Mathematics (Syllabus D) 4024 / 11 — May/June 2024

DifficultyEasy

Worked solution

Approach

We examine each letter in the word $G R A N T$ individually to determine if it possesses a line of symmetry (reflectional symmetry). A line of symmetry divides a shape into two identical halves that are mirror images of each other.

Working

Let's check each letter:

$G$ : The standard capital letter $G$ does not have any line of symmetry. No vertical or horizontal fold produces matching halves.
$R$ : The standard capital letter $R$ does not have any line of symmetry.
$A$ : The capital letter $A$ has one vertical line of symmetry down the centre. If you fold it vertically along the middle, the left side mirrors the right side.
$N$ : The capital letter $N$ does not have a line of symmetry (it has rotational symmetry, but not reflectional symmetry).
$T$ : The capital letter $T$ has one vertical line of symmetry down the centre. Folding it vertically aligns the two halves perfectly.

Thus, the letters with a line of symmetry are $A$ and $T$ .

Answer

A \text{ and } T

Final answer

A and T

Detailed explanation

Walkthrough

The question asks for letters from the word "GRANT" that have a line of symmetry. We look at each letter:

G: Has no line of symmetry.
R: Has no line of symmetry.
A: Has a vertical line of symmetry through the peak and base.
N: Has no line of symmetry (rotational only).
T: Has a vertical line of symmetry through the stem.

Therefore, the correct letters are A and T.

Key Takeaways

Understanding what a line of symmetry is: a line that divides a figure into two congruent parts that are mirror images.
Familiarity with the symmetries of capital letters.

Common Mistakes

Identifying rotational symmetry instead of line symmetry (e.g., choosing N).
Missing one of the symmetric letters.

Things to Be Careful About

Ensure you are looking for line (reflectional) symmetry, not rotational symmetry.
Standard capital letter forms are assumed unless specified otherwise.

Techniques used

identify letters with reflectional symmetry

(b)

rotational symmetry

______

Q22MEasyMensurationNumber

At the start of the day the mass of a bird is $4.628\,\text{kg}$ .
Later in the day the mass of this bird is $4.693\,\text{kg}$ .

Calculate the increase in the mass of the bird.
Give your answer in grams.

______ $\text{g}$

DifficultyEasy

Worked solution

Approach

First, determine the increase in mass in kilograms by subtracting the initial mass from the later mass. Then, convert this difference from kilograms to grams.

Working

\text{Initial mass} = 4.628 \, \text{kg}

\text{Later mass} = 4.693 \, \text{kg}

Calculate the difference:

\begin{aligned} \text{Increase in kg} &= 4.693 - 4.628 \\ &= 0.065 \, \text{kg} \end{aligned}

Convert the increase from kilograms to grams. We know that $1 \, \text{kg} = 1000 \, \text{g}$ .

\text{Increase in g} = 0.065 \times 1000

\text{Increase in g} = 65 \, \text{g}

Alternatively, we could convert both masses to grams first:

4.628 \, \text{kg} = 4628 \, \text{g}

4.693 \, \text{kg} = 4693 \, \text{g}

\text{Increase} = 4693 - 4628 = 65 \, \text{g}

Answer

65

Final answer

65

Detailed explanation

Walkthrough

The question asks for the increase in the bird's mass in grams. The masses are given in kilograms (kg). There are two main ways to solve this: finding the difference in kg first and then converting, or converting to grams first and then finding the difference.

Method 1: Subtract first, then convert

Subtract: Find the difference between the final mass and the initial mass.
$4.693 - 4.628 = 0.065 \, \text{kg}$
Convert: To change kilograms to grams, multiply by 1000 because there are 1000 grams in 1 kilogram.
$0.065 \times 1000 = 65 \, \text{g}$

Method 2: Convert first, then subtract

Convert: Change both masses into grams immediately.
$4.628 \, \text{kg} = 4628 \, \text{g}$
$4.693 \, \text{kg} = 4693 \, \text{g}$
Subtract: Calculate the difference.
$4693 - 4628 = 65 \, \text{g}$

Both methods lead to the same result. Method 2 avoids decimal arithmetic which can sometimes be prone to error, but Method 1 is often quicker if you are comfortable with decimals.

Key Takeaways

Always check the units required for the answer. If the question asks for grams but gives kilograms, a conversion step is necessary.
$1 \, \text{kg} = 1000 \, \text{g}$ . Multiplying by 1000 moves the decimal point three places to the right.
When subtracting decimals, align the decimal points vertically to ensure accuracy.

Common Mistakes

Forgetting to convert the final answer to grams.
Incorrect decimal subtraction (e.g., misaligning digits).
Dividing by 1000 instead of multiplying when converting kg to g.
Rounding errors (though none are needed here as the answer is exact).

Things to Be Careful About

Ensure you read the unit in the question carefully ('grams' not 'kilograms').
In non-calculator papers, showing the multiplication by 1000 clearly helps secure method marks if the final calculation is wrong.
The mark scheme accepts answers where the student converts to grams first ( $4628$ and $4693$ seen), so either approach is valid.

Techniques used

convert mass from kilograms to gramssubtract two decimal numberscalculate the difference between two quantities

Q3EasyNumber

Work out.

(a)

$0.3 \times 0.02$

______

(b)

$15\%$ of $40$

______

(c)

$7 - (-25)$

______

Q4Medium-EasyGeometry

(a)

$ABC$ is a triangle with $AC = 5\,\text{cm}$ and $BC = 10\,\text{cm}$ .

Using a ruler and compasses only, construct triangle $ABC$ .
$AB$ has been drawn for you.

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

To construct triangle $ABC$ with $AC = 5\,\text{cm}$ and $BC = 10\,\text{cm}$ given the base $AB$ , use a compass to draw arcs from $A$ and $B$ with radii equal to the required side lengths. The intersection of these arcs above $AB$ will be vertex $C$ .

Working

Place the compass point on $A$ and set the radius to $5\,\text{cm}$ (the length of $AC$ ). Draw an arc above the line $AB$ .
Place the compass point on $B$ and set the radius to $10\,\text{cm}$ (the length of $BC$ ). Draw another arc above the line $AB$ that intersects the first arc.
Label the intersection point of the two arcs as $C$ .
Use a ruler to draw straight lines $AC$ and $BC$ to complete the triangle.

Answer

Triangle $ABC$ is constructed with vertex $C$ at the intersection of the arc of radius $5\,\text{cm}$ from $A$ and the arc of radius $10\,\text{cm}$ from $B$ .

Final answer

Triangle constructed with vertex C at the intersection of arcs of radius 5 cm from A and 10 cm from B

Detailed explanation

Walkthrough

The question asks for a construction using ruler and compasses only, given two side lengths $AC = 5\,\text{cm}$ and $BC = 10\,\text{cm}$ and the base $AB$ already drawn.

Arc from A: Set the compass to $5\,\text{cm}$ . Place the needle at $A$ and draw an arc. This arc represents all possible locations for $C$ that are exactly $5\,\text{cm}$ from $A$ .
Arc from B: Set the compass to $10\,\text{cm}$ . Place the needle at $B$ and draw an arc that crosses the first arc. This arc represents all possible locations for $C$ that are exactly $10\,\text{cm}$ from $B$ .
Intersection: The point where the two arcs cross is the only point that is simultaneously $5\,\text{cm}$ from $A$ and $10\,\text{cm}$ from $B$ . Label this point $C$ .
Complete the triangle: Draw straight lines from $A$ to $C$ and from $B$ to $C$ using the ruler.

Key Takeaways

In an SSS (side-side-side) construction, two arcs are drawn from the endpoints of the given base line, with radii equal to the other two side lengths.
The intersection of these arcs locates the third vertex.
Construction arcs should be drawn lightly and clearly, as they are part of the working that earns marks.

Common Mistakes

Forgetting to draw the construction arcs; the mark scheme requires "intersecting arcs" to be visible to award the method mark.
Drawing arcs with the wrong radii (e.g., using $10\,\text{cm}$ from $A$ instead of $5\,\text{cm}$ ).
Not extending the arcs sufficiently to ensure a clear intersection point.
Drawing the final triangle lines too thick, obscuring the construction arcs.

Things to Be Careful About

Use a sharp pencil for construction work to ensure precision.
The mark scheme awards B1 for an acceptable vertex clearly indicated with no or incorrect arcs, but full marks require the correct intersecting arcs.
Always draw construction arcs lightly so they can be distinguished from the final triangle edges.

Techniques used

draw arcs from A with radius 5 cmdraw arcs from B with radius 10 cmlocate vertex C at intersection of arcs

(b)

Measure angle $BAC$ .

Angle $BAC$ = ______

DifficultyEasy

Worked solution

Approach

Place the centre of a protractor at vertex $A$ , with the baseline along $AB$ . Read the angle where the line $AC$ crosses the protractor scale.

Working

The angle $BAC$ is measured directly from the constructed triangle. The value will depend on the exact length of $AB$ printed in the original question paper. For example, if $AB = 10\,\text{cm}$ , the theoretical angle is $\cos^{-1}(0.25) \approx 75.5^\circ$ .

Answer

Angle $BAC$ = measured value from construction (accept $75^\circ$ to $77^\circ$ if $AB = 10\,\text{cm}$ )

Final answer

Measured from construction (accept 75° to 77° if AB = 10 cm)

Detailed explanation

Walkthrough

After constructing the triangle in part (a), the candidate must measure angle $BAC$ using a protractor.

Place the centre hole of the protractor exactly on vertex $A$ .
Align the baseline of the protractor with the line segment $AB$ .
Read the angle where the line $AC$ crosses the protractor scale. Ensure you use the correct scale (the one that starts at $0^\circ$ at $AB$ and increases towards $AC$ ).

Key Takeaways

Protractor measurement requires careful alignment of the centre and baseline.
In construction questions, small measurement errors are expected and accepted within a tolerance (usually $\pm 2^\circ$ ).

Common Mistakes

Placing the protractor centre on the wrong point (e.g., on $B$ instead of $A$ ).
Reading the wrong scale on the protractor (inner vs outer).
Not aligning the baseline exactly along $AB$ .

Things to Be Careful About

The mark scheme states "Their angle $BAC \pm 2^\circ$ ", meaning the answer is accepted based on the candidate's own construction. If the construction in part (a) was slightly inaccurate, the measured angle in part (b) will still be marked correct as long as it is within $2^\circ$ of the true value for that construction.
Always state the measured value clearly; do not leave the answer blank.

Techniques used

measure angle with a protractor

Q5EasyStatisticsProbability

The table shows information about a class of 28 students and the distances they live from their school.

	Boys	Girls
Distance of $1\,\text{km}$ or less	11
Distance of more than $1\,\text{km}$	3	6

(a)

Complete the table.

DifficultyEasy

Worked solution

Approach

The table provides information about the distances boys and girls live from school. We know the total number of students is 28. We can calculate the number of girls who live $1\,\text{km}$ or less by subtracting the known counts from the total.

Working

First, let's sum up all the students whose positions are already filled in the table:

Boys living $1\,\text{km}$ or less: $11$
Boys living more than $1\,\text{km}$ : $3$
Girls living more than $1\,\text{km}$ : $6$

Total known students = $11 + 3 + 6 = 20$ .

The total number of students in the class is given as $28$ . Let $x$ be the number of girls living $1\,\text{km}$ or less.

x + 20 = 28

Solving for $x$ :

x = 28 - 20 = 8

So, there are $8$ girls who live $1\,\text{km}$ or less.

Answer

The missing value in the table is $8$ .

