Mathematics (Syllabus D) 4024 / 12 — May/June 2025

DifficultyMedium-Easy

Worked solution

Approach

Evaluate the expression $6 - 2 \times (-4)$ using the order of operations. Multiplication takes precedence over subtraction.

Working

First, perform the multiplication: $2 \times (-4) = -8$ .

Substitute this back into the expression:

6 - (-8)

Subtracting a negative is equivalent to adding a positive:

6 + 8 = 14

Answer

14

Final answer

14

Detailed explanation

Walkthrough

To solve $6 - 2 \times (-4)$ , we must follow the standard order of operations (often remembered by acronyms like BODMAS or PEMDAS), which dictates that multiplication is performed before addition or subtraction.

Identify the multiplication part: $2 \times (-4)$ .
Calculate the product: A positive times a negative gives a negative, so $2 \times -4 = -8$ .
Replace the multiplication in the original expression with the result: The expression becomes $6 - (-8)$ .
Simplify the double negative: Subtracting a negative number is the same as adding its positive counterpart. So, $6 - (-8)$ becomes $6 + 8$ .
Perform the final addition: $6 + 8 = 14$ .

Key Takeaways

Always check for multiplication or division before doing addition or subtraction.
Remember the sign rules for multiplication: positive × negative = negative.
Remember the sign rule for subtraction: minus a negative equals plus a positive.

Common Mistakes

Performing the subtraction first ( $6 - 2 = 4$ ) and then multiplying ( $4 \times -4 = -16$ ). This violates the order of operations.
Getting the sign wrong in the final step, such as thinking $6 - (-8) = 6 - 8 = -2$ .

Things to Be Careful About

Ensure you do not ignore the brackets around the $-4$ . It represents a negative number being multiplied, not just a subtraction of 4 later on.

Techniques used

apply order of operations (BODMAS/PEMDAS)multiply a positive integer by a negative integersubtract a negative number from a positive number

(b)

$80 \times 0.2$

______

(c)

$\frac{2}{9} \div \frac{5}{6}$

______

DifficultyMedium-Easy

Worked solution

Approach

To divide by a fraction, multiply by its reciprocal. The expression is $\frac{2}{9} \div \frac{5}{6}$ .

Working

Keep the first fraction, change division to multiplication, and flip the second fraction (find the reciprocal):

\frac{2}{9} \times \frac{6}{5}

Multiply the numerators together and the denominators together:

\text{Numerator: } 2 \times 6 = 12

\text{Denominator: } 9 \times 5 = 45

So the unsimplified result is:

\frac{12}{45}

Simplify the fraction by finding the highest common factor (HCF) of 12 and 45. Both are divisible by 3:

12 \div 3 = 4

45 \div 3 = 15

The simplified fraction is:

\frac{4}{15}

Answer

\frac{4}{15}

Final answer

4/15

Detailed explanation

Walkthrough

We are asked to calculate $\frac{2}{9} \div \frac{5}{6}$ .

Reciprocal Rule: Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{5}{6}$ is $\frac{6}{5}$ . So, rewrite the problem as $\frac{2}{9} \times \frac{6}{5}$ .
Multiply Across: Multiply the top numbers (numerators) and the bottom numbers (denominators).
- Numerator: $2 \times 6 = 12$
- Denominator: $9 \times 5 = 45$
- Result: $\frac{12}{45}$
Simplify: Check if the fraction can be reduced. The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 45 are 1, 3, 5, 9, 15, 45. The largest common factor is 3. Divide both the numerator and denominator by 3.
- $12 \div 3 = 4$
- $45 \div 3 = 15$
- Final answer: $\frac{4}{15}$ .

Note: You could also simplify before multiplying by cancelling common factors between diagonals. Here, 6 and 9 share a factor of 3 ( $6\to2$ , $9\to3$ ), leading directly to $\frac{2 \times 2}{3 \times 5} = \frac{4}{15}$ . Both methods are valid.

Key Takeaways

To divide fractions, invert the divisor and multiply.
Always simplify your final answer to its lowest terms.
Cross-cancellation before multiplying can keep numbers smaller and reduce errors.

Common Mistakes

Multiplying straight across without flipping the second fraction (i.e., calculating $\frac{2}{9} \times \frac{5}{6}$ ).
Forgetting to simplify the final fraction (leaving it as $\frac{12}{45}$ ).
Incorrectly multiplying mixed numbers or whole numbers with fractions (not applicable here but common in similar questions).

