Mathematics (Syllabus D) 4024 / 13 — May/June 2025

(b)

Write $\frac{16}{25}$ as a decimal.

______

DifficultyEasy

Worked solution

Approach

To convert a fraction to a decimal, you can either divide the numerator by the denominator or multiply the numerator and denominator by a number that makes the denominator a power of 10 ( $10, 100, 1000, \dots$ ).

Working

The fraction is $\frac{16}{25}$ .
We want to change the denominator 25 into a power of 10. Since $25 \times 4 = 100$ , we multiply both the numerator and the denominator by 4.

\frac{16}{25} = \frac{16 \times 4}{25 \times 4}

Calculate the numerator:

16 \times 4 = 64

Calculate the denominator:

25 \times 4 = 100

So,

\frac{16}{25} = \frac{64}{100}

Convert the fraction $\frac{64}{100}$ to a decimal by placing the decimal point two places from the right:

\frac{64}{100} = 0.64

Answer

0.64

Final answer

0.64

Detailed explanation

Walkthrough

The question asks us to write the fraction $\frac{16}{25}$ as a decimal.

Look at the denominator, 25. We know that multiplying 25 by 4 gives 100, which is a power of 10. This is easier than performing long division ( $16 \div 25$ ).
Multiply both the top (numerator) and bottom (denominator) of the fraction by 4 to keep the value equivalent.
$\frac{16 \times 4}{25 \times 4} = \frac{64}{100}$
Now, convert $\frac{64}{100}$ to a decimal. Dividing by 100 moves the decimal point two places to the left.
$64 \rightarrow 6.4 \rightarrow 0.64$

Key Takeaways

Fractions with denominators of 2, 4, 5, 8, 10, 20, 25, 50, etc., can often be easily converted to decimals by scaling the denominator to a power of 10.
Specifically, $\times 4$ turns 25 into 100; $\times 2$ turns 5 into 10; $\times 125$ turns 8 into 1000.

Common Mistakes

Multiplying only the numerator or only the denominator.
Incorrect multiplication (e.g., $16 \times 4$ calculated as 54 or 66).
Misplacing the decimal point.

Things to Be Careful About

Ensure the final answer is in decimal form, not fraction form.

Techniques used

convert fraction to decimalmultiply numerator and denominator to reach a power of 10

Q22MEasyStatistics

The favourite ice cream flavours of 20 children are shown.


Vanilla	Vanilla	Strawberry	Chocolate	Chocolate
Strawberry	Chocolate	Chocolate	Vanilla	Strawberry
Chocolate	Vanilla	Vanilla	Strawberry	Chocolate
Strawberry	Chocolate	Vanilla	Chocolate	Vanilla

Complete the frequency table.

Flavour	Tally	Frequency
Vanilla
Strawberry
Chocolate

DifficultyEasy

Worked solution

Approach

The question provides a list of 20 ice cream flavours. We need to count how many times each specific flavour appears in the list to complete the frequency table.

Working

We go through the grid row by row and tally each flavour:

Vanilla:

Row 1: Vanilla (2nd), Vanilla (5th)
Row 2: Vanilla (4th)
Row 3: Vanilla (2nd), Vanilla (3rd)
Row 4: Vanilla (3rd), Vanilla (5th)
Total Vanilla = $2 + 1 + 2 + 2 = 7$

Strawberry:

Row 1: Strawberry (3rd)
Row 2: Strawberry (1st), Strawberry (5th)
Row 3: Strawberry (4th)
Row 4: Strawberry (1st)
Total Strawberry = $1 + 2 + 1 + 1 = 5$

Chocolate:

Row 1: Chocolate (4th), Chocolate (5th)
Row 2: Chocolate (2nd), Chocolate (3rd)
Row 3: Chocolate (1st), Chocolate (5th)
Row 4: Chocolate (2nd), Chocolate (4th)
Total Chocolate = $2 + 2 + 2 + 2 = 8$

Check: Total children = $7 + 5 + 8 = 20$ . This matches the number of children mentioned in the question.

Answer

\begin{array}{|c|c|c|} \hline \text{Flavour} & \text{Tally} & \text{Frequency} \\ \hline \text{Vanilla} & ||||\ || & 7 \\ \hline \text{Strawberry} & ||||\ | & 5 \\ \hline \text{Chocolate} & ||||\ ||| & 8 \\ \hline \end{array}

Final answer

Vanilla: 7, Strawberry: 5, Chocolate: 8

Detailed explanation

Walkthrough

To solve this problem, we must process the raw data provided in the text description of the grid. The grid lists the favourite ice cream flavour for each of the 20 children. The goal is to convert this unstructured list into a structured frequency table.

Identify the categories: The flavours listed are Vanilla, Strawberry, and Chocolate.
Tally the data: Systematically scan the list (row by row or column by column) and mark a tally for each occurrence of a flavour.
- For Vanilla, we find it appears 7 times.
- For Strawberry, we find it appears 5 times.
- For Chocolate, we find it appears 8 times.
Verify: Sum the frequencies ( $7 + 5 + 8$ ) to ensure they equal the total number of children (20). If they match, the counting is likely correct.
Complete the table: Write the final counts in the 'Frequency' column.

Key Takeaways

Tallying is an efficient way to count occurrences in a large set of data.
A frequency table summarizes categorical data by showing how often each category occurs.
Always check that the sum of frequencies equals the total sample size.

Common Mistakes

Miscounting due to skipping an item or double-counting one.
Incorrectly adding the final numbers (e.g., getting a total other than 20).
Leaving the 'Tally' column blank if required (though marks are usually for frequency).

Things to Be Careful About

Ensure you read all rows and columns carefully; missing a single entry changes the answer.
The mark scheme awards partial credit (B1) for getting two out of three frequencies correct, so accuracy in each count is important.

Techniques used

tally raw datacount frequencies from a tallyverify total frequency

Q3Medium-EasyGeometry

(a)

Fig. 1

$AB$ is a straight line.

Work out the value of $x$ .

$x$ = ______

(b)

Fig. 2

$PQR$ is an isosceles triangle with $PQ = PR$ .
Angle $QPR = 46^\circ$ .

Work out the value of $y$ .

$y$ = ______

DifficultyMedium-Easy

Worked solution

Approach

In an isosceles triangle, the angles opposite the equal sides (the base angles) are equal. The sum of all angles in any triangle is $180^\circ$ . We use these two properties to find $y$ .

Working

Since $PQ = PR$ , the base angles are equal:

\angle PQR = \angle PRQ = y^\circ

The sum of angles in triangle $PQR$ is $180^\circ$ :

y + y + 46 = 180

2y + 46 = 180

Subtract $46$ from both sides:

2y = 134

Divide by $2$ :

y = \frac{134}{2}

y = 67

Answer

67

Final answer

67

Detailed explanation

Walkthrough

The diagram shows an isosceles triangle $PQR$ where the sides $PQ$ and $PR$ are marked as equal with tick marks. A key property of isosceles triangles is that the angles opposite the equal sides (the base angles) are equal. Therefore, $\angle PQR = \angle PRQ = y^\circ$ . The sum of all interior angles in any triangle is always $180^\circ$ . We can write the equation $y + y + 46 = 180$ . Solving this gives $2y = 134$ , so $y = 67$ . Alternatively, you can subtract the vertex angle from $180^\circ$ and divide by $2$ directly: $\frac{180 - 46}{2} = 67$ .

Key Takeaways

Isosceles triangles have two equal sides and two equal base angles. The sum of angles in a triangle is always $180^\circ$ . Combining these two facts allows you to find unknown angles quickly and efficiently.

Common Mistakes

Assuming all three angles are equal (that would be an equilateral triangle, which requires all three sides to be equal).
Forgetting to divide the remaining angle sum by $2$ to find a single base angle, giving $134$ as the answer instead of $67$ .
Adding $46$ to $180$ instead of subtracting it.
Matching the wrong angles to the equal sides; the angles opposite $PQ$ and $PR$ are at $R$ and $Q$ respectively, not at $P$ .

Things to Be Careful About

The tick marks on $PQ$ and $PR$ indicate they are equal, which means the angles opposite them (at $Q$ and $R$ ) are equal, NOT the angle at $P$ . Always match the equal sides to the opposite base angles. The diagram is NOT TO SCALE, so do not estimate angles by eye. The mark scheme accepts the direct calculation $\frac{180 - 46}{2}$ as a valid method.

Techniques used

identify base angles in an isosceles triangleapply the triangle angle sum propertysolve a simple linear equation

Q42MEasyAlgebra and Graphs

Find the value of $5x + 3y$ when $x = 4$ and $y = -2$ .

______

DifficultyEasy

Worked solution

Approach

Substitute the given values $x = 4$ and $y = -2$ into the expression $5x + 3y$ and evaluate.

Working

Replace $x$ with $4$ and $y$ with $-2$ :

5(4) + 3(-2)

Perform the multiplications. Remember that multiplying a positive number by a negative number gives a negative result:

20 + (-6)

Simplify the addition:

20 - 6

Calculate the final result:

14

Answer

14

Final answer

14

Detailed explanation

Walkthrough

The problem asks for the numerical value of the expression $5x + 3y$ when specific numbers are assigned to the variables $x$ and $y$ .

First, we substitute the value of $x$ , which is $4$ , into the term $5x$ . This becomes $5 \times 4$ . Next, we substitute the value of $y$ , which is $-2$ , into the term $3y$ . It is crucial to use brackets here to handle the negative sign correctly: $3 \times (-2)$ .

Next, we perform the arithmetic operations. Multiplication comes before addition in the order of operations (BODMAS/PEMDAS).
$5 \times 4 = 20$ .
$3 \times (-2) = -6$ .

Finally, we add these two results together: $20 + (-6)$ , which is equivalent to $20 - 6$ . The result is $14$ .

Key Takeaways

Substitution: Always replace the variable with its value, using brackets if the value is negative.
Order of Operations: Perform multiplication before addition/subtraction.
Signed Numbers: Be careful with signs. A positive times a negative is negative ( $+ \times - = -$ ).

Common Mistakes

Sign Errors: Forgetting that $y$ is negative, leading to calculation of $20 + 6 = 26$ instead of $20 - 6$ .
Order of Operations: Adding $5$ and $4$ first (getting $9$ ), then multiplying by $3$ or similar incorrect groupings.
Arithmetic: Simple multiplication errors like $3 \times 2 = 7$ or $20 - 6 = 13$ .

Things to Be Careful About

Ensure you write $3(-2)$ or $3 \times (-2)$ rather than just $3-2$ . The coefficient is $3$ , not $-1$ .
Double-check the final subtraction step.

Techniques used

substitute given values into an algebraic expressionperform multiplication with signed integersadd integers to find the final value

Q5EasyNumber

Write down

(a)

the reciprocal of 7

______

(b)

an irrational number.

______

Q6Medium-EasyCoordinate Geometry

Point $A$ , point $B$ and point $C$ are plotted on the grid.

Fig. 3

(a)

On the grid, plot and label the point $D (3, 6)$ .

(b)

Find the equation of the line

3M

(i)

$BC$

______

(ii)

$AB$ .