	Boys	Girls
Distance of $1\,\text{km}$ or less	11	8
Distance of more than $1\,\text{km}$	3	6

Final answer

8

Detailed explanation

Walkthrough

The problem gives a two-way table with one missing entry. The key piece of information is that the total number of students in the class is 28. A two-way table aggregates data such that the sum of all cells equals the grand total.

Step 1: Identify the known values. We have 11 boys within 1 km, 3 boys further away, and 6 girls further away.
Step 2: Sum these known values. $11 + 3 + 6 = 20$ . This represents all the students we currently know about in the table.
Step 3: Use the grand total to find the missing part. Since the total must be 28, the remaining students (the girls within 1 km) must make up the difference between 20 and 28.
Step 4: Calculate the difference. $28 - 20 = 8$ . This is the number we insert into the table.

Key Takeaways

In any two-way table or contingency table, the sum of all individual category counts must equal the total population size provided. If one cell is missing, you can always find it by subtracting the sum of the other cells from the total.

Common Mistakes

Adding the wrong combination of numbers (e.g., adding only the column totals if they were complete, but here rows/columns are incomplete).
Misreading the total number of students (e.g., using the number of boys instead of the total class size).

Things to Be Careful About

Ensure you include ALL given numbers in your initial sum. In this case, the 'Girls' column was incomplete, so you had to use the row-wise breakdown plus the one known girl count to reach the total.

Techniques used

calculate total number of studentsdetermine missing frequency by subtraction

(b)

A student is chosen at random from the class.

Write down the probability that the student lives more than $1\,\text{km}$ from the school.

______

DifficultyEasy

Worked solution

Approach

The probability of an event is defined as the number of favorable outcomes divided by the total number of possible outcomes. Here, the event is choosing a student who lives more than $1\,\text{km}$ from the school.

Working

Identify the total number of outcomes:
The problem states there are 28 students in the class. So, the denominator is $28$ .
Identify the number of favorable outcomes:
We need the number of students who live more than $1\,\text{km}$ from the school. Looking at the table row for "Distance of more than $1\,\text{km}$ ":

Boys: $3$
Girls: $6$

Total students living more than $1\,\text{km}$ = $3 + 6 = 9$ .

Calculate the probability:

P(\text{more than } 1\,\text{km}) = \frac{\text{Number of students > 1 km}}{\text{Total number of students}} = \frac{9}{28}

The fraction $\frac{9}{28}$ cannot be simplified further because $9 = 3^2$ and $28 = 2^2 \times 7$ , so they share no common factors.

Answer

\frac{9}{28}

Final answer

9/28

Detailed explanation

Walkthrough

Probability is calculated as $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$ .

First, determine the denominator. The question asks for a probability based on choosing a student from the whole class, so the total number of outcomes is the total class size, which is 28.

Next, determine the numerator. We are interested in students who live "more than $1\,\text{km}$ ". We look at the specific row in the table labeled "Distance of more than $1\,\text{km}$ ". This row contains two groups: Boys (3) and Girls (6). To get the total number of favorable students, we add these together: $3 + 6 = 9$ .

Finally, write the ratio as a fraction: $\frac{9}{28}$ . Check if it simplifies. The factors of 9 are 1, 3, 9. None of these divide evenly into 28. Thus, the fraction is in its simplest form.

Key Takeaways

When calculating probability from a table, ensure you sum across the relevant categories (rows or columns) to get the correct count for the event. Do not just pick a single number from the table unless the event refers to exactly that cell.

Common Mistakes

Using only the number of boys or only the number of girls who live far away, ignoring the other gender.
Using the total number of boys or girls as the denominator instead of the total class size.
Failing to simplify the fraction if possible (though $\frac{9}{28}$ is already simple, this is a common error in similar problems).

Things to Be Careful About

Read the condition carefully: "more than $1\,\text{km}$ " excludes those who live $1\,\text{km}$ or less. Make sure you select the correct row from the table.

Techniques used

identify favorable outcomesidentify total outcomesform the probability fraction

Q6EasyAlgebra and Graphs

(a)

Simplify $6r + 7s - r + 3s$ .

______

(b)

Bananas cost $x$ cents each.
Apples cost $y$ cents each.

Write an expression for the total cost of 7 bananas and 5 apples.

______ $\text{cents}$

Q73MMedium-EasyGeometry

Fig. 2

A square and a regular pentagon are joined along one edge as shown in the diagram.

Calculate the value of $x$ .

$x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

The angle $x$ is at a vertex where a square and a regular pentagon meet. The three angles at this vertex — the interior angle of the square, the interior angle of the pentagon, and $x$ — must sum to $360^\circ$ because they are angles around a point. We calculate the interior angles of the square and the regular pentagon, then subtract their sum from $360^\circ$ .

Working

The interior angle of a square is $90^\circ$ .

The interior angle of a regular pentagon with $n = 5$ sides is:

\frac{(n-2) \times 180}{n} = \frac{(5-2) \times 180}{5} = \frac{3 \times 180}{5} = 108^\circ

The angles around the shared vertex sum to $360^\circ$ :

x + 90 + 108 = 360

x + 198 = 360

x = 360 - 198 = 162

Answer

x = 162

Final answer

162

Detailed explanation

Walkthrough

The problem asks for the value of $x$ , which is the angle between a side of a square and a side of a regular pentagon that share a common vertex. At this vertex, three angles meet: the interior angle of the square, the interior angle of the regular pentagon, and the angle $x$ . Because these three angles go all the way around the point, they must add up to $360^\circ$ .

First, we identify the interior angle of the square, which is a standard value of $90^\circ$ .

Next, we find the interior angle of the regular pentagon. A regular pentagon has $n = 5$ sides. The formula for the interior angle of a regular $n$ -sided polygon is $\frac{(n-2) \times 180}{n}$ . Substituting $n = 5$ gives $\frac{3 \times 180}{5} = 108^\circ$ .

Finally, we set up the equation for the angles around the point: $x + 90 + 108 = 360$ . Solving for $x$ gives $x = 360 - 198 = 162$ .

Key Takeaways

The interior angle of a regular $n$ -sided polygon is given by $\frac{(n-2) \times 180}{n}$ .
Angles around a single point always sum to $360^\circ$ .

Common Mistakes

Using the exterior angle formula ( $\frac{360}{n}$ ) and forgetting to subtract it from $180^\circ$ to get the interior angle.
Forgetting that angles around a point sum to $360^\circ$ and instead adding the angles to $180^\circ$ .
Miscalculating the interior angle of the pentagon (e.g., using $\frac{360}{5} = 72^\circ$ as the interior angle instead of the exterior angle).

Things to Be Careful About

Ensure you are using the interior angle of the pentagon ( $108^\circ$ ), not the exterior angle ( $72^\circ$ ), when summing the angles around the vertex.
The diagram is marked NOT TO SCALE, so do not estimate $x$ visually; rely entirely on the geometric properties and calculations.

Techniques used

calculate interior angle of a regular polygonapply angles around a point summing to 360

Q82MEasyNumber

Ahmed invests $4000 at a rate of $1.5\%$ per year simple interest.

Calculate the value of the investment after 2 years.

$ ______

DifficultyEasy

Worked solution

Approach

The problem asks for the total value of an investment after a certain period with simple interest. We use the formula $I = \frac{P \times r \times t}{100}$ to find the interest earned, where $P$ is the principal amount, $r$ is the annual interest rate in percent, and $t$ is the time in years. The final value is the sum of the principal and the interest.

Working

Given:
Principal ( $P$ ) = $4000
Rate ( $r$ ) = $1.5\%$
Time ( $t$ ) = $2$ years

First, calculate the simple interest earned per year or total over the period. Since it is simple interest, the interest is constant each year.

\text{Interest} = 4000 \times \frac{1.5}{100} \times 2

Calculate the yearly interest:

4000 \times \frac{1.5}{100} = 40 \times 1.5 = 60

Multiply by the number of years (2):

60 \times 2 = 120

So, the total interest earned is $120.

Now, add the interest to the original principal to get the final value:

\text{Final Value} = 4000 + 120 = 4120

Answer

4120

Final answer

4120

Detailed explanation

Walkthrough

The problem involves calculating the future value of an investment under simple interest conditions. Simple interest means that the interest is calculated only on the original principal amount, not on any accumulated interest from previous periods.

Identify the variables:
- Principal ( $P$ ): The initial amount invested, which is $4000.
- Rate ( $r$ ): The annual interest rate, which is $1.5\%$ .
- Time ( $t$ ): The duration of the investment, which is $2$ years.
Calculate the Interest:
The formula for simple interest is $I = P \times \frac{r}{100} \times t$ .
Substitute the values: $I = 4000 \times \frac{1.5}{100} \times 2$ .
First, find $1.5\%$ of $4000$ . To do this, divide $4000$ by $100$ to get $40$ , then multiply by $1.5$ . $40 \times 1.5 = 60$ . This is the interest for one year.
Since the investment is for $2$ years, multiply the yearly interest by $2$ : $60 \times 2 = 120$ . So, the total interest is $120.
Calculate the Final Value:
The question asks for the value of the investment, which includes both the original money and the interest earned.
Add the interest to the principal: $4000 + 120 = 4120$ .

Key Takeaways

Simple interest is calculated only on the principal amount.
The formula for simple interest is $I = \frac{Prt}{100}$ .
The total value of the investment is Principal + Interest.
Be careful to distinguish between "interest earned" and "total value".

Common Mistakes

Calculating compound interest instead of simple interest (using $P(1+r/100)^t$ ).
Forgetting to add the interest back to the principal to find the final value.
Arithmetic errors when calculating percentages.

Things to Be Careful About

Ensure you use the correct formula for simple interest, not compound interest.
Check if the question asks for the interest amount or the total accumulated value.
In non-calculator papers, ensure you can perform the percentage calculations manually (e.g., finding $1.5\%$ of $4000$ ).

Techniques used

calculate simple interest using I = Prtadd the interest to the principal

Q9Medium-EasyNumber

(a)

Fig. 3

The diagram shows the distance–time graph of a cyclist for the first two stages of a race, $AB$ and $BC$ .

Work out the average speed of the cyclist for the second stage, $BC$ .

______ $\text{km/h}$

DifficultyMedium-Easy

Worked solution

Approach

Identify the distance travelled and the time taken for stage $BC$ from the graph, then apply the formula $\text{speed} = \frac{\text{distance}}{\text{time}}$ .

Working

From the graph, point $B$ is at $20\,\text{km}$ at $09:00$ and point $C$ is at $26\,\text{km}$ at $09:30$ .

Distance for stage $BC$ :

26 - 20 = 6\,\text{km}

Time for stage $BC$ :

09:30 - 09:00 = 30\,\text{minutes} = 0.5\,\text{hours}

Average speed for stage $BC$ :

\text{Speed} = \frac{6}{0.5} = 12\,\text{km/h}

Answer

12

Final answer

12

Detailed explanation

Walkthrough

The question asks for the average speed during stage $BC$ . Average speed is calculated by dividing the total distance travelled by the total time taken. First, read the coordinates of points $B$ and $C$ from the distance-time graph. Point $B$ is at $20\,\text{km}$ and $09:00$ , while point $C$ is at $26\,\text{km}$ and $09:30$ . The distance covered between $B$ and $C$ is the difference in their vertical coordinates: $26 - 20 = 6\,\text{km}$ . The time taken is the difference in their horizontal coordinates: from $09:00$ to $09:30$ , which is $30$ minutes. Since speed is required in $\text{km/h}$ , convert $30$ minutes to $0.5$ hours. Finally, divide the distance by the time: $6 \div 0.5 = 12\,\text{km/h}$ .

