Mathematics (Syllabus D) 4024 / 11 — October/November 2025

Q2Medium-EasyStatisticsProbability

Some children record the number of hours they each spend watching television one day.
These are the results.

1 \quad 2 \quad 2 \quad 2 \quad 6 \quad 7 \quad 7 \quad 8 \quad 10

(a)

Write down the mode.

______

(b)

One of the 9 children is chosen at random.

Find the probability that this child spends more than 6 hours watching television on this day.

______

DifficultyMedium-Easy

Worked solution

Approach

Calculate the probability by dividing the number of children who satisfy the condition by the total number of children. The condition is spending more than 6 hours watching television.

Working

First, determine the total number of children recorded in the data set. Counting the entries:

1, 2, 2, 2, 6, 7, 7, 8, 10

There are $9$ values in total, so the denominator is $9$ .

Next, identify the values that are strictly greater than $6$ . Looking at the list:

$1$ is not $> 6$
$2$ is not $> 6$
$6$ is not $> 6$ (it is equal to 6)
$7$ is $> 6$
$7$ is $> 6$
$8$ is $> 6$
$10$ is $> 6$

The values satisfying the condition are $7, 7, 8,$ and $10$ . There are $4$ such values.

Therefore, the probability $P$ is:

P(\text{hours} > 6) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{9}

Answer

\frac{4}{9}

Final answer

4/9

Detailed explanation

Walkthrough

The probability of an event is calculated as the ratio of favorable outcomes to total possible outcomes.

Total outcomes: The problem states "One of the 9 children is chosen", confirming there are 9 items in our sample space. We can also verify this by counting the numbers in the list provided.
Favorable outcomes: The condition is "more than 6 hours". This means we must exclude any child who watched exactly 6 hours. We scan the list for numbers strictly larger than 6. The numbers are 7, 7, 8, and 10. Counting these gives us 4 children.
Form the fraction: Place the count of favorable outcomes (4) over the total count (9).

Key Takeaways

Probability is always $\frac{\text{favorable}}{\text{total}}$ .
Read inequality conditions carefully: "more than" means $>$ , while "at least" means $\geq$ .

Common Mistakes

Including the value '6' in the count. The question asks for more than 6, so 6 is excluded. If included, the numerator would incorrectly become 5.
Miscounting the total number of children.
Writing the decimal approximation instead of the exact fraction (though usually acceptable unless specified otherwise, fractions are preferred in 4024 non-calculator papers).

Things to Be Careful About

The phrase "more than" excludes the boundary value. If the question said "6 hours or more", the answer would be $\frac{5}{9}$ .

Techniques used

count total number of outcomescount favorable outcomesform a probability fraction

Q32MMedium-EasyNumber

A shirt costs $24.
In a sale the cost is reduced by 15%.

Work out the cost of the shirt in the sale.

$ ______

DifficultyMedium-Easy

Worked solution

Approach

The shirt originally costs $24. It is reduced by 15%. We need to calculate the sale price.

There are two common methods:

Calculate the discount amount (15% of $24) and subtract it from the original price.
Calculate the remaining percentage (100% - 15% = 85%) and multiply the original price by this decimal/fraction.

We will use Method 1 as it aligns directly with the mark scheme's primary method route.

Working

Method 1: Subtract the discount

First, find 15% of $24.

15\% \text{ of } 24 = \frac{15}{100} \times 24

Calculate the value:

\frac{15}{100} \times 24 = 0.15 \times 24

Break down the multiplication for clarity:

0.10 \times 24 = 2.40

0.05 \times 24 = 1.20

2.40 + 1.20 = 3.60

So the discount is $3.60.

Now, subtract the discount from the original cost:

24 - 3.60 = 20.40

Method 2: Multiply by the remaining percentage

The sale price is $100\% - 15\% = 85\%$ of the original price.

0.85 \times 24

Calculation:

0.85 \times 24 = 20.40

Answer

20.40

Final answer

20.40

Detailed explanation

Walkthrough

The problem asks for the new price of an item after a percentage reduction. The original price is $24, and the reduction is 15%.

Step 1: Understand what "reduced by 15%" means.
It means we take away 15% of the original $24 from the $24. Alternatively, it means we keep 85% of the original price.

Step 2: Calculate the discount amount.
Using the fraction method: $15\% = \frac{15}{100}$ .
Discount = $\frac{15}{100} \times 24$ .
$15 \times 24 = 360$ , so $360 / 100 = 3.60$ . The discount is $3.60.

Step 3: Calculate the final sale price.
Original Price - Discount = Sale Price.
$24 - 3.60 = 20.40$ .

Alternatively, using the multiplier method:
Remaining percentage = $100\% - 15\% = 85\% = 0.85$ .
Sale Price = $0.85 \times 24$ .
$0.85 \times 24 = 20.40$ .

Both methods yield $20.40.

Key Takeaways

To reduce a quantity by a percentage, you can either subtract the calculated percentage amount from the original, or multiply the original by $(1 - \text{decimal percentage})$ .
$15\%$ can be easily calculated as $10\% + 5\%$ . $10\%$ of 24 is 2.4, and $5\%$ is half of that, which is 1.2. Summing them gives 3.6.
Always include the currency symbol and correct decimal places in the final answer when dealing with money.

Common Mistakes

Adding 15% instead of subtracting it (calculating an increase).
Calculating 15% correctly but forgetting to subtract it from the original price (stopping at 3.60).
Arithmetic errors in multiplying decimals (e.g., $0.15 \times 24$ ).
Rounding errors (though here the answer is exact).

Things to Be Careful About

The question involves money, so the answer should ideally be given to two decimal places ($20.40), although $20.4 is mathematically equivalent. The mark scheme accepts $20.4[0]$ .
Ensure you are calculating the sale price, not just the discount amount.
In non-calculator exams, show your working clearly for the percentage calculation (e.g., $24 \times \frac{15}{100}$ ) to secure method marks if the final answer is wrong.

Techniques used

calculate a percentage of a quantitysubtract the reduction from the original amount

Q4Medium-EasyGeometry

The diagram shows five straight lines.

Fig. 1

The diagram shows three horizontal parallel lines and two sloping parallel lines.

(a)

Find the value of $p$ .
Give a geometrical reason to explain your answer.

$p$ = ______ because ______

DifficultyMedium-Easy

Worked solution

Approach

The top and bottom horizontal lines are parallel, and the right-hand sloping line acts as a transversal cutting through them. The angle marked $112^\circ$ and the angle marked $p^\circ$ lie between these two parallel lines and on the same side of the transversal, making them co-interior (or allied) angles. Co-interior angles between parallel lines sum to $180^\circ$ .

Working

p + 112 = 180

Subtract $112$ from both sides:

p = 180 - 112

p = 68

Answer

$p = 68$ because co-interior angles sum to $180^\circ$ .

Final answer

p = 68, co-interior angles sum to 180

Detailed explanation

Walkthrough

The question provides a diagram with three horizontal parallel lines and two sloping parallel lines. For part (a), we focus on the top horizontal line, the bottom horizontal line, and the right-hand sloping line that intersects both. These two horizontal lines are parallel, and the sloping line is a transversal. The angle $112^\circ$ is below the top line and to the left of the transversal. The angle $p^\circ$ is above the bottom line and also to the left of the transversal. Because they are on the same side of the transversal and between the parallel lines, they are co-interior angles (also called allied angles or consecutive interior angles). A key property of parallel lines cut by a transversal is that co-interior angles add up to $180^\circ$ . Therefore, we set up the equation $p + 112 = 180$ and solve for $p$ to get $68$ .

Key Takeaways

Co-interior angles (angles on the same side of a transversal and between parallel lines) always sum to $180^\circ$ .
When applying parallel line rules, clearly identify which two lines are parallel and which line is the transversal cutting through them.

Common Mistakes

Confusing co-interior angles with alternate interior angles (which are equal) or corresponding angles (which are also equal). Always check the position: if they are on the same side and between the parallels, they sum to $180$ .
Forgetting to state the geometrical reason clearly; the mark scheme awards one mark for the value and one mark for the correct reason ("co-interior angles sum to 180").

Things to Be Careful About

Ensure the reason is stated precisely. Saying "angles on a straight line" or "angles in a triangle" will not earn the method mark. The exact reason required is that co-interior angles sum to $180^\circ$ .
The diagram is marked "NOT TO SCALE", so do not estimate angles by eye; rely only on the given values and parallel line properties.

Techniques used

identify co-interior angles between parallel linesapply the co-interior angle sum property

(b)

Find the value of $q$ .
Give a geometrical reason to explain your answer.

$q$ = ______ because ______

DifficultyMedium-Easy

Worked solution

Approach

We now use the value $p = 68$ found in part (a). Consider the two sloping lines, which are parallel to each other, and the bottom horizontal line as a transversal. The angle $p^\circ$ is above the bottom horizontal line and to the left of the right-hand sloping line. By the property of corresponding angles between parallel lines, the angle in the same position at the intersection of the left-hand sloping line and the bottom horizontal line is also $68^\circ$ . Alternatively, and more directly, consider the two parallel sloping lines cut by the middle horizontal line. The angle at the right-hand sloping line and middle horizontal line (above the middle line, left of the transversal) is equal to $p^\circ$ because they are corresponding angles between the top and bottom parallel lines cut by the right-hand sloping transversal. Then, $q^\circ$ and this angle are corresponding angles between the parallel sloping lines cut by the middle horizontal transversal. Thus, $q = p = 68$ .

A simpler route accepted by the mark scheme: The angle $p^\circ$ and the angle $q^\circ$ are corresponding angles. To see this, note that the angle corresponding to $p^\circ$ on the left-hand sloping line (above the bottom horizontal line, left of the left-hand sloping line) is equal to $p^\circ$ . Then, using the middle horizontal line as a transversal cutting the parallel sloping lines, $q^\circ$ is in the corresponding position to that angle. Ultimately, $q = p$ .

Working

q = p

Substitute $p = 68$ :

q = 68

Answer

$q = 68$ because corresponding angles are equal.

Final answer

q = 68, corresponding angles are equal

Detailed explanation

Walkthrough

For part (b), we need to find $q$ . The mark scheme indicates that $q$ is equal to $p$ by corresponding angles, with follow-through from the value $68$ found in part (a). To justify this geometrically: the two sloping lines are parallel. The horizontal lines act as transversals. The angle $p^\circ$ is at the intersection of the right-hand sloping line and the bottom horizontal line (above the horizontal, left of the sloping line). The angle $q^\circ$ is at the intersection of the left-hand sloping line and the middle horizontal line (above the horizontal, left of the sloping line). Because the sloping lines are parallel and the horizontal lines are parallel, the angle $q^\circ$ is in the exact same relative position to its intersection as $p^\circ$ is to its intersection. Therefore, they are corresponding angles and are equal. Substituting the value from part (a), $q = 68$ .

Key Takeaways

Corresponding angles are equal when two parallel lines are cut by a transversal. They are in the same relative position at each intersection.
When a problem has multiple parts, always check if an earlier result can be used directly in a later part (follow-through).

Common Mistakes

Not using the value from part (a); the mark scheme explicitly allows follow-through (FT) on their $68$ , so if a student got $p$ wrong in part (a), they can still get the mark for $q$ if they correctly state $q = p$ and use their own $p$ value. However, since $p$ is correctly $68$ , $q$ must be $68$ .
Stating the wrong reason, such as "alternate angles are equal" or "co-interior angles sum to 180". The angles $p$ and $q$ are on the same side of the transversal and in the same position relative to the parallel lines, making them corresponding.

Things to Be Careful About

The geometrical reason must be specific: "corresponding angles are equal". Vague reasons like "angles are equal" will not earn the mark.
Ensure the follow-through from part (a) is clearly stated if using it, though here the value is simply $68$ .

Techniques used

use the result from part (a)identify corresponding angles between parallel sloping lines

Q5Medium-EasyNumberStatistics

The table shows how some people travel to work.

Type of travel	Number of people	Pie chart angle
Cycle	20	$120^\circ$
Walk	12	$72^\circ$
Car
Bus	5

(a)

Complete the table.

DifficultyMedium-Easy

Worked solution

Approach

Find the angle per person using the Cycle row, then find the total number of people, and use these to calculate the missing values for Car and Bus.

Working

From the Cycle row, 20 people correspond to $120^\circ$ .

\text{Angle per person} = \frac{120^\circ}{20} = 6^\circ

The total angle in a pie chart is $360^\circ$ , so the total number of people is:

\text{Total people} = \frac{360^\circ}{6^\circ} = 60

For the Bus row, there are 5 people:

\text{Bus angle} = 5 \times 6^\circ = 30^\circ

For the Car row, the number of people is the total minus the others:

\text{Car people} = 60 - 20 - 12 - 5 = 23

The angle for Car is:

\text{Car angle} = 23 \times 6^\circ = 138^\circ

Answer

Type of travel	Number of people	Pie chart angle
Cycle	20	$120^\circ$
Walk	12	$72^\circ$
Car	23	$138^\circ$
Bus	5	$30^\circ$

Final answer

Car: 23 people, 138°; Bus: 30°

Detailed explanation

Walkthrough

First, determine the scale of the pie chart by dividing the given angle by the given frequency. For Cycle, $120^\circ$ represents 20 people, so $1$ person represents $6^\circ$ . Next, find the total number of people by dividing the full circle angle ( $360^\circ$ ) by the angle per person ( $6^\circ$ ), giving $60$ people. Then, calculate the missing angle for Bus by multiplying its frequency ( $5$ ) by $6^\circ$ , which gives $30^\circ$ . Finally, find the missing frequency for Car by subtracting the known frequencies from the total ( $60 - 20 - 12 - 5 = 23$ ), and calculate its angle by multiplying $23$ by $6^\circ$ , giving $138^\circ$ .