Things to Be Careful About

Ensure the question asks for the simplest form. While $\frac{12}{45}$ is mathematically equal, exams usually require the simplest form for full marks unless stated otherwise.

Techniques used

divide fractions by multiplying by the reciprocalmultiply numerators and denominatorssimplify the resulting fraction

Q2EasyProbability

A bag contains 11 balls.
There are 5 blue balls and 4 yellow balls.
The rest of the balls are green.
A ball is taken from the bag at random.

Find the probability that the ball is

(a)

yellow

______

DifficultyEasy

Worked solution

Approach

The probability of an event is given by the ratio:

P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}

We need to find the probability that a randomly selected ball is yellow.

Working

From the question stem:

Total number of balls = $11$
Number of yellow balls = $4$

The number of favourable outcomes (picking a yellow ball) is $4$ .
The total number of possible outcomes is $11$ .

Substituting these values into the formula:

P(\text{yellow}) = \frac{4}{11}

Answer

\frac{4}{11}

Final answer

4/11

Detailed explanation

Walkthrough

To find the probability of picking a yellow ball, we use the basic definition of probability for a single event. This is the number of balls that are yellow divided by the total number of balls in the bag. The problem states there are 4 yellow balls out of a total of 11 balls. Therefore, the probability is simply the fraction $\frac{4}{11}$ .

Key Takeaways

Probability is always a number between 0 and 1.
For equally likely outcomes, $P(A) = \frac{n(A)}{n(S)}$ , where $n(A)$ is the number of outcomes in event A and $n(S)$ is the total number of outcomes.
The denominator represents the total sample space.

Common Mistakes

Swapping the numerator and denominator (writing $\frac{11}{4}$ ).
Using the wrong count for the numerator (e.g., using the number of green or blue balls instead of yellow).
Not reducing the fraction if it can be simplified (though $\frac{4}{11}$ is already in its simplest form).

Things to Be Careful About

Ensure you use the correct total from the stem ( $11$ ), not just the sum of the parts mentioned in the specific part question.
The answer should be left as a fraction unless decimals are requested.

Techniques used

identify total number of outcomescount favorable outcomesform the probability fraction

(b)

not blue.

______

DifficultyEasy

Worked solution

Approach

We need to find the probability that the ball is not blue. There are two ways to do this:

Count the number of balls that are not blue and divide by the total.
Use the complement rule: $P(\text{not blue}) = 1 - P(\text{blue})$ .

Working

Method 1: Counting favourable outcomes
The balls that are not blue are the yellow ones and the green ones.

Number of yellow balls = $4$
Number of green balls: The rest of the 11 balls are green.
$\text{Green} = 11 - 5 - 4 = 2$
Total non-blue balls = Yellow + Green = $4 + 2 = 6$ .

Alternatively, simply subtract the blue balls from the total:
$\text{Non-blue balls} = 11 - 5 = 6$

Total outcomes = $11$ .

P(\text{not blue}) = \frac{6}{11}

Method 2: Complement Rule

P(\text{blue}) = \frac{5}{11}

P(\text{not blue}) = 1 - \frac{5}{11} = \frac{11}{11} - \frac{5}{11} = \frac{6}{11}

Answer

\frac{6}{11}

Final answer

6/11

Detailed explanation

Walkthrough

The event "not blue" includes all outcomes except those where the ball is blue. Since the only colours are blue, yellow, and green, the non-blue balls are the yellow and green ones. We know there are 5 blue balls out of 11. So, the number of non-blue balls is $11 - 5 = 6$ . The probability is therefore $\frac{6}{11}$ . Alternatively, one could calculate $1 - P(\text{blue}) = 1 - \frac{5}{11} = \frac{6}{11}$ .

Key Takeaways

The probability of an event NOT happening is $1$ minus the probability of it happening.
Mutually exclusive events (like being blue vs being not blue) cover the entire sample space, so their probabilities sum to 1.

Common Mistakes

Calculating the probability of being green instead of not blue.
Forgetting that "not blue" includes both yellow AND green balls.
Arithmetic errors when subtracting fractions from 1.

Things to Be Careful About

Read the question carefully: "not blue" is different from "green".
Verify that the counts add up: $5 (\text{blue}) + 4 (\text{yellow}) + 2 (\text{green}) = 11$ . This confirms the total is correct.

