/ˈdɪstəns/
parabola/pəˈræbələ/
vertex/ˈvɜːteks/
roots/ruːts/
graphically/ˈɡræfɪkli/
1.4

Geometry: plane, solid and Euclidean geometry

Syllabus

Unit 4 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC004.4: Analyze simple problems involving planar shapes and solids and solve problems in Euclidean geometry.

Source: Cambridge International syllabus

Plane geometry is about flat shapes, solid geometry about three-dimensional ones, and Euclidean geometry is the reasoning that connects them.

Angles on a straight line add to $180°$, angles around a point to $360°$, and the interior angles of a plane Euclidean triangle to $180°$. In a simple polygon 多边形 of $n$ sides the interior angles add to $(n - 2) \times 180°$.

  • Congruent 全等 shapes are identical in size and shape; similar 相似 shapes have the same shape with all lengths in one ratio.
  • Pythagoras' theorem 勾股定理 holds in a right-angled triangle: $a^2 + b^2 = c^2$, where $c$ is the hypotenuse 斜边.
  • Area 面积 is measured in square units and volume 体积 in cubic units. A cylinder has volume $\pi r^2 h$; a sphere has volume $\tfrac{4}{3}\pi r^3$ and surface area $4\pi r^2$.
  • ⚠ Scaling is not linear. If every length of a solid is doubled, its area is multiplied by $2^2 = 4$ and its volume by $2^3 = 8$.

Worked example. a triangular prism has a constant cross-section

A triangular prism has a right-triangle cross-section with perpendicular sides 3 and 4 cm and length 10 cm; perspective is not to scale.

Known: perpendicular triangle sides are 3 and 4 cm; prism length is 10 cm. Find the cross-sectional area, then multiply by the prism length.

$$A=\frac12 bh=\frac12(3)(4)=6\ \text{cm}^2$$
$$V=AL=6(10)=60\ \text{cm}^3$$

The sloping triangle side is not its perpendicular height. Volume is not the area of the triangular end.

Continue with practice sheet 1.4. Solve before opening the solutions, and check both the method and the final units.

Vocabulary Train
English
polygon/ˈpɒlɪɡən/
Congruent/ˈkɒŋɡruːənt/
similar/ˈsɪmɪlə/
Pythagoras' theorem/paɪˈθæɡərəs ˈθɪərəm/
hypotenuse/haɪˈpɒtənjuːs/
Area/ˈeərɪə/
volume/ˈvɒljuːm/
1.5

Trigonometry

Syllabus

Unit 5 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC004.5: Calculate solutions to various problems using trigonometric methods.

Source: Cambridge International syllabus

Trigonometry connects the angles of a triangle to its sides.

In a right-angled triangle, with $\theta$ one of the acute angles, the three ratios are $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ and $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$.

For a nondegenerate plane Euclidean triangle, the sine rule 正弦定理 and the cosine rule 余弦定理 apply:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, \qquad a^2 = b^2 + c^2 - 2bc\cos A$$
  • An angle can be measured in degrees 度 or in radians 弧度, where $\pi$ radians is $180°$. Check which mode your calculator is in before every question.
  • The unit circle 单位圆 extends the ratios beyond $90°$ and explains why $\sin$ and $\cos$ repeat every $360°$: they are periodic 周期的.
  • A bearing 方位角 is measured clockwise from north and always written with three figures, such as $075°$.
  • Angles of elevation and depression 仰角与俯角 are measured from the horizontal, upwards and downwards respectively. Add the observer or instrument height if the answer needs height above the ground. With side-side-angle data, the sine rule may allow two triangles: check both supplementary angles against the remaining angle sum.

Worked example. height from a horizontal distance and angle

A right triangle has a 12 m horizontal base and an angle of elevation of 30 degrees; the unknown height is opposite the angle.

Known: horizontal distance $d=12$ m and elevation $\theta=30^\circ$. Height h is opposite and d is adjacent, so use tangent in degree mode.

$$\tan\theta=\frac{h}{d}$$
$$h=d\tan\theta=12\tan30^\circ\approx6.93\ \text{m}$$

This is the height above the observer's horizontal sight level, not automatically the height above the ground.

Continue with practice sheet 1.5. Solve before opening the solutions, and check both the method and the final units.

Vocabulary Train
English
sine rule/saɪn ruːl/
cosine rule/ˈkəʊsaɪn ruːl/
degrees/dɪˈɡriːz/
radians/ˈreɪdɪənz/
unit circle/ˈjuːnɪt ˈsɜːkl/
periodic/ˌpɪərɪˈɒdɪk/
bearing/ˈbeərɪŋ/
Angles of elevation and depression/ˈæŋɡlz ɒv ˌelɪˈveɪʃn ænd dɪˈpreʃn/
1.6

Exponential and logarithmic functions

Syllabus

Unit 6 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC004.6: Solve and graph exponential and logarithmic functions and equations.