Key Takeaways

In a pie chart, the total angle is always $360^\circ$ and represents the total frequency.
You can find the scale (angle per unit) by dividing a known angle by its corresponding frequency.
Missing frequencies and angles can be found by using the scale and the total frequency.

Common Mistakes

Forgetting that the total angle in a pie chart is $360^\circ$ .
Calculating the total frequency incorrectly by not using the scale.
Forgetting to subtract all known frequencies from the total to find the missing frequency.
Not showing working for the calculated values (mark scheme requires "www" or at least the total/ scale to be visible for method marks).

Things to Be Careful About

Ensure all angles sum to exactly $360^\circ$ ( $120 + 72 + 138 + 30 = 360$ ).
Ensure all frequencies sum to the total ( $20 + 12 + 23 + 5 = 60$ ).
The mark scheme awards method marks for finding the total number of people ( $60$ ) or the scale ( $6^\circ$ represents $1$ person), so show these intermediate steps clearly.

Techniques used

find the value of one unit in degreescalculate the total number of peoplefind the missing frequency and angle

(b)

Complete the pie chart to show the information in the table.

Fig. 2

DifficultyMedium-Easy

Worked solution

Approach

Use the angles calculated in part (a) to draw the remaining sectors on the pie chart, starting from the existing Cycle sector.

Working

The angles for each sector are:

Cycle: $120^\circ$ (already drawn)
Walk: $72^\circ$
Car: $138^\circ$
Bus: $30^\circ$

Total angle: $120^\circ + 72^\circ + 138^\circ + 30^\circ = 360^\circ$ .

Draw the sectors in order, labeling each with the type of travel.

Answer

Pie chart with sectors: Cycle ( $120^\circ$ ), Walk ( $72^\circ$ ), Car ( $138^\circ$ ), Bus ( $30^\circ$ ).

Final answer

Pie chart with sectors Cycle (120°), Walk (72°), Car (138°), Bus (30°)

Detailed explanation

Walkthrough

The pie chart already has the Cycle sector drawn ( $120^\circ$ ). Starting from the end of the Cycle sector, draw the Walk sector with an angle of $72^\circ$ . Next, draw the Car sector with an angle of $138^\circ$ . Finally, draw the Bus sector with an angle of $30^\circ$ to complete the circle. Label each sector with its corresponding type of travel. Ensure the angles sum to $360^\circ$ .

Key Takeaways

A pie chart must have sectors that sum to $360^\circ$ .
Sectors should be drawn in order and labeled clearly.
The size of each sector is proportional to the frequency it represents.

Common Mistakes

Drawing sectors with incorrect angles.
Forgetting to label the sectors.
Not starting from the correct position or drawing sectors in the wrong order, leading to a messy diagram.
Not completing the circle (remaining angle should be exactly $0^\circ$ ).

Things to Be Careful About

Use a protractor to measure the angles accurately.
The mark scheme awards follow-through marks (B1 FT) if one sector is drawn correctly based on part (a), so ensure the angles from part (a) are correct.
Labels should be clear and not overlap with the sector lines.

Techniques used

calculate sector anglesdraw sectors on a pie chartlabel sectors

Q62MMedium-EasyNumber

Sam buys some chocolate.
He eats $\frac{2}{7}$ of the chocolate.
The mass of the remaining chocolate is $350\,\text{g}$ .

Work out the mass of the chocolate Sam buys.

______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Approach

Sam eats $\frac{2}{7}$ of the chocolate. We first determine what fraction of the chocolate remains. Then, using the fact that this remaining fraction equals $350\,\text{g}$ , we calculate the total mass by dividing by the numerator and multiplying by the denominator (the unitary method).

Working

The fraction of chocolate eaten is:

\frac{2}{7}

The fraction of chocolate remaining is:

1 - \frac{2}{7} = \frac{7}{7} - \frac{2}{7} = \frac{5}{7}

We are given that the mass of the remaining chocolate is $350\,\text{g}$ . Therefore:

\frac{5}{7} \times \text{Total Mass} = 350\,\text{g}

To find the total mass, we divide $350$ by $5$ to find the value of $\frac{1}{7}$ :

\frac{1}{7} \times \text{Total Mass} = \frac{350}{5} = 70\,\text{g}

Now, multiply by $7$ to find the full mass ( $\frac{7}{7}$ ):

\text{Total Mass} = 70 \times 7 = 490\,\text{g}

Answer

490

Final answer

490

Detailed explanation

Walkthrough

First, identify the portion of the chocolate that was not eaten. Since Sam ate $\frac{2}{7}$ of the total, the remaining part is the complement of this fraction with respect to the whole ( $1$ or $\frac{7}{7}$ ). Subtracting $\frac{2}{7}$ from $1$ gives $\frac{5}{7}$ .

Next, link this fraction to the physical quantity provided. The problem states that the remaining mass is $350\,\text{g}$ . This means that $5$ parts out of every $7$ parts of the total mass equal $350\,\text{g}$ .

To find the total mass, use the unitary method: divide the known mass by the number of parts it represents ( $5$ ) to find the mass of a single part ( $1/7$ ), then multiply by the total number of parts ( $7$ ).

Calculation steps:

Remaining fraction: $\frac{5}{7}$
Value of one part ( $\frac{1}{7}$ ): $350 \div 5 = 70$
Total value ( $\frac{7}{7}$ ): $70 \times 7 = 490$

Key Takeaways

When a fraction of a quantity is used or removed, subtract that fraction from $1$ to find the remainder.
To find a total from a fractional part, divide the part's value by the fraction's numerator, then multiply by the denominator.

Common Mistakes

Incorrect subtraction: Calculating $1 - \frac{2}{7}$ as $\frac{5}{7}$ incorrectly (e.g., $7-2=5$ , but forgetting the denominator stays the same, or getting the wrong sum).
Multiplying instead of dividing: Multiplying $350$ by $\frac{5}{7}$ or $\frac{2}{7}$ instead of finding the total. This leads to a smaller number than the part itself.
Scaling error: Dividing $350$ by $5$ correctly but then multiplying by the wrong number (e.g., multiplying by $5$ again or by $2$ ).
Confusing 'eaten' with 'remaining': Using $\frac{2}{7}$ directly with the $350\,\text{g}$ figure, assuming $350\,\text{g}$ is the amount eaten.

Things to Be Careful About

Ensure you answer for the total mass bought, not just the remaining mass or the eaten mass.
Check your arithmetic: $350 / 5 = 70$ and $70 \times 7 = 490$ .
The mark scheme awards a specific shortcut mark (B1) for writing $\frac{350}{5}$ , so showing this division explicitly helps secure partial credit if the final multiplication is incorrect.

Techniques used

calculate the remaining fraction of a wholeuse unitary method to find the total from a known part

Q73MMedium-EasyNumber

Car hire
$40 per day plus $0.30 per kilometre

Amy hires a car for 10 days.
She pays a total of $670.

Work out the number of kilometres Amy travels in the car.

______ $\text{km}$

DifficultyMedium-Easy

Worked solution

Approach

The total cost consists of two parts: a fixed daily hire fee and a variable mileage fee. We need to isolate the mileage fee by removing the fixed cost from the total amount paid, then divide by the cost per kilometre to find the distance.

Working

First, calculate the total fixed cost for hiring the car for 10 days at $40 per day:

\text{Fixed Cost} = 40 \times 10 = 400

Next, determine how much of the total $670 was spent on the mileage fee by subtracting the fixed cost:

\text{Mileage Fee} = 670 - 400 = 270

Finally, since the cost is $0.30 per kilometre, divide the total mileage fee by the rate per kilometre to find the number of kilometres travelled:

\text{Kilometres} = \frac{270}{0.3}

To perform this division easily without a calculator, multiply both numerator and denominator by 10:

\frac{270 \times 10}{0.3 \times 10} = \frac{2700}{3} = 900

Answer

900 \text{ km}

Final answer

900 km

Detailed explanation

Walkthrough

This question describes a linear relationship between cost and distance: Total Cost = (Daily Rate × Days) + (Rate per km × Kilometres).

Calculate the fixed portion: The car hire charges $40 for every day. Amy hires the car for 10 days. So, the base cost is $40 \times 10 = 400$ . This amount is paid regardless of how far she drives.
Isolate the variable portion: Amy pays a total of $670. Since $400 of that covers the daily hire, the rest must cover the distance driven. We subtract the fixed cost from the total: $670 - 400 = 270$ . This means $270 was spent solely on mileage.
Calculate the distance: The mileage charge is $0.30 per kilometre. To find out how many kilometres correspond to $270, we divide the mileage money by the cost per kilometre: $270 \div 0.30$ .
On a non-calculator paper, dividing by a decimal like 0.3 can be tricky. A good method is to convert the divisor into a whole number. Multiply 0.3 by 10 to get 3. Do the same to the dividend (270 becomes 2700). Now the calculation is simply $2700 \div 3$ , which is 900.

Key Takeaways

Structure of Costs: Many real-world problems involve a 'fixed cost' plus a 'variable cost'. Understanding this structure allows you to separate the components using subtraction.
Division by Decimals: When dividing by a decimal (like 0.3), converting it to an integer (multiplying top and bottom by 10, 100, etc.) makes the arithmetic much simpler and less prone to error.

Common Mistakes

Incorrect Order of Operations: Subtracting 40 from 670 first ( $670 - 40 = 630$ ) ignores that the $40 is a daily rate for 10 days. You must multiply the daily rate by the number of days before subtracting.
Multiplication Error: Calculating $40 \times 10$ incorrectly.
Decimal Division: Struggling with $270 \div 0.3$ or misplacing the decimal point in the final answer.

Things to Be Careful About

Units: Ensure you are calculating kilometres, not dollars. The question asks for "number of kilometres".
Non-Calculator Component: Remember to show your working for the decimal division clearly (e.g., converting $270 \div 0.3$ to $2700 \div 3$ ) rather than just stating the result.
Total vs Daily: Always check if a rate given in the question is per item/day/unit or a lump sum. Here, $40 is 'per day', so it must be scaled by the number of days.

Techniques used

calculate the fixed daily cost componentsubtract fixed costs from total paymentdivide remaining amount by variable rate per unit

Q8Medium-EasyTransformations and Vectors

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

Fig. 3

(a)

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

______

DifficultyMedium-Easy

Worked solution

Approach

Compare corresponding vertices of triangle $A$ and triangle $B$ to determine the type of transformation and its parameters.

Working

Triangle $A$ has vertices at $(-4, 2)$ , $(-3, 2)$ and $(-4, 5)$ . Triangle $B$ has vertices at $(5, 2)$ , $(6, 2)$ and $(6, 5)$ .

The $y$ -coordinates are unchanged, and the $x$ -coordinates change from negative to positive, indicating a reflection in a vertical line.

The midpoint of the $x$ -coordinates of corresponding points gives the line of reflection:

\frac{-4 + 6}{2} = 1

Checking with another pair:

\frac{-3 + 5}{2} = 1

The transformation is a reflection in the line $x = 1$ .

Answer

Reflection in the line $x = 1$ .

Final answer

Reflection in the line x = 1

Detailed explanation

Walkthrough

To describe a transformation fully, we must identify the type of transformation and provide the specific parameter (the line of reflection, centre of rotation, vector of translation, or scale factor and centre of enlargement). Comparing the coordinates of the vertices of triangle $A$ and triangle $B$ , we see that the $y$ -coordinates remain the same while the $x$ -coordinates change sign and shift. This indicates a reflection in a vertical line. To find the line of reflection, we calculate the midpoint between the $x$ -coordinates of any pair of corresponding vertices. The midpoint of $-4$ and $6$ is $1$ , and the midpoint of $-3$ and $5$ is also $1$ . Therefore, the line of reflection is $x = 1$ .

Key Takeaways

A reflection in a vertical line $x = k$ maps $(x, y)$ to $(2k - x, y)$ . The $y$ -coordinates are unchanged, and the $x$ -coordinates are equidistant from the line of reflection.

Common Mistakes

Describing the transformation as a translation or rotation instead of a reflection.
Giving only the type of transformation without specifying the line, or vice versa. Both are required for full marks.
Calculating the midpoint incorrectly or using the wrong coordinates.

Things to Be Careful About

A full description of a reflection requires both the type of transformation and the equation of the line of reflection.
Ensure the coordinates read from the grid are correct before performing any calculations.

Techniques used

identify the type of transformationfind the line of reflection

(b)

Draw the image of triangle $A$ after a rotation of $180^\circ$ about centre $(0, 0)$ .

DifficultyEasy

Worked solution

Approach

Apply the rule for a $180^\circ$ rotation about the origin $(0, 0)$ , which maps $(x, y)$ to $(-x, -y)$ , to each vertex of triangle $A$ .

Working

The vertices of triangle $A$ are $(-4, 2)$ , $(-3, 2)$ and $(-4, 5)$ .