Techniques used

calculate complementary outcomessum probabilities of mutually exclusive eventsform the probability fraction

Q33MMedium-EasyGeometryNumber

The area of one face of a cube is $9\,\text{cm}^2$ .

On the $1\,\text{cm}$ grid, draw an accurate net of the cube.

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

First, determine the side length of each face of the cube from the given area. Then, use the $1\,\text{cm}$ grid scale to draw a valid net consisting of six $3\,\text{cm} \times 3\,\text{cm}$ squares.

Working

The area of one face (a square) is given as $9\,\text{cm}^2$ .

\text{Side length} = \sqrt{9} = 3\,\text{cm}

Since the grid is $1\,\text{cm}$ per square, each face of the cube must be drawn as a $3 \times 3$ block of grid squares.

A cube has 6 faces. A valid net must consist of 6 squares connected edge-to-edge in a pattern that can fold into a cube. A common valid net is a cross shape: a horizontal row of 4 squares, with one square attached above the second square and one square attached below the second square.

Answer

A net of a cube drawn on the grid, consisting of 6 squares each measuring $3 \times 3$ grid units, arranged in a valid net configuration (e.g., a cross shape), with internal fold lines drawn.

Final answer

Net of a cube with 6 squares of size 3x3 grid units, e.g., a cross shape (4 in a row, 1 above and 1 below the second square), with internal fold lines.

Detailed explanation

Walkthrough

Find the side length: The problem states the area of one face is $9\,\text{cm}^2$ . Since a face of a cube is a square, its area is $s^2$ . Solving $s^2 = 9$ gives $s = 3\,\text{cm}$ .
Determine the grid scale: The grid provided is $1\,\text{cm}$ per square. Therefore, each $3\,\text{cm}$ side of the cube's face must span exactly 3 grid squares. Each face is a $3 \times 3$ block.
Construct the net: A cube has 6 identical square faces. A net is a 2D pattern that can be folded to form the 3D shape. There are 11 distinct nets for a cube. A simple and common one is the 'cross' or 'T' shape. For example, draw a horizontal row of four $3 \times 3$ squares. Then, draw a fifth $3 \times 3$ square attached to the top edge of the second square in the row, and a sixth $3 \times 3$ square attached to the bottom edge of the same second square.
Internal lines: The marking scheme notes that 'no/incorrect internal lines' reduces the mark to 2. Internal lines represent the folds where the net will be joined. These should be drawn clearly between the adjacent squares.

Key Takeaways

The side length of a square is the square root of its area: $s = \sqrt{A}$ .
A net of a cube always consists of exactly 6 squares.
When drawing on a grid, always convert the required dimensions into grid units using the given scale.

Common Mistakes

Wrong scale: Drawing squares that are $1 \times 1$ (ignoring the area) or $9 \times 9$ (confusing area with side length).
Invalid net: Drawing 6 squares that do not connect in a way that can fold into a cube (e.g., a $2 \times 3$ block is not a valid net, nor is a row of 6 squares).
Missing internal lines: Forgetting to draw the lines between the squares that indicate folds, which the mark scheme penalises.
Inaccuracy: Not aligning the squares perfectly to the grid lines, resulting in a net that is not 'accurate'.

Things to Be Careful About

Accuracy: The question asks for an 'accurate' net. Ensure all lines are straight and follow the grid exactly. Each face must be exactly $3 \times 3$ grid squares.
Mark scheme guidance: Full marks (3) require a correct net with edge length $3\,\text{cm}$ and correct internal lines. An open net (5 squares) or missing internal lines only scores 2 marks. A single square of the correct size scores 1 mark.
Units: The grid is in cm, and the calculated side is in cm, so no unit conversion is needed for the drawing, but be aware if the grid scale were different.

Techniques used

calculate side length from area of a squareidentify a valid net configuration for a cubedraw on grid at correct scale

Q43MMedium-EasyGeometry

Fig. 2

$AEC$ and $BED$ are straight lines.
$ED = EC$ .

Find the value of $x$ .

$x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Use the properties of an isosceles triangle to find the angles in triangle DEC. Then use vertically opposite angles to find an angle in triangle AEB, and finally use the angle sum of a triangle to find $x$ .