Key Takeaways

The gradient of a distance-time graph represents speed.
Always ensure the units of time and distance are consistent before calculating speed.
Reading values from a graph requires identifying the correct coordinates on both axes.

Common Mistakes

Reading the coordinates of $C$ incorrectly (e.g., using $28\,\text{km}$ instead of $26\,\text{km}$ ), which gives a distance of $8\,\text{km}$ and an incorrect speed of $16\,\text{km/h}$ .
Forgetting to convert minutes to hours, resulting in a speed of $12\,\text{km/min}$ or calculating $6 \div 30 = 0.2$ .
Calculating the speed for the entire journey $AC$ instead of just stage $BC$ .

Things to Be Careful About

The mark scheme awards a method mark for stating either the correct distance ( $6\,\text{km}$ ) or the correct time ( $30$ mins) before applying the formula.
The answer must be in $\text{km/h}$ , so the time in minutes must be converted to hours by dividing by $60$ .
When reading graphs, ensure you are reading the exact values at the intersection of the grid lines.

Techniques used

read distance and time from a distance-time graphconvert minutes to hourscalculate average speed using distance over time

(b)

A bus travels at an average speed of $20\,\text{km/h}$ .

Work out the time the bus takes to travel $50\,\text{km}$ .

______ $\text{hours}$

Q10Medium-EasyNumber

(a)

Fig. 4

The diagram shows the fuel gauge of a car.
This car has 40 litres of fuel in the tank.

Calculate the amount of fuel that the tank contains when it is full.

______ $\text{litres}$

DifficultyMedium-Easy

Worked solution

Approach

The fuel gauge has 8 equal divisions from empty to full. The needle points to the 5th mark from empty, meaning the tank is $\frac{5}{8}$ full. We are given that this amount is $40$ litres. We need to find the total capacity of the full tank.

Working

\frac{5}{8} \times \text{Total Capacity} = 40

To find the total capacity, divide $40$ by $\frac{5}{8}$ (or multiply by the reciprocal $\frac{8}{5}$ ):

\text{Total Capacity} = 40 \div \frac{5}{8}

\text{Total Capacity} = 40 \times \frac{8}{5}

\text{Total Capacity} = \frac{320}{5} = 64

Answer

64

Final answer

64

Detailed explanation

Walkthrough

First, examine the fuel gauge diagram. The scale runs from 'empty' to 'full' and is divided into 8 equal sections. The needle is pointing to the 5th mark from the 'empty' end, which tells us that the tank is currently $\frac{5}{8}$ full. The problem states that the current amount of fuel is $40$ litres. This means $\frac{5}{8}$ of the total tank capacity equals $40$ litres. To find the full capacity, we divide the known amount ( $40$ ) by the fraction it represents ( $\frac{5}{8}$ ), which is the same as multiplying by $\frac{8}{5}$ . Calculating $40 \times \frac{8}{5}$ gives $64$ litres.

Key Takeaways

How to read a fraction from a segmented diagram.
How to find a whole quantity when given a fractional part of it.

Common Mistakes

Counting the divisions incorrectly (e.g., counting from 'full' instead of 'empty', or miscounting the total number of divisions).
Multiplying $40$ by $\frac{5}{8}$ instead of dividing, which would give the amount for $\frac{5}{8}$ of a smaller tank.
Forgetting that the answer must be in litres.

Things to Be Careful About

Always verify the total number of divisions on the gauge; here it is $8$ , not $10$ . The fraction is $\frac{5}{8}$ , not $\frac{5}{10}$ .
The mark scheme requires the exact integer value $64$ ; do not round or add unnecessary decimal places.

Techniques used

read the fraction from the gauge diagramset up a proportion equationsolve for the total capacity

(b)

The car uses $5.4$ litres of fuel for every $100\,\text{km}$ travelled.

Calculate the amount of fuel that the car will use on a journey of $300\,\text{km}$ .

______ $\text{litres}$

Q112MMedium-EasyNumber

Tom’s pet eats $\frac{3}{5}$ of a tin of food each day.
Tom needs food for his pet for 12 days.

Calculate the number of tins of food Tom needs to buy.

______

DifficultyMedium-Easy

Worked solution

Approach

Calculate the total amount of food needed in tins by multiplying the daily consumption by the number of days. Then convert the improper fraction to a whole number or mixed number to determine how many tins must be purchased.

Working

Tom's pet eats $\frac{3}{5}$ of a tin each day. The number of days is $12$ .

Total food required:

\text{Total} = \frac{3}{5} \times 12

Multiply the numerator by $12$ :

\text{Total} = \frac{3 \times 12}{5} = \frac{36}{5}

Convert the improper fraction $\frac{36}{5}$ into a mixed number:

\frac{36}{5} = 7 \frac{1}{5}

This means Tom needs $7.2$ tins of food. Since he cannot buy partial tins, he must buy enough to cover the full amount. Therefore, he needs to buy $8$ tins.

Answer

8

Final answer

8

Detailed explanation

Walkthrough

First, we need to determine the total quantity of food required for the 12-day period. We are given that the pet eats $\frac{3}{5}$ of a tin per day. To find the total, we multiply the daily rate by the number of days:

\text{Total tins} = \frac{3}{5} \times 12

Performing the multiplication involves multiplying the integer $12$ by the numerator $3$ , while keeping the denominator $5$ unchanged:

\frac{3 \times 12}{5} = \frac{36}{5}

Next, we interpret this result. The fraction $\frac{36}{5}$ can be converted to a mixed number or decimal. Dividing $36$ by $5$ gives $7$ with a remainder of $1$ , so:

\frac{36}{5} = 7 \frac{1}{5} = 7.2

Since Tom needs to buy whole tins, and $7$ tins would only provide $7$ tins worth of food (which is less than the required $7.2$ ), he must purchase the next whole number of tins. Thus, he buys $8$ tins.

Key Takeaways

Multiplying a fraction by an integer: Multiply the numerator by the integer and keep the denominator the same ( $\frac{a}{b} \times n = \frac{an}{b}$ ).
Interpreting fractional results in real-world contexts: If the result represents discrete items that cannot be bought in fractions, round up to the next whole number if there is any remainder.

Common Mistakes

Forgetting to multiply the numerator by the days (e.g., writing $\frac{3}{60}$ instead of $\frac{36}{5}$ ).
Rounding incorrectly: Simply rounding $7.2$ down to $7$ would leave the pet short of food. The mark scheme awards M1 for $\frac{36}{5}$ , so showing the exact value is crucial even if the final answer requires rounding up.
Arithmetic errors in $3 \times 12$ .

Things to Be Careful About

Ensure you show the intermediate step $\frac{36}{5}$ as it corresponds to Method Mark 1 (M1) in the marking scheme.
The question asks for the number of tins to buy, implying a practical constraint (you can't buy 0.2 of a tin). Always check if the context requires rounding up, down, or to the nearest whole number.

Techniques used

multiply daily consumption by number of daysevaluate the resulting fraction to a mixed number or decimal

Q123MMedium-EasyAlgebra and Graphs

Solve.

4(2x - 3) + 5(x + 5) = 20

$x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Expand the two sets of brackets, combine the $x$ terms and the constant terms separately to simplify the equation into the form $ax = b$ , then divide by $a$ to find $x$ .

Working

Start with the given equation:

4(2x - 3) + 5(x + 5) = 20

Expand the brackets. Multiply every term inside each bracket by the number outside:

8x - 12 + 5x + 25 = 20

Collect the $x$ terms together ( $8x$ and $5x$ ) and the constant terms together ( $-12$ and $+25$ ):

(8x + 5x) + (-12 + 25) = 20

Simplify the sums:

13x + 13 = 20

Subtract $13$ from both sides to isolate the $x$ term:

13x = 20 - 13

13x = 7

Divide both sides by $13$ :

x = \frac{7}{13}

Answer

x = \frac{7}{13}

Final answer

7/13

Detailed explanation

Walkthrough

The problem asks us to solve a linear equation where the variable $x$ is contained within two different sets of brackets. The first step in such problems is always to remove the brackets (expand). We multiply the terms inside the first bracket $(2x - 3)$ by $4$ , giving $8x - 12$ . Then we multiply the terms inside the second bracket $(x + 5)$ by $5$ , giving $5x + 25$ .

Once expanded, the equation becomes $8x - 12 + 5x + 25 = 20$ . At this stage, we must group 'like terms'. This means putting all terms containing $x$ on one side and all plain numbers (constants) on the other. $8x$ and $5x$ are like terms; adding them gives $13x$ . $-12$ and $+25$ are constants; adding them gives $+13$ . The simplified equation is now $13x + 13 = 20$ .

To find $x$ , we need it to stand alone. Since $13$ is currently being added to $13x$ , we subtract $13$ from both sides of the equals sign. $20 - 13$ leaves $7$ on the right, so we have $13x = 7$ . Finally, because $x$ is multiplied by $13$ , we divide both sides by $13$ to get the final answer.

Key Takeaways

Expansion: Always distribute the multiplier to every term inside the bracket, paying close attention to signs (e.g., $4 \times -3 = -12$ ).
Like Terms: Only add or subtract terms that are identical (e.g., $x$ with $x$ , constants with constants). You cannot add $8x$ and $5$ directly.
Inverse Operations: To isolate a variable, do the opposite of what is happening to it. Addition is undone by subtraction; multiplication is undone by division.

Common Mistakes

Incomplete Expansion: Forgetting to multiply the constant inside the bracket (e.g., writing $4(2x - 3)$ as $8x - 3$ instead of $8x - 12$ ).
Sign Errors: Making mistakes with negative numbers during expansion (e.g., $5 \times -3 = -15$ ) or when collecting constants ( $-12 + 25$ ).
Adding Across the Equals Sign: Adding $13$ to the left side but forgetting to subtract it from the right side.
Incorrect Division: Writing $x = 13/7$ instead of $7/13$ (flipping the numerator and denominator).

Things to Be Careful About

Accuracy: Ensure you expand both brackets correctly before simplifying.
Form: The mark scheme expects the answer as an exact fraction ( $\frac{7}{13}$ ). Do not round this to a decimal unless explicitly asked to do so.
Non-Calculator Component: On Paper 1, arithmetic must be shown clearly. Steps like $20 - 13 = 7$ should be visible.

Techniques used

expand bracketscollect like termssolve the resulting linear equation

Q132MMedium-EasyNumber

Fig. 5

The pie chart shows the proportion of junior members and senior members at a gym.
There are 120 more senior members than junior members.

Calculate the total number of junior and senior members at the gym.

______

DifficultyMedium-Easy

Worked solution

Approach

The total angle in a pie chart is $360^\circ$ . The junior sector is $60^\circ$ , so the senior sector is $300^\circ$ . The difference between these two sectors corresponds to the 120 more senior members than junior members. We can use this to find the number of members per degree, then calculate the total.

Working

Angle for senior members:

360^\circ - 60^\circ = 300^\circ

Difference in angle between senior and junior members:

300^\circ - 60^\circ = 240^\circ

This $240^\circ$ represents 120 members. The number of members per degree is:

\frac{120}{240} = 0.5 \text{ members per } ^\circ

Total number of members (corresponding to $360^\circ$ ):

360 \times 0.5 = 180

Answer

180

Final answer

180

Detailed explanation

Walkthrough

A pie chart represents a whole as $360^\circ$ . The junior sector is given as $60^\circ$ , so the senior sector must be $360^\circ - 60^\circ = 300^\circ$ . The problem states there are 120 more senior members than junior members. The difference in their sector angles is $300^\circ - 60^\circ = 240^\circ$ . This angular difference of $240^\circ$ directly corresponds to the numerical difference of 120 members. By finding how many members each degree represents ( $120 \div 240 = 0.5$ ), we can scale up to the full $360^\circ$ of the pie chart to find the total number of members: $360 \times 0.5 = 180$ .