Applying the rotation rule $(x, y) \to (-x, -y)$ :

(-4, 2) \to (4, -2)

(-3, 2) \to (3, -2)

(-4, 5) \to (4, -5)

Draw the triangle with vertices at $(3, -2)$ , $(4, -2)$ and $(4, -5)$ .

Answer

Triangle with vertices at $(3, -2)$ , $(4, -2)$ and $(4, -5)$ .

Final answer

Triangle with vertices at (3, -2), (4, -2) and (4, -5)

Detailed explanation

Walkthrough

A rotation of $180^\circ$ about the origin $(0, 0)$ maps any point $(x, y)$ to $(-x, -y)$ . This means both the $x$ -coordinate and the $y$ -coordinate change sign. Applying this to the vertices of triangle $A$ : $(-4, 2)$ becomes $(4, -2)$ , $(-3, 2)$ becomes $(3, -2)$ , and $(-4, 5)$ becomes $(4, -5)$ . These new coordinates are plotted on the grid to draw the image of triangle $A$ .

Key Takeaways

A $180^\circ$ rotation about the origin is equivalent to a point reflection through the origin, where $(x, y) \to (-x, -y)$ .

Common Mistakes

Forgetting to change the sign of both coordinates.
Plotting the points incorrectly on the grid.
Drawing a triangle that is not congruent to the original.

Things to Be Careful About

Ensure the drawn triangle is the correct size and orientation. The mark scheme awards partial credit for correct size and orientation even if the position is wrong, but full marks require the correct position.
When drawing on a grid, verify that each vertex is placed exactly at the calculated coordinates.

Techniques used

apply rotation formulaplot the image vertices

Q92MMedium-EasyNumber

By writing each number correct to 1 significant figure, work out an estimate for the value of

\frac{\sqrt{102.5} \times 8.7}{27}

______

DifficultyMedium-Easy

Worked solution

Approach

To estimate the value of the expression, we first round each number in the calculation to 1 significant figure. Then we substitute these rounded values back into the expression and evaluate.

Working

The expression is:

\frac{\sqrt{102.5} \times 8.7}{27}

Step 1: Round $102.5$ to 1 significant figure.
The first digit is $1$ . The next digit is $0$ , so we round down to:

102.5 \approx 100

Step 2: Estimate $\sqrt{102.5}$ .
Since $102.5$ has been rounded to $100$ , we use:

\sqrt{102.5} \approx \sqrt{100} = 10

Note: While $\sqrt{102.5}$ is slightly larger than $10$ , for an estimate to 1 significant figure based on the rounded input, $10$ is the standard approach. Alternatively, one might just round the final result of the surd if calculated directly, but using the rounded integer is clearer for estimation steps. Let's look at the marking scheme. It expects $100, 9, 30$ . So $\sqrt{100}=10$ is implied as part of the process or the numerator becomes $10 \times 9$ .

Step 3: Round $8.7$ to 1 significant figure.
The first digit is $8$ . The next digit is $7$ , so we round up to:

8.7 \approx 9

Step 4: Round $27$ to 1 significant figure.
The first digit is $2$ . The next digit is $7$ , so we round up to:

27 \approx 30

Step 5: Substitute the estimated values into the expression.

\frac{10 \times 9}{30}

Step 6: Calculate the final value.

\frac{90}{30} = 3

Answer

3

Final answer

3

Detailed explanation

Walkthrough

The question asks for an estimate using 1 significant figure (s.f.) for each number involved in the calculation. This simplifies complex numbers into easy-to-calculate integers.

Round $102.5$ : To 1 s.f., we look at the first non-zero digit ( $1$ ). The digit after it is $0$ , which is less than $5$ , so we keep the $1$ and replace subsequent digits with zeros. Result: $100$ . Consequently, $\sqrt{102.5}$ is estimated as $\sqrt{100} = 10$ .
Round $8.7$ : The first digit is $8$ . The next digit is $7$ , which is $5$ or greater, so we round up. Result: $9$ .
Round $27$ : The first digit is $2$ . The next digit is $7$ , so we round up. Result: $30$ .
Calculate: Substitute these values into the fraction: $\frac{10 \times 9}{30}$ Multiply the numerator: $10 \times 9 = 90$ .
Divide by the denominator: $90 \div 30 = 3$ .

Key Takeaways

Estimation Strategy: When asked to estimate, always round intermediate values to 1 significant figure to make mental math easier.
Square Roots of Rounded Numbers: If you round a number under a square root to a perfect square (like $102.5 \to 100$ ), take the square root of that perfect square ( $10$ ).
Rounding Rules: Identify the first significant digit. Look at the next digit to decide whether to round up or down.

Common Mistakes

Rounding to 1 decimal place instead of 1 significant figure (e.g., $102.5 \to 102.5$ , $8.7 \to 8.7$ , $27 \to 27.0$ ).
Incorrectly rounding $27$ to $20$ (forgetting that $7 > 5$ , so it rounds up to $30$ ).
Incorrectly rounding $8.7$ to $8$ .
Failing to round $102.5$ before taking the square root, leading to a messy calculation rather than an estimate.
Calculating the exact value instead of the estimate.

Things to Be Careful About

Significant Figures vs Decimal Places: Ensure you are counting from the first non-zero digit, not just the first digit overall or decimal place.
Intermediate Rounding: The mark scheme awards marks for seeing the specific rounded values ( $100, 9, 30$ ). Even if your final arithmetic is slightly different due to choosing different approximations (e.g. keeping $\sqrt{102.5} \approx 10.1$ ), you must show the 1 s.f. rounded values to get full credit.
Accuracy: The final answer should be consistent with the estimates used. Here, the estimate is exactly $3$ .

Techniques used

round numbers to one significant figureestimate the value of a surdcalculate the estimated result

Q102MEasyMensuration

A circle has diameter $12\,\text{cm}$ .

Find the area of the circle.
Give your answer in terms of $\pi$ .

______ $\text{cm}^2$

DifficultyEasy

Worked solution

Approach

The area of a circle is $A = \pi r^2$ . The question gives the diameter, so first find the radius by halving the diameter, then substitute into the area formula.

Working

The diameter is $12\,\text{cm}$ , so the radius is

r = \frac{12}{2} = 6\,\text{cm}

Substitute $r = 6$ into the area formula:

A = \pi r^2 = \pi \times 6^2 = 36\pi\,\text{cm}^2

Answer

36\pi\,\text{cm}^2

Final answer

36π cm^2

Detailed explanation

Walkthrough

The question gives the diameter of the circle, not the radius. The area formula uses the radius, so first halve the diameter to get $r = 6\,\text{cm}$ . Then substitute into $A = \pi r^2$ :

A = \pi \times 6^2 = 36\pi

Because the question asks for the answer in terms of $\pi$ , leave the $\pi$ symbol in the answer rather than multiplying by a decimal approximation. The units are $\text{cm}^2$ because area is measured in square units.

Key Takeaways

The area of a circle is $A = \pi r^2$ .
The radius is half the diameter.
"In terms of $\pi$ " means the final answer keeps the $\pi$ symbol and is not rounded to a decimal.

Common Mistakes

Using the diameter $12$ directly in the area formula: $\pi \times 12^2$ gives $144\pi$ , which is wrong.
Giving a decimal such as $113.1$ instead of $36\pi$ . The mark scheme requires $36\pi$ exactly.
Mixing up area and circumference: the circumference would be $2\pi r = 12\pi$ , but the question asks for area.

Things to Be Careful About

Always halve the diameter before substituting into the area formula.
Include the correct units: $\text{cm}^2$ .
This is a non-calculator paper, so the working should be shown by hand; the exact form $36\pi$ is the required answer.

Techniques used

halve the diameter to find the radiussubstitute the radius into the area formula for a circleleave the answer in terms of pi

Q113MMedium-EasyStatistics

A bag contains 10 plums.
The mean mass of 6 of the plums is $50\,\text{g}$ .
The mean mass of the remaining 4 plums is $40\,\text{g}$ .

Calculate the mean mass of all 10 plums in the bag.

______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Approach

The mean mass is defined as the total mass divided by the number of items. We can calculate the total mass for each group of plums separately using their respective means, add these together to get the grand total mass, and then divide by the total number of plums (10) to find the overall mean.

Working

First, calculate the total mass of the first group of 6 plums:

\text{Total mass}_1 = \text{mean} \times \text{count} = 50 \times 6 = 300\,\text{g}

Next, calculate the total mass of the remaining 4 plums:

\text{Total mass}_2 = \text{mean} \times \text{count} = 40 \times 4 = 160\,\text{g}

Now, find the total mass of all 10 plums by adding the two sub-totals:

\text{Grand Total Mass} = 300 + 160 = 460\,\text{g}

Finally, calculate the mean mass of all 10 plums:

\text{Overall Mean} = \frac{\text{Grand Total Mass}}{\text{Total Count}} = \frac{460}{10} = 46\,\text{g}

Alternatively, this can be set up in one expression:

\frac{(6 \times 50) + (4 \times 40)}{10} = \frac{300 + 160}{10} = \frac{460}{10} = 46

Answer

46

Final answer

46

Detailed explanation

Walkthrough

The problem asks for the mean of a combined set of data, given the means of its subsets. The fundamental formula for the mean is:

\text{Mean} = \frac{\text{Sum of values}}{\text{Number of values}}

Rearranging this gives us the tool we need: $\text{Sum of values} = \text{Mean} \times \text{Number of values}$ .

Step 1: Find the sum of the masses for the first group. We are told there are 6 plums with a mean mass of $50\,\text{g}$ . So, the total mass of these 6 plums is $6 \times 50 = 300\,\text{g}$ .

Step 2: Find the sum of the masses for the second group. There are 4 plums with a mean mass of $40\,\text{g}$ . The total mass here is $4 \times 40 = 160\,\text{g}$ .

Step 3: Combine the groups. The bag contains all these plums, so the total number of plums is $6 + 4 = 10$ , and the total mass is $300 + 160 = 460\,\text{g}$ .

Step 4: Calculate the new mean. Divide the combined total mass by the combined total number of plums: $\frac{460}{10} = 46\,\text{g}$ .

Key Takeaways

The mean of a combined group is NOT the average of the means of the subgroups (i.e., do not just calculate $(50+40)/2$ ). You must weight the means by the number of items in each group.
Always convert means back to totals before combining them if the group sizes differ.

Common Mistakes

Averaging the means: Calculating $\frac{50 + 40}{2} = 45$ . This is incorrect because the groups have different numbers of items ( $6$ vs $4$ ).
Adding the counts incorrectly: Using $8$ or another number instead of $10$ as the divisor.
Arithmetic errors in multiplication or addition.

Things to Be Careful About

Ensure you use the correct counts ( $6$ and $4$ ) for the respective means.
Check that the final answer includes the correct unit if required, though the question prompt already provides ' $\text{g}$ ' outside the blank line.

Techniques used

calculate total mass from mean and countsum the partial totalsdivide by the overall count to find the combined mean

Q123MMedium-EasyMensurationNumber

Fig. 4

The diagram shows an empty container.
The container is a cuboid of width $10\,\text{cm}$ , length $8\,\text{cm}$ and height $9\,\text{cm}$ .
Water drips into the container at a rate of $6$ millilitres per second.

Find the time, in seconds, it takes to fill the container.

______ $\text{seconds}$

DifficultyMedium-Easy

Worked solution

Approach

Find the volume of the cuboid, convert it to millilitres, and divide by the dripping rate to obtain the time in seconds.

Working

The volume of a cuboid is given by the product of its width, length, and height:

\text{Volume} = 10 \times 8 \times 9

\text{Volume} = 720 \, \text{cm}^3

Since $1 \, \text{cm}^3$ is equivalent to $1 \, \text{ml}$ , the container holds $720 \, \text{ml}$ of water.

The water drips in at a rate of $6 \, \text{ml}$ per second, so the time to fill the container is:

\text{Time} = \frac{720}{6}

\text{Time} = 120 \, \text{seconds}

Answer

120

Final answer

120

Detailed explanation

Walkthrough

The problem asks for the time required to fill a cuboid-shaped container with water. First, we calculate the capacity of the container by multiplying its three dimensions: width, length, and height. This gives the volume in cubic centimetres. Because $1 \text{ cm}^3$ is exactly equal to $1 \text{ ml}$ , the numerical value of the volume in $\text{cm}^3$ is also the volume in millilitres. Finally, we divide the total volume in millilitres by the rate at which water enters the container (in $\text{ml}$ per second) to find the total time in seconds.

Key Takeaways

The volume of a cuboid is calculated as $\text{length} \times \text{width} \times \text{height}$ .
$1 \text{ cm}^3$ is equivalent to $1 \text{ ml}$ , so volume in cubic centimetres can be directly used as volume in millilitres.
Time = $\frac{\text{Total Volume}}{\text{Rate}}$ .

Common Mistakes

Forgetting to convert cubic centimetres to millilitres (though numerically they are the same, the units must be consistent with the given rate).
Multiplying the rate by the volume instead of dividing.
Using the wrong dimensions or omitting one dimension when calculating the volume.

Things to Be Careful About

Ensure the units of volume match the units of the rate. Here, $\text{cm}^3$ and $\text{ml}$ are directly interchangeable.
The final answer must be in seconds, as requested.
The diagram is marked "NOT TO SCALE", so do not attempt to measure lengths from the drawing; use only the given numerical values.