______

DifficultyMedium-Easy

Worked solution

Approach

Find the gradient $m$ of line $AB$ using the coordinates of $A(3, -3)$ and $B(0, 3)$ , then use the $y$ -intercept (where the line crosses the $y$ -axis) to write the equation in the form $y = mx + c$ .

Working

The gradient $m$ of line $AB$ is:

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 3}{3 - 0} = \frac{-6}{3} = -2

The line passes through $B(0, 3)$ , which is on the $y$ -axis, so the $y$ -intercept is $c = 3$ .

Substituting $m = -2$ and $c = 3$ into $y = mx + c$ :

y = -2x + 3

This can also be written as $y = 3 - 2x$ .

Answer

y = 3 - 2x

Final answer

y = 3 - 2x

Detailed explanation

Walkthrough

Step 1: Calculate the gradient. The gradient formula is $m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$ . Using $A(3, -3)$ and $B(0, 3)$ :

m = \frac{-3 - 3}{3 - 0} = \frac{-6}{3} = -2

This tells us the line falls $2$ units for every $1$ unit it moves to the right.

Step 2: Find the $y$ -intercept. The $y$ -intercept is where the line crosses the $y$ -axis, i.e., where $x = 0$ . Point $B$ is at $(0, 3)$ , which is on the $y$ -axis, so the $y$ -intercept is $c = 3$ .

Step 3: Write the equation. Using $y = mx + c$ with $m = -2$ and $c = 3$ :

y = -2x + 3

This is equivalent to $y = 3 - 2x$ , both of which are accepted.

Key Takeaways

The gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ can be applied in either order of the two points.
If one of the given points lies on the $y$ -axis (i.e., has $x = 0$ ), that point directly gives the $y$ -intercept.
The equation $y = mx + c$ is the standard form for a non-vertical straight line.

Common Mistakes

Calculating the gradient as $\frac{3 - (-3)}{3 - 0} = \frac{6}{3} = 2$ (getting the sign wrong) — this happens when the numerator is computed as $y_1 - y_2$ instead of $y_2 - y_1$ while keeping the denominator as $x_2 - x_1$ .
Forgetting that the $y$ -intercept is the $y$ -value when $x = 0$ , not just any $y$ -value on the line.
Writing the equation as $x = -2y + 3$ by swapping $x$ and $y$ .

Things to Be Careful About

The mark scheme accepts $y = 3 - 2x$ as an equivalent form to $y = -2x + 3$ .
Working must be shown (marked as "www" in the scheme) to earn the method mark for the gradient calculation.
The gradient must be computed correctly: $\frac{-3 - 3}{3 - 0} = \frac{-6}{3} = -2$ .

Techniques used

calculate the gradient from two pointsidentify the y-interceptwrite the equation in slope-intercept form

Q73MMedium-EasyNumber

Asif is making a rectangular lawn $14\,\text{m}$ by $19\,\text{m}$ .
He uses grass seed to make the lawn.
The grass seed costs $0.34 for $1\,\text{m}^2$ .

By writing each number correct to 1 significant figure, estimate the cost of seed needed.

$ ______

DifficultyMedium-Easy

Worked solution

Approach

To estimate the cost, we first round each number involved in the calculation ( $14$ , $19$ , and $0.34$ ) to one significant figure. Then, we calculate the area using the rounded dimensions and multiply by the rounded cost per square metre.

Working

The dimensions of the lawn are $14\,\text{m}$ and $19\,\text{m}$ . The cost is $0.34 per $\text{m}^2$ .

Step 1: Round each number to 1 significant figure (1 s.f.)

$14$ : The first digit is $1$ . The next digit is $4$ , which is less than $5$ , so we round down.
$14 \approx 10$
$19$ : The first digit is $1$ . The next digit is $9$ , which is $5$ or more, so we round up.
$19 \approx 20$
$0.34$ : The first significant digit is $3$ . The next digit is $4$ , which is less than $5$ , so we round down.
$0.34 \approx 0.3$

Step 2: Calculate the estimated cost
The cost is given by:
$\text{Cost} = \text{Area} \times \text{Cost per m}^2$
$\text{Area} \approx 10 \times 20$
$\text{Cost} \approx (10 \times 20) \times 0.3$

Calculate the product:
$10 \times 20 = 200$
$200 \times 0.3 = 60$

So the estimated cost is $60.

Answer

$60$

Final answer

60

Detailed explanation

Walkthrough

The question asks for an estimate of the cost of grass seed. To do this efficiently without a calculator, we use the method of estimation by rounding.

Identify the numbers: We have two lengths ( $14$ and $19$ ) and a unit price ( $0.34$ ).
Round to 1 significant figure: This simplifies the arithmetic significantly.
- For $14$ , the leading digit is $1$ . Since the following digit ( $4$ ) is small, $14$ rounds to $10$ .
- For $19$ , the leading digit is $1$ . Since the following digit ( $9$ ) is large, $19$ rounds up to $20$ .
- For $0.34$ , the leading non-zero digit is $3$ . Since the following digit ( $4$ ) is small, $0.34$ rounds to $0.3$ .
Perform the calculation: Multiply the rounded numbers together.
- Area $\approx 10 \times 20 = 200$ .
- Cost $\approx 200 \times 0.3 = 60$ .

This gives a quick approximation of the total cost.

Key Takeaways

Rounding to 1 significant figure is the standard technique for quick estimation in Cambridge O Level Mathematics.
Remember that for numbers like $19$ , rounding to 1 s.f. moves it to the next multiple of ten ( $20$ ), not just truncating it.
For decimals like $0.34$ , the leading zeros are not significant. The first significant digit is the first non-zero digit ( $3$ ).

Common Mistakes

Rounding $14$ to $15$ (rounding to nearest 5 instead of 1 s.f.).
Rounding $0.34$ to $0.4$ (incorrectly rounding up when the next digit is 4).
Forgetting to round all three numbers before multiplying.
Calculating the exact answer first and then rounding the final result, which defeats the purpose of the estimation exercise and may not follow the mark scheme's requirement for "rounded values seen".

Things to Be Careful About

Ensure you round to one significant figure, not two or to the nearest whole number.
When rounding $0.34$ to $0.3$ , keep the decimal place correct. It becomes $0.3$ , not $3$ .
The mark scheme awards marks for seeing the rounded values ( $10, 20, 0.3$ ) and the final product. Make sure your working shows these intermediate rounded values clearly.

Techniques used

round each value to one significant figuremultiply the rounded values to estimate the total cost

Q8MediumAlgebra and Graphs

(a)

Solve.

4x - 7 = 9

$x$ = ______

(b)

Solve the simultaneous equations.
You must show all your working.

\begin{aligned} 5y - 8x &= -5 \\ 3y + 2x &= 14 \end{aligned}

$x$ = ______
$y$ = ______

DifficultyMedium

Worked solution

Approach

We have two simultaneous equations:

$5y - 8x = -5$
$3y + 2x = 14$

Notice that the coefficient of $x$ in equation (2) is $+2$ , while in equation (1) it is $-8$ . We can eliminate $x$ by multiplying equation (2) by 4 so that the $x$ terms become $+8x$ and $-8x$ . Then we add the two equations.

Working

Multiply equation (2) by 4:

4(3y + 2x) = 4(14)

12y + 8x = 56 \quad \text{(Equation 3)}

Now add Equation (1) and Equation (3):

(5y - 8x) + (12y + 8x) = -5 + 56

17y = 51

Solve for $y$ :

y = \frac{51}{17}

y = 3

Substitute $y = 3$ into Equation (2) to find $x$ :

3(3) + 2x = 14

9 + 2x = 14

Subtract 9 from both sides:

2x = 14 - 9

2x = 5

Divide by 2:

x = \frac{5}{2}

x = 2\frac{1}{2} \quad \text{or} \quad 2.5

Answer

x = 2\frac{1}{2}, \quad y = 3

Final answer

x = 2.5, y = 3

Detailed explanation

Walkthrough

To solve simultaneous equations by elimination, we want to remove one variable. Looking at the coefficients of $x$ ( $-8$ and $+2$ ), we see that if we multiply the second equation by 4, the $x$ term becomes $+8x$ . Adding this to the first equation (which has $-8x$ ) will cancel out $x$ completely.

After eliminating $x$ , we are left with an equation in terms of $y$ only: $17y = 51$ . Dividing by 17 gives $y = 3$ .

Once we have the value of $y$ , we substitute it back into one of the original equations (the simpler one, usually) to find $x$ . Substituting $y=3$ into $3y + 2x = 14$ gives $9 + 2x = 14$ . Solving this simple linear equation yields $2x = 5$ , so $x = 2.5$ or $2\frac{1}{2}$ .

Key Takeaways

Look for coefficients that are multiples of each other to make elimination easy.
When adding/subtracting equations, ensure signs are handled correctly.
Always check your solution by substituting values into the other original equation if time permits.

Common Mistakes

Incorrectly multiplying all terms in the equation by the scaling factor (e.g., forgetting to multiply the constant term 14 by 4).
Sign errors when adding the equations (e.g., subtracting instead of adding, leading to $-16y = -51$ which is also valid but requires care).
Arithmetic errors in division ( $51 / 17$ ).
Substitution errors when finding $x$ (e.g., using the wrong value for $y$ ).
Failing to convert improper fractions to mixed numbers or decimals if preferred, though both are accepted (oe).

Things to Be Careful About

The question asks to "show all your working". You must demonstrate the elimination step clearly. The mark scheme awards A1 for either correct value, so getting one right partially helps, but full marks require both. The answer for $x$ can be given as a fraction ( $\frac{5}{2}$ ), mixed number ( $2\frac{1}{2}$ ), or decimal ( $2.5$ ).

Techniques used

form equations from a worded contexteliminate one variablesubstitute back to find the second variable

Q9Medium-EasyNumber

Work out.

(a)

$8 \div 0.1$

______

(b)

$\frac{4}{5} - \frac{1}{2} \times \frac{3}{4}$

______

DifficultyMedium-Easy

Worked solution

Approach

Follow the order of operations (BODMAS/PEMDAS). Multiplication comes before subtraction. First, multiply $\frac{1}{2}$ by $\frac{3}{4}$ . Then, subtract this result from $\frac{4}{5}$ by finding a common denominator.

Working

Step 1: Perform the multiplication.

\frac{1}{2} \times \frac{3}{4} = \frac{1 \times 3}{2 \times 4} = \frac{3}{8}

Now substitute this back into the expression:

\frac{4}{5} - \frac{3}{8}

Step 2: Find a common denominator for 5 and 8. The lowest common multiple (LCM) of 5 and 8 is 40.
Convert each fraction:

\frac{4}{5} = \frac{4 \times 8}{5 \times 8} = \frac{32}{40}

\frac{3}{8} = \frac{3 \times 5}{8 \times 5} = \frac{15}{40}

Step 3: Subtract the numerators.