Key Takeaways

The total angle in a pie chart is always $360^\circ$ .
The ratio of quantities in a pie chart is equal to the ratio of their central angles.
Differences in quantities can be found by taking the difference in their corresponding angles.

Common Mistakes

Subtracting $60^\circ$ from 100 instead of 360 to find the senior angle.
Assuming the $60^\circ$ sector represents 60 members instead of using the ratio.
Forgetting to add the junior and senior members together at the end, giving only the junior or senior count.

Things to Be Careful About

Always remember a full circle is $360^\circ$ , not 100 (which is for percentages).
Ensure you are using the difference in angles ( $240^\circ$ ) to match the difference in numbers (120), not the junior angle ( $60^\circ$ ) to the senior number (120).
The final answer is the TOTAL number of members, not just the number of junior or senior members.

Techniques used

find the angle for the senior sectoruse the difference in angles to find the value per unitcalculate the total from the total angle

Q143MMedium-EasyAlgebra and Graphs

Solve the simultaneous equations.
Show all your working.

\begin{aligned} 2x + 3y &= 7 \\ x - 6y &= 6 \end{aligned}

$x$ = ______
$y$ = ______

DifficultyMedium-Easy

Worked solution

Approach

We have two simultaneous equations:

\begin{aligned} 2x + 3y &= 7 \quad (1) \\ x - 6y &= 6 \quad (2) \end{aligned}

To eliminate $y$ , we can multiply equation (1) by $2$ so that the coefficient of $y$ becomes $+6$ , matching the magnitude in equation (2).

Working

Multiply equation (1) by $2$ :

2(2x + 3y) = 2(7)

4x + 6y = 14 \quad (3)

Now add equation (2) and equation (3) to eliminate $y$ :

(x - 6y) + (4x + 6y) = 6 + 14

5x = 20

Divide by $5$ :

x = 4

Substitute $x = 4$ into equation (2):

4 - 6y = 6

Rearrange to solve for $y$ :

-6y = 6 - 4

-6y = 2

y = -\frac{2}{6} = -\frac{1}{3}

Answer

x = 4, \quad y = -\frac{1}{3}

Final answer

x = 4, y = -1/3

Detailed explanation

Walkthrough

The problem asks us to find the values of $x$ and $y$ that satisfy both equations simultaneously. The most direct method here is elimination.

Aligning coefficients: We look at the coefficients of $y$ . In equation (1), it is $3$ , and in equation (2), it is $-6$ . If we multiply equation (1) by $2$ , the $y$ term becomes $6y$ . This allows us to cancel out the $y$ terms when we add the modified equation (1) to equation (2).
Eliminating: Adding $(x - 6y)$ and $(4x + 6y)$ results in $5x$ . The constants $6$ and $14$ sum to $20$ . This leaves a simple single-variable equation: $5x = 20$ .
Solving: Dividing by $5$ gives $x = 4$ .
Substitution: To find $y$ , we plug $x=4$ back into one of the original equations (equation (2) was chosen as it looked slightly simpler). Solving $4 - 6y = 6$ involves subtracting $4$ from both sides to get $-6y = 2$ , then dividing by $-6$ to get $y = -1/3$ .

Key Takeaways

When solving simultaneous equations by elimination, ensure the coefficients of the variable you want to remove are equal in magnitude but opposite in sign (for addition) or identical in sign (for subtraction).
Always substitute your answer back into an original equation to verify it works (e.g., check with equation (1): $2(4) + 3(-1/3) = 8 - 1 = 7$ ).

Common Mistakes

Arithmetic errors: Forgetting to multiply the entire right-hand side of the equation when scaling (e.g., getting $4x + 6y = 7$ instead of $14$ ).
Sign errors: Incorrectly adding or subtracting negative numbers, especially when dealing with the constant terms or the final division for $y$ .
Incorrect substitution: Substituting the value of $x$ into the wrong equation or making algebraic mistakes during the rearrangement.

Things to Be Careful About

Accuracy: The mark scheme accepts 'oe' (other equivalent) forms like $-2/6$ , but simplifying to $-1/3$ is standard practice.
Working: Show all steps clearly ('nfww' - no follow-through without working). The marks are awarded for the method (M1) and the correct answers (A2). A0 SC1 is available only for correct answers with no working shown.

Techniques used

form a new equation by eliminating one variablesolve the resulting linear equationsubstitute back to find the second variable

Q152MMedium-EasyNumber

By writing each number correct to one significant figure, estimate the value of

\frac{2.87 \times \sqrt{396.5}}{1.92^2}

______

DifficultyMedium-Easy

Worked solution

Approach

To estimate the value of the expression $\frac{2.87 \times \sqrt{396.5}}{1.92^2}$ , we first round each number in the expression to one significant figure. Then, we substitute these rounded values back into the expression and calculate the result.

Working

Step 1: Round each number to one significant figure.

The number $2.87$ starts with the digit 2. The next digit is 8, so we round up. To one significant figure, $2.87 \approx 3$ .
The number $396.5$ starts with the digit 3. The next digit is 9, so we round up. To one significant figure, $396.5 \approx 400$ .
The number $1.92$ starts with the digit 1. The next digit is 9, so we round up. To one significant figure, $1.92 \approx 2$ .

Step 2: Substitute the rounded values into the original expression.
The expression becomes:

\frac{3 \times \sqrt{400}}{2^2}

Step 3: Evaluate the components.

Calculate the square root: $\sqrt{400} = 20$ .
Calculate the square: $2^2 = 4$ .

Step 4: Perform the multiplication and division.
Substitute these values back into the fraction:

\frac{3 \times 20}{4}

Calculate the numerator:

3 \times 20 = 60

Now divide by the denominator:

\frac{60}{4} = 15

Answer

The estimated value is 15.

15

Final answer

15

Detailed explanation

Walkthrough

The problem asks for an estimate of a numerical expression. Estimation in mathematics often involves simplifying complex or precise numbers into easier-to-handle values to get a quick approximate answer. The standard method here is to round each component to one significant figure.

First, look at the numerator's factors:

$2.87$ : We keep only the first non-zero digit (2) and look at the second digit (8). Since 8 is 5 or greater, we round the 2 up to 3. So, $2.87 \rightarrow 3$ .
$\sqrt{396.5}$ : We cannot easily take the square root of 396.5 mentally. Instead, we first round the number inside the square root. The first digit is 3, and the second is 9. Rounding 396.5 to one significant figure gives 400. Now, taking the square root is easy because 400 is a perfect square ( $20^2 = 400$ ). So, $\sqrt{396.5} \approx \sqrt{400} = 20$ .

Next, look at the denominator:
3. $1.92^2$ : We first round the base number 1.92. The first digit is 1, and the second is 9. Rounding to one significant figure gives 2. Then we square this rounded value: $2^2 = 4$ .

Finally, put it all together into the fraction:

\frac{3 \times 20}{4}

Multiply the top numbers: $3 \times 20 = 60$ . Divide by the bottom number: $60 \div 4 = 15$ .

The mark scheme awards marks for identifying the correct rounded values (B1 for two out of three correct rounds) and the final calculated estimate (A1 for 15).

Key Takeaways

One Significant Figure: This means keeping only the first non-zero digit and rounding the rest based on the next digit. Zeros used as placeholders (like in 400) are not counted as significant figures in this context; they just show the magnitude.
Estimation Strategy: Simplify the numbers before doing the operations. It is much easier to calculate $\sqrt{400}$ than $\sqrt{396.5}$ .
Order of Operations: When estimating a power like $1.92^2$ , you can either round the base first ( $2^2=4$ ) or evaluate then round ( $1.92^2 \approx 3.68 \rightarrow 4$ ). Both lead to the same result here, but rounding the base is generally the intended method for simple estimation unless specified otherwise.

Common Mistakes

Incorrect Rounding: Rounding 2.87 to 2 instead of 3, or 1.92 to 1 instead of 2. Remember that if the digit after the significant figure is 5 or more, round up.
Rounding the Square Root: Trying to round $\sqrt{396.5}$ directly. You must round the radicand (the number inside) first to find a compatible perfect square.
Calculation Errors: Making arithmetic mistakes when multiplying or dividing the simplified numbers (e.g., calculating $60/4$ incorrectly).
Using Too Many Figures: Carrying forward too many decimal places defeats the purpose of estimation and may lead to a different answer than expected by the "one significant figure" rule.

Things to Be Careful About

Significant Figures vs Decimal Places: Ensure you are rounding to significant figures, not decimal places. For example, 2.87 to one decimal place is 2.9, but to one significant figure is 3.
Placeholder Zeros: When rounding 396.5 to one significant figure, the answer is 400, not 4. The zeros are crucial to maintain the correct order of magnitude.
Mark Scheme Specifics: The mark scheme explicitly looks for the intermediate rounded values ( $3$ , $400$ , $2$ ). Showing these steps clearly is important for earning partial credit (B1) even if the final calculation is wrong.

Techniques used

round numbers to one significant figureestimate square roots of rounded valuescalculate the estimated value using rounded numbers

Q16MediumGeometry

Fig. 6

$A, B, C$ and $D$ are points on the circumference of a circle centre $O$ .
Angle $BCD = 58^\circ$ and angle $DBC = 72^\circ$ .

(a)

Complete the statement below.

Angle $DAB$ = ______ because ______

DifficultyMedium-Easy

Worked solution

Approach

Points $A$ , $B$ , $C$ , and $D$ all lie on the circumference of the circle, making $ABCD$ a cyclic quadrilateral. In a cyclic quadrilateral, opposite angles are supplementary.

Working

\angle DAB + \angle BCD = 180^\circ

Substitute the given value $\angle BCD = 58^\circ$ :

\angle DAB + 58^\circ = 180^\circ

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

Answer

Angle $DAB = 122^\circ$ because angles in opposite segments are supplementary.

Final answer

122°, angles in opposite segments are supplementary

Detailed explanation

Walkthrough

Since all four points $A$ , $B$ , $C$ , and $D$ lie on the circle, the quadrilateral $ABCD$ is cyclic. A fundamental circle theorem states that opposite angles in a cyclic quadrilateral are supplementary, meaning they add up to $180^\circ$ . The angles $\angle DAB$ and $\angle BCD$ are opposite angles. We are given $\angle BCD = 58^\circ$ , so we subtract this from $180^\circ$ to find $\angle DAB$ .

Key Takeaways

A cyclic quadrilateral is a four-sided shape whose vertices all lie on a single circle.
Opposite angles in a cyclic quadrilateral always sum to $180^\circ$ .

Common Mistakes

Forgetting that the angles must sum to $180^\circ$ and instead adding or subtracting incorrectly.
Using the wrong pair of angles; only diagonally opposite angles (like $\angle DAB$ and $\angle BCD$ , or $\angle ABC$ and $\angle ADC$ ) are supplementary.

Things to Be Careful About

The mark scheme accepts the reason as "angles in opposite segments are supplementary". Ensure the reason is stated clearly and matches the theorem exactly. No working is required for this part, but the calculation must be correct.

Techniques used

identify a cyclic quadrilateralapply the opposite angles supplementary theorem

(b)

3M

(i)

Find angle $DOC$ .

Angle $DOC$ = ______

(ii)

Find angle $BCO$ .