Techniques used

calculate the volume of a cuboidconvert cubic centimetres to millilitresdivide volume by flow rate

Q133MMedium-EasyNumber

Work out $2\frac{2}{3} + 3\frac{3}{4}$ .

Give your answer as a mixed number in its simplest form.

______

DifficultyMedium-Easy

Worked solution

Approach

Convert both mixed numbers into improper fractions, find a common denominator for the two fractions, add them together, and finally convert the resulting improper fraction back into a mixed number in its simplest form.

Working

First, convert $2\frac{2}{3}$ to an improper fraction:

2\frac{2}{3} = \frac{(2 \times 3) + 2}{3} = \frac{8}{3}

Next, convert $3\frac{3}{4}$ to an improper fraction:

3\frac{3}{4} = \frac{(3 \times 4) + 3}{4} = \frac{15}{4}

Now we need to add these two fractions:

\frac{8}{3} + \frac{15}{4}

Find a common denominator for 3 and 4. The lowest common multiple is 12. Convert both fractions:

\frac{8}{3} = \frac{8 \times 4}{3 \times 4} = \frac{32}{12}

\frac{15}{4} = \frac{15 \times 3}{4 \times 3} = \frac{45}{12}

Add the two fractions:

\frac{32}{12} + \frac{45}{12} = \frac{77}{12}

Finally, convert $\frac{77}{12}$ back into a mixed number. Divide 77 by 12:

77 \div 12 = 6 \text{ remainder } 5

So the answer is:

6\frac{5}{12}

The fraction $\frac{5}{12}$ cannot be simplified further as 5 and 12 have no common factors.

Answer

6\frac{5}{12}

Final answer

6 5/12

Detailed explanation

Walkthrough

We are asked to add two mixed numbers: $2\frac{2}{3}$ and $3\frac{3}{4}$ . Mixed numbers can be tricky to add directly because you have to deal with whole parts and fractional parts simultaneously. A reliable method is to first turn them into improper fractions (where the numerator is larger than the denominator).

To turn $2\frac{2}{3}$ into an improper fraction, multiply the whole number ( $2$ ) by the denominator ( $3$ ) and add the numerator ( $2$ ). This gives $(2 \times 3) + 2 = 8$ . Place this over the original denominator to get $\frac{8}{3}$ .

Similarly, for $3\frac{3}{4}$ , multiply the whole number ( $3$ ) by the denominator ( $4$ ) and add the numerator ( $3$ ). This gives $(3 \times 4) + 3 = 15$ . Placing this over the original denominator gives $\frac{15}{4}$ .

Now we must add $\frac{8}{3}$ and $\frac{15}{4}$ . We cannot simply add the top and bottom numbers; we must make the denominators (the bottom numbers) the same. The smallest number that both 3 and 4 divide into is 12.

To change $\frac{8}{3}$ to something over 12, we multiply the top and bottom by 4, giving $\frac{32}{12}$ . To change $\frac{15}{4}$ to something over 12, we multiply the top and bottom by 3, giving $\frac{45}{12}$ .

Now we can just add the top numbers (numerators): $32 + 45 = 77$ . The bottom number stays 12, so we have $\frac{77}{12}$ .

The question asks for a mixed number, so we convert $\frac{77}{12}$ back. How many times does 12 go into 77? It goes in 6 times ( $12 \times 6 = 72$ ), with 5 left over ( $77 - 72 = 5$ ). So the mixed number is $6\frac{5}{12}$ . Since 5 and 12 share no common factors, this is in its simplest form.

Key Takeaways

Converting mixed numbers to improper fractions makes addition much easier.
Always find a common denominator before adding or subtracting fractions.
If the final answer needs to be a mixed number, remember to divide the numerator by the denominator to find the whole part and the remainder.

Common Mistakes

Adding whole numbers and fractions separately without converting properly (e.g., getting $5\frac{5}{7}$ by adding tops and bottoms directly).
Forgetting to change BOTH the numerator and the denominator when finding equivalent fractions (e.g., writing $\frac{8}{12}$ instead of $\frac{32}{12}$ ).
Calculation errors when multiplying the numerator and denominator to find the common denominator.
Failing to simplify the final improper fraction back into a mixed number if requested.

Things to Be Careful About

Make sure your final answer is in the correct format. The question specifically asks for a "mixed number in its simplest form". An improper fraction like $\frac{77}{12}$ might earn partial marks (M1) but not full credit (B2 cao).
Ensure the fractional part of your mixed number is fully simplified (e.g., don't leave it as $\frac{10}{24}$ if it can be reduced to $\frac{5}{12}$ ).
Double-check your multiplication: $32 + 45$ is indeed $77$ , and $12 \times 6$ is $72$ .

Techniques used

convert mixed numbers to improper fractionsfind a common denominatoradd the numeratorssimplify the result to a mixed number

Q14Medium-EasyAlgebra and GraphsMensurationCoordinate Geometry

Lia cycles $800\,\text{m}$ in 105 seconds.
The diagram shows the speed–time graph for this journey.

Fig. 5

(a)

Work out the value of $v$ .

$v$ = ______

DifficultyMedium-Easy

Worked solution

Approach

The total distance travelled is equal to the area under the speed–time graph. The graph consists of a triangular section from $t = 0$ to $t = 10$ and a rectangular section from $t = 10$ to $t = 105$ . Together, these form a trapezium with parallel sides along the time axis and the constant speed line.

Working

The area of a trapezium is given by:

\text{Area} = \frac{1}{2}(a + b)h

where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular height between them.

From the graph:

The bottom parallel side (along the time axis) has length $105$ .
The top parallel side (the constant speed portion) has length $105 - 10 = 95$ .
The height is the constant speed $v$ .

Equating the area to the total distance of $800\,\text{m}$ :

\frac{1}{2}(105 + 95) \times v = 800

\frac{1}{2}(200) \times v = 800

100v = 800

v = \frac{800}{100}

v = 8

Answer

v = 8

Final answer

8

Detailed explanation

Walkthrough

The problem gives the total distance ( $800\,\text{m}$ ) and a speed-time graph. A fundamental principle in kinematics is that the distance travelled is the area under the speed-time graph. The graph starts at the origin, rises linearly to a speed $v$ at $t = 10\,\text{s}$ , and then stays constant at $v$ until $t = 105\,\text{s}$ . The region under this graph is a trapezium. The parallel sides of the trapezium are horizontal: the bottom side is the total time, $105\,\text{s}$ , and the top side is the duration at constant speed, which is $105 - 10 = 95\,\text{s}$ . The height of the trapezium is the maximum speed, $v$ . We set the area formula $\frac{1}{2}(a+b)h$ equal to $800$ and solve for $v$ .

Key Takeaways

The area under a speed-time graph represents the distance travelled.
Composite shapes under graphs (like a triangle plus a rectangle) can often be treated as a single trapezium for area calculation.
Identifying the parallel sides and height of a trapezium on a graph requires reading the axis values correctly.

Common Mistakes

Calculating the area as a triangle only (forgetting the rectangular part at the end), which would give $\frac{1}{2} \times 105 \times v = 800$ .
Using the wrong parallel side lengths, such as using $10$ instead of $95$ for the top side.
Forgetting to subtract $10$ from $105$ to find the length of the constant-speed segment.

Things to Be Careful About

Ensure the area units match the distance units (metres here).
The mark scheme accepts the equation $800 = \frac{1}{2}(105 + 105 - 10) \times v$ , showing that identifying the top side as $(105 - 10)$ is the key step.
Working must be shown (nfww) to earn the method marks.

Techniques used

recognise that distance equals the area under a speed-time graphidentify the shape under the graph as a trapeziumapply the trapezium area formulasolve the resulting linear equation for the unknown speed

(b)

Work out Lia’s acceleration during the first 10 seconds.

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

DifficultyEasy

Worked solution

Approach

Acceleration is defined as the rate of change of speed. On a speed-time graph, this corresponds to the gradient (slope) of the line. During the first $10$ seconds, the graph is a straight line from $(0, 0)$ to $(10, v)$ .

Working

Using the value $v = 8$ from part (a), the coordinates are $(0, 0)$ and $(10, 8)$ .

The gradient is:

\text{Gradient} = \frac{\text{change in speed}}{\text{change in time}} = \frac{8 - 0}{10 - 0}

\text{Acceleration} = \frac{8}{10}

\text{Acceleration} = 0.8

Answer

0.8

Final answer

0.8

Detailed explanation

Walkthrough

The question asks for the acceleration during the first $10$ seconds. Acceleration is the rate of change of speed, which is represented by the gradient of the speed-time graph. The graph is a straight line from the origin $(0,0)$ to $(10, v)$ . We use the value $v = 8$ calculated in part (a). The change in speed is $8 - 0 = 8\,\text{m/s}$ , and the change in time is $10 - 0 = 10\,\text{s}$ . Dividing the change in speed by the change in time gives the acceleration.

Key Takeaways

On a speed-time graph, the gradient of a straight line segment represents the acceleration during that time interval.
Acceleration can be calculated as $\frac{\Delta \text{speed}}{\Delta \text{time}}$ .

Common Mistakes

Using the total time ( $105\,\text{s}$ ) instead of the time interval ( $10\,\text{s}$ ) for the acceleration calculation.
Forgetting to use the value of $v$ from part (a) and trying to calculate acceleration without it.
Confusing acceleration with the total distance or average speed.

Things to Be Careful About

The mark scheme allows $\frac{8}{10}$ as an answer, but $0.8$ is the standard decimal form. Both are acceptable.
Follow-through (FT) marks are awarded if a different value for $v$ was used in part (a), as long as the correct method (their $v \div 10$ ) is applied.

Techniques used

recognise that acceleration is the gradient of a speed-time graphcalculate the gradient using the coordinates of the first segment

Q15Medium-EasyStatistics

The cumulative frequency diagram shows information about journey times.

Fig. 6

Use the cumulative frequency diagram to find an estimate for

(a)

the median

______ $\text{minutes}$

(b)

the number of journeys that took 20 minutes or more.

______

DifficultyMedium-Easy

Worked solution

Approach

To find the number of journeys that took 20 minutes or more, find the cumulative frequency at 20 minutes and subtract it from the total frequency.

Working

Total frequency is $60$ .

From the graph, at time $= 20$ minutes, the cumulative frequency is $23$ .

Number of journeys taking 20 minutes or more $= 60 - 23 = 37$ .

60 - 23 = 37

Answer

37

Final answer

37

Detailed explanation

Walkthrough

The question asks for the number of journeys that took $20$ minutes or more. The cumulative frequency at $20$ minutes gives the number of journeys that took $20$ minutes or less. To find the number of journeys that took $20$ minutes or more, we subtract the cumulative frequency at $20$ minutes from the total frequency. From the graph, at time $= 20$ minutes, the cumulative frequency is approximately $23$ . The total frequency is $60$ . Therefore, the number of journeys is $60 - 23 = 37$ .

Key Takeaways

Cumulative frequency at a given value represents the number of observations up to that value.
The number of observations greater than a value is found by subtracting the cumulative frequency at that value from the total frequency.

Common Mistakes

Reading the frequency at $20$ minutes instead of the cumulative frequency.
Forgetting to subtract from the total frequency to find the number of journeys above the threshold.
Misreading the cumulative frequency value from the graph.

Things to Be Careful About

The mark scheme awards a method mark (B1) for stating the cumulative frequency at $20$ minutes (e.g., $23$ ), even if the final subtraction is incorrect.
Ensure you subtract the cumulative frequency at the given time from the total frequency, not the other way around.
Readings from graphs are estimates, so a value like $23$ is expected to earn the method mark.

Techniques used

read cumulative frequency at a given timesubtract from total frequency to find the upper tail

Q16MediumGeometry

Fig. 7

Points $Q$ , $R$ , $S$ and $T$ lie on a circle, centre $O$ .
$UV$ is a tangent to the circle at $T$ .
$SOQV$ is a straight line.
Angle $QST = 32^\circ$ .

(a)

Find angle $QRT$ .
Give a geometrical reason to explain your answer.

Angle $QRT$ = ______ because ______

(b)

Find angle $TQS$ .

Angle $TQS$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Recognise that $SQ$ is a diameter since the straight line $SOQV$ passes through the centre $O$ . Use the angle in a semicircle theorem to find $\angle STQ$ , then use the angle sum of $\triangle STQ$ to find $\angle TQS$ .

Working

Since $SOQV$ is a straight line passing through the centre $O$ , $SQ$ is a diameter of the circle.
The angle subtended by a diameter at the circumference is $90^\circ$ , so:

\angle STQ = 90^\circ

In $\triangle STQ$ , the sum of the interior angles is $180^\circ$ :

\angle TQS + \angle STQ + \angle TSQ = 180^\circ

Substitute the known values ( $\angle TSQ = 32^\circ$ and $\angle STQ = 90^\circ$ ):

\angle TQS = 180^\circ - 90^\circ - 32^\circ = 58^\circ

Answer

Angle $TQS = 58^\circ$ .

Final answer

58

Detailed explanation

Walkthrough

First, observe that the line $SOQV$ passes through the centre $O$ , meaning $SQ$ is a diameter. A key circle theorem states that the angle subtended by a diameter at any point on the circumference is a right angle. Therefore, $\angle STQ = 90^\circ$ .