\frac{32}{40} - \frac{15}{40} = \frac{32 - 15}{40} = \frac{17}{40}

Answer

\frac{17}{40}

Final answer

17/40

Detailed explanation

Walkthrough

This problem tests the ability to handle mixed operations with fractions. The critical first step is recognizing that multiplication must be performed before subtraction. If you subtracted first ( $\frac{4}{5} - \frac{1}{2}$ ), you would get the wrong answer. After getting $\frac{3}{8}$ , you are left with $\frac{4}{5} - \frac{3}{8}$ . To subtract these, you need a common denominator. Since 5 and 8 share no factors, their least common multiple is $5 \times 8 = 40$ . You scale the top and bottom of each fraction appropriately ( $\times 8$ for the first, $\times 5$ for the second) so they have the same denominator, allowing you to simply subtract the numerators: $32 - 15 = 17$ .

Key Takeaways

Order of Operations: Multiplication and division happen before addition and subtraction.
Fraction Subtraction: Requires a common denominator (lowest common multiple of the denominators).
Scaling Fractions: Multiply numerator and denominator by the same number to create an equivalent fraction.

Common Mistakes

Subtracting numerators and denominators directly (e.g., $\frac{4-3}{5-8}$ ). This is incorrect.
Forgetting to perform the multiplication $\frac{1}{2} \times \frac{3}{4}$ first.
Using the wrong common denominator or making arithmetic errors when scaling the numerators ( $4 \times 8$ vs $5 \times 8$ ).

Things to Be Careful About

Check if the final fraction can be simplified. In this case, 17 is prime and does not divide 40, so $\frac{17}{40}$ is in its simplest form.
Ensure you copy the question correctly; mixing up the signs (+ vs -) changes the entire problem.

Techniques used

multiply two fractionsfind a common denominator for subtractionsubtract numerators while keeping the denominator constant

Q10Medium-EasyProbability

A spinner can land on red, green, yellow or blue.
The table shows the probability of each outcome.

Colour	Red	Green	Yellow	Blue
Probability	0.35	0.2	0.15	0.4

(a)

One of the values in the table is incorrect.

Explain how you know this.

(b)

The probability of the spinner landing on blue is incorrect.

Work out the correct probability of the spinner landing on blue.

______

DifficultyMedium-Easy

Worked solution

Approach

We are told the blue probability ( $0.4$ ) is the incorrect one. The probabilities for Red, Green, and Yellow are assumed correct. Since the total probability must be $1$ , we can find the correct Blue probability by subtracting the sum of the other three from $1$ .

Working

First, sum the correct probabilities:

P(\text{Red}) + P(\text{Green}) + P(\text{Yellow}) = 0.35 + 0.2 + 0.15

= 0.7

Now, subtract this sum from $1$ to find $P(\text{Blue})$ :

P(\text{Blue}) = 1 - 0.7

P(\text{Blue}) = 0.3

Answer

0.3

Final answer

0.3

Detailed explanation

Walkthrough

Part (a) established that the total was $1.1$ , meaning the probabilities were too high. Part (b) specifies that the error lies specifically with the 'Blue' entry ( $0.4$ ). This implies the entries for Red ( $0.35$ ), Green ( $0.2$ ), and Yellow ( $0.15$ ) are correct.

The fundamental rule is:

P(\text{Red}) + P(\text{Green}) + P(\text{Yellow}) + P(\text{Blue}) = 1

Substituting the known correct values:

0.35 + 0.2 + 0.15 + P(\text{Blue}) = 1

0.7 + P(\text{Blue}) = 1

Solving for $P(\text{Blue})$ :

P(\text{Blue}) = 1 - 0.7 = 0.3

Key Takeaways

If one probability in a complete set is unknown or identified as erroneous, it can be found by subtracting the sum of the others from $1$ .

Common Mistakes

Subtracting the incorrect value ( $0.4$ ) from something, rather than using the correct values.
Failing to realize that the total MUST be $1$ .
Calculation errors in the subtraction $1 - 0.7$ .

Things to Be Careful About

Read the question carefully: it explicitly states WHICH probability is incorrect. Do not try to guess which one is wrong; use the ones stated to be correct. Ensure the final answer is in decimal form consistent with the question ( $0.3$ , not $3/10$ unless asked, though both are mathematically equivalent, the context uses decimals).

Techniques used

subtract known probabilities from 1calculate the remaining probability

Q113MMedium-EasyNumber

A green light flashes every 12 minutes.
A red light flashes every 45 minutes.
The two lights flash together at 9 am.

Find the next time when the two lights will flash together.

______

DifficultyMedium-Easy

Worked solution

Approach

The lights flash together again after a time that is a common multiple of both flashing intervals. The first such time is the lowest common multiple (LCM) of 12 minutes and 45 minutes. Find the LCM by prime factorisation, convert it to hours, then add it to 9 am.

Working

Prime factorise each interval:

12 = 2^2 \times 3

45 = 3^2 \times 5

Take the highest power of each prime factor:

\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180

So the lights flash together every 180 minutes. Convert to hours:

180 \text{ minutes} = 3 \text{ hours}

Add this to the starting time:

9 \text{ am} + 3 \text{ hours} = 12 \text{ pm}

Answer

12:00 \text{ (12 pm)}

Final answer

12:00 (12 pm)

Detailed explanation

Walkthrough

The two lights flash together at 9 am. For them to flash together again, the number of minutes that has passed must be divisible by both 12 and 45. The smallest positive number divisible by both is the lowest common multiple (LCM).

To find the LCM, write each interval as a product of prime factors:

12 = 2^2 \times 3, \qquad 45 = 3^2 \times 5

For each prime that appears, take the highest power that appears in either factorisation: $2^2$ , $3^2$ and $5$ . Multiplying these gives:

2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180

So the lights coincide every 180 minutes. Since 180 minutes is exactly 3 hours, adding 3 hours to 9 am gives 12 pm, written as 12:00.

Key Takeaways

This question tests finding the LCM using prime factorisation and then applying it to a real-life time problem. The key skill is recognising that 'flash together again' means a common multiple of the two intervals, and that the next time uses the smallest such multiple.

Common Mistakes

A common error is to add the two intervals and use 57 minutes; this is not a common multiple of 12 and 45. Another mistake is to use the HCF (3) instead of the LCM. Some students also convert 180 minutes incorrectly, or write the final time as 12 am instead of 12 pm.

Things to Be Careful About

The answer must be the next time, so use the LCM, not just any common multiple. 180 minutes is exactly 3 hours, so the time is 12 noon. The mark scheme accepts 12:00 or 12 pm. If using 24-hour time, 12:00 also means noon, but the question gives 9 am, so 12 pm is clearest.

Techniques used

find the LCM by prime factorisationconvert minutes to hoursadd the elapsed time to the starting time

Q12EasyTransformations and Vectors

$\overrightarrow{PQ} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$

(a)

The position vector of point $Q$ is $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ .

Find the position vector of point $P$ .

$\overrightarrow{OP}$ = ______

DifficultyEasy

Worked solution

Approach

We are given the displacement vector $\overrightarrow{PQ}$ and the position vector of point $Q$ ( $\overrightarrow{OQ}$ ). We need to find the position vector of point $P$ ( $\overrightarrow{OP}$ ).

The relationship between these vectors is:

\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}

Rearranging this formula to solve for $\overrightarrow{OP}$ :

\overrightarrow{OP} = \overrightarrow{OQ} - \overrightarrow{PQ}

Working

Substitute the given vectors into the equation. Let $\overrightarrow{OQ} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\overrightarrow{PQ} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ .

\overrightarrow{OP} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ -2 \end{pmatrix}

Perform the subtraction component-wise (top minus top, bottom minus bottom):

Top component: $2 - 4 = -2$
Bottom component: $5 - (-2) = 5 + 2 = 7$

So,

\overrightarrow{OP} = \begin{pmatrix} -2 \\ 7 \end{pmatrix}

Answer

\overrightarrow{OP} = \begin{pmatrix} -2 \\ 7 \end{pmatrix}

Final answer

(-2, 7)

Detailed explanation

Walkthrough

To find the position vector of $P$ , we must understand how position vectors relate to the vector connecting two points. The vector $\overrightarrow{PQ}$ tells us how to get from $P$ to $Q$ . Mathematically, this is defined as the position of $Q$ minus the position of $P$ :

\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}

Since we want to find $\overrightarrow{OP}$ , we rearrange the equation by adding $\overrightarrow{OP}$ to both sides and subtracting $\overrightarrow{PQ}$ from both sides:

\overrightarrow{OP} = \overrightarrow{OQ} - \overrightarrow{PQ}

Now we substitute the numbers given in the question. Remember that subtracting a negative number is the same as adding a positive one. So, $5 - (-2)$ becomes $5 + 2$ .

Key Takeaways

Position Vector Rule: To go from point A to point B, you subtract the position vector of A from the position vector of B ( $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ ).
Vector Subtraction: Subtract the corresponding components individually. Be careful with signs, especially when subtracting a negative value.

Common Mistakes

Adding instead of subtracting: Calculating $\overrightarrow{OQ} + \overrightarrow{PQ}$ gives the wrong result.
Sign errors: Forgetting that subtracting $-2$ means adding $2$ (e.g., calculating $5 - 2 = 3$ ).
Reversing the order: Calculating $\overrightarrow{PQ} - \overrightarrow{OQ}$ .

Things to Be Careful About

Ensure you identify which vector is the "from" vector and which is the "to" vector in your head. $\overrightarrow{PQ}$ starts at $P$ and ends at $Q$ . Therefore $P$ is the starting point (position vector being subtracted) and $Q$ is the ending point (position vector being kept).

Techniques used

relate position vectors to displacement vectorssubtract a column vector

(b)

$|\overrightarrow{PQ}| = \sqrt{a}$

Find the value of $a$ .

$a$ = ______

DifficultyEasy

Worked solution

Approach

The magnitude (or length) of a column vector $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ is given by the formula:

|\mathbf{v}| = \sqrt{x^2 + y^2}

We are given $\overrightarrow{PQ} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ and asked to express its magnitude in the form $\sqrt{a}$ .

Working

Identify the components of $\overrightarrow{PQ}$ :
$x = 4$
$y = -2$

Substitute these values into the magnitude formula:

|\overrightarrow{PQ}| = \sqrt{4^2 + (-2)^2}

Calculate the squares:
$4^2 = 16$
$(-2)^2 = 4$

Add the results together:

|\overrightarrow{PQ}| = \sqrt{16 + 4} = \sqrt{20}

Comparing this to the required form $\sqrt{a}$ , we can see that:
$a = 20$

Answer

a = 20

Final answer

20

Detailed explanation

Walkthrough

The magnitude of a vector represents its length. Geometrically, if you draw the vector on a grid, it forms the hypotenuse of a right-angled triangle where the horizontal side is the x-component and the vertical side is the y-component. By Pythagoras' theorem, the length squared is equal to the sum of the squares of the components. Thus, Length = $\sqrt{x^2 + y^2}$ .

Here, the x-component is 4 and the y-component is -2. Squaring them removes any negative signs ( $(-2)^2 = 4$ ). Adding $16 + 4$ gives 20. The question asks for the answer in the form $\sqrt{a}$ , so we leave the 20 inside the square root rather than simplifying it to $2\sqrt{5}$ or approximating it.

Key Takeaways

Magnitude Formula: Always square both components, add them, and then take the square root.
Squaring Negatives: Remember that squaring a negative number yields a positive result.