Angle $BCO$ = ______

DifficultyMedium

Worked solution

Approach

Triangle $OCD$ is formed by two radii $OC$ and $OD$ , making it an isosceles triangle. We can find the base angles $\angle OCD$ and $\angle ODC$ using the sum of angles in a triangle. Then, $\angle BCO$ is found by subtracting $\angle OCD$ from the given $\angle BCD$ .

Working

In triangle $OCD$ , $OC = OD$ (radii of the circle), so $\angle OCD = \angle ODC$ .

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

\angle OCD + \angle ODC + \angle DOC = 180^\circ

2 \times \angle OCD + 144^\circ = 180^\circ

2 \times \angle OCD = 180^\circ - 144^\circ = 36^\circ

\angle OCD = 18^\circ

We are given $\angle BCD = 58^\circ$ . From the diagram, $\angle BCD = \angle BCO + \angle OCD$ :

\angle BCO = \angle BCD - \angle OCD

\angle BCO = 58^\circ - 18^\circ = 40^\circ

Answer

Angle $BCO = 40^\circ$ .

Final answer

40

Detailed explanation

Walkthrough

First, look at triangle $OCD$ . Since $O$ is the centre and $C$ and $D$ are on the circumference, $OC$ and $OD$ are both radii. This makes triangle $OCD$ isosceles, meaning the base angles $\angle OCD$ and $\angle ODC$ are equal. We know the angle at the top, $\angle DOC = 144^\circ$ (from part b(i)). Subtract this from $180^\circ$ and divide by $2$ to find $\angle OCD = 18^\circ$ . Finally, the given angle $\angle BCD$ is split into $\angle BCO$ and $\angle OCD$ . Subtract $18^\circ$ from $58^\circ$ to get $\angle BCO = 40^\circ$ .

Key Takeaways

Lines from the centre to the circumference are radii and are equal in length, creating isosceles triangles.
Base angles of an isosceles triangle are equal, and the sum of angles in any triangle is $180^\circ$ .
Angles can be split into parts and subtracted to find unknown angles.

Common Mistakes

Forgetting that $OC = OD$ and trying to solve triangle $OCD$ without recognizing it is isosceles.
Subtracting the wrong angle or adding instead of subtracting when splitting $\angle BCD$ .

Things to Be Careful About

Follow through with the value of $\angle DOC$ from part b(i). If a candidate gets $\angle DOC$ wrong in part b(i), they can still earn method marks for correctly calculating $\angle OCD$ using their incorrect value. The mark scheme awards $M1$ for the correct method: $(180 - \text{their } 144) / 2$ .

Techniques used

recognize an isosceles triangle formed by radiicalculate base angles of an isosceles trianglesubtract angles to find the required angle

Q17MediumStatistics

The table shows the time that each of 60 children spends in a play area.

Time ( $t$ mins)	$0 < t \leq 10$	$10 < t \leq 40$	$40 < t \leq 60$	$60 < t \leq 90$	$90 < t \leq 120$
Frequency	4	7	8	24	17

(a)

Draw a cumulative frequency diagram to show this information.

Fig. 7

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

To draw a cumulative frequency diagram, first calculate the cumulative frequency at the upper boundary of each class interval by adding frequencies successively. Then plot these points on the given grid and join them with a smooth increasing curve.

Working

Step 1: Calculate cumulative frequencies

Time ( $t$ mins)	Upper boundary	Frequency	Cumulative frequency
$0 < t \leq 10$	10	4	4
$10 < t \leq 40$	40	7	4 + 7 = 11
$40 < t \leq 60$	60	8	11 + 8 = 19
$60 < t \leq 90$	90	24	19 + 24 = 43
$90 < t \leq 120$	120	17	43 + 17 = 60

The curve should also start at $(0, 0)$ since no children spend 0 minutes in the play area.

Step 2: Plot points and draw the curve

Plot the following coordinates on the grid:

(0, 0), \quad (10, 4), \quad (40, 11), \quad (60, 19), \quad (90, 43), \quad (120, 60)

Join these points with a smooth, increasing curve (an ogive). Ensure the curve does not have any sharp corners or decrease at any point.

Answer

A cumulative frequency diagram with the points $(0, 0)$ , $(10, 4)$ , $(40, 11)$ , $(60, 19)$ , $(90, 43)$ , $(120, 60)$ plotted and joined by a smooth increasing curve.

Final answer

Cumulative frequency curve through (0,0), (10,4), (40,11), (60,19), (90,43), (120,60)

Detailed explanation

Walkthrough

The cumulative frequency diagram is built by accumulating the frequencies from the lowest class to the highest. We start at 0 cumulative frequency at time 0. At the end of the first class ( $t=10$ ), 4 children have finished, so we plot $(10, 4)$ . At $t=40$ , we add the next 7 children, giving $4+7=11$ , so we plot $(40, 11)$ . Continuing this process: at $t=60$ , $11+8=19$ ; at $t=90$ , $19+24=43$ ; at $t=120$ , $43+17=60$ . These five points, plus the origin $(0,0)$ , are plotted on the grid. A smooth curve is drawn through them to represent the cumulative distribution.

Key Takeaways

Cumulative frequency is the running total of frequencies up to a given value.
Points are plotted at the upper class boundary against the cumulative frequency.
The curve must be smooth and always increasing (non-decreasing).

Common Mistakes

Plotting at the lower class boundary instead of the upper class boundary.
Forgetting to start the curve at $(0, 0)$ .
Drawing a straight line graph (polygon) instead of a smooth curve.
Making the curve decrease at any point.
Misreading the cumulative frequency axis (e.g., reading frequency instead of cumulative frequency).

Things to Be Careful About

Ensure the curve is smooth; sharp corners at the plotted points are incorrect.
The final cumulative frequency must match the total number of children ( $60$ ).
When reading values from the diagram later, use horizontal lines from the cumulative frequency axis and vertical lines down to the time axis.

Techniques used

calculate cumulative frequencies from grouped dataplot upper class boundaries against cumulative frequenciesdraw a smooth increasing cumulative frequency curve (ogive)

(b)

Use your diagram to estimate

5M

(i)

the median

______ $\text{minutes}$

DifficultyMedium-Easy

Worked solution

Approach

The median is the value at the middle of the data set. For $N = 60$ children, the median position is $\frac{N}{2} = 30$ . Draw a horizontal line from cumulative frequency 30 to the curve, then read the corresponding time on the horizontal axis.

Working

Median position:

\frac{60}{2} = 30

Locate $30$ on the cumulative frequency axis. Move horizontally to the curve and then vertically down to the time axis.

From the diagram, at cumulative frequency $30$ , the time $t$ is approximately $74$ minutes.

(Interpolation check: between $t=60$ (CF=19) and $t=90$ (CF=43), the CF range is $24$ over $30$ minutes. We need $30 - 19 = 11$ more cumulative frequency. $\frac{11}{24} \times 30 = 13.75$ . So $t = 60 + 13.75 = 73.75 \approx 74$ .)

Answer

74 \text{ minutes}

(Accept any value between 73 and 75 minutes based on diagram reading.)

Final answer

74

Detailed explanation

Walkthrough

The median splits the data into two equal halves. With 60 children total, the median is at the 30th child. On the cumulative frequency graph, we find 30 on the vertical axis, draw a horizontal line to intersect the curve, and then drop a vertical line to the horizontal axis to read the time. Reading from the curve, this falls at approximately 74 minutes.

Key Takeaways

The median position for $N$ data points is $\frac{N}{2}$ .
Read the median value from the cumulative frequency curve by finding the time corresponding to the median cumulative frequency.

Common Mistakes

Using the wrong position for the median (e.g., using $N$ instead of $\frac{N}{2}$ ).
Reading the cumulative frequency value instead of the time value at the end of the horizontal line.
Not allowing for the natural variation in diagram reading (mark schemes accept a range).

Things to Be Careful About

The question asks to 'estimate' using the diagram, so an exact interpolated value is not required, but must be consistent with the drawn curve.
Ensure the reading is taken from the correct axis (time, not cumulative frequency).

Techniques used

determine the median position from total frequencyread the median value from the cumulative frequency curve

(ii)

the interquartile range

______ $\text{minutes}$

DifficultyMedium

Worked solution

Approach

The interquartile range (IQR) is the difference between the upper quartile (UQ) and the lower quartile (LQ). For $N = 60$ , the LQ is at position $\frac{N}{4} = 15$ and the UQ is at position $\frac{3N}{4} = 45$ . Read these values from the curve and subtract.

Working

Lower Quartile (LQ):

Position: $\frac{60}{4} = 15$

Locate 15 on the cumulative frequency axis. Move horizontally to the curve and vertically down to the time axis.

At CF = 15, the time is $t = 50$ minutes.
(Interpolation check: between $t=40$ (CF=11) and $t=60$ (CF=19), range is 8 over 20 mins. Need $15-11=4$ . $\frac{4}{8} \times 20 = 10$ . $t = 40 + 10 = 50$ .)

Upper Quartile (UQ):

Position: $\frac{3 \times 60}{4} = 45$

Locate 45 on the cumulative frequency axis. Move horizontally to the curve and vertically down to the time axis.

At CF = 45, the time is approximately $t = 93.5$ minutes.
(Interpolation check: between $t=90$ (CF=43) and $t=120$ (CF=60), range is 17 over 30 mins. Need $45-43=2$ . $\frac{2}{17} \times 30 \approx 3.53$ . $t = 90 + 3.53 = 93.53$ .)

Interquartile Range:

\text{IQR} = \text{UQ} - \text{LQ} = 93.5 - 50 = 43.5 \text{ minutes}

Answer

43.5 \text{ minutes}

(Accept any value between 42 and 45 minutes based on diagram reading.)

Final answer

43.5

Detailed explanation

Walkthrough

The interquartile range measures the spread of the middle 50% of the data. We find the lower quartile at the 15th value (25% of 60) and the upper quartile at the 45th value (75% of 60). Reading from the cumulative frequency curve, the 15th value occurs at 50 minutes, and the 45th value occurs at approximately 93.5 minutes. Subtracting the lower quartile from the upper quartile gives the IQR.

Key Takeaways

LQ position is $\frac{N}{4}$ and UQ position is $\frac{3N}{4}$ .
IQR is calculated as UQ minus LQ.
Both quartiles must be read from the same cumulative frequency curve.

Common Mistakes

Confusing the positions of LQ and UQ.
Reading the cumulative frequency value instead of the time value.
Forgetting to subtract LQ from UQ (giving just one quartile value).
Using incorrect quartile positions (e.g., using $\frac{N}{2}$ for LQ).

Things to Be Careful About

Allow for diagram reading variation; mark schemes accept a range (typically $\pm 1$ or $\pm 2$ minutes).
Ensure follow-through marks are applied if earlier readings are incorrect but consistent with the drawn curve.

Techniques used

determine lower and upper quartile positionsread quartile values from the cumulative frequency curvecalculate the interquartile range as the difference

(iii)

the number of children who spend more than 80 minutes in the play area.

______

DifficultyMedium

Worked solution

Approach

To find the number of children who spend more than 80 minutes, first read the cumulative frequency at $t = 80$ minutes from the diagram. Then subtract this value from the total number of children (60).

Working

Step 1: Read cumulative frequency at $t = 80$

Locate 80 on the time axis. Move vertically up to the curve and horizontally to the cumulative frequency axis.

At $t = 80$ , the cumulative frequency is approximately 35.
(Interpolation check: between $t=60$ (CF=19) and $t=90$ (CF=43), range is 24 over 30 mins. From 60 to 80 is 20 mins. $\frac{20}{30} \times 24 = 16$ . CF = $19 + 16 = 35$ .)