Next, look at $\triangle STQ$ . We now know two of its angles: $\angle STQ = 90^\circ$ and $\angle TSQ = 32^\circ$ (given as $\angle QST$ ). The sum of angles in any triangle is $180^\circ$ , so we can find the third angle $\angle TQS$ by subtracting the other two from $180^\circ$ :

\angle TQS = 180^\circ - 90^\circ - 32^\circ = 58^\circ

Key Takeaways

A straight line through the centre of a circle is a diameter, and it subtends a $90^\circ$ angle at the circumference.
Once a right angle is established in a triangle, the other two acute angles must sum to $90^\circ$ .

Common Mistakes

Forgetting that $SQ$ is a diameter and not using the $90^\circ$ angle.
Rounding or writing the answer to an incorrect number of significant figures; angles in this context are typically exact integers.

Things to Be Careful About

The mark scheme specifically looks for the step identifying $\angle STQ = 90^\circ$ . Make sure this is shown clearly in the working.

Techniques used

recognise angle in a semicircle is 90use angle sum of triangle

(c)

Find angle $TVQ$ .

Angle $TVQ$ = ______

DifficultyMedium

Worked solution

Approach

Use the alternate segment theorem to find $\angle QTV$ , then use angles on a straight line and the angle sum of $\triangle TVQ$ to find $\angle TVQ$ .

Working

$UV$ is a tangent to the circle at $T$ , and $TQ$ is a chord.
By the alternate segment theorem, the angle between the tangent and the chord equals the angle in the alternate segment:

\angle QTV = \angle TSQ = 32^\circ

Since $SQV$ is a straight line, $\angle TQS$ and $\angle TQV$ are supplementary:

\angle TQV = 180^\circ - \angle TQS = 180^\circ - 58^\circ = 122^\circ

In $\triangle TVQ$ , the sum of the interior angles is $180^\circ$ :

\angle TVQ = 180^\circ - \angle QTV - \angle TQV

Substitute the known values:

\angle TVQ = 180^\circ - 32^\circ - 122^\circ = 26^\circ

Answer

Angle $TVQ = 26^\circ$ .

Final answer

26

Detailed explanation

Walkthrough

The tangent $UV$ touches the circle at $T$ , and $TQ$ is a chord drawn from the point of tangency. The alternate segment theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Here, the angle between tangent $TV$ and chord $TQ$ is $\angle QTV$ , and the angle in the alternate segment is $\angle TSQ$ (or $\angle QST$ ). Thus, $\angle QTV = 32^\circ$ .

Next, we need another angle in $\triangle TVQ$ . We know $\angle TQS = 58^\circ$ from part (b). Since $SQV$ is a straight line, $\angle TQS$ and $\angle TQV$ form a linear pair and add to $180^\circ$ :

\angle TQV = 180^\circ - 58^\circ = 122^\circ

Finally, use the angle sum of $\triangle TVQ$ to find $\angle TVQ$ :

\angle TVQ = 180^\circ - 32^\circ - 122^\circ = 26^\circ

Key Takeaways

The alternate segment theorem links the angle between a tangent and a chord to an angle on the circumference.
Angles on a straight line add to $180^\circ$ , which is useful for finding exterior angles of triangles.

Common Mistakes

Applying the alternate segment theorem to the wrong chord or tangent angle.
Forgetting that $SQV$ is a straight line and trying to use $\angle TQS$ directly in $\triangle TVQ$ without finding the supplementary angle $\angle TQV$ .
Arithmetic errors when subtracting from $180^\circ$ .

Things to Be Careful About

The mark scheme accepts alternative routes, such as finding $\angle STV = 90^\circ + 32^\circ = 122^\circ$ and then using $\triangle STV$ to find $\angle TVQ = 180^\circ - 32^\circ - 122^\circ = 26^\circ$ . Both methods are valid.
Ensure all angle values used in the final triangle calculation are clearly derived.

Techniques used

apply alternate segment theoremuse angle sum of triangle

Q17MediumMensurationTrigonometryGeometry

Fig. 8

The diagram shows a prism.
The cross-section of the prism is an isosceles triangle with sides $13\,\text{cm}$ , $13\,\text{cm}$ and base $10\,\text{cm}$ .
The length of the prism is $3\,\text{cm}$ .

(a)

Show that the volume of the prism is $180\,\text{cm}^3$ .

DifficultyMedium-Easy

Worked solution

Approach

To find the volume of the prism, we first need the area of its triangular cross-section. The cross-section is an isosceles triangle with sides $13$ , $13$ and $10$ . We can find its height by splitting it into two right-angled triangles and using Pythagoras' theorem. Then we multiply the cross-sectional area by the length of the prism.

Working

The base of the isosceles triangle is $10\,\text{cm}$ , so the perpendicular height bisects the base into two segments of $5\,\text{cm}$ .
Let $h$ be the height of the triangle. By Pythagoras' theorem:

\begin{aligned} h^2 + 5^2 &= 13^2 \\ h^2 + 25 &= 169 \\ h^2 &= 144 \\ h &= 12\,\text{cm} \end{aligned}

The area of the triangular cross-section is:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 12 = 60\,\text{cm}^2

The volume of the prism is the cross-sectional area multiplied by its length ( $3\,\text{cm}$ ):

\text{Volume} = 60 \times 3 = 180\,\text{cm}^3

Answer

The volume of the prism is $180\,\text{cm}^3$ , as required.

Final answer

180 cm^3

Detailed explanation

Walkthrough

The volume of any prism is given by the formula: Volume = Area of cross-section × length. The cross-section here is an isosceles triangle with sides 13 cm, 13 cm and base 10 cm. To find its area, we need its height. Because the triangle is isosceles, dropping a perpendicular from the apex to the base bisects the base, creating two right-angled triangles with hypotenuse 13 cm and base 5 cm. Using Pythagoras' theorem ( $h^2 + 5^2 = 13^2$ ), we find the height is 12 cm. The area of the triangle is then $\frac{1}{2} \times 10 \times 12 = 60\,\text{cm}^2$ . Multiplying by the prism length of 3 cm gives a volume of 180 cm³.

Key Takeaways

The volume of a prism is the area of its uniform cross-section multiplied by its length.
In an isosceles triangle, the perpendicular height from the apex bisects the base, allowing the use of Pythagoras' theorem to find the height.

Common Mistakes

Forgetting to halve the base before applying Pythagoras' theorem (using 10 instead of 5).
Using the wrong formula for the area of a triangle (e.g., forgetting the $\frac{1}{2}$ factor).
Multiplying the triangle's perimeter or side lengths instead of its area by the prism's length.

Things to Be Careful About

Ensure all lengths are in the same units (cm) before calculating area and volume.
The question asks to 'show that', so the final line of working must explicitly state or conclude with the target value of $180\,\text{cm}^3$ .
Pythagoras' theorem requires the square of the hypotenuse (the longest side, 13) to be equal to the sum of the squares of the other two sides.

Techniques used

apply Pythagoras' theorem to find the height of the trianglecalculate the area of the triangular cross-sectionmultiply by the length to find the volume of the prism

(b)

A mathematically similar prism has a volume of $1440\,\text{cm}^3$ .
The cross-section of this prism is an isosceles triangle with base $x\,\text{cm}$ .

Calculate the value of $x$ .

$x$ = ______

DifficultyMedium

Worked solution

Approach

For mathematically similar solids, the ratio of their volumes is equal to the cube of the linear scale factor. We can find the volume ratio, take the cube root to get the linear scale factor, and then use it to find the unknown base length $x$ .

Working

The volume of the original prism is $180\,\text{cm}^3$ and the volume of the similar prism is $1440\,\text{cm}^3$ .
The ratio of the volumes is:

\frac{V_{\text{new}}}{V_{\text{original}}} = \frac{1440}{180} = 8

The linear scale factor $k$ is the cube root of the volume ratio:

k = \sqrt[3]{8} = 2

This means all lengths in the new prism are $2$ times the corresponding lengths in the original prism. The original base is $10\,\text{cm}$ , so:

x = 10 \times 2 = 20

Alternatively, setting up the ratio directly:

\frac{10}{x} = \sqrt[3]{\frac{180}{1440}} = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}

Solving for $x$ :

x = 10 \times 2 = 20

Answer

x = 20

Final answer

20

Detailed explanation

Walkthrough

When two solids are mathematically similar, their corresponding lengths are in the ratio $k : 1$ (the linear scale factor), their areas are in the ratio $k^2 : 1$ , and their volumes are in the ratio $k^3 : 1$ . Here, the volume ratio is $\frac{1440}{180} = 8$ . To find the linear scale factor, we take the cube root of 8, which is 2. This means every length in the new prism is twice the corresponding length in the original prism. Since the original base is 10 cm, the new base $x$ is $10 \times 2 = 20$ cm.

Key Takeaways

For similar solids, Volume ratio = (Linear scale factor) $^3$ .
To find a length from a volume ratio, take the cube root of the volume ratio to get the linear scale factor, then multiply the original length by this factor.

Common Mistakes

Confusing the volume ratio with the linear ratio (e.g., assuming $x = 10 \times 8 = 80$ ).
Taking the square root instead of the cube root when finding the linear scale factor from a volume ratio.
Setting up the ratio incorrectly, such as $\frac{x}{10} = \frac{1440}{180}$ instead of using the cube root.

Things to Be Careful About

Always verify whether the given ratio is for lengths, areas, or volumes, as the relationship to the linear scale factor changes accordingly ( $k$ , $k^2$ , or $k^3$ ).
Ensure the cube root is calculated correctly; $\sqrt[3]{8} = 2$ , not 4.
The question asks for the value of $x$ , so the final answer should be just the number (20), though including the unit (cm) is good practice in working.

Techniques used

calculate the ratio of volumestake the cube root to find the linear scale factorapply the linear scale factor to the base length

Q182MMedium-EasyNumber

Write $0.1\dot{4}$ as a fraction.

______

DifficultyMedium-Easy

Worked solution

Approach

To convert the recurring decimal $0.1\dot{4}$ into a fraction, we can use the algebraic method. Let $x$ equal the given number. We then multiply $x$ by powers of 10 to shift the decimal point so that the recurring parts align, allowing us to subtract one equation from the other to remove the recurring digit.

Working

Let $x = 0.1\dot{4}$ . This means:

x = 0.1444\ldots \quad (1)

We want to create another equation where the repeating part starts at the same position as in (1), but shifted. Multiplying (1) by 10 moves the decimal point just before the repeating '4':

10x = 1.444\ldots \quad (2)

Now, multiply (1) by 100 to move the decimal point past the first '4' as well (or simply notice that multiplying (2) by 10 gives $100x$ ):

100x = 14.444\ldots \quad (3)

Subtract equation (2) from equation (3). The recurring parts ( $0.444\ldots$ ) cancel out:

100x - 10x = 14.444\ldots - 1.444\ldots

90x = 13

Solve for $x$ :

x = \frac{13}{90}

The fraction $\frac{13}{90}$ cannot be simplified further because 13 is a prime number and does not divide 90.

Answer

\frac{13}{90}

Final answer

13/90

Detailed explanation

Walkthrough

The problem asks to write the recurring decimal $0.1\dot{4}$ as a fraction. This notation means $0.14444...$ , where the digit 4 repeats infinitely.

Define the variable: Let $x = 0.1444...$
Shift the decimal point: We need to create a second equation such that when we subtract it from the first (or vice versa), the infinite repeating tail disappears.
- Multiply $x$ by 10 to get $10x = 1.444...$
- Multiply $x$ by 100 to get $100x = 14.444...$
Subtract: Subtracting $10x$ from $100x$ eliminates the recurring $.444...$ part.
$100x - 10x = 14.444... - 1.444...$
$90x = 13$
Solve: Divide by 90 to isolate $x$ .
$x = \frac{13}{90}$
Check for simplification: 13 is prime and does not divide 90, so the fraction is in its simplest form.

An alternative approach mentioned in some mark schemes is splitting the decimal: $0.1\dot{4} = 0.1 + 0.0444... = \frac{1}{10} + \frac{4}{90}$ . Finding a common denominator gives $\frac{9}{90} + \frac{4}{90} = \frac{13}{90}$ .

Key Takeaways

The algebraic method for converting recurring decimals involves multiplying by powers of 10 to align the repeating digits.
Subtracting the aligned equations removes the recurring part, leaving a linear equation.
Always check if the final fraction can be simplified.

Common Mistakes

Forgetting to subtract correctly (e.g., doing $100x - x$ instead of $100x - 10x$ ).
Misinterpreting the scope of the dot notation (e.g., treating it as $0.141414...$ ).
Failing to simplify the fraction if possible (though $\frac{13}{90}$ is already simple).
Arithmetic errors when calculating $100 - 10$ or $14 - 1$ .

Things to Be Careful About

Ensure you multiply by the correct power of 10. If only one digit repeats after the decimal, multiplying by 10 once puts the repeat immediately after the decimal, and multiplying by 100 puts it again. The difference between these multipliers is $100 - 10 = 90$ , which becomes the denominator.
The question requires an exact fractional answer, not a decimal approximation.
Mark scheme note: "oe" means any equivalent form is accepted, but usually this implies the simplified fraction.

Techniques used

convert recurring decimal to fraction using algebraic methodeliminate the recurring part by subtractionsimplify the resulting fraction

Q193MMediumAlgebra and Graphs

$g(x) = \frac{x}{2} + 4 \quad h(x) = 3x - 1$

Solve $hg(x) = 5$ .