Common Mistakes

Forgetting to square the components (e.g., $\sqrt{4 + (-2)}$ ).
Adding the absolute values without squaring first.
Simplifying the surd unnecessarily (the question specifically asks for $\sqrt{a}$ , so the final answer for 'a' is just the number under the root).

Things to Be Careful About

Check the question format carefully. It asks for the value of $a$ where the magnitude is $\sqrt{a}$ . Do not write $2\sqrt{5}$ as the final answer; write $a=20$ .

Techniques used

calculate the magnitude of a column vectorevaluate a square root expression

Q13MediumMensurationTrigonometry

The diagram shows a trapezium with lengths in centimetres.

Fig. 4

(a)

Work out the area of the trapezium.

______ $\text{cm}^2$

(b)

Work out the perimeter of the trapezium.

______ $\text{cm}$

DifficultyMedium

Worked solution

Approach

To find the perimeter, we need the lengths of all four sides. Three are given ( $6$ , $9$ , and $4$ ). The fourth is the slanted right side. We can find it by dropping a perpendicular from the top-right vertex to the bottom side, creating a right-angled triangle.

Working

The horizontal base of this right-angled triangle is the difference between the bottom and top parallel sides:

9 - 6 = 3

The vertical height is $4$ . By Pythagoras' theorem, the length of the slanted side $s$ is:

\begin{aligned} s^2 &= 3^2 + 4^2 \\ s^2 &= 9 + 16 \\ s^2 &= 25 \\ s &= 5 \end{aligned}

The perimeter is the sum of all four side lengths:

\begin{aligned} \text{Perimeter} &= 6 + 9 + 4 + 5 \\ &= 24 \end{aligned}

Answer

24

Final answer

24

Detailed explanation

Walkthrough

To calculate the perimeter, we must add the lengths of all four sides of the trapezium. We are given the top side ( $6$ ), the bottom side ( $9$ ), and the perpendicular left side ( $4$ ). The right side is slanted, so we need to find its length.

By dropping an imaginary perpendicular line from the top-right vertex down to the bottom side, we form a right-angled triangle on the right side of the trapezium. The vertical side of this triangle is the same as the left side of the trapezium, which is $4$ . The horizontal base of this triangle is the difference between the bottom and top parallel sides: $9 - 6 = 3$ .

Using Pythagoras' theorem on this right-angled triangle:

s^2 = 3^2 + 4^2 = 9 + 16 = 25

Taking the square root gives $s = 5$ . Finally, we add all four sides together to find the perimeter: $6 + 9 + 4 + 5 = 24$ .

Key Takeaways

When a trapezium has a slanted side, you can often find its length by constructing a right-angled triangle using the difference between the parallel sides and the perpendicular height. Pythagoras' theorem is then used to find the hypotenuse.

Common Mistakes

Forgetting to subtract the top side length from the bottom side length to find the base of the right-angled triangle (using $9$ instead of $3$ ).
Using the wrong sides in Pythagoras' theorem (e.g., adding $6$ and $4$ instead of $3$ and $4$ ).
Forgetting to include all four sides when calculating the perimeter.
Rounding errors if using a calculator for the square root, though $\sqrt{25}$ is exact here.

Things to Be Careful About

The diagram is marked NOT TO SCALE, so do not estimate lengths visually. Always use the given numerical values. The right-angle symbol at the bottom-left confirms that the left side is the perpendicular height, making the construction of the auxiliary right-angled triangle valid. Ensure the final answer is in $\text{cm}$ .

Techniques used

construct a right-angled triangle to find the slanted sideapply Pythagoras' theoremsum all four side lengths for the perimeter

Q14Medium-EasyNumberAlgebra and Graphs

(a)

Find the value of $3^3$ .

______

(b)

$2^x = 32$

Find the value of $x$ .

$x$ = ______

(c)

Simplify.

3a^{-3} \times 6a

______

DifficultyMedium-Easy

Worked solution

Approach

To simplify the expression $3a^{-3} \times 6a$ , we group the numerical coefficients together and the variable terms together, then apply the rules of indices.

Working

Group the numbers and the variables:

(3 \times 6) \times (a^{-3} \times a)

Multiply the coefficients:

3 \times 6 = 18

For the variables, recall that $a$ is the same as $a^1$ . When multiplying terms with the same base, add the exponents:

a^{-3} \times a^1 = a^{-3 + 1} = a^{-2}

Combine these results:

18a^{-2}

The mark scheme also accepts the answer with a positive index. A negative exponent indicates a reciprocal:

18a^{-2} = \frac{18}{a^2}

Answer

\frac{18}{a^2}

Final answer

18/a^2

Detailed explanation

Walkthrough

The expression is a product of two algebraic terms: $3a^{-3}$ and $6a$ .

Coefficients: Multiply the numbers $3$ and $6$ to get $18$ .
Variables: Multiply $a^{-3}$ by $a$ . Since there is no visible exponent on the second $a$ , it is understood to be $a^1$ . The law of indices states that $x^m \times x^n = x^{m+n}$ . So, we add the exponents: $-3 + 1 = -2$ . This gives $a^{-2}$ .
Final Form: The simplified term is $18a^{-2}$ . It is standard practice to write answers with positive indices if possible. Using the rule $x^{-n} = \frac{1}{x^n}$ , we move $a^{-2}$ to the denominator to get $\frac{18}{a^2}$ .

Key Takeaways

Always multiply coefficients separately from variables.
Remember that a variable written alone has an implicit exponent of $1$ (e.g., $a = a^1$ ).
Negative indices can be converted to positive indices by taking the reciprocal.

Common Mistakes

Adding the coefficients instead of multiplying them ( $3+6=9$ ).
Subtracting exponents instead of adding them when multiplying.
Forgetting that $a$ has an exponent of $1$ (treating it as $a^0$ or just dropping it).
Leaving the answer with a negative index when a positive one is preferred or required by specific instructions (though here both forms are accepted).

Things to Be Careful About

Check the sign of the resulting exponent carefully. $-3 + 1$ is $-2$ , not $-4$ or $2$ .

Techniques used

multiply numerical coefficientsapply the product rule for indicesrewrite negative exponents as fractions

Q15MediumNumber

(a)

Work out 70% of 120.

______

(b)

Find 8 as a percentage of 25.

______ %

(c)

In a sale, the original price of a jacket is reduced by 12%.
The sale price of the jacket is $66.

Calculate the original price of the jacket.

$ ______

DifficultyMedium

Worked solution

Approach

Let the original price be $x$ . The jacket was reduced by $12\%$ , so the sale price represents $(100 - 12)\%$ of the original price. We can set up an equation where $88\%$ of $x$ equals the sale price of $66, and solve for $x$ .

Working

Original price $= x$
Percentage reduction $= 12\%$
Percentage of original price paid $= 100\% - 12\% = 88\%$

Convert $88\%$ to a fraction:

88\% = \frac{88}{100}

Set up the equation using the sale price of $66:

\frac{88}{100} x = 66

To isolate $x$ , multiply both sides by $100$ and divide by $88$ :

x = 66 \times \frac{100}{88}

Simplify the fraction $\frac{66}{88}$ . Both numbers are divisible by $22$ :

66 \div 22 = 3

88 \div 22 = 4

So,

x = 3 \times \frac{100}{4}

Divide $100$ by $4$ :

\frac{100}{4} = 25

Multiply by $3$ :

x = 3 \times 25 = 75

Answer

\$75

Final answer

75

Detailed explanation

Walkthrough

This is a reverse percentage problem. We know the final price after a discount, but we need to find the starting price.

Step 1: Understand the percentages. The original price is always $100\%$ . Since there was a $12\%$ reduction, the customer paid $100\% - 12\% = 88\%$ of the original price.

Step 2: Relate the percentage to the money. We know that this $88\%$ corresponds to $66. Therefore, $88\%$ of the Original Price $= 66$ .

Step 3: Solve for the Original Price. We can write this algebraically as $0.88 \times x = 66$ or $\frac{88}{100}x = 66$ . To find $x$ , we divide $66$ by $0.88$ (or multiply by $\frac{100}{88}$ ).

Using fractions is often safer for non-calculator exams. $x = 66 \times \frac{100}{88}$ . We can simplify $\frac{66}{88}$ by dividing top and bottom by their highest common factor, which is $22$ . This leaves $\frac{3}{4}$ . So $x = 3 \times \frac{100}{4} = 3 \times 25 = 75$ .

Key Takeaways

In a discount/sale problem, the sale price is $(100 - \text{discount}\%)$ of the original price.
To find the original price from a discounted price, divide the sale price by the remaining percentage (as a decimal or fraction).

Common Mistakes

Subtracting $12\%$ of $66$ from $66$ . This is incorrect because the $12\%$ reduction was taken from the original price, not the sale price.
Adding $12\%$ to $66$ . This assumes the sale price was $112\%$ of the original, which is wrong.
Calculating $1\%$ correctly but making arithmetic errors when multiplying by $100$ .

Things to Be Careful About

Always identify which value represents $100\%$ . Here, the unknown original price is $100\%$ , not the $66.
Show working clearly. The mark scheme requires seeing the logic of equating the reduced percentage to the sale price.

Techniques used

identify original value as 100%calculate remaining percentagedivide sale price by remaining percentage factor

Q16Medium-EasyStatistics

The cumulative frequency diagram gives information about the marks scored by 80 students in an exam.

Fig. 5

(a)

Use the cumulative frequency diagram to complete the frequency table.
The first two frequencies have been completed for you.

Mark ( $m$ )	$0 < m \leq 20$	$20 < m \leq 40$	$40 < m \leq 55$	$55 < m \leq 70$	$70 < m \leq 90$	$90 < m \leq 100$
Frequency	5	10

DifficultyMedium-Easy

Worked solution

Approach

Read the cumulative frequency at each class boundary from the diagram, then find each frequency by subtracting consecutive cumulative frequencies.

Working

From the diagram, read the cumulative frequency at each upper class boundary:

\begin{aligned} C(20) &= 5 \\ C(40) &= 15 \\ C(55) &= 35 \\ C(70) &= 60 \\ C(90) &= 77 \\ C(100) &= 80 \end{aligned}

Each frequency is the difference between consecutive cumulative frequencies:

\begin{aligned} \text{Frequency for } 40 < m \leq 55 &= C(55) - C(40) = 35 - 15 = 20 \\ \text{Frequency for } 55 < m \leq 70 &= C(70) - C(55) = 60 - 35 = 25 \\ \text{Frequency for } 70 < m \leq 90 &= C(90) - C(70) = 77 - 60 = 17 \\ \text{Frequency for } 90 < m \leq 100 &= C(100) - C(90) = 80 - 77 = 3 \end{aligned}

Answer

\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Mark } (m) & 0 < m \leq 20 & 20 < m \leq 40 & 40 < m \leq 55 & 55 < m \leq 70 & 70 < m \leq 90 & 90 < m \leq 100 \\ \hline \text{Frequency} & 5 & 10 & 20 & 25 & 17 & 3 \\ \hline \end{array}

The four missing frequencies are $20$ , $25$ , $17$ , $3$ .

Final answer

20, 25, 17, 3

Detailed explanation

Walkthrough

A cumulative frequency diagram plots the running total of frequencies against the upper boundary of each class. To recover the individual frequency for a class, we read the cumulative frequency at its upper boundary and subtract the cumulative frequency at its lower boundary.