Step 2: Calculate number of children > 80 minutes

Total children = 60

\text{Number} > 80 \text{ mins} = 60 - 35 = 25

Answer

25 \text{ children}

(Accept any value between 23 and 27 based on diagram reading.)

Final answer

25

Detailed explanation

Walkthrough

The cumulative frequency at a given time tells us how many children spent up to that time in the play area. To find how many spent more than 80 minutes, we find the cumulative frequency at 80 minutes and subtract it from the total number of children. Reading from the graph at $t=80$ , the cumulative frequency is about 35. Subtracting this from 60 gives 25 children who spent more than 80 minutes.

Key Takeaways

Cumulative frequency at a value gives the count of observations less than or equal to that value.
To find the number greater than a value, subtract the cumulative frequency at that value from the total frequency.

Common Mistakes

Reading the cumulative frequency and using it as the final answer (forgetting to subtract from the total).
Reading the wrong axis (reading time from the vertical axis or cumulative frequency from the horizontal axis).
Misreading the graph at $t=80$ (e.g., reading 30 or 40 instead of 35).

Things to Be Careful About

The question asks for the number of children, not a percentage or proportion.
Ensure the reading at $t=80$ is consistent with the curve drawn in part (a). Mark schemes apply follow-through if the reading is consistent with the student's own curve.

Techniques used

read cumulative frequency at a specific time valuesubtract from total frequency to find the number above that value

Q183MMedium-EasyAlgebra and Graphs

In this table, $p$ is directly proportional to $q^2$ .

$p$	12	$a$	48
$q$	2	7	$b$

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

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Since $p$ is directly proportional to $q^2$ , we can write the relationship as $p = kq^2$ for some constant $k$ . We use the first column of values ( $p=12, q=2$ ) to calculate $k$ , then substitute $k$ into the equation to find $a$ (when $q=7$ ) and $b$ (when $p=48$ ).

Working

Step 1: Find the constant of proportionality, $k$
$p = kq^2$
Substitute $p = 12$ and $q = 2$ :
$12 = k(2)^2$
$12 = 4k$
$k = \frac{12}{4} = 3$
So the formula connecting $p$ and $q$ is $p = 3q^2$ .

Step 2: Calculate $a$
In the second column, $p = a$ and $q = 7$ . Substitute these into the formula:
$a = 3 \times (7)^2$
$a = 3 \times 49$
$a = 147$

Step 3: Calculate $b$
In the third column, $p = 48$ and $q = b$ . Substitute these into the formula:
$48 = 3(b)^2$
Divide both sides by 3:
$16 = b^2$
Take the square root of both sides:
$b = \pm \sqrt{16}$
$b = \pm 4$
(Note: In the context of many syllabi, the positive root is often accepted unless specified otherwise, but mathematically both are valid. The mark scheme accepts $\pm 4$ or just $4$ .)

Answer

$a = 147$
$b = 4$ (or $\pm 4$ )

Final answer

a = 147, b = 4

Detailed explanation

Walkthrough

The question states that $p$ is directly proportional to $q^2$ . This means there is a constant multiplier $k$ such that $p = kq^2$ .

First, we need to find this constant $k$ . We look at the table for a complete row where both $p$ and $q$ are known numbers. The first column gives $p = 12$ when $q = 2$ . Substituting these into our formula:
$12 = k \times 2^2$
$12 = 4k$
Dividing by 4 gives $k = 3$ . Now we know the specific rule for this table is $p = 3q^2$ .

Next, we find $a$ . The second column tells us that when $q = 7$ , $p = a$ . Using our rule:
$a = 3 \times 7^2$
$a = 3 \times 49$
$a = 147$

Finally, we find $b$ . The third column tells us that when $p = 48$ , $q = b$ . Using the rule again:
$48 = 3 \times b^2$
Divide by 3:
$16 = b^2$
Square root both sides:
$b = 4$
(Strictly $b = \pm 4$ , but typically the positive value is the expected answer in this context unless domain restrictions apply).

Key Takeaways

Direct Proportion: If $y$ is directly proportional to $x^n$ , write $y = kx^n$ .
Finding $k$ : Always use a complete pair of values from the data to solve for $k$ .
Using the Formula: Once you have the formula with the known $k$ , substitute any other given value to find the unknown.

Common Mistakes

Incorrect Proportionality Statement: Writing $p = kq$ instead of $p = kq^2$ . This leads to wrong values for everything.
Order of Operations: For calculating $a$ , forgetting to square $q$ before multiplying by $k$ (e.g., doing $3 \times 7$ instead of $3 \times 7^2$ ).
Arithmetic Errors: Multiplication errors like $3 \times 49$ or square root errors for $b$ .
Ignoring the Square Root: Forgetting that $b^2 = 16$ implies $b$ could be negative, though usually the positive root is sufficient in basic tables.

Things to Be Careful About

Ensure you square the $q$ value before multiplying by $k$ .
When solving for $b$ from $b^2 = 16$ , remember to take the square root.
Check your arithmetic carefully, especially large multiplications like $3 \times 49$ .

Techniques used

identify the constant of proportionality from a known pairuse the constant to find missing values in a proportional relationship

Q19MediumAlgebra and Graphs

$f(x) = 2x - 5$

(a)

Find $f(11)$ .

______

(b)

Find $f^{-1}(x)$ .

$f^{-1}(x)$ = ______

DifficultyMedium-Easy

Worked solution

Approach

To find the inverse function $f^{-1}(x)$ , we follow these steps:

Replace $f(x)$ with $y$ .
Swap the roles of $x$ and $y$ (replace $x$ with $y$ and $y$ with $x$ ).
Rearrange the resulting equation to make $y$ the subject.
Replace $y$ with $f^{-1}(x)$ .

Working

Step 1: Write $f(x)$ as $y$ .

y = 2x - 5

Step 2: Swap $x$ and $y$ .

x = 2y - 5

Step 3: Make $y$ the subject.
Add 5 to both sides:

x + 5 = 2y

Divide by 2:

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

Step 4: Write in inverse function notation.

f^{-1}(x) = \frac{x + 5}{2}

(Note: This can also be written as $\frac{1}{2}x + \frac{5}{2}$ )

Answer

f^{-1}(x) = \frac{x + 5}{2}

Final answer

(x + 5)/2

Detailed explanation

Walkthrough

An inverse function reverses the operation of the original function. If $f$ maps $x$ to $y$ , then $f^{-1}$ maps $y$ back to $x$ . The standard algebraic method is to treat $f(x)$ as $y$ , swap $x$ and $y$ to reflect the reversal of input/output, and then solve for the new $y$ .
Starting with $y = 2x - 5$ , swapping gives $x = 2y - 5$ . We then isolate $y$ by adding 5 to get $x+5 = 2y$ , and dividing by 2 to get $y = \frac{x+5}{2}$ . Finally, we label this result $f^{-1}(x)$ .

Key Takeaways

The inverse function $f^{-1}(x)$ undoes what $f(x)$ does.
Swapping $x$ and $y$ is the key geometric/algebraic step that reflects the function across the line $y=x$ .

Common Mistakes

Forgetting to swap $x$ and $y$ . Just solving $y = 2x - 5$ for $x$ without swapping first yields $x = \frac{y+5}{2}$ , which looks similar but isn't formally derived via the standard variable swap method (though mathematically equivalent if you just rename variables at the end). The mark scheme explicitly awards a mark for seeing $x = 2y - 5$ or equivalent intermediate forms.
Incorrectly isolating $y$ . For example, dividing only the 2x by 2 and forgetting the 5.

Things to Be Careful About

The final answer must be in terms of $x$ . Ensure you replace $y$ with $f^{-1}(x)$ at the very end. The form $\frac{x+5}{2}$ is preferred but $\frac{1}{2}x + \frac{5}{2}$ is also acceptable (oe).

Techniques used

swap x and y variablesrearrange a linear equation to isolate the new x

(c)

Solve $f(x) = x^2 + x - 11$ .

$x$ = ______ or $x$ = ______

3M Similar

DifficultyMedium

Worked solution

Approach

We are asked to solve $f(x) = x^2 + x - 11$ . Since $f(x) = 2x - 5$ , we substitute this expression into the equation, collect all terms on one side to form a quadratic equation equal to zero, and then solve for $x$ .

Working

Set the expressions equal:

2x - 5 = x^2 + x - 11

Rearrange to form a standard quadratic equation $ax^2 + bx + c = 0$ . Subtract $2x$ and add 5 to both sides:

0 = x^2 + x - 2x - 11 + 5

Simplify:

x^2 - x - 6 = 0

Factorise the quadratic. We need two numbers that multiply to $-6$ and add to $-1$ . These numbers are $-3$ and $2$ .

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

Set each factor to zero to find the solutions:

x - 3 = 0 \implies x = 3

x + 2 = 0 \implies x = -2

Answer

x = 3 \text{ or } x = -2

Final answer

3 or -2

Detailed explanation

Walkthrough

This question combines function evaluation with solving quadratic equations. First, we use the definition of $f(x)$ to create an equation involving only $x$ : $2x - 5 = x^2 + x - 11$ . The goal is to solve for $x$ . We move all terms to one side to set the equation to zero. Subtracting $2x$ and adding 5 from both sides simplifies the left-hand side to 0 and transforms the right-hand side into $x^2 - x - 6$ . Now we have a standard quadratic equation. We factorise $x^2 - x - 6$ by finding factors of -6 that sum to -1, which are -3 and +2. Setting each bracket to zero gives the two possible values for $x$ .

Key Takeaways

When solving $f(x) = g(x)$ , substitute the expression for $f(x)$ and rearrange into a polynomial equation.
Always check your sign changes when moving terms across the equals sign.
Factorisation requires finding two numbers with the correct product and sum.

Common Mistakes

Errors in rearranging the equation (e.g., getting $x^2 + 3x - 16 = 0$ ).
Incorrect factorisation (e.g., signs mixed up like $(x+3)(x-2)$ ).
Forgetting the negative root ( $x=-2$ ).

Things to Be Careful About

The mark scheme awards marks for the correct quadratic form ( $x^2 - x - 6 = 0$ ) and for the method used to solve their quadratic. Ensure the final equation is set to 0 before factorising. The answers are exact integers, so no rounding is needed.

Techniques used

form a quadratic equation by equating two functionsfactorise a quadratic expressionsolve for roots

Q202MMedium-Easy[Legacy] Matrices

The matrix $\mathbf{N}$ satisfies the equation

3\mathbf{N} = \mathbf{N} + 5 \begin{pmatrix} 4 & 0 \\ 6 & -2 \end{pmatrix}

Find $\mathbf{N}$ .

$\mathbf{N}$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Treat the matrix equation like an ordinary linear equation: collect all $\mathbf{N}$ terms on one side, then divide by the scalar coefficient to find $\mathbf{N}$ .

Working

Start with

3\mathbf{N} = \mathbf{N} + 5 \begin{pmatrix} 4 & 0 \\ 6 & -2 \end{pmatrix}

Subtract $\mathbf{N}$ from both sides:

2\mathbf{N} = 5 \begin{pmatrix} 4 & 0 \\ 6 & -2 \end{pmatrix}

Multiply the matrix by $5$ :

2\mathbf{N} = \begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix}

Divide both sides by $2$ :

\mathbf{N} = \frac{1}{2} \begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix} = \begin{pmatrix} 10 & 0 \\ 15 & -5 \end{pmatrix}

Answer

\mathbf{N} = \begin{pmatrix} 10 & 0 \\ 15 & -5 \end{pmatrix}

Final answer

[[10, 0], [15, -5]]

Detailed explanation

Walkthrough

The equation is a matrix equation, but the same rules of algebra apply as for numbers. We want to isolate $\mathbf{N}$ .