$x$ = ______

DifficultyMedium

Worked solution

Approach

To solve $hg(x) = 5$ , we first determine the expression for the composite function $hg(x)$ by substituting $g(x)$ into $h(x)$ . Then we set this expression equal to 5 and solve for $x$ .

Working

The functions are given as:

g(x) = \frac{x}{2} + 4 \quad h(x) = 3x - 1

The composite function $hg(x)$ means $h(g(x))$ . We substitute the entire expression for $g(x)$ wherever there is an $x$ in $h(x)$ :

hg(x) = 3\left( \frac{x}{2} + 4 \right) - 1

Expand the brackets:

hg(x) = \frac{3x}{2} + 12 - 1

Simplify:

hg(x) = \frac{3x}{2} + 11

Now set $hg(x) = 5$ :

\frac{3x}{2} + 11 = 5

Subtract 11 from both sides:

\frac{3x}{2} = 5 - 11

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

Multiply both sides by 2:

3x = -12

Divide by 3:

x = -4

Alternatively, using the inverse method shown in the mark scheme:

\frac{x}{2} + 4 = \frac{5 + 1}{3}

\frac{x}{2} + 4 = \frac{6}{3}

\frac{x}{2} + 4 = 2

\frac{x}{2} = 2 - 4

\frac{x}{2} = -2

x = -4

Answer

x = -4

Final answer

-4

Detailed explanation

Walkthrough

The problem asks us to find the value of $x$ that satisfies the condition where the composite function $hg(x)$ equals 5.

First, recall that $hg(x)$ is read as "h of g of x" or $h(g(x))$ . This means we take the output of the inner function $g(x)$ and use it as the input for the outer function $h(x)$ .

Given $g(x) = \frac{x}{2} + 4$ and $h(x) = 3x - 1$ , we substitute $g(x)$ into $h(x)$ :

hg(x) = 3(g(x)) - 1 = 3\left( \frac{x}{2} + 4 \right) - 1

Next, we simplify this expression. Distribute the 3 across the terms inside the parentheses:

3 \times \frac{x}{2} = \frac{3x}{2}

3 \times 4 = 12

So, $hg(x) = \frac{3x}{2} + 12 - 1$ . Combining the constants gives:

hg(x) = \frac{3x}{2} + 11

We are told that $hg(x) = 5$ , so we set up the equation:

\frac{3x}{2} + 11 = 5

To isolate $x$ , subtract 11 from both sides:

\frac{3x}{2} = 5 - 11

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

Multiply both sides by 2 to remove the denominator:

3x = -12

Finally, divide by 3:

x = -4

Key Takeaways

Composite Function Notation: $hg(x)$ always means $h(g(x))$ . The function written second (on the left) is applied last.
Substitution: When substituting an expression into another, use brackets if the substituted expression has more than one term to avoid sign errors.
Linear Equations: Solving involves isolating the variable through inverse operations (subtracting constants, dividing coefficients).

Common Mistakes

Order of Composition: Students sometimes calculate $gh(x)$ instead of $hg(x)$ , leading to the wrong equation ( $3x-1$ plugged into $x/2+4$ ).
Bracket Errors: Forgetting to bracket $g(x)$ when substituting into $h(x)$ , e.g., writing $3x/2 + 4 - 1$ instead of $3(x/2 + 4) - 1$ . This results in incorrect constant terms.
Arithmetic with Negatives: Subtracting 11 from 5 incorrectly as positive 6 or failing to carry the negative sign through the steps.
Inconsistent Method: Trying to solve without simplifying the composite function first can lead to messy algebra.

Things to Be Careful About

Check Your Work: Substitute $x = -4$ back into the original functions. $g(-4) = -4/2 + 4 = 2$ . $h(2) = 3(2) - 1 = 5$ . The result matches.
Fraction Arithmetic: Ensure you handle the fraction $\frac{3x}{2}$ correctly during multiplication and division steps.
Sign Errors: Pay close attention to signs when expanding brackets or moving terms across the equals sign.

Techniques used

find the composite function hg(x)form an equation from the composite functionsolve the resulting linear equation

Q202MMedium-EasyAlgebra and Graphs

Write as a single fraction in its simplest form.

\frac{5}{2x} - \frac{4}{7x}

______

DifficultyMedium-Easy

Worked solution

Approach

Find a common denominator for the two algebraic fractions, rewrite each fraction with this denominator, then subtract the numerators and simplify.

Working

The denominators are $2x$ and $7x$ . Their lowest common denominator is $14x$ .

Rewrite each fraction with denominator $14x$ :

\frac{5}{2x} = \frac{5 \times 7}{2x \times 7} = \frac{35}{14x}

\frac{4}{7x} = \frac{4 \times 2}{7x \times 2} = \frac{8}{14x}

Subtract the numerators, keeping the common denominator:

\frac{35}{14x} - \frac{8}{14x} = \frac{35 - 8}{14x} = \frac{27}{14x}

Answer

\frac{27}{14x}

Final answer

27/(14x)

Detailed explanation

Walkthrough

We start with two fractions with different denominators, $2x$ and $7x$ . Before subtracting, both fractions must be written over the same denominator. The smallest denominator that both $2x$ and $7x$ divide into is $14x$ .

Multiply the first fraction by $\frac{7}{7}$ and the second fraction by $\frac{2}{2}$ . These multiplications do not change the value of either fraction, but they give both fractions the common denominator $14x$ :

\frac{5}{2x} = \frac{35}{14x}, \quad \frac{4}{7x} = \frac{8}{14x}

Now subtract the numerators, keeping the denominator unchanged:

\frac{35}{14x} - \frac{8}{14x} = \frac{27}{14x}

The fraction is already in simplest form because $27$ and $14$ have no common factor.

Key Takeaways

Algebraic fractions are combined in the same way as numerical fractions: find a common denominator first.
The variable $x$ is part of the denominator, so the lowest common denominator of $2x$ and $7x$ is $14x$ .
When subtracting fractions with the same denominator, subtract only the numerators and keep the denominator unchanged.
A fraction is in simplest form when the numerator and denominator have no common factor.

Common Mistakes

Trying to subtract the fractions without first writing them with a common denominator.
Forgetting to multiply the numerator when scaling the denominator, for example writing $\frac{5}{14x}$ instead of $\frac{35}{14x}$ .
Subtracting the denominators as well as the numerators.
Giving an unsupported final answer: the mark scheme awards a method mark for showing the equivalent fractions $\frac{35}{14x}$ and $\frac{8}{14x}$ , so working must be shown.

Things to Be Careful About

The value $x = 0$ is not allowed, since the original fractions would have zero denominators.
Keep the denominator as one term, $14x$ , throughout the subtraction.
This is a non-calculator question, so show the by-hand scaling clearly: multiply numerator and denominator by $7$ for the first fraction and by $2$ for the second.
The final answer must be a single fraction in simplest form; $\frac{27}{14x}$ is the required exact form.

Techniques used

find the lowest common denominatorrewrite each fraction with the common denominatorsubtract the numeratorssimplify the resulting fraction

Q21Medium-EasyNumber

(a)

Evaluate.

27^{-\frac{2}{3}}

______

DifficultyMedium-Easy

Worked solution

Approach

Write $27$ as a power of the prime base $3$ , then apply the fractional index using the power rule $(a^m)^n = a^{mn}$ . Finally, convert the resulting negative index into a reciprocal.

Working

Since $27 = 3^3$ ,

27^{-\frac{2}{3}} = \left(3^3\right)^{-\frac{2}{3}}

Using $\left(a^m\right)^n = a^{mn}$ :

= 3^{3 \times \left(-\frac{2}{3}\right)} = 3^{-2}

Using $a^{-n} = \frac{1}{a^n}$ :

= \frac{1}{3^2} = \frac{1}{9}

Answer

\frac{1}{9}

Final answer

1/9

Detailed explanation

Walkthrough

We are asked to evaluate $27^{-\frac{2}{3}}$ . The key is to connect the base $27$ with the denominator $3$ of the fractional index: $27$ is $3^3$ . Then apply the law

(a^m)^n = a^{mn}

so $(3^3)^{-\frac{2}{3}} = 3^{3 \times -\frac{2}{3}}$ . Multiplying $3$ by $-\frac{2}{3}$ gives $-2$ , leaving $3^{-2}$ . A negative index means the reciprocal: $a^{-n} = \frac{1}{a^n}$ , so $3^{-2} = \frac{1}{9}$ . This is the exact fractional value, not a rounded decimal.

Key Takeaways

This question tests the index laws for negative and fractional powers. The most useful steps are always: rewrite a base as a prime power, then multiply the exponents, then apply any negative index by taking the reciprocal.

Common Mistakes

Forgetting the negative sign and giving $9$ , since $27^{\frac{2}{3}} = 9$ .
Taking a square root before a cube root, or mixing up $27^{\frac{1}{3}}$ and $27^{\frac{1}{2}}$ .
Giving a decimal such as $0.111$ instead of the exact fraction $\frac{1}{9}$ .

Things to Be Careful About

This is the non-calculator paper, so the answer must be presented exactly as a fraction. The mark scheme gives a method mark for reaching $3^{-2}$ or $\frac{1}{27^{\frac{2}{3}}}$ , so show that intermediate line. Be especially careful with the negative index: it turns the entire result into its reciprocal.

Techniques used

rewrite an integer as a power of its prime baseapply a fractional index using the power ruleconvert a negative index into a reciprocal

(b)

Simplify.

\left(\frac{1}{100}x^{100}\right)^{\frac{1}{2}}

______

DifficultyMedium-Easy

Worked solution

Approach

Apply the index $\frac{1}{2}$ separately to the numerical factor and to the $x$ term, using $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$ .

Working

\left(\frac{1}{100}x^{100}\right)^{\frac{1}{2}} = \left(\frac{1}{100}\right)^{\frac{1}{2}}\cdot\left(x^{100}\right)^{\frac{1}{2}}

First:

\left(\frac{1}{100}\right)^{\frac{1}{2}} = \frac{1}{\sqrt{100}} = \frac{1}{10}

Then:

\left(x^{100}\right)^{\frac{1}{2}} = x^{100 \times \frac{1}{2}} = x^{50}

Combining these:

\left(\frac{1}{100}x^{100}\right)^{\frac{1}{2}} = \frac{1}{10}x^{50}

Answer

\frac{1}{10}x^{50}

Final answer

1/10 x^50

Detailed explanation

Walkthrough

The expression is $\left(\frac{1}{100}x^{100}\right)^{\frac{1}{2}}$ . We can apply the exponent to each factor separately using

(ab)^n = a^n b^n

so it becomes $\left(\frac{1}{100}\right)^{\frac{1}{2}}\left(x^{100}\right)^{\frac{1}{2}}$ . For the constant factor, raising to the power $\frac{1}{2}$ is the same as taking the square root, and $\sqrt{100} = 10$ , so we get $\frac{1}{10}$ . For the variable factor, use $(a^m)^n = a^{mn}$ , multiplying $100$ by $\frac{1}{2}$ to obtain $x^{50}$ . The simplified final answer is $\frac{1}{10}x^{50}$ .

Key Takeaways

This question checks two fundamental index laws: powers in a product can each be raised to the exponent separately, and a power raised to another power is found by multiplying exponents. It also connects fractional powers to roots, particularly $\frac{1}{2}$ being the square root.

Common Mistakes

Applying the exponent only to $x^{100}$ and leaving the coefficient as $\frac{1}{100}$ .
Treating $x^{100}$ after taking the power as $x^{50}$ correctly but forgetting to convert $\frac{1}{100}$ to $\frac{1}{10}$ .
Writing $x^{100 \times \frac{1}{2}} = x^{100}$ .
Writing the coefficient as $\frac{1}{\sqrt{100}} = \frac{1}{10}$ is correct; writing $\frac{1}{100} \times \frac{1}{2} = \frac{1}{200}$ would be wrong because the power does not multiply the coefficient.

Things to Be Careful About

The final answer should be given in exact form, prefer $\frac{1}{10}x^{50}$ to $0.1x^{50}$ . The coefficient and the exponent must be simplified correctly. The mark scheme awards one method mark for arriving at a form like $kx^{50}$ or $\frac{1}{10}x^k$ with non-zero $k$ , so keep the $\frac{1}{10}$ and $x^{50}$ parts visible.

Techniques used

apply a fractional power to each factor of a productmultiply the numerator and denominator powerstake the square root of a perfect square constant

Q22MediumAlgebra and Graphs

The table shows some values for $y = x^3 - 3x + 2$ .

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$-16$		$4$	$2$	$0$		$20$

(a)

Complete the table.

DifficultyEasy

Worked solution

Approach

Substitute $x = -2$ and $x = 2$ into the given expression $y = x^3 - 3x + 2$ to find the missing table values.

Working

For $x = -2$ :

\begin{aligned} y &= (-2)^3 - 3(-2) + 2 \\ &= -8 + 6 + 2 \\ &= 0 \end{aligned}

For $x = 2$ :

\begin{aligned} y &= (2)^3 - 3(2) + 2 \\ &= 8 - 6 + 2 \\ &= 4 \end{aligned}

Answer

The missing values in the table are $0$ (for $x = -2$ ) and $4$ (for $x = 2$ ).