First, read the cumulative frequency at each class boundary from the graph:

At $m = 20$ , $C = 5$ (matches the given frequency of 5 for the first class).
At $m = 40$ , $C = 15$ (matches $5 + 10 = 15$ for the first two classes).
At $m = 55$ , $C = 35$ .
At $m = 70$ , $C = 60$ .
At $m = 90$ , $C = 77$ .
At $m = 100$ , $C = 80$ (matches the total number of students).

Then compute each missing frequency:

For $40 < m \leq 55$ : $35 - 15 = 20$ .
For $55 < m \leq 70$ : $60 - 35 = 25$ .
For $70 < m \leq 90$ : $77 - 60 = 17$ .
For $90 < m \leq 100$ : $80 - 77 = 3$ .

Check: $5 + 10 + 20 + 25 + 17 + 3 = 80$ ✓

Key Takeaways

The cumulative frequency at a class boundary is the sum of all frequencies up to and including that class.
Individual frequency = cumulative frequency at upper boundary minus cumulative frequency at lower boundary.
The final cumulative frequency must equal the total number of data points.

Common Mistakes

Reading the wrong axis or misreading grid values (the graph has fine gridlines; read carefully).
Subtracting in the wrong order (lower from upper, not upper from lower).
Forgetting that the first two frequencies are already given and should be verified against the graph.

Things to Be Careful About

Read cumulative frequency values to the nearest gridline. Values like $C(90) = 77$ require careful interpolation between gridlines.
The class boundaries are not evenly spaced ( $55$ , $70$ , $90$ ), so read the graph at the exact $m$ -values given.
Always verify the total frequency equals $80$ .

Techniques used

read cumulative frequency values from a graph at class boundariessubtract consecutive cumulative frequencies to find individual frequencies

(b)

Use the cumulative frequency diagram to find an estimate of

5M

(i)

the median mark

______

(ii)

the 30th percentile

______

(iii)

the number of students who scored 76 marks or more.

______

Q173MMedium-EasyAlgebra and Graphs

Rearrange the formula to make $x$ the subject.

ax = \frac{3x + 2}{5}

$x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

To make $x$ the subject, we must first eliminate the fraction, then collect all terms involving $x$ on one side of the equation, factorise $x$ , and finally divide to isolate $x$ .

Working

Start with the given equation:

ax = \frac{3x + 2}{5}

Multiply both sides by 5 to remove the denominator (M1):

5(ax) = 3x + 2

Simplify the left side:

5ax = 3x + 2

Subtract $3x$ from both sides to group the $x$ terms together (M1FT):

5ax - 3x = 2

Factorise the left side by taking out the common factor $x$ (M1FT):

x(5a - 3) = 2

Divide both sides by $(5a - 3)$ to make $x$ the subject:

x = \frac{2}{5a - 3}

Answer

x = \frac{2}{5a - 3}

Final answer

2/(5a-3)

Detailed explanation

Walkthrough

The goal is to rearrange the formula so that $x$ stands alone on one side.

Clear the fraction: The right-hand side has a denominator of 5. To get rid of it, multiply every term in the equation by 5. On the right, the 5 cancels with the denominator, leaving just $3x + 2$ . On the left, $5 \times ax$ becomes $5ax$ . This gives us $5ax = 3x + 2$ .
Group the $x$ terms: We want all parts of the equation containing $x$ on one side. Currently, we have $5ax$ on the left and $3x$ on the right. Subtract $3x$ from both sides. This moves the $3x$ to the left side, resulting in $5ax - 3x = 2$ . The constant term $2$ remains on the right.
Factorise: Now we have two terms on the left, $5ax$ and $-3x$ , that share a common factor of $x$ . We can write this as $x$ multiplied by what remains: $x(5a - 3)$ . So the equation is $x(5a - 3) = 2$ .
Isolate $x$ : To leave $x$ by itself, we need to remove the $(5a - 3)$ multiplying it. We do this by dividing both sides by $(5a - 3)$ . This yields $x = \frac{2}{5a - 3}$ .

Key Takeaways

When the subject appears in multiple terms, you cannot simply divide or subtract to isolate it; you must factorise first.
Clearing fractions early often simplifies the algebra significantly.
Be careful with signs when moving terms across the equals sign (e.g., $+3x$ becomes $-3x$ ).

Common Mistakes

Incorrect grouping: Students might try to combine $5ax$ and $3x$ into $2ax$ or similar errors. Remember, you can only combine 'like terms' (terms with exactly the same variable factors). Here, the variables are effectively different until factored.
Sign errors: Forgetting to change the sign of $3x$ when moving it to the other side, leading to $5ax + 3x = 2$ .
Order of operations: Attempting to subtract 3 before clearing the fraction, which leads to messy fractional coefficients.

Things to Be Careful About

Final Form: Ensure the answer is explicitly in the form $x = \dots$ . Do not leave it as $x(5a-3) = 2$ .
Variable vs Constant: Treat $a$ as a constant coefficient. It stays with $x$ during the subtraction step but becomes part of the divisor in the final step.
Accuracy: As per the mark scheme, an incorrect final answer after correct method steps will limit marks to M1/M2 only (Max 2 marks). Precision in the final division is key.

Techniques used

clear the denominator by multiplicationgroup terms containing the subject variablefactorise to isolate the subject

Q18MediumAlgebra and Graphs

The diagram shows the speed–time graph for part of a car’s journey.

Fig. 6

(a)

The car travels a total distance of $1.01\,\text{km}$ in the 90 seconds.

Calculate the value of $V$ .

$V$ = ______

4M Similar

DifficultyMedium

Worked solution

Approach

The distance travelled by the car is equal to the area under the speed-time graph. We are given the total distance as $1.01\,\text{km}$ , which must be converted to metres ( $1010\,\text{m}$ ) to match the speed units ( $\text{m/s}$ ). The area under the graph can be split into three geometric shapes: a trapezium from $t=0$ to $t=30$ , a rectangle from $t=30$ to $t=70$ , and a triangle from $t=70$ to $t=90$ . We sum these areas, set the total equal to $1010$ , and solve for $V$ .

Working

Convert the total distance to metres:

1.01\,\text{km} = 1010\,\text{m}

Calculate the area under the graph in terms of $V$ :

Area from $t=0$ to $t=30$ (trapezium with parallel sides $24$ and $V$ , width $30$ ):

\text{Area}_1 = \frac{1}{2} \times 30 \times (24 + V) = 15(24 + V) = 360 + 15V

Area from $t=30$ to $t=70$ (rectangle with width $40$ and height $V$ ):

\text{Area}_2 = 40 \times V = 40V

Area from $t=70$ to $t=90$ (triangle with base $20$ and height $V$ ):

\text{Area}_3 = \frac{1}{2} \times 20 \times V = 10V

Total area under the graph:

\text{Total Area} = (360 + 15V) + 40V + 10V = 360 + 65V

Set the total area equal to the total distance in metres:

360 + 65V = 1010

Solve for $V$ :

65V = 1010 - 360

65V = 650

V = \frac{650}{65} = 10

Answer

V = 10

Final answer

10

Detailed explanation

Walkthrough

The problem gives a speed-time graph and asks for the value of an unknown constant speed $V$ . The key principle here is that the distance travelled is the area under the speed-time graph.

First, we must ensure units are consistent. The speed is in $\text{m/s}$ and time is in seconds, so distance must be in metres. We convert $1.01\,\text{km}$ to $1010\,\text{m}$ .

Next, we calculate the area under the graph by splitting it into simple shapes:

From $t=0$ to $t=30$ , the graph is a straight line from $(0, 24)$ to $(30, V)$ . This forms a trapezium with parallel vertical sides of length $24$ and $V$ , and a horizontal width of $30$ . The area is $\frac{1}{2} \times 30 \times (24 + V) = 360 + 15V$ .
From $t=30$ to $t=70$ , the speed is constant at $V$ . This forms a rectangle with width $70 - 30 = 40$ and height $V$ . The area is $40V$ .
From $t=70$ to $t=90$ , the graph is a straight line from $(70, V)$ to $(90, 0)$ . This forms a right-angled triangle with base $90 - 70 = 20$ and height $V$ . The area is $\frac{1}{2} \times 20 \times V = 10V$ .

Adding these areas together gives the total distance: $360 + 15V + 40V + 10V = 360 + 65V$ .

We set this equal to the given total distance of $1010\,\text{m}$ : $360 + 65V = 1010$ . Subtracting $360$ from both sides gives $65V = 650$ , and dividing by $65$ yields $V = 10$ .

Key Takeaways

The area under a speed-time graph represents the distance travelled.
Always check that units are consistent before setting up equations (e.g., convert km to m when speed is in m/s).
Complex shapes under graphs can be split into simpler geometric figures (trapeziums, rectangles, triangles) to calculate the total area.

Common Mistakes

Forgetting to convert the distance from kilometres to metres, leading to setting the area equal to $1.01$ instead of $1010$ .
Miscalculating the area of the trapezium by using the wrong formula or misidentifying the parallel sides.
Using the wrong width for the rectangular section (e.g., using $70$ instead of $70 - 30 = 40$ ).

Things to Be Careful About

The question specifies the distance in km ( $1.01\,\text{km}$ ), but the graph axes use m/s and seconds. Converting to $1010\,\text{m}$ is essential.
The answer must be given as a number; $V = 10$ is exact and requires no rounding.
When calculating areas, ensure you use the correct dimensions from the graph (e.g., the width of the last triangle is $90 - 70 = 20$ , not $90$ ).

Techniques used

convert distance to consistent unitscalculate area under a speed-time graph as total distanceform and solve a linear equation for V

(b)

Calculate the deceleration of the car in the last 20 seconds.

______ $\text{m/s}^2$

DifficultyEasy

Worked solution

Approach

Deceleration is the rate at which speed decreases. On a speed-time graph, the acceleration (or deceleration) is given by the gradient of the line. For the last $20$ seconds (from $t=70$ to $t=90$ ), the speed drops from $V$ to $0$ . We use the value of $V$ found in part (a) to calculate this gradient.

Working

From part (a), $V = 10\,\text{m/s}$ .

The car's speed changes from $10\,\text{m/s}$ at $t=70\,\text{s}$ to $0\,\text{m/s}$ at $t=90\,\text{s}$ .

The change in speed is:

\Delta v = 0 - 10 = -10\,\text{m/s}

The time interval is:

\Delta t = 90 - 70 = 20\,\text{s}

The acceleration is:

\text{Acceleration} = \frac{\Delta v}{\Delta t} = \frac{-10}{20} = -0.5\,\text{m/s}^2

Since the question asks for deceleration, we take the magnitude of the negative acceleration:

\text{Deceleration} = 0.5\,\text{m/s}^2

Answer

0.5

Final answer

0.5

Detailed explanation

Walkthrough

Part (b) asks for the deceleration during the last $20$ seconds of the journey. On a speed-time graph, the gradient of the line represents the acceleration. A negative gradient indicates deceleration.