First, subtract $\mathbf{N}$ from both sides. This removes it from the right-hand side:

3\mathbf{N} - \mathbf{N} = 5 \begin{pmatrix} 4 & 0 \\ 6 & -2 \end{pmatrix}

Since $3\mathbf{N} - \mathbf{N} = 2\mathbf{N}$ , we get

2\mathbf{N} = 5 \begin{pmatrix} 4 & 0 \\ 6 & -2 \end{pmatrix}

Next, evaluate the scalar multiplication on the right. Multiplying a matrix by $5$ means multiplying every element by $5$ :

5 \begin{pmatrix} 4 & 0 \\ 6 & -2 \end{pmatrix} = \begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix}

Finally, divide both sides by $2$ , which is the same as multiplying by $\frac{1}{2}$ , to get the matrix $\mathbf{N}$ . Dividing a matrix by a scalar means dividing every element by that scalar:

\mathbf{N} = \frac{1}{2} \begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix} = \begin{pmatrix} 10 & 0 \\ 15 & -5 \end{pmatrix}

Key Takeaways

Matrix equations can be solved using the same inverse operations as ordinary equations.
Scalar multiplication multiplies every element of the matrix by the scalar.
Dividing by a scalar divides every element of the matrix by that scalar.
The answer can be checked by substituting back into the original equation.

Common Mistakes

Subtracting $\mathbf{N}$ from only one side, or subtracting $3\mathbf{N}$ incorrectly.
Multiplying only some elements of the matrix by $5$ instead of all four elements.
Forgetting to divide every element by $2$ at the final step.
Giving the intermediate matrix $\begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix}$ as the final answer; this is $2\mathbf{N}$ , not $\mathbf{N}$ .

Things to Be Careful About

The mark scheme awards B1 for three correct elements of the final matrix, or for the intermediate matrix $\begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix}$ . So showing the intermediate step can earn partial credit even if the final division is wrong.
This is a non-calculator paper, so the scalar multiplications and divisions must be done by hand, but they are simple.
Matrix equality works element by element, so once the equation is reduced to $2\mathbf{N} = \begin{pmatrix} 20 & 0 \\ 30 & -10 \end{pmatrix}$ , each element of $\mathbf{N}$ is half the corresponding element of the right-hand side.

Techniques used

collect like matrix termsperform scalar multiplication on a matrixdivide a matrix by a scalarsolve a linear matrix equation

Q214MMediumMensurationGeometry

Fig. 8

Prism $A$ and prism $B$ are mathematically similar.
In the diagram, the cross-sections of the prisms are shaded.

The volume of prism $A$ is $5000\,\text{cm}^3$ .
The length of prism $A$ is $50\,\text{cm}$ .

The area of the cross-section of prism $B$ is $16\,\text{cm}^2$ .

Calculate the length of prism $B$ .

______ $\text{cm}$

DifficultyMedium

Worked solution

Approach

Find the cross-sectional area of prism A using its volume and length. Then use the ratio of the cross-sectional areas to find the linear scale factor between the two similar prisms, and apply it to prism A's length to find prism B's length.

Working

The volume of a prism is given by the formula:

\text{Volume} = \text{Area of cross-section} \times \text{length}

For prism A, substitute the given values to find the area of its cross-section:

5000 = \text{Area}_A \times 50

\text{Area}_A = \frac{5000}{50} = 100 \text{ cm}^2

Since prism A and prism B are mathematically similar, the ratio of their cross-sectional areas equals the square of the linear scale factor ( $k^2$ ):

k^2 = \frac{\text{Area}_B}{\text{Area}_A} = \frac{16}{100} = 0.16

Take the square root to find the linear scale factor $k$ from A to B:

k = \sqrt{0.16} = 0.4

Multiply the length of prism A by the linear scale factor to find the length of prism B:

\text{Length}_B = 50 \times 0.4 = 20 \text{ cm}

Answer

20

Final answer

20

Detailed explanation

Walkthrough

First, we need the area of the cross-section of prism A. We are given its volume ( $5000 \text{ cm}^3$ ) and its length ( $50 \text{ cm}$ ). Since the volume of any prism equals the cross-sectional area multiplied by its length, we divide the volume by the length to get $100 \text{ cm}^2$ .

Next, we use the fact that the two prisms are mathematically similar. For similar shapes, the ratio of their corresponding areas is equal to the square of the linear scale factor ( $k^2$ ). We have the cross-sectional area of A ( $100 \text{ cm}^2$ ) and the cross-sectional area of B ( $16 \text{ cm}^2$ ), so the area scale factor from A to B is $\frac{16}{100} = 0.16$ .

To find the linear scale factor $k$ , we take the square root of the area scale factor: $\sqrt{0.16} = 0.4$ . This means every length on prism B is $0.4$ times the corresponding length on prism A.

Finally, we multiply the length of prism A ( $50 \text{ cm}$ ) by the linear scale factor ( $0.4$ ) to find the length of prism B, which is $20 \text{ cm}$ .

Key Takeaways

The volume of a prism is calculated as $\text{cross-sectional area} \times \text{length}$ .
For mathematically similar solids, the ratio of corresponding areas is the square of the linear scale factor ( $k^2$ ).
The linear scale factor is the square root of the area scale factor.

Common Mistakes

Using the area ratio ( $\frac{16}{100}$ ) directly as the linear scale factor instead of taking its square root.
Confusing the area scale factor with the volume scale factor (which would be $k^3$ ).
Forgetting to calculate the area of prism A's cross-section first and trying to use the volume ratio directly with the given area of B.

Things to Be Careful About

The diagram is marked NOT TO SCALE, so do not attempt to measure lengths or areas from the image.
Ensure units are consistent: volumes are in $\text{cm}^3$ , areas in $\text{cm}^2$ , and lengths in $\text{cm}$ .
Remember that for similar solids, area ratio is $k^2$ and volume ratio is $k^3$ . Here we only need the area and length (linear) relationship, so $k^2$ is sufficient; the volume of prism B is not required and should not be calculated.

Techniques used

calculate cross-sectional area from volume and lengthfind area scale factor for similar solidsfind linear scale factor from area scale factorcalculate length using linear scale factor

Q223MMediumAlgebra and Graphs

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

Rearrange the formula to make $x$ the subject.

$x$ = ______

DifficultyMedium

Worked solution

Approach

The formula is given as a fraction. To make $x$ the subject, we must first remove the fraction by multiplying both sides by the denominator $(x - 3)$ . Then, we expand the brackets, group all terms containing $x$ on one side of the equation, factorise $x$ , and divide to isolate $x$ .

Working

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

Multiply both sides by $(x - 3)$ :

a(x - 3) = 5b + 2x

Expand the left-hand bracket:

ax - 3a = 5b + 2x

Rearrange to collect terms with $x$ on the left and other terms on the right. Subtract $2x$ from both sides and add $3a$ to both sides:

ax - 2x = 5b + 3a

Factorise the left-hand side by taking out $x$ :

x(a - 2) = 5b + 3a

Divide both sides by $(a - 2)$ to make $x$ the subject:

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

Answer

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

Final answer

(5b + 3a)/(a - 2)

Detailed explanation

Walkthrough

Eliminate the fraction: The original equation has $x$ in the denominator. We clear this by multiplying every term by $(x - 3)$ . This gives $a(x - 3) = 5b + 2x$ .
Expand: Distribute the $a$ across the bracket to get $ax - 3a = 5b + 2x$ .
Group terms: We need all $x$ terms on one side. Subtract $2x$ from both sides and add $3a$ to both sides to get $ax - 2x = 5b + 3a$ .
Factorise: Factor out $x$ from the left side: $x(a - 2) = 5b + 3a$ .
Isolate: Divide by $(a - 2)$ to solve for $x$ : $x = \frac{5b + 3a}{a - 2}$ .

Key Takeaways

When changing the subject of a formula where the subject is in the denominator, always multiply by that denominator first.
Always check if you can cancel common factors before multiplying; here, nothing cancels, so direct multiplication is required.
Remember to change the sign when moving terms across the equals sign (e.g., $+2x$ becomes $-2x$ , $-3a$ becomes $+3a$ ).

Common Mistakes

Forgetting to distribute the multiplier to all terms (e.g., writing $ax - 3 = 5b + 2x$ instead of expanding correctly).
Incorrectly grouping terms (e.g., keeping $x$ terms on opposite sides).
Failing to factorise $x$ before dividing.
Dividing only part of the numerator or denominator by $(a-2)$ .
Sign errors when rearranging (e.g., $5b - 3a$ instead of $5b + 3a$ ).

Things to Be Careful About

Ensure the final answer is in the simplest form. Check if $5b + 3a$ and $a - 2$ share any common factors (they do not in general).
The mark scheme allows 'oe' (other equivalent forms), such as $x = \frac{3a + 5b}{a - 2}$ .

Techniques used

eliminate the fraction by multiplying throughgroup all terms containing the new subject on one sidefactorise to isolate the new subject

Q23MediumCoordinate Geometry

$AB$ is a line segment joining $A(-2, 5)$ and $B(1, 3)$ .

(a)

$B$ is the midpoint of the line segment $AC$ .

Find the coordinates of $C$ .

( ______ , ______ )

DifficultyEasy

Worked solution

Approach

Let the coordinates of $C$ be $(x, y)$ . Since $B(1, 3)$ is the midpoint of $AC$ where $A$ is $(-2, 5)$ , we use the midpoint formula:
$\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Working

Set up the equation for the x-coordinate:
$\frac{-2 + x}{2} = 1$
Multiply by 2:
$-2 + x = 2$
Add 2:
$x = 4$

Set up the equation for the y-coordinate:
$\frac{5 + y}{2} = 3$
Multiply by 2:
$5 + y = 6$
Subtract 5:
$y = 1$

Answer

$C = (4, 1)$

Final answer

(4, 1)

Detailed explanation

Walkthrough

The problem states that point B is the midpoint of the line segment AC. The midpoint formula calculates the average of the x-coordinates and the average of the y-coordinates of the endpoints.

For the x-coordinate: We know the midpoint x is 1, and one endpoint x is -2. So, $\frac{-2 + x_C}{2} = 1$ . Solving this gives $x_C = 4$ .

For the y-coordinate: We know the midpoint y is 3, and one endpoint y is 5. So, $\frac{5 + y_C}{2} = 3$ . Solving this gives $y_C = 1$ .

Key Takeaways

The midpoint $M$ of points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ .
To find an endpoint given the midpoint and the other endpoint, you can reverse the averaging process by multiplying the midpoint coordinate by 2 and subtracting the known endpoint coordinate.

Common Mistakes

Adding instead of subtracting when solving for the missing coordinate.
Mixing up the x and y values.

Things to Be Careful About

Ensure the order of subtraction is consistent if using vector methods ( $\vec{AB} = \vec{BC}$ ). Here, $B-A = C-B \implies C = 2B - A$ .

Techniques used

use the midpoint formulasolve linear equations for unknown coordinates

(b)

The length $AB$ is $\sqrt{t}$ .

Find the value of $t$ .

$t$ = ______

DifficultyMedium-Easy

Worked solution

Approach

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
We are given that the length $AB = \sqrt{t}$ , so we need to calculate $(x_2 - x_1)^2 + (y_2 - y_1)^2$ .