Working

In triangle DEC, $ED = EC$ , so it is an isosceles triangle. The base angles are equal:

\angle DCE = \angle CDE = 35^\circ

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

\angle DEC = 180^\circ - (35^\circ + 35^\circ) = 180^\circ - 70^\circ = 110^\circ

Since AEC and BED are straight lines intersecting at $E$ , $\angle AEB$ and $\angle DEC$ are vertically opposite angles. Vertically opposite angles are equal:

\angle AEB = \angle DEC = 110^\circ

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

x^\circ + 48^\circ + 110^\circ = 180^\circ

x + 158 = 180

x = 22

Answer

x = 22

Final answer

22

Detailed explanation

Walkthrough

First, we look at triangle DEC. We are given that $ED = EC$ , which means triangle DEC is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, $\angle DCE = \angle CDE = 35^\circ$ .

Next, we use the fact that the angles in any triangle add up to $180^\circ$ . So in triangle DEC, $\angle DEC = 180^\circ - 35^\circ - 35^\circ = 110^\circ$ .

Now we move to the intersection at $E$ . The lines AEC and BED are straight lines, which means $\angle AEB$ and $\angle DEC$ are vertically opposite angles. Vertically opposite angles are equal, so $\angle AEB = 110^\circ$ . (Alternatively, $\angle BEC$ and $\angle DEC$ are angles on a straight line BED, so $\angle BEC = 180^\circ - 110^\circ = 70^\circ$ , and then $\angle AEB = 180^\circ - 70^\circ = 110^\circ$ using angles on straight line AEC).

Finally, we look at triangle AEB. We know two of its angles: $\angle EBA = 48^\circ$ and $\angle AEB = 110^\circ$ . The third angle is $\angle EAB = x^\circ$ . Since the angles in a triangle add up to $180^\circ$ , we have $x + 48 + 110 = 180$ , which gives $x = 22$ .

Key Takeaways

Isosceles triangles have two equal base angles opposite the equal sides.
Vertically opposite angles formed by two intersecting straight lines are equal.
The sum of angles in any triangle is always $180^\circ$ .

Common Mistakes

Forgetting that the base angles of an isosceles triangle are equal, and instead assuming the apex angle is $35^\circ$ .
Not recognising vertically opposite angles and instead trying to use angles on a straight line incorrectly.
Forgetting to subtract the sum of the two known angles from $180^\circ$ to find $x$ .

Things to Be Careful About

The diagram is marked "NOT TO SCALE", so do not measure angles with a protractor.
Ensure you correctly identify which angles are equal in the isosceles triangle (the base angles, not the apex angle).
Keep track of degrees and do not forget the degree symbol when writing angle values, though $x$ itself is just a number.

Techniques used

identify base angles in an isosceles triangleuse vertically opposite anglesapply the angle sum property of a triangle

Q52MEasyAlgebra and Graphs

Solve.

5(4 - x) = 35

$x$ = ______

DifficultyEasy

Worked solution

Approach

Expand the brackets to remove the parentheses, then rearrange the equation to isolate $x$ .

Working

Start with the given equation:

5(4 - x) = 35

Expand the brackets (multiply each term inside by 5):

20 - 5x = 35

Subtract 20 from both sides to move the constant term:

-5x = 35 - 20

Simplify the right-hand side:

-5x = 15

Divide both sides by -5 to solve for $x$ :

x = \frac{15}{-5}

Calculate the final value:

x = -3

Answer

x = -3

Final answer

-3

Detailed explanation

Walkthrough

The problem asks us to solve for $x$ in the equation $5(4 - x) = 35$ . The first step is to simplify the left-hand side. We can do this by 'expanding the brackets', which means multiplying the number outside ( $5$ ) by each term inside the brackets ( $4$ and $-x$ ). This gives $20 - 5x$ . Now the equation is $20 - 5x = 35$ . To get $x$ on its own, we need to move the $20$ to the other side. We do this by subtracting $20$ from both sides, resulting in $-5x = 15$ . Finally, since $x$ is multiplied by $-5$ , we divide both sides by $-5$ to find the value of $x$ . Dividing a positive number ( $15$ ) by a negative number ( $-5$ ) results in a negative answer ( $-3$ ).

Key Takeaways

Expanding Brackets: Always distribute the multiplier to every term inside the parentheses.
Inverse Operations: Use addition/subtraction to move constants and multiplication/division to isolate variables.
Sign Rules: Remember that a positive divided by a negative yields a negative result.