We focus on the time interval from $t=70$ to $t=90$ . At $t=70$ , the speed is $V$ , which we found to be $10\,\text{m/s}$ in part (a). At $t=90$ , the speed is $0\,\text{m/s}$ .

The change in speed is $0 - 10 = -10\,\text{m/s}$ . The time taken for this change is $90 - 70 = 20\,\text{s}$ .

Acceleration is calculated as $\frac{\text{change in speed}}{\text{time}} = \frac{-10}{20} = -0.5\,\text{m/s}^2$ . The negative sign indicates the car is slowing down. Deceleration is the positive value of this rate, so the deceleration is $0.5\,\text{m/s}^2$ .

Key Takeaways

The gradient of a speed-time graph gives the acceleration.
Deceleration is simply the magnitude of negative acceleration.
Always use the correct value from previous parts (here, $V=10$ ) when calculating subsequent quantities.

Common Mistakes

Forgetting to use the value of $V$ from part (a) and trying to calculate the gradient without it.
Calculating the gradient as $\frac{10}{20} = 0.5$ and then incorrectly stating the acceleration is $0.5$ without recognizing it as deceleration (though the question asks for deceleration, so the positive value is correct, the reasoning must be sound).
Using the wrong time interval (e.g., using $90$ instead of $90 - 70 = 20$ ).

Things to Be Careful About

The question asks for deceleration, which is a positive quantity representing the rate of slowing down. Ensure the final answer is positive.
Units are already provided in the question ( $\text{m/s}^2$ ), so only the numerical value is needed in the answer box.
Follow-through from part (a) is expected; using an incorrect value of $V$ will lead to an incorrect deceleration.

Techniques used

identify the gradient of the speed-time graph as accelerationcalculate deceleration as the negative of acceleration

Q193MMedium-EasyTransformations and Vectors

The diagram shows triangle $A$ and triangle $B$ .

Fig. 7

Describe fully the single transformation that maps triangle $A$ onto triangle $B$ .

______

DifficultyMedium-Easy

Worked solution

Approach

Join corresponding vertices of the two triangles. The lines intersect at the centre of enlargement. The scale factor is found by comparing the lengths of corresponding sides or the distances of corresponding vertices from the centre.

Working

Lines through corresponding vertices:
Vertex $(-5, -2)$ of $A$ maps to $(4, 1)$ of $B$ .
Vertex $(-1, -2)$ of $A$ maps to $(2, 1)$ of $B$ .
Vertex $(-5, -4)$ of $A$ maps to $(4, 2)$ of $B$ .

Equation of line through $(-5, -2)$ and $(4, 1)$ :

m = \frac{1 - (-2)}{4 - (-5)} = \frac{3}{9} = \frac{1}{3}

y - 1 = \frac{1}{3}(x - 4) \implies x - 3y = 1

Equation of line through $(-1, -2)$ and $(2, 1)$ :

m = \frac{1 - (-2)}{2 - (-1)} = \frac{3}{3} = 1

y - 1 = 1(x - 2) \implies x - y = 1

Finding the intersection (centre of enlargement):

\begin{aligned} x - 3y &= 1 \\ x - y &= 1 \end{aligned}

Subtracting the second equation from the first:

-2y = 0 \implies y = 0

Substituting $y = 0$ into $x - y = 1$ :

x = 1

The centre of enlargement is $(1, 0)$ .

Finding the scale factor:

The length of the horizontal side of $A$ from $(-5, -2)$ to $(-1, -2)$ is $4$ units.

The corresponding horizontal side of $B$ from $(4, 1)$ to $(2, 1)$ is $2$ units.

\text{Scale factor} = \frac{2}{4} = \frac{1}{2}

Since the triangles are on opposite sides of the centre (the image is inverted), the scale factor is negative.

\text{Scale factor} = -\frac{1}{2}

Answer

Enlargement with centre $(1, 0)$ and scale factor $-\frac{1}{2}$ .

Final answer

Enlargement, centre (1, 0), scale factor -1/2

Detailed explanation

Walkthrough

To describe the transformation mapping triangle $A$ to triangle $B$ , we first observe that the triangles are similar but oriented in opposite directions. This indicates an enlargement with a negative scale factor.

We find the centre of enlargement by drawing lines through corresponding vertices of the two triangles. The vertex $(-5, -2)$ on $A$ maps to $(4, 1)$ on $B$ . The line through these points has gradient $\frac{1}{3}$ and equation $x - 3y = 1$ . The vertex $(-1, -2)$ on $A$ maps to $(2, 1)$ on $B$ . The line through these points has gradient $1$ and equation $x - y = 1$ . Solving these two equations simultaneously gives the intersection point $(1, 0)$ , which is the centre of enlargement.

To find the scale factor, we compare the lengths of corresponding sides. The horizontal side of $A$ has length $4$ , and the corresponding horizontal side of $B$ has length $2$ . The ratio of the image length to the original length is $\frac{2}{4} = \frac{1}{2}$ . Because the image is inverted relative to the centre, the scale factor is negative, giving $-\frac{1}{2}$ .

Key Takeaways

An enlargement with a negative scale factor produces an image on the opposite side of the centre and inverted.
The centre of enlargement is the fixed point where lines joining corresponding vertices of the original and image shapes intersect.
The scale factor is the ratio of a length in the image to the corresponding length in the original shape, with a negative sign if the image is inverted.

Common Mistakes

Forgetting to state the type of transformation (enlargement) or giving an incomplete description (must include type, centre, and scale factor for full marks).
Giving a positive scale factor when the image is inverted; forgetting the negative sign for enlargements with the image on the opposite side of the centre.
Reading the centre coordinates incorrectly from the grid.
Describing the transformation as a rotation or reflection; the change in size (from side length 4 to side length 2) confirms it is an enlargement.

Things to Be Careful About

A full description of an enlargement requires all three elements: the type of transformation, the centre coordinates, and the scale factor. Missing any one of these costs a mark.
Ensure coordinates are read accurately from the grid axes.
The scale factor must include the negative sign when the image is on the opposite side of the centre of enlargement.

Techniques used

identify the type of transformation from the orientation of the shapesfind the centre of enlargement by intersecting lines through corresponding verticescalculate the scale factor from the ratio of corresponding lengths

Q20Medium-EasyMensuration

Fig. 8

$OAB$ and $OCD$ are sectors of circles, each with centre $O$ and angle $60^\circ$ .
Sector $OAB$ has radius $6\,\text{cm}$ .
Sector $OCD$ has radius $9\,\text{cm}$ .

(a)

The perimeter of the shaded region is $(a\pi + b)\,\text{cm}$ .

Find the value of $a$ and the value of $b$ .

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Approach

The perimeter of the shaded region consists of two arc lengths (arc CD and arc AB) and two straight line segments (AC and BD). Calculate each arc length using the formula $\frac{\theta}{360} \times 2\pi r$ , and find the straight lengths by subtracting the inner radius from the outer radius.

Working

The angle at the centre is $60^\circ$ .

Length of arc CD:

\frac{60}{360} \times 2\pi \times 9 = 3\pi \text{ cm}

Length of arc AB:

\frac{60}{360} \times 2\pi \times 6 = 2\pi \text{ cm}

Length of AC:

OC - OA = 9 - 6 = 3 \text{ cm}

Length of BD:

OD - OB = 9 - 6 = 3 \text{ cm}

Perimeter of shaded region:

3\pi + 2\pi + 3 + 3 = 5\pi + 6 \text{ cm}

Comparing with $(a\pi + b)$ , we have $a = 5$ and $b = 6$ .

Answer

a = 5, \quad b = 6

Final answer

a = 5, b = 6

Detailed explanation

Walkthrough

The perimeter of the shaded region is the total length of its boundary. This boundary is made up of four pieces: the outer arc CD, the inner arc AB, and the two straight line segments AC and BD that connect the ends of the arcs.

First, we calculate the length of arc CD. The formula for arc length is $\frac{\theta}{360} \times 2\pi r$ , where $\theta$ is the angle at the centre and $r$ is the radius. For arc CD, $\theta = 60^\circ$ and $r = 9$ , giving $\frac{60}{360} \times 2\pi \times 9 = 3\pi$ cm.

Next, we calculate the length of arc AB using the same formula with $r = 6$ , giving $\frac{60}{360} \times 2\pi \times 6 = 2\pi$ cm.

The straight segments AC and BD are parts of the radii. Since $OC = 9$ and $OA = 6$ , the length $AC = 9 - 6 = 3$ cm. Similarly, $BD = 9 - 6 = 3$ cm.

Adding all four pieces together gives the total perimeter: $3\pi + 2\pi + 3 + 3 = 5\pi + 6$ cm. Matching this to the form $a\pi + b$ gives $a = 5$ and $b = 6$ .

Key Takeaways

The perimeter of a shaded region bounded by arcs and straight lines is the sum of the lengths of all those boundary pieces.
Arc length is found using $\frac{\theta}{360} \times 2\pi r$ .
Straight segments connecting concentric arcs are found by subtracting the inner radius from the outer radius.

Common Mistakes

Forgetting to include the two straight line segments (AC and BD) in the perimeter calculation.
Using the full circle circumference $2\pi r$ instead of the arc length formula $\frac{\theta}{360} \times 2\pi r$ .
Adding the radii together instead of subtracting to find the length of AC and BD.

Things to Be Careful About

Ensure the angle is in degrees when using the $\frac{\theta}{360}$ formula.
The answer must be in the exact form $a\pi + b$ ; do not convert $\pi$ to a decimal.
Keep track of which radius corresponds to which arc (9 cm for the outer arc, 6 cm for the inner arc).

Techniques used

calculate arc length of a sectorfind straight line segment lengths by subtractionsum arc lengths and straight segments for perimeter

(b)

Calculate the area of the shaded region.
Give your answer in its simplest form in terms of $\pi$ .

______ $\text{cm}^2$

DifficultyMedium-Easy

Worked solution

Approach

The area of the shaded region is the area of the larger sector OCD minus the area of the smaller sector OAB. Use the formula $\frac{\theta}{360} \times \pi r^2$ for each sector.

Working

Area of sector OCD:

\frac{60}{360} \times \pi \times 9^2 = \frac{1}{6} \times 81\pi = \frac{81\pi}{6} = \frac{27\pi}{2} \text{ cm}^2

Area of sector OAB:

\frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times 36\pi = 6\pi \text{ cm}^2

Area of shaded region:

\frac{27\pi}{2} - 6\pi = \frac{27\pi}{2} - \frac{12\pi}{2} = \frac{15\pi}{2} \text{ cm}^2

This can also be written as $7.5\pi$ .

Answer

\frac{15\pi}{2} \text{ cm}^2

Final answer

15pi/2

Detailed explanation

Walkthrough

The shaded region is the area between two concentric sectors with the same central angle. To find its area, we calculate the area of the larger sector (OCD) and subtract the area of the smaller sector (OAB).

The formula for the area of a sector is $\frac{\theta}{360} \times \pi r^2$ . For sector OCD, $\theta = 60^\circ$ and $r = 9$ , so the area is $\frac{60}{360} \times \pi \times 9^2 = \frac{1}{6} \times 81\pi = \frac{27\pi}{2}$ cm $^2$ .