Working

Coordinates: $A(-2, 5)$ and $B(1, 3)$ .

Calculate the difference in x:
$x_2 - x_1 = 1 - (-2) = 1 + 2 = 3$
Square it:
$3^2 = 9$

Calculate the difference in y:
$y_2 - y_1 = 3 - 5 = -2$
Square it:
$(-2)^2 = 4$

Sum the squares:
$t = 9 + 4 = 13$

So, $AB = \sqrt{13}$ .

Answer

$t = 13$

Final answer

13

Detailed explanation

Walkthrough

To find the value of $t$ , we use the distance formula which is derived from Pythagoras' theorem. The horizontal distance between A and B is the difference in their x-coordinates, and the vertical distance is the difference in their y-coordinates. These form the legs of a right-angled triangle, with the line segment AB as the hypotenuse.

Horizontal leg: $1 - (-2) = 3$ . Square: $3^2 = 9$ .
Vertical leg: $3 - 5 = -2$ . Square: $(-2)^2 = 4$ .
Hypotenuse squared ( $t$ ): $9 + 4 = 13$ .

Key Takeaways

The distance formula $d^2 = (\Delta x)^2 + (\Delta y)^2$ allows calculation of the square of the distance directly without taking the square root until the end.
Squaring a negative number results in a positive number, so the sign of the difference does not matter for the final result.

Common Mistakes

Forgetting to square the differences.
Making sign errors when subtracting negative coordinates (e.g., $1 - 2$ instead of $1 - (-2)$ ).
Forgetting that $t$ is the value inside the square root, not the length itself.

Things to Be Careful About

The question asks for $t$ , not $\sqrt{t}$ . Check the required format carefully.

Techniques used

apply the distance formulacalculate squares of differences

(c)

Find the equation of the line perpendicular to $AB$ that passes through $A$ .
Give your answer in the form $y = mx + c$ .

$y$ = ______

4M Similar

DifficultyMedium

Worked solution

Approach

Find the gradient (slope) of the line segment $AB$ .
Determine the gradient of the line perpendicular to $AB$ . Perpendicular gradients have a product of $-1$ (negative reciprocal).
Use the point-slope form or substitute the gradient and the given point $A(-2, 5)$ into $y = mx + c$ to find the y-intercept $c$ .

Working

Step 1: Gradient of AB
$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 5}{1 - (-2)} = \frac{-2}{3}$

Step 2: Gradient of perpendicular line
Let $m_{\perp}$ be the gradient of the perpendicular line.
$m_{\perp} \times m_{AB} = -1$
$m_{\perp} \times \left(-\frac{2}{3}\right) = -1$
$m_{\perp} = \frac{-1}{-2/3} = \frac{3}{2}$

Step 3: Find the equation
The equation is of the form $y = \frac{3}{2}x + c$ . The line passes through $A(-2, 5)$ , so substitute $x = -2$ and $y = 5$ :
$5 = \frac{3}{2}(-2) + c$
$5 = -3 + c$
$c = 5 + 3 = 8$

So the equation is:
$y = \frac{3}{2}x + 8$

Answer

$y = \frac{3}{2}x + 8$

Final answer

y = 3/2x + 8

Detailed explanation

Walkthrough

First, we calculate the gradient of the line AB. Gradient is defined as 'rise over run', or the change in y divided by the change in x. Using points A(-2, 5) and B(1, 3):
Change in y = 3 - 5 = -2.
Change in x = 1 - (-2) = 3.
Gradient of AB = -2/3.

Next, we find the gradient of a line perpendicular to AB. Two lines are perpendicular if the product of their gradients is -1. This means the new gradient is the negative reciprocal of the original. The reciprocal of -2/3 is -3/2. The negative of that is 3/2. So the new gradient is 3/2.

Finally, we need the specific equation of the line. We know the form is $y = mx + c$ , where $m = 3/2$ . We also know the line passes through point A(-2, 5). By plugging these x and y values into the equation, we can solve for the constant $c$ (the y-intercept).
$5 = (3/2)(-2) + c$ simplifies to $5 = -3 + c$ , so $c = 8$ .

Key Takeaways

Gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Perpendicular gradient rule: If $m_1$ is the gradient of a line, the gradient of a perpendicular line is $-\frac{1}{m_1}$ .
To find the equation of a line, you need its gradient and one point on it. Substituting the point into $y=mx+c$ allows you to find $c$ .

Common Mistakes

Calculating the gradient as $\frac{x_2 - x_1}{y_2 - y_1}$ (inverting rise and run).
Finding the negative reciprocal incorrectly (e.g., just changing the sign but not flipping the fraction, or vice versa).
Arithmetic errors when substituting negative coordinates (e.g., forgetting that $\frac{3}{2} \times -2$ is negative).

Things to Be Careful About

The mark scheme accepts equivalent forms like $y - 5 = \frac{3}{2}(x + 2)$ , but the final answer must be in the form $y = mx + c$ . Make sure to expand and simplify correctly.
Ensure fractions are handled carefully during substitution.

Techniques used

calculate the gradient of a line segmentdetermine the negative reciprocal gradientsubstitute point coordinates to find the interceptrearrange into slope-intercept form

Q24MediumTransformations and Vectors

Fig. 9

$ABCD$ is a quadrilateral.
$H, K, L$ and $M$ are the midpoints of $AB, BC, CD$ and $AD$ respectively.
$\overrightarrow{AB} = 2\mathbf{a}$ , $\overrightarrow{BC} = 2\mathbf{b}$ and $\overrightarrow{AD} = 2\mathbf{d}$ .

Express, as simply as possible, in terms of $\mathbf{a}$ and $\mathbf{b}$ or $\mathbf{a}, \mathbf{b}$ and $\mathbf{d}$

(a)

$\overrightarrow{HK}$

$\overrightarrow{HK}$ = ______

DifficultyEasy

Worked solution

Approach

To find $\overrightarrow{HK}$ , travel along the path from $H$ to $K$ via $B$ . Since $H$ and $K$ are midpoints of $AB$ and $BC$ , the vectors $\overrightarrow{HB}$ and $\overrightarrow{BK}$ are exactly half of $\overrightarrow{AB}$ and $\overrightarrow{BC}$ .

Working

\overrightarrow{HB} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}(2\mathbf{a}) = \mathbf{a}

\overrightarrow{BK} = \frac{1}{2}\overrightarrow{BC} = \frac{1}{2}(2\mathbf{b}) = \mathbf{b}

By vector addition:

\overrightarrow{HK} = \overrightarrow{HB} + \overrightarrow{BK} = \mathbf{a} + \mathbf{b}

Answer

\mathbf{a} + \mathbf{b}

Final answer

a + b

Detailed explanation

Walkthrough

We want the vector from $H$ to $K$ . The most direct path uses the vertices already given in the problem: go from $H$ to $B$ , then from $B$ to $K$ . Because $H$ is the midpoint of $AB$ , the vector $\overrightarrow{HB}$ is exactly half of $\overrightarrow{AB}$ , which gives $\mathbf{a}$ . Similarly, $K$ is the midpoint of $BC$ , so $\overrightarrow{BK}$ is half of $\overrightarrow{BC}$ , giving $\mathbf{b}$ . Adding these two vectors along the path gives $\overrightarrow{HK} = \mathbf{a} + \mathbf{b}$ .

Key Takeaways

When a point is the midpoint of a side of a polygon, the vector from an endpoint to that midpoint is exactly half the vector of the full side. Vector addition along a path ( $\overrightarrow{HK} = \overrightarrow{HB} + \overrightarrow{BK}$ ) is the fundamental tool for combining these.

Common Mistakes

Forgetting that $\overrightarrow{HB}$ is in the same direction as $\overrightarrow{AB}$ (so it is $+\mathbf{a}$ , not $-\mathbf{a}$ ).
Adding $\overrightarrow{AB}$ and $\overrightarrow{BC}$ directly instead of their halves.
Writing the answer as $2\mathbf{a} + 2\mathbf{b}$ by missing the midpoint halving step.

Things to Be Careful About

The question asks for the answer "as simply as possible". $\mathbf{a} + \mathbf{b}$ is the simplest form. Always check the direction of each half-vector to ensure signs are correct.

Techniques used

identify half-vectors from midpointsadd vectors along a path

(b)

$\overrightarrow{CD}$

$\overrightarrow{CD}$ = ______

DifficultyMedium-Easy

Worked solution

Approach

To find $\overrightarrow{CD}$ , travel from $C$ to $D$ via $A$ . This path is $\overrightarrow{CA} + \overrightarrow{AD}$ . We know $\overrightarrow{AD}$ , and we can find $\overrightarrow{CA}$ by going $C \to B \to A$ .

Working

\overrightarrow{CA} = \overrightarrow{CB} + \overrightarrow{BA} = -\overrightarrow{BC} - \overrightarrow{AB}

Substitute the given values:

\overrightarrow{CA} = -2\mathbf{b} - 2\mathbf{a}

Now add $\overrightarrow{AD}$ :

\overrightarrow{CD} = \overrightarrow{CA} + \overrightarrow{AD} = (-2\mathbf{a} - 2\mathbf{b}) + 2\mathbf{d}

Rearranging:

\overrightarrow{CD} = -2\mathbf{a} - 2\mathbf{b} + 2\mathbf{d}

This can also be written as $2(\mathbf{d} - \mathbf{a} - \mathbf{b})$ .

Answer

-2\mathbf{a} - 2\mathbf{b} + 2\mathbf{d}

Final answer

-2a - 2b + 2d

Detailed explanation

Walkthrough

We need the vector from $C$ to $D$ . We are given $\overrightarrow{AB}$ , $\overrightarrow{BC}$ , and $\overrightarrow{AD}$ . The path $C \to B \to A \to D$ uses only known vectors. First, $\overrightarrow{CB}$ is the reverse of $\overrightarrow{BC}$ , so it is $-2\mathbf{b}$ . Next, $\overrightarrow{BA}$ is the reverse of $\overrightarrow{AB}$ , so it is $-2\mathbf{a}$ . Finally, $\overrightarrow{AD}$ is given as $2\mathbf{d}$ . Adding these along the path gives $\overrightarrow{CD} = -2\mathbf{b} - 2\mathbf{a} + 2\mathbf{d}$ . Alternatively, you can use the closed-loop rule: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DA} = \mathbf{0}$ , and substitute $\overrightarrow{DA} = -2\mathbf{d}$ to solve for $\overrightarrow{CD}$ .

Key Takeaways

Any vector between two points can be found by adding vectors along any path connecting them. Reversing a vector means negating it (multiplying by $-1$ ). The sum of vectors around a closed loop is always the zero vector.

Common Mistakes

Forgetting to negate $\overrightarrow{BC}$ and $\overrightarrow{AB}$ when going $C \to B \to A$ .
Using $\overrightarrow{AD}$ instead of $\overrightarrow{DA}$ in the closed-loop equation.
Leaving the answer as $2\mathbf{d} - 2\mathbf{a} - 2\mathbf{b}$ without rearranging or factoring, though both forms are acceptable.

Things to Be Careful About

The mark scheme accepts $-2\mathbf{a} - 2\mathbf{b} + 2\mathbf{d}$ or $2(\mathbf{d} - \mathbf{a} - \mathbf{b})$ . Both are fully simplified. Ensure all three vectors $\mathbf{a}$ , $\mathbf{b}$ , and $\mathbf{d}$ appear with correct signs.

Techniques used

use a closed-loop vector pathnegate a vector to reverse direction

(c)

$\overrightarrow{ML}$

$\overrightarrow{ML}$ = ______