Source: Cambridge International syllabus

These two functions describe growth and decay, and they undo each other.

An exponential function 指数函数 has the form $y=a^x$ with $a>0$ and $a\ne1$: the variable is in the exponent 指数. Each unit step multiplies its value by a. It grows when $a>1$ and decays when $0<a<1$. Its y-intercept is 1 and its values stay positive; the x-axis is a horizontal asymptote.

A logarithm 对数 answers the reverse question. $\log_a y = x$ means exactly $a^x = y$, so $\log_{10}1000=3$. For real logarithms the argument must be positive, and the base must be positive and different from 1. The graph $y=\log_a x$ passes through $(1,0)$ and has vertical asymptote $x=0$.

$$\log(mn) = \log m + \log n, \qquad \log\!\left(\frac{m}{n}\right) = \log m - \log n, \qquad \log(m^k) = k \log m$$
  • For positive arguments and the same valid base, the laws turn multiplication into addition, which is what makes a logarithm useful for solving an equation with the unknown in the exponent.
  • Natural logarithms 自然对数 use the base $e \approx 2.718$ and are written $\ln$.
  • Compound interest 复利 is exponential: $A = P(1 + r)^n$ after $n$ periods.
  • Exponential decay 指数衰减 has a base between 0 and 1, and can describe a retained proportion each period under stated assumptions. It does not guarantee that a real process follows that model indefinitely.

Worked example. repeated reductions multiply rather than subtract a fixed amount

A model starting at 100 is multiplied by 0.8 each period, giving 80 then 64.

Known: an invented quantity starts at 100 units and decreases by 20% per whole period. The retained proportion is 0.8, so use a multiplicative model.

$$Q(n)=Q_0(1-r)^n$$
$$Q(2)=100(1-0.20)^2=64$$

The first decrease is 20 units and the second is 16. A fixed subtraction of 20 each period would be a different model.

Continue with practice sheet 1.6. Solve before opening the solutions, and check both the method and the final units.

Vocabulary Train
English
exponential function/ˌekspəˈnenʃl ˈfʌŋkʃn/
exponent/ekˈspəʊnənt/
logarithm/ˈlɒɡərɪθəm/
Natural logarithms/ˈnætʃərəl ˈlɒɡərɪθəmz/
Compound interest/ˈkɒmpaʊnd ˈɪntrest/
Exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/
Additional notes PDF

Extra practice: four questions with worked answers

This pack adds practice to three units of GAC004 Mathematics I. It covers arithmetic review (sheet 1.1), introductory algebra (sheet 1.2) and geometry (sheet 1.4). The questions practise basic arithmetic, algebraic methods and plane geometry. Use this pack beside the GAC004 handout and its practice sheets.

It is original practice material. It does not change your centre's assessment brief, and it is not an official paper.

Answer the questions in “Practice questions” first. Write your full method. Then compare your method with each worked answer, not only your final number. A clear, stated method shows your reasoning better than a long report.

Each answer shows the same structure: what is known, why the rule applies, the rule in symbols, then the numbers. One calculation stage sits on each line. The check at the end is part of the answer, so run it yourself on a fresh copy.

Question Sheet What it practises
Q1 1.1 Two price changes in a row
Q2 1.1 Undo one increase
Q3 1.2 Divide by a negative number
Q4 1.4 From lengths to areas

Practice questions

Answer these first. Do not look ahead.

Q1. A price of 200 units rises by 10%, then falls by 10%.

  • Find the final price.
  • Explain why the final price is not 200 units.

Q2. After a 20% increase, a figure is 84. Find the original figure.

Q3. Solve $9-2x\le3$. Show your solution on a number line 数轴.

Q4. Two similar 相似 models have a length ratio 长度比 of $3:5$. The smaller model has surface area 表面积 $36\ \text{cm}^2$. Find the larger surface area.

Worked answers

Answer to Q1: each change uses the current price

Known: $P_0=200$ units, rise $r=10\%$, fall $r=10\%$. Each percentage 百分比 uses the current price. So apply a multiplier 乘数 at each stage.

The increase equation 方程, in symbols:

$$P_{new}=P_{old}(1+r/100)$$
$$P_1=P_0(1+r/100)=200(1+10/100)=220$$

The decrease, in symbols:

$$P_{new}=P_{old}(1-r/100)$$
$$P_2=P_1(1-r/100)=220(1-10/100)=198$$

The final price is $\mathbf{198}$ units.