For sector OAB, $\theta = 60^\circ$ and $r = 6$ , so the area is $\frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times 36\pi = 6\pi$ cm $^2$ .

Subtracting the smaller area from the larger area gives $\frac{27\pi}{2} - 6\pi$ . To subtract, we write $6\pi$ as $\frac{12\pi}{2}$ , resulting in $\frac{27\pi - 12\pi}{2} = \frac{15\pi}{2}$ cm $^2$ . The question asks for the simplest form in terms of $\pi$ , so $\frac{15\pi}{2}$ (or $7.5\pi$ ) is the final answer.

Key Takeaways

The area of a shaded region between two concentric sectors is the difference of their individual areas.
Sector area is found using $\frac{\theta}{360} \times \pi r^2$ .
When subtracting fractions involving $\pi$ , find a common denominator before combining.

Common Mistakes

Using the circumference formula $2\pi r$ instead of the area formula $\pi r^2$ .
Forgetting to square the radius in the area formula.
Subtracting the areas incorrectly or failing to find a common denominator when combining the $\pi$ terms.
Giving a decimal approximation for $\pi$ when the question asks for the answer in terms of $\pi$ .

Things to Be Careful About

The question requires the answer in its simplest form in terms of $\pi$ . Do not evaluate $\pi$ as 3.14159... or leave it as an unsimplified fraction like $\frac{81\pi}{6} - 6\pi$ .
Ensure the final answer is in cm $^2$ , not cm.
$\frac{15\pi}{2}$ and $7.5\pi$ are both acceptable final forms.

Techniques used

calculate area of a sectorsubtract smaller sector area from larger sector areasimplify the resulting fraction

Q214MMediumCoordinate Geometry

Line $L$ has equation $2y = 5x + 3$ .

Find the equation of the line perpendicular to $L$ which passes through the point $(1, 4)$ .

______

DifficultyMedium

Worked solution

Approach

First, determine the gradient of the given line $L$ by rearranging its equation into the form $y = mx + c$ . Then, calculate the gradient of the required perpendicular line using the negative reciprocal property. Finally, substitute the coordinates of the given point $(1, 4)$ into the equation of the perpendicular line to find the y-intercept (or use the point-slope form) and write the final equation.

Working

Step 1: Find the gradient of line $L$
The equation of line $L$ is:

2y = 5x + 3

Rearrange to make $y$ the subject:

y = \frac{5}{2}x + \frac{3}{2}

The gradient ( $m_L$ ) is the coefficient of $x$ :

m_L = \frac{5}{2}

Step 2: Find the gradient of the perpendicular line
Let $m_{\perp}$ be the gradient of the line perpendicular to $L$ . The product of the gradients of two perpendicular lines is $-1$ :

m_L \times m_{\perp} = -1

Substituting $m_L = \frac{5}{2}$ :

\frac{5}{2} \times m_{\perp} = -1

m_{\perp} = -\frac{2}{5}

Step 3: Find the equation of the perpendicular line
The equation of the perpendicular line is of the form:

y = -\frac{2}{5}x + c

This line passes through the point $(1, 4)$ . Substitute $x = 1$ and $y = 4$ into the equation:

4 = -\frac{2}{5}(1) + c

Solve for $c$ :

4 = -\frac{2}{5} + c

c = 4 + \frac{2}{5}

c = \frac{20}{5} + \frac{2}{5} = \frac{22}{5}

So the equation in slope-intercept form is:

y = -\frac{2}{5}x + \frac{22}{5}

Alternatively, this can be written in general form by multiplying the entire equation by 5:

5y = -2x + 22

5y + 2x = 22

Answer

The equation of the line is:

y = -\frac{2}{5}x + \frac{22}{5}

or equivalently,

5y + 2x = 22

Final answer

y = -2/5 x + 22/5

Detailed explanation

Walkthrough

The problem asks for the equation of a line that satisfies two conditions: it is perpendicular to a given line $L$ , and it passes through a specific point $(1, 4)$ .

Identify the gradient of the original line: The given equation is $2y = 5x + 3$ . To find the gradient easily, we rearrange this into the slope-intercept form $y = mx + c$ , where $m$ is the gradient. Dividing both sides by 2 gives $y = \frac{5}{2}x + \frac{3}{2}$ . Thus, the gradient of line $L$ is $\frac{5}{2}$ . This corresponds to the first method mark (M1) in the scheme.
Determine the perpendicular gradient: Two lines are perpendicular if the product of their gradients is $-1$ . If the gradient of one line is $m$ , the gradient of the perpendicular line is $-\frac{1}{m}$ (the negative reciprocal). Therefore, the gradient of our new line is the negative reciprocal of $\frac{5}{2}$ , which is $-\frac{2}{5}$ . This corresponds to the second method mark (M1FT).
Find the specific equation: We now know the gradient of the new line is $-\frac{2}{5}$ , so its equation looks like $y = -\frac{2}{5}x + c$ . Since the line passes through the point $(1, 4)$ , these coordinates must satisfy the equation. Substituting $x=1$ and $y=4$ gives $4 = -\frac{2}{5}(1) + c$ . Solving for $c$ : add $\frac{2}{5}$ to both sides to get $c = 4 + \frac{2}{5} = \frac{22}{5}$ . This substitution step earns the third method mark (M1FT).
Final Form: The equation is $y = -\frac{2}{5}x + \frac{22}{5}$ . It is often acceptable to present the answer in integer form by clearing the denominator. Multiplying by 5 yields $5y = -2x + 22$ , or $5y + 2x = 22$ . Both forms are correct.

Key Takeaways

Gradient from Equation: Always rearrange linear equations to $y = mx + c$ to quickly identify the gradient $m$ .
Perpendicular Gradients: Remember the rule $m_1 \times m_2 = -1$ . The gradient of a perpendicular line is the negative reciprocal ( $-\frac{1}{m}$ ).
Finding the Intercept: Once you have the gradient, use a known point on the line to solve for the y-intercept $c$ .

Common Mistakes

Incorrect Perpendicular Gradient: Students often forget the negative sign (giving $\frac{2}{5}$ instead of $-\frac{2}{5}$ ) or only take the reciprocal without negating it.
Algebraic Errors: When solving $4 = -\frac{2}{5} + c$ , students might subtract $\frac{2}{5}$ instead of adding it, leading to an incorrect intercept.
Formatting: Failing to simplify the fraction or leaving the answer in an unsimplified fractional form when an integer form is preferred (though usually both are accepted unless specified).

Things to Be Careful About

Sign Errors: Be very careful with negative signs when dealing with reciprocals and subtraction of fractions.
Fraction Arithmetic: Adding integers and fractions requires converting the integer to a fraction with a common denominator (e.g., $4 = \frac{20}{5}$ ).
Method Marks: Ensure you show the rearrangement of the original line and the calculation of the negative reciprocal clearly, as these are key steps for earning method marks.

Techniques used

rearrange the given equation to find the gradientdetermine the negative reciprocal for the perpendicular gradientsubstitute the given point into the line equation to find the constantsimplify the equation into standard or slope-intercept form

Q224MMediumAlgebra and Graphs

Simplify.

\frac{x^2 - 9}{5x^2 - 11x - 12}

______

DifficultyMedium

Worked solution

Approach

To simplify the algebraic fraction $\frac{x^2 - 9}{5x^2 - 11x - 12}$ , we must factorise both the numerator and the denominator completely. Once factorised, we cancel any common binomial factors.

Working

Step 1: Factorise the numerator.
The numerator is $x^2 - 9$ . This is a difference of two perfect squares ( $a^2 - b^2 = (a-b)(a+b)$ ):

x^2 - 9 = (x - 3)(x + 3)

(B1 mark awarded for seeing this correct factorisation)

Step 2: Factorise the denominator.
The denominator is $5x^2 - 11x - 12$ . We look for two numbers that multiply to $5 \times (-12) = -60$ and add to $-11$ . These numbers are $-15$ and $4$ .
We split the middle term:

5x^2 - 15x + 4x - 12

Group terms:

5x(x - 3) + 4(x - 3)

Extract the common bracket:

(5x + 4)(x - 3)

(B2 mark awarded for seeing this full factorisation, or B1 for partial progress like the grouped form above)

Step 3: Simplify the fraction.
Rewrite the original expression with the factorised forms:

\frac{(x - 3)(x + 3)}{(5x + 4)(x - 3)}

Cancel the common factor $(x - 3)$ from the top and bottom:

\frac{x + 3}{5x + 4}

Answer

\frac{x + 3}{5x + 4}

Final answer

(x + 3)/(5x + 4)

Detailed explanation

Walkthrough

The goal is to reduce an algebraic fraction to its simplest form. This involves two main tasks: factorising the numerator and factorising the denominator.

First, look at the numerator $x^2 - 9$ . This fits the pattern of the difference of two squares, which states that $a^2 - b^2 = (a - b)(a + b)$ . Here, $a=x$ and $b=3$ , so it becomes $(x - 3)(x + 3)$ . This is a straightforward recall fact.

Next, look at the denominator $5x^2 - 11x - 12$ . This is a non-monic quadratic (the coefficient of $x^2$ is not 1). The most reliable method is to find two numbers that multiply to give $ac$ ( $5 \times -12 = -60$ ) and add to give $b$ ( $-11$ ). The pair $-15$ and $4$ works because $(-15) \times 4 = -60$ and $(-15) + 4 = -11$ . We use these to rewrite the middle term: $5x^2 - 15x + 4x - 12$ . Then we factorise by grouping: take out $5x$ from the first two terms to get $5x(x - 3)$ , and take out $4$ from the last two to get $4(x - 3)$ . Since $(x - 3)$ is now a common factor, we pull it out to get $(5x + 4)(x - 3)$ .

Finally, we write the fraction as $\frac{(x - 3)(x + 3)}{(5x + 4)(x - 3)}$ . We can see that $(x - 3)$ appears in both the numerator and the denominator, so they cancel each other out, leaving the final simplified answer $\frac{x + 3}{5x + 4}$ .

Key Takeaways

Always check for special products like the difference of two squares before attempting more complex methods.
For quadratics where the leading coefficient is not 1, splitting the middle term (using the product-sum method) is a robust technique.
Simplifying fractions requires identifying and removing common factors across the entire numerator and denominator.

Common Mistakes

Incorrect factorisation: Getting the signs wrong in the denominator (e.g., writing $(5x - 4)(x + 3)$ ) results in losing the B2 mark. Note that $(5x - 4)(x + 3)$ expands to $5x^2 + 11x - 12$ , which has the wrong sign for the middle term.
Incomplete cancellation: Canceling only parts of terms, such as canceling the $x$ 's or the $3$ 's individually without treating $(x-3)$ as a single unit.
Failing to factorise: Leaving the answer as the original unsimplified fraction.

Things to Be Careful About

Check your work: Expand your factors back out to ensure they match the original expressions. For the denominator, $(5x+4)(x-3) = 5x^2 - 15x + 4x - 12 = 5x^2 - 11x - 12$ . If this doesn't match, the factorisation is wrong.
Domain restrictions: While usually not required for simple simplification questions at this level, note that the original expression is undefined for $x = 3$ and $x = -4/5$ . The simplified form removes the hole at $x=3$ but does not change the domain restriction implicitly. However, for 'Simplify' questions, just providing the reduced algebraic form is sufficient.