Final answer

0, 4

Detailed explanation

Walkthrough

The table provides several $(x, y)$ pairs for the function $y = x^3 - 3x + 2$ , but two $y$ -values are missing. To find them, we simply replace $x$ with the given value in the formula and calculate. For $x = -2$ , we compute $(-2)^3 = -8$ , then $-3 \times (-2) = +6$ , and finally $-8 + 6 + 2 = 0$ . For $x = 2$ , we compute $2^3 = 8$ , then $-3 \times 2 = -6$ , and finally $8 - 6 + 2 = 4$ . These complete the set of points needed to sketch the graph in part (b).

Key Takeaways

Substituting numerical values into an algebraic expression and correctly handling negative numbers and indices. Evaluating $x^3$ when $x$ is negative yields a negative result.

Common Mistakes

Forgetting that $(-2)^3 = -8$ and instead writing $8$ . This is a sign error with odd powers of negative numbers.
Mistaking $-3(-2)$ as $-6$ instead of $+6$ . Multiplying two negative numbers gives a positive result.
Arithmetic errors when adding the three terms together.

Things to Be Careful About

Always show the substitution step clearly so method marks are awarded. Pay close attention to the sign of $x^3$ when $x$ is negative; an odd power preserves the negative sign. The mark scheme awards one mark for each correct value (B1 each), so both $0$ and $4$ are required for full credit.

Techniques used

substitute x = -2 into the cubic expressionsubstitute x = 2 into the cubic expressionevaluate powers and simplify

(b)

Draw the graph of $y = x^3 - 3x + 2$ for $-3 \leq x \leq 3$ .

Fig. 9

DifficultyMedium-Easy

Worked solution

Approach

Use the completed table from part (a) to obtain seven $(x, y)$ coordinate pairs. Plot each point on the provided grid, then join them with a smooth, continuous curve that reflects the shape of a cubic function.

Working

The seven points to plot are:
$(-3, -16)$ , $(-2, 0)$ , $(-1, 4)$ , $(0, 2)$ , $(1, 0)$ , $(2, 4)$ , $(3, 20)$ .

The curve rises from the bottom left, passes through $(-3, -16)$ , crosses the $x$ -axis at $(-2, 0)$ , reaches a local maximum near $(-1, 4)$ , dips to a local minimum near $(1, 0)$ , rises again through $(2, 4)$ , and ends at $(3, 20)$ .

Answer

A smooth cubic curve passing through all seven plotted points.

Final answer

Graph drawn with points (-3, -16), (-2, 0), (-1, 4), (0, 2), (1, 0), (2, 4), (3, 20) joined by a smooth curve

Detailed explanation

Walkthrough

Part (b) asks you to draw the graph of $y = x^3 - 3x + 2$ over the range $-3 \leq x \leq 3$ . The grid provided has the $x$ -axis from $-3$ to $3$ and the $y$ -axis from $-20$ to $20$ , which perfectly accommodates all the values in the table. Plot each of the seven $(x, y)$ pairs: $(-3, -16)$ , $(-2, 0)$ , $(-1, 4)$ , $(0, 2)$ , $(1, 0)$ , $(2, 4)$ , and $(3, 20)$ . Once the points are plotted, draw a smooth curve through them. A cubic graph has a characteristic 'S' shape; here it rises from the bottom left, curves up to a local maximum around $x = -1$ , curves down to a local minimum around $x = 1$ , and then rises steeply to the top right. The mark scheme gives follow-through marks based on how many points are correctly plotted, so accuracy in plotting is essential.

Key Takeaways

Drawing a graph from a table of values requires accurate plotting of coordinates and joining them with a smooth curve that matches the expected function type. For cubics, look for the local maximum and minimum to guide the curve's shape.

Common Mistakes

Plotting points incorrectly, especially negative $y$ -values or large values like $-16$ and $20$ . Ensure you read the axes correctly; the $y$ -axis has major gridlines every $5$ units.
Joining the points with straight line segments instead of a smooth curve. The graph of a cubic function is smooth and continuous.
Not extending the curve to the boundaries of the given range $x = -3$ and $x = 3$ .

Things to Be Careful About

The mark scheme awards marks based on the number of correctly plotted points (B3FT for 6 points, B2FT for 4 points, B1FT for 2 points) and a correct curve. Even if the curve is slightly off, correct points will still earn marks. Ensure your curve is smooth and does not have sharp corners at the plotted points. The $y$ -axis scale is every $5$ units, so $-16$ is slightly below the $-15$ line, and $20$ is exactly on the top gridline.

Techniques used

plot the seven points from the completed tabledraw a smooth cubic curve through the plotted points

(c)

By drawing a suitable line on the grid, find the solutions of the equation $x^3 - 7x + 1 = 0$ .

$x$ = ______ or $x$ = ______ or $x$ = ______

DifficultyMedium

Worked solution

Approach

We are given the curve $y = x^3 - 3x + 2$ and asked to solve $x^3 - 7x + 1 = 0$ . We can rewrite the equation to match the form of the curve by subtracting the two expressions to find a linear equation $y = mx + c$ whose intersections with the curve give the solutions.

Working

Subtract the target equation from the curve equation:

\begin{aligned} (x^3 - 3x + 2) - (x^3 - 7x + 1) &= y - 0 \\ x^3 - 3x + 2 - x^3 + 7x - 1 &= y \\ 4x + 1 &= y \end{aligned}

So we draw the straight line $y = 4x + 1$ on the same grid.

Points on the line $y = 4x + 1$ :

When $x = 0$ , $y = 1$ (0, 1)
When $x = 1$ , $y = 5$ (1, 5)
When $x = -1$ , $y = -3$ (-1, -3)
When $x = -2$ , $y = -7$ (-2, -7)
When $x = 2$ , $y = 9$ (2, 9)
When $x = -3$ , $y = -11$ (-3, -11)

The line $y = 4x + 1$ intersects the curve $y = x^3 - 3x + 2$ at three points. Reading the $x$ -coordinates of these intersections from the graph:

The leftmost intersection is between $x = -3$ and $x = -2$ , approximately at $x \approx -2.7$ .
The middle intersection is between $x = 0$ and $x = 1$ , approximately at $x \approx 0.1$ .
The rightmost intersection is between $x = 2$ and $x = 3$ , approximately at $x \approx 2.6$ .

Answer

The solutions are $x = -2.7$ or $x = 0.1$ or $x = 2.6$ .

Acceptable ranges from the mark scheme:

$-2.8$ to $-2.6$
$0.1$ to $0.2$
$2.4$ to $2.7$

Final answer

x = -2.7, x = 0.1, x = 2.6

Detailed explanation

Walkthrough

Part (c) asks to solve $x^3 - 7x + 1 = 0$ using the graph drawn in part (b). Since we already have the curve $y = x^3 - 3x + 2$ , we need to find a straight line that, when intersected with this curve, gives the solutions to the target equation. We do this by setting the curve equation equal to a line equation $y = mx + c$ and matching terms.

If $y = x^3 - 3x + 2$ and we want to solve $x^3 - 7x + 1 = 0$ , we can subtract the second from the first:
$(x^3 - 3x + 2) - (x^3 - 7x + 1) = y - 0 \implies 4x + 1 = y$ .

So we draw the line $y = 4x + 1$ . We find a few points on this line to draw it accurately: $(0, 1)$ , $(1, 5)$ , $(-1, -3)$ , etc. When we draw this line on the grid, it crosses the cubic curve at three points. The $x$ -coordinates of these intersection points are the solutions to the equation $x^3 - 7x + 1 = 0$ .

Reading from the graph:

The first intersection is near $x = -2.7$ (between $-2.8$ and $-2.6$ ).
The second intersection is near $x = 0.1$ (between $0.1$ and $0.2$ ).
The third intersection is near $x = 2.6$ (between $2.4$ and $2.7$ ).

Key Takeaways

When asked to solve an equation using a given graph, rearrange the equation to express it as the intersection of the given curve and a simple straight line. The $x$ -coordinates of the intersection points are the solutions.

Common Mistakes

Drawing the wrong line. For example, drawing $y = 4x - 1$ or $y = -4x + 1$ . The sign and constant must be correct: $y = 4x + 1$ .
Reading the intersections incorrectly. Ensure you are reading the $x$ -coordinate (horizontal axis) at the point where the line crosses the curve, not the $y$ -coordinate.
Not drawing the line long enough to show all three intersections. The line $y = 4x + 1$ extends from the bottom left to the top right of the grid.

Things to Be Careful About

The mark scheme allows a range of acceptable values for each intersection: $-2.8$ to $-2.6$ , $0.1$ to $0.2$ , and $2.4$ to $2.7$ . This is because graphical reading is inherently approximate. Ensure your line is drawn with a ruler and is truly straight. The mark scheme gives M2 for a ruled line $y = 4x + 1$ , M1 for a short or unruled line, and A2 for three correct values (or A1 for two). If only M1 is scored, SC1 is awarded for three correct solutions. Always use a ruler for the line to maximize method marks.

Techniques used

rearrange the equation to find the intersection of the curve and a straight linedraw the straight line y = 4x + 1 on the gridread the x-coordinates of the intersection points

Q23Medium-EasyNumber

(a)

Simplify.

\sqrt{8} \times \sqrt{6} + \sqrt{3}

______

(b)

Rationalise the denominator and simplify.

\frac{3}{2 - \sqrt{5}}

______

DifficultyMedium-Easy

Worked solution

Approach

Multiply the numerator and denominator by the conjugate $2+\sqrt{5}$ to rationalise the denominator, then simplify.

Working

\frac{3}{2-\sqrt{5}}\times\frac{2+\sqrt{5}}{2+\sqrt{5}}=\frac{3(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})}

Expand the denominator using the difference of two squares:

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

So

\frac{3(2+\sqrt{5})}{-1}=-3(2+\sqrt{5})

Expanding gives the equivalent form

-3(2+\sqrt{5})=-6-3\sqrt{5}

Answer

-3(2+\sqrt{5})\quad\text{or}\quad -6-3\sqrt{5}

Final answer

-3(2 + √5)

Detailed explanation

Walkthrough

To rationalise a denominator of the form $a-\sqrt{b}$ , multiply by the conjugate $a+\sqrt{b}$ . Here the conjugate is $2+\sqrt{5}$ . Multiplying by $\frac{2+\sqrt{5}}{2+\sqrt{5}}$ does not change the value because it is equal to $1$ . The numerator becomes $3(2+\sqrt{5})$ . The denominator is a difference of two squares: $(2-\sqrt{5})(2+\sqrt{5})=4-5=-1$ . Therefore the whole fraction is $\frac{3(2+\sqrt{5})}{-1}=-3(2+\sqrt{5})$ , which may also be written as $-6-3\sqrt{5}$ .

Key Takeaways

Rationalising a denominator removes the surd from the denominator. The conjugate turns the denominator into a difference of two squares, so the surd terms cancel. A negative denominator must be handled by changing all signs in the numerator.

Common Mistakes

Forgetting to multiply the numerator by the same conjugate.
Expanding the denominator incorrectly; the cross terms $2\sqrt{5}$ and $-2\sqrt{5}$ cancel, leaving $4-5=-1$ .
Leaving the answer as $\frac{3(2+\sqrt{5})}{-1}$ without simplifying the division by $-1$ .
Giving a decimal approximation instead of the exact surd form.
Not realising that $-3(2+\sqrt{5})$ , $-(6+3\sqrt{5})$ and $-6-3\sqrt{5}$ are all accepted.

Things to Be Careful About

The mark scheme awards a method mark for multiplying by the conjugate, a mark for the expanded denominator $4+2\sqrt{5}-2\sqrt{5}-5$ , and a final mark for the simplified answer. On the non-calculator paper, show the expansion by hand. Any of the three equivalent final forms is accepted.

Techniques used

multiply by the conjugateexpand a difference of two squaressimplify a fraction with a negative denominator

Q24MediumMensurationNumberAlgebra and Graphs

The diagram shows a rectangle and a triangle.

Fig. 10

The length of the rectangle is $x\,\text{cm}$ and the area of the rectangle is $30\,\text{cm}^2$ .
The triangle has sides of length $(x - 1)\,\text{cm}$ , $(x + 2)\,\text{cm}$ and $(2x - 3)\,\text{cm}$ .

(a)

Write down an expression, in terms of $x$ , for the width of the rectangle.

______ $\text{cm}$

(b)

The perimeter of the rectangle is equal to the perimeter of the triangle.

Form an equation in $x$ and show that it simplifies to $x^2 - x - 30 = 0$ .

DifficultyMedium

Worked solution

Approach

Write an expression for the perimeter of the rectangle using the width found in part (a), and an expression for the perimeter of the triangle by summing its three sides. Equate the two perimeters and simplify to obtain the required quadratic equation.

Working

Perimeter of the rectangle:

P_{\text{rect}} = 2 \left( \text{width} + \text{length} \right) = 2 \left( \frac{30}{x} + x \right)

Perimeter of the triangle:

P_{\text{tri}} = (x - 1) + (x + 2) + (2x - 3) = 4x - 2

Equate the two perimeters:

2 \left( \frac{30}{x} + x \right) = 4x - 2

Expand the left-hand side:

\frac{60}{x} + 2x = 4x - 2

Multiply the entire equation by $x$ to remove the algebraic fraction:

60 + 2x^2 = 4x^2 - 2x

Rearrange all terms to one side:

4x^2 - 2x^2 - 2x - 60 = 0

2x^2 - 2x - 60 = 0

Divide the entire equation by $2$ :

x^2 - x - 30 = 0

This matches the required form.