Why is it not 200? The two multipliers form one combined multiplier:

$$\frac{P_2}{P_0}=(1+r/100)(1-r/100)$$
$$\frac{P_2}{P_0}=(1+r/100)(1-r/100)=(1+10/100)(1-10/100)=0.99$$

So the final price keeps $99\%$ of the starting price. The overall loss is $1\%$, not zero.

The rise and the fall also started from different amounts:

$$\Delta_{rise}=P_1-P_0=220-200=20$$
$$\Delta_{fall}=P_1-P_2=220-198=22$$

10% of 220 is larger than 10% of 200. The fall removes more than the rise added.

Check: the overall loss is $1\%$ of the starting price.

$$1-0.99=0.01$$
$$P_0\times0.01=200\times0.01=2$$
$$P_0-P_2=200-198=2$$

Both give 2 units, so the two checks agree.

Answer to Q2: undo the increase by dividing

Known: new figure $N=84$, increase $r=20\%$. The increase multiplied the original figure by $1+r/100$. To reverse a percentage change, divide by this multiplier. Never subtract the same percentage back.

The increase equation, in symbols:

$$N=P_0(1+r/100)$$

Rearrange for the original figure:

$$P_0=\frac{N}{1+r/100}$$

Substitute:

$$P_0=\frac{N}{1+r/100}=\frac{84}{1+20/100}$$

Work out the multiplier:

$$1+\frac{r}{100}=1+\frac{20}{100}=1.2$$

Divide:

$$P_0=\frac{N}{1+r/100}=\frac{84}{1.2}=70$$

The original figure is $\mathbf{70}$.

Reverse check: run the increase forward again.

$$N=P_0(1+r/100)=70(1+20/100)=84$$

The original 70 returns the stated 84, so the answer is correct.

The trap: subtracting 20% of the new figure.

$$84-84\times0.20=67.2$$

That is wrong. The increase was calculated on 70, not on 84.

Answer to Q3: solve, then check both sides

Known: the inequality 不等式 $9-2x\le3$. Solve it like an equation. Dividing by a negative number reverses 反向 the inequality sign.

Subtract 9 from both sides:

$$9-2x\le3$$
$$-2x\le3-9$$
$$-2x\le-6$$

Divide both sides by $-2$, and reverse the sign:

$$x\ge\frac{-6}{-2}=3$$

The solution is $x\ge3$.

Boundary 边界 check: at $x=3$ the two sides are equal.

$$9-2x=9-2(3)=9-6=3$$

So 3 is included. On the number line, mark a closed circle 实心圆点 at 3 and shade to the right.

Inside check: $x=4$.

$$9-2x=9-2(4)=9-8=1\le3$$

True.

Outside check: $x=2$.

$$9-2x=9-2(2)=9-4=5>3$$

False. Both checks agree with $x\ge3$.

Answer to Q4: from lengths to areas

Known: length ratio $3:5$; smaller surface area $A_s=36\ \text{cm}^2$. Let $k$ be the scale factor 比例因子 from a smaller length $L_s$ to the matching larger length $L_L$. Each area is a product of two lengths. So areas scale as the square 平方 of $k$.

The general relation between the two areas:

$$\frac{A_L}{A_s}=k^2$$

The length ratio gives the scale factor:

$$k=\frac{L_L}{L_s}=\frac{5}{3}$$

The area ratio:

$$\frac{A_L}{A_s}=k^2=\left(\frac{5}{3}\right)^2=\frac{25}{9}$$

Rearrange for the larger area:

$$A_L=A_s k^2$$

Substitute:

$$A_L=A_s k^2=36\times\frac{25}{9}=100$$

The larger surface area is $\mathbf{100}\ \text{cm}^2$.

Ratio check:

$$\frac{A_L}{A_s}=\frac{100}{36}=\frac{25}{9}$$
$$k^2=\left(\frac{5}{3}\right)^2=\frac{25}{9}$$

The area ratio $25:9$ is the square of the length ratio $5:3$. Only areas scale by $\frac{25}{9}$. A length scales by $k=\frac{5}{3}$.

What to do next

  • Try the matching exercises on practice sheets 1.1, 1.2 and 1.4.
  • Run every check yourself on a fresh copy.
  • Record the step where your method differed, then try a new question.

A stated method with a check shows your reasoning clearly. A bare number does not.

Interactive lessons on this topic

Work through it step by step, with instant-check exercises.

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