Techniques used

factorise a quadratic expression by difference of two squaresfactorise a quadratic trinomial by grouping or inspectioncancel common factors in algebraic fractions

Q23MediumAlgebra and Graphs

(a)

Expand and simplify.

(x - 3)(2x + 5)(x + 2)

______

DifficultyMedium-Easy

Worked solution

Approach

Expand two of the brackets first to obtain a quadratic, then multiply the quadratic by the third bracket and collect like terms.

Working

First, expand $(x - 3)(2x + 5)$ :

\begin{aligned} (x - 3)(2x + 5) &= 2x^2 + 5x - 6x - 15 \\ &= 2x^2 - x - 15 \end{aligned}

Now multiply this result by $(x + 2)$ :

\begin{aligned} (2x^2 - x - 15)(x + 2) &= 2x^2(x) + 2x^2(2) - x(x) - x(2) - 15(x) - 15(2) \\ &= 2x^3 + 4x^2 - x^2 - 2x - 15x - 30 \\ &= 2x^3 + 3x^2 - 17x - 30 \end{aligned}

Answer

2x^3 + 3x^2 - 17x - 30

Final answer

2x^3 + 3x^2 - 17x - 30

Detailed explanation

Walkthrough

To expand three linear factors, we multiply them in pairs. First, we expand $(x - 3)(2x + 5)$ by multiplying each term in the first bracket by each term in the second: $x \times 2x = 2x^2$ , $x \times 5 = 5x$ , $-3 \times 2x = -6x$ , and $-3 \times 5 = -15$ . Combining the like terms $5x$ and $-6x$ gives $2x^2 - x - 15$ . Next, we multiply this quadratic by the remaining factor $(x + 2)$ , distributing each term: $2x^2 \times x = 2x^3$ , $2x^2 \times 2 = 4x^2$ , $-x \times x = -x^2$ , $-x \times 2 = -2x$ , $-15 \times x = -15x$ , and $-15 \times 2 = -30$ . Finally, we collect like terms: $4x^2 - x^2 = 3x^2$ and $-2x - 15x = -17x$ , yielding the simplified expression $2x^3 + 3x^2 - 17x - 30$ .

Key Takeaways

When expanding three or more linear factors, expand them sequentially in pairs. Always collect like terms carefully at each stage to avoid sign errors.

Common Mistakes

Forgetting to multiply every term in one bracket by every term in the other (e.g. missing the $-15 \times 2 = -30$ term).
Sign errors when combining negative and positive like terms, such as writing $-2x - 15x = 13x$ instead of $-17x$ .
Not simplifying the final answer to a four-term expression as required.

Things to Be Careful About

Ensure all intermediate multiplication steps are shown clearly, as partial marks are awarded for correct unsimplified expressions or for having three out of four terms correct.
Keep track of negative signs throughout the expansion; a single sign error will cascade and invalidate the final answer.

Techniques used

expand the first pair of bracketsmultiply the resulting quadratic by the remaining linear factorcollect like terms

(b)

Sketch the graph of $y = (x - 3)(2x + 5)(x + 2)$ .
On the sketch, label the values where the graph crosses the axes.

Fig. 9

4M Similar

DifficultyMedium

Worked solution

Approach

Find the x-intercepts by setting each factor equal to zero, find the y-intercept by substituting $x = 0$ , and use the leading term to determine the end behaviour of the cubic curve.

Working

The x-intercepts occur where $y = 0$ :

(x - 3)(2x + 5)(x + 2) = 0

Setting each factor to zero:

x - 3 = 0 \implies x = 3

2x + 5 = 0 \implies x = -2.5

x + 2 = 0 \implies x = -2

The y-intercept occurs where $x = 0$ :

y = (0 - 3)(2(0) + 5)(0 + 2) = (-3)(5)(2) = -30

The leading term of the expanded expression is $2x^3$ . Since the coefficient is positive, the graph starts in the bottom-left (as $x \to -\infty$ , $y \to -\infty$ ) and ends in the top-right (as $x \to \infty$ , $y \to \infty$ ).

Answer

Sketch a positive cubic curve crossing the x-axis at $-2.5$ , $-2$ , and $3$ , and crossing the y-axis at $-30$ .

Final answer

Positive cubic curve crossing the x-axis at -2.5, -2, and 3, and the y-axis at -30.

Detailed explanation

Walkthrough

To sketch the graph, we first identify the key points where the curve crosses the axes. The x-intercepts are found by setting $y = 0$ , which means at least one of the factors $(x - 3)$ , $(2x + 5)$ , or $(x + 2)$ must be zero. Solving each gives $x = 3$ , $x = -2.5$ , and $x = -2$ . The y-intercept is found by substituting $x = 0$ into the equation, giving $y = (-3)(5)(2) = -30$ . Next, we determine the shape of the curve. The expanded form is $2x^3 + 3x^2 - 17x - 30$ , and the leading term is $2x^3$ . Because the coefficient of $x^3$ is positive, the curve has the standard positive cubic shape: it comes from the bottom-left quadrant, crosses the x-axis at $-2.5$ , rises to a local maximum between $-2.5$ and $-2$ , crosses the x-axis again at $-2$ , falls to cross the y-axis at $-30$ (which is the local minimum), and then rises to cross the x-axis a third time at $3$ before continuing to the top-right.

Key Takeaways

For polynomial graphs given in factored form, the x-intercepts are directly readable from the factors. The y-intercept is found by evaluating the function at $x = 0$ . The leading term determines the end behaviour (top-right to bottom-left for negative odd powers, bottom-left to top-right for positive odd powers).

Common Mistakes

Forgetting to solve $2x + 5 = 0$ correctly and writing $x = -5$ instead of $x = -2.5$ .
Drawing a negative cubic curve (top-left to bottom-right) by ignoring the positive leading coefficient $2$ .
Drawing a straight line or a quadratic shape instead of a cubic curve with the appropriate turning points.
Labeling the y-intercept as $30$ instead of $-30$ , or drawing a horizontal line at $y = -30$ instead of a single intercept point.

Things to Be Careful About

The sketch does not require a scale, but the relative positions of the intercepts and the overall cubic shape must be correct.
All three x-intercepts and the y-intercept must be labelled with their exact values; partial marks are awarded for fewer correctly labelled intercepts.
Ensure the curve is smooth and passes through all four labelled points without sharp corners.

Techniques used

find x-intercepts by setting each linear factor to zerofind y-intercept by substituting x = 0determine end behaviour from the leading termdraw a positive cubic curve through the intercepts

Q24MediumTransformations and VectorsNumber

The diagram shows the quadrilateral $OACB$ .

Fig. 10

$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .
$OB$ is parallel to $AC$ and $AC = 2OB$ .

(a)

Find, in its simplest form, in terms of $\mathbf{a}$ and $\mathbf{b}$

2M

(i)

$\overrightarrow{AB}$

$\overrightarrow{AB}$ = ______

(ii)

$\overrightarrow{OC}$ .

$\overrightarrow{OC}$ = ______

(b)

$X$ is the point on $OC$ such that $OX : XC = 4 : 1$ .

Find $\overrightarrow{BX}$ .
Give your answer in its simplest form in terms of $\mathbf{a}$ and $\mathbf{b}$ .

$\overrightarrow{BX}$ = ______

DifficultyMedium

Worked solution

Approach

Find the position of point $X$ on $OC$ using the ratio $OX : XC = 4 : 1$ , then write a vector path for $\overrightarrow{BX}$ and simplify in terms of $\mathbf{a}$ and $\mathbf{b}$ .

Working

Since $OX : XC = 4 : 1$ , the total number of parts along $OC$ is $4 + 1 = 5$ .
Therefore:

\overrightarrow{OX} = \frac{4}{5}\overrightarrow{OC} = \frac{4}{5}(\mathbf{a} + 2\mathbf{b}) = \frac{4}{5}\mathbf{a} + \frac{8}{5}\mathbf{b}

Now find $\overrightarrow{BX}$ by choosing the path via $O$ :

\begin{aligned} \overrightarrow{BX} &= \overrightarrow{BO} + \overrightarrow{OX} \\ &= -\mathbf{b} + \left(\frac{4}{5}\mathbf{a} + \frac{8}{5}\mathbf{b}\right) \\ &= \frac{4}{5}\mathbf{a} + \left(\frac{8}{5} - 1\right)\mathbf{b} \\ &= \frac{4}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \end{aligned}

Answer

\frac{4}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}

Final answer

4/5a + 3/5b

Detailed explanation

Walkthrough

Point $X$ lies on the line segment $OC$ dividing it in the ratio $OX : XC = 4 : 1$ . The total parts are $4 + 1 = 5$ , which means $OX$ is $\frac{4}{5}$ of the entire length $OC$ .
Express $\overrightarrow{OX}$ in vector form: $\overrightarrow{OX} = \frac{4}{5}\overrightarrow{OC} = \frac{4}{5}(\mathbf{a} + 2\mathbf{b}) = \frac{4}{5}\mathbf{a} + \frac{8}{5}\mathbf{b}$ .
To find $\overrightarrow{BX}$ , choose a convenient path from $B$ to $X$ . A direct path is from $B$ to $O$ , and then from $O$ to $X$ :

\overrightarrow{BX} = \overrightarrow{BO} + \overrightarrow{OX} = -\mathbf{b} + \overrightarrow{OX}

Substitute the expression for $\overrightarrow{OX}$ into the path equation and collect like terms:

-\mathbf{b} + \frac{4}{5}\mathbf{a} + \frac{8}{5}\mathbf{b} = \frac{4}{5}\mathbf{a} + \left(\frac{8}{5} - 1\right)\mathbf{b} = \frac{4}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}

Alternatively, this can be written with a single denominator as $\frac{4\mathbf{a} + 3\mathbf{b}}{5}$ .

Key Takeaways

When a line is divided in ratio $m : n$ , the fraction from the start to the division point is $\frac{m}{m+n}$ .
Follow a clear path using known vectors: $\overrightarrow{BX} = \overrightarrow{BO} + \overrightarrow{OX}$ .
Simplify vector components by grouping like vectors $\mathbf{a}$ and $\mathbf{b}$ .

Common Mistakes

Using $\frac{4}{1}$ or $\frac{1}{4}$ instead of $\frac{4}{5}$ for the fraction representing $OX$ .
Using $+\mathbf{b}$ instead of $-\mathbf{b}$ for $\overrightarrow{BO}$ .
Miscalculating the fraction arithmetic: $\frac{8}{5} - 1 = \frac{3}{5}$ .

Things to Be Careful About

Ensure the final answer is simplified and clearly expressed in terms of $\mathbf{a}$ and $\mathbf{b}$ . Both $\frac{4}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}$ and $\frac{4\mathbf{a} + 3\mathbf{b}}{5}$ are acceptable forms.

Techniques used

divide a vector by a given ratioform a vector path between two pointssimplify algebraic vector expressions

Q25Medium-EasyNumber

(a)

Simplify.

\sqrt{300} - \sqrt{48}

______

(b)

Rationalise the denominator.
Give your answer in its simplest form.

\frac{6}{\sqrt{7} + 2}

______