Answer

x^2 - x - 30 = 0

Final answer

x^2 - x - 30 = 0

Detailed explanation

Walkthrough

First, calculate the perimeter of the rectangle. Using the width $\frac{30}{x}$ and length $x$ , the perimeter is $2\left(\frac{30}{x} + x\right)$ . Next, calculate the perimeter of the triangle by adding its three side lengths: $(x - 1) + (x + 2) + (2x - 3) = 4x - 2$ . Equating these gives $2\left(\frac{30}{x} + x\right) = 4x - 2$ . Expanding the brackets yields $\frac{60}{x} + 2x = 4x - 2$ . To clear the fraction, multiply every term by $x$ , resulting in $60 + 2x^2 = 4x^2 - 2x$ . Rearranging all terms to the right-hand side gives $2x^2 - 2x - 60 = 0$ . Finally, dividing by $2$ simplifies this to $x^2 - x - 30 = 0$ .

Key Takeaways

When forming equations from geometry, carefully write out all perimeter or area expressions before equating them. Clearing algebraic fractions by multiplying through by the denominator is a standard technique to convert rational equations into polynomials.

Common Mistakes

Forgetting to multiply every term by $x$ when clearing the fraction, which leaves a stray $\frac{60}{x}$ term. Expanding $2\left(\frac{30}{x} + x\right)$ incorrectly as $\frac{60}{x} + x$ instead of $\frac{60}{x} + 2x$ . Sign errors when rearranging terms to one side of the equation.

Things to Be Careful About

The mark scheme requires full working to be shown; do not skip steps when clearing the fraction or rearranging the equation. Ensure the final equation is exactly $x^2 - x - 30 = 0$ with no common factors remaining.

Techniques used

write the perimeter of the rectangle as an expressionwrite the perimeter of the triangle as an expressionform an equation equating the two perimeterssimplify the equation to a quadratic

(c)

Solve the equation $x^2 - x - 30 = 0$ .

$x$ = ______ or $x$ = ______

(d)

Work out the perimeter of the rectangle.

______ $\text{cm}$

DifficultyMedium-Easy

Worked solution

Approach

Since $x$ represents a length in the diagram, $x$ must be positive. Discard the negative root $x = -5$ and use $x = 6$ . Substitute this value into the perimeter expression for the rectangle (or the triangle) to find the numerical perimeter.

Working

From part (c), $x = 6$ or $x = -5$ . Since $x$ is a length, $x = 6$ .

Perimeter of the rectangle:

P_{\text{rect}} = 2 \left( \frac{30}{x} + x \right)

Substitute $x = 6$ :

P_{\text{rect}} = 2 \left( \frac{30}{6} + 6 \right)

P_{\text{rect}} = 2 (5 + 6)

P_{\text{rect}} = 2 \times 11 = 22

(Alternatively, using the triangle perimeter: $4(6) - 2 = 24 - 2 = 22$ .)

Answer

22

Final answer

22

Detailed explanation

Walkthrough

The equation in part (b) was based on the geometric shapes, so $x$ must represent a valid physical length. This means $x > 0$ , so we reject $x = -5$ and use $x = 6$ . Substitute $x = 6$ into the perimeter expression for the rectangle: $2\left(\frac{30}{6} + 6\right) = 2(5 + 6) = 2(11) = 22\,\text{cm}$ . You can verify this using the triangle's perimeter expression: $4(6) - 2 = 22\,\text{cm}$ .

Key Takeaways

When solving equations derived from geometric contexts, always check whether both algebraic solutions are physically meaningful. Lengths, areas, and other physical quantities must be positive.

Common Mistakes

Forgetting to reject the negative root $x = -5$ , or substituting $x = -5$ into the perimeter formula, which would give an incorrect or nonsensical result. Calculation errors when evaluating $2(5 + 6)$ .

Things to Be Careful About

Include the unit in the final answer if the question asks for it (the blank is followed by $\text{cm}$ , so just the number is needed, but be aware of units). Ensure you use the correct value of $x$ from the valid physical context.

Techniques used

select the valid positive value of xsubstitute the positive value of x into the perimeter expressioncalculate the numerical perimeter

Q253MMediumAlgebra and Graphs

Factorise.

6x^3 + 5x^2 - 4x

______

DifficultyMedium

Worked solution

Approach

The expression is a cubic polynomial. The first step is to look for a common factor in all three terms. After extracting the common factor, we are left with a quadratic expression which can be factorised by finding two numbers that multiply to give the product of the coefficient of $x^2$ and the constant term, and add to give the coefficient of $x$ .

Working

Step 1: Extract the common factor.
All three terms contain an $x$ . We factor out $x$ :

6x^3 + 5x^2 - 4x = x(6x^2 + 5x - 4)

Step 2: Factorise the quadratic $6x^2 + 5x - 4$ .
We need to split the middle term ( $5x$ ) into two terms. To do this, we multiply the coefficient of $x^2$ (which is $6$ ) by the constant term (which is $-4$ ):

6 \times (-4) = -24

We look for two numbers that multiply to $-24$ and add to $+5$ (the coefficient of the middle term). The pairs of factors of $-24$ are:

$1$ and $-24$ (sum $-23$ )
$-1$ and $24$ (sum $23$ )
$2$ and $-12$ (sum $-10$ )
$-2$ and $12$ (sum $10$ )
$3$ and $-8$ (sum $-5$ )
$-3$ and $8$ (sum $5$ ) ✓

The numbers are $-3$ and $8$ . So we rewrite $5x$ as $-3x + 8x$ :

6x^2 + 5x - 4 = 6x^2 - 3x + 8x - 4

Now we group the terms in pairs and factorise each pair:

= 3x(2x - 1) + 4(2x - 1)

Notice that $(2x - 1)$ is a common bracket. We factor this out:

= (2x - 1)(3x + 4)

Step 3: Combine with the initial common factor.
Don't forget the $x$ we took out at the very beginning:

x(2x - 1)(3x + 4)

Answer

x(2x - 1)(3x + 4)

Final answer

x(2x - 1)(3x + 4)

Detailed explanation

Walkthrough

This question requires factorising a cubic expression. Since there is no single formula for factorising cubics directly, we break the problem down into smaller, standard algebraic tasks.

First, we examine the three terms: $6x^3$ , $5x^2$ , and $-4x$ . We look for a 'common factor'—a number or variable that divides into every term evenly. Here, $x$ is present in all three terms, so we pull it out to the front of the bracket. This simplifies the expression inside the bracket from a degree-3 polynomial to a degree-2 polynomial (a quadratic): $6x^2 + 5x - 4$ . This corresponds to the M1 mark for extracting $x$ .

Next, we must factorise the quadratic $6x^2 + 5x - 4$ . A reliable method for quadratics where the leading coefficient is not 1 is 'splitting the middle term'. We calculate the product of the outer coefficients ( $6$ and $-4$ ), which is $-24$ . We then find two numbers that multiply to $-24$ but add up to the middle coefficient, $5$ . These numbers are $8$ and $-3$ . We replace $5x$ with $8x - 3x$ .

Then, we use 'grouping'. We put the first two terms in one bracket and the last two in another: $(6x^2 - 3x)$ and $(8x - 4)$ . We factorise these individually. From the first group, we take out $3x$ to get $3x(2x - 1)$ . From the second group, we take out $4$ to get $4(2x - 1)$ . Because both groups now share the identical bracket $(2x - 1)$ , we can pull that bracket out as well, leaving $(3x + 4)$ . Finally, we attach the original $x$ to the front.

Key Takeaways

Always check for a common factor first. It makes the remaining polynomial simpler and easier to handle.
Splitting the middle term is a robust technique for factorising quadratics like $ax^2 + bx + c$ when $a \neq 1$ . Find factors of $ac$ that sum to $b$ .
Grouping allows you to reveal hidden common brackets within a four-term polynomial.
Verification: You can check your answer by expanding the brackets. $(2x-1)(3x+4) = 6x^2 + 8x - 3x - 4 = 6x^2 + 5x - 4$ . Multiplying by $x$ gives the original expression.

Common Mistakes

Forgetting the common factor: Students often factorise the quadratic part correctly to get $(2x - 1)(3x + 4)$ but forget to include the $x$ they pulled out at the start. This usually results in only partial credit (M1).
Incorrect splitting: Choosing the wrong pair of factors for $-24$ (e.g., using $3$ and $-8$ which sum to $-5$ instead of $5$ ) leads to incorrect grouping and final brackets.
Sign errors: When factoring out negative numbers or handling the signs during grouping (e.g., writing $-4(-2x + 1)$ instead of $+4(2x - 1)$ ), which can make the common bracket disappear.
Arithmetic errors: Miscalculating the product $6 \times -4$ or the sum of the factors.

Things to Be Careful About

Show your working clearly. The mark scheme awards marks for intermediate steps (M1 for extracting $x$ , M1 for the correct split/grouping setup). Writing just the final answer without showing the split might result in lost method marks if the final answer is wrong.
Non-calculator component: Ensure you perform the multiplication and addition of integers mentally or on paper carefully, as calculators are not permitted.
Final form: Ensure the answer is fully factorised into linear brackets multiplied by the common factor.

Techniques used

extract common factorfactorise quadratic trinomial by splitting the middle termgroup factors to complete factorisation

Q26MediumTransformations and Vectors

Fig. 11

The diagram shows triangle $OAB$ .
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = 4\mathbf{b}$ .
$T$ is a point on $AB$ such that $AT : TB = 3 : 2$ .

(a)

Find $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

$\overrightarrow{AB}$ = ______

(b)

Show that the position vector of $T$ is $\frac{2}{5}(\mathbf{a} + 6\mathbf{b})$ .

DifficultyMedium-Easy

Worked solution

Approach

Point $T$ lies on $AB$ such that $AT : TB = 3 : 2$ . This means $\overrightarrow{AT} = \frac{3}{5}\overrightarrow{AB}$ . We can find the position vector $\overrightarrow{OT}$ by adding $\overrightarrow{OA}$ and $\overrightarrow{AT}$ .

Working

\overrightarrow{AT} = \frac{3}{5}\overrightarrow{AB}

Substitute $\overrightarrow{AB} = 4\mathbf{b} - \mathbf{a}$ from part (a):

\overrightarrow{AT} = \frac{3}{5}(4\mathbf{b} - \mathbf{a})

Now find $\overrightarrow{OT}$ :

\overrightarrow{OT} = \overrightarrow{OA} + \overrightarrow{AT}

\overrightarrow{OT} = \mathbf{a} + \frac{3}{5}(4\mathbf{b} - \mathbf{a})

Expand the bracket:

\overrightarrow{OT} = \mathbf{a} + \frac{12}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}

Collect the $\mathbf{a}$ terms ( $\mathbf{a} - \frac{3}{5}\mathbf{a} = \frac{2}{5}\mathbf{a}$ ):

\overrightarrow{OT} = \frac{2}{5}\mathbf{a} + \frac{12}{5}\mathbf{b}

Factor out $\frac{2}{5}$ :

\overrightarrow{OT} = \frac{2}{5}(\mathbf{a} + 6\mathbf{b})

This matches the required expression.

Answer

Shown as required.

Final answer

Shown as required

Detailed explanation

Walkthrough

Since $T$ divides $AB$ in the ratio $3:2$ , the segment $AT$ is $\frac{3}{3+2} = \frac{3}{5}$ of the total segment $AB$ . Therefore, $\overrightarrow{AT} = \frac{3}{5}\overrightarrow{AB}$ . We already know $\overrightarrow{AB}$ from part (a). To find the position vector of $T$ (which is $\overrightarrow{OT}$ ), we start at $O$ , go to $A$ ( $\overrightarrow{OA} = \mathbf{a}$ ), and then go to $T$ ( $\overrightarrow{AT}$ ). Adding these gives $\overrightarrow{OT} = \mathbf{a} + \frac{3}{5}(4\mathbf{b} - \mathbf{a})$ . Expanding and collecting like terms yields $\frac{2}{5}\mathbf{a} + \frac{12}{5}\mathbf{b}$ , which factors neatly to $\frac{2}{5}(\mathbf{a} + 6\mathbf{b})$ .

Key Takeaways

When a point divides a line segment in a ratio $m:n$ , the vector from the start of the segment to that point is $\frac{m}{m+n}$ of the total vector. Always add this to the position vector of the starting point.

Common Mistakes

Using the wrong fraction (e.g., $\frac{2}{5}$ instead of $\frac{3}{5}$ for $\overrightarrow{AT}$ ).
Algebra errors when collecting $\mathbf{a}$ terms: remembering that $\mathbf{a} = \frac{5}{5}\mathbf{a}$ .
Forgetting to factor out the common fraction at the end to match the required form.

Things to Be Careful About

The question asks to "show that", so you must finish at exactly the printed target $\frac{2}{5}(\mathbf{a} + 6\mathbf{b})$ . Do not start from the target and work backwards. Show all algebraic steps clearly to earn the method marks.

Techniques used

divide a vector in a given ratioadd position vectors using the triangle rule

(c)

$Q$ is a point on $OB$ and $\overrightarrow{QT} = \frac{1}{5}(2\mathbf{a} - 3\mathbf{b})$ .

Find $OQ : QB$ .

______ : ______