Common Mistakes

Forgetting to multiply the second term inside the bracket (e.g., writing $20 - x$ instead of $20 - 5x$ ).
Sign errors when moving terms across the equals sign (e.g., adding 20 instead of subtracting).
Incorrectly dividing signs (e.g., thinking $15 / -5 = 3$ ).

Things to Be Careful About

Ensure you maintain the negative sign attached to the $5x$ term throughout the working.
Check your answer by substituting $x = -3$ back into the original equation: $5(4 - (-3)) = 5(7) = 35$ . This confirms the solution is correct.

Techniques used

expand bracketsisolate the variable termsolve a linear equation

Q6Medium-EasyMensurationGeometry

The scale drawing shows the positions of two villages, $A$ and $B$ .
The scale is $1\,\text{cm}$ to $5\,\text{km}$ .

Fig. 3

(a)

Find the actual distance between village $A$ and village $B$ .

______ $\text{km}$

DifficultyMedium-Easy

Worked solution

Approach

Measure the length of the line segment $AB$ on the scale drawing using a ruler. Then multiply this length by the scale factor to find the actual distance in kilometres.

Working

Measure the length of $AB$ on the drawing. The acceptable measured length is between $5.8\,\text{cm}$ and $6.2\,\text{cm}$ . Assume a measured length of $6.0\,\text{cm}$ for this working.

\text{Actual distance} = 6.0 \times 5 = 30\,\text{km}

Any answer in the range $29\,\text{km}$ to $31\,\text{km}$ is accepted based on the measured length.

Answer

$30\,\text{km}$ (accept $29$ to $31\,\text{km}$ )

Final answer

30

Detailed explanation

Walkthrough

First, use a ruler to measure the length of the line segment connecting $A$ and $B$ on the provided scale drawing. Because the drawing is printed at a specific size, small measurement variations are expected, so any value between $5.8\,\text{cm}$ and $6.2\,\text{cm}$ is considered correct. Next, apply the scale ratio. The scale is $1\,\text{cm}$ to $5\,\text{km}$ , meaning every centimetre on the drawing represents $5\,\text{km}$ in reality. Multiply the measured length by $5$ to convert to kilometres. For example, $6.0\,\text{cm} \times 5 = 30\,\text{km}$ .

Key Takeaways

Always measure lengths on scale drawings with a ruler.
To convert a measured distance to an actual distance, multiply by the scale factor (the real-world distance per unit of drawing distance).
Acceptable measurement tolerances are given in the mark scheme; answers within this range are valid.

Common Mistakes

Forgetting to multiply by the scale factor and giving the answer in centimetres.
Multiplying by the wrong number (e.g., dividing by 5 instead of multiplying).
Not accepting the tolerance range for the measured length.

Things to Be Careful About

Ensure the ruler is aligned exactly with the line segment $AB$ and read the measurement to the nearest millimetre.
The final answer must be in kilometres, not centimetres.
The mark scheme allows a range of answers ( $29$ to $31$ ) due to measurement tolerance; any value in this range earns full marks.

Techniques used

measure the length of a line segment on a scale drawingconvert units using the given scale ratio

(b)

Village $C$ is on a bearing of $060^\circ$ from village $A$ .
Village $C$ is on a bearing of $320^\circ$ from village $B$ .

Find and label the position of village $C$ on the scale drawing.

DifficultyMedium-Easy

Worked solution

Approach

To find the position of village $C$ , draw two rays from villages $A$ and $B$ using the given bearings. The intersection of these two rays will be the location of village $C$ .

Working

At village $A$ , the North arrow is already drawn. Measure and draw a ray at a bearing of $060^\circ$ (which is $60^\circ$ clockwise from the North line at $A$ ).
At village $B$ , the North arrow is already drawn. Measure and draw a ray at a bearing of $320^\circ$ (which is $320^\circ$ clockwise from the North line at $B$ , or equivalently $40^\circ$ counter-clockwise from North).
Mark the point where these two rays intersect and label it $C$ .

Answer

Village $C$ is located at the intersection of the ray from $A$ at a bearing of $060^\circ$ and the ray from $B$ at a bearing of $320^\circ$ . See diagram.

Final answer

Point C at the intersection of the 060° bearing from A and the 320° bearing from B

Detailed explanation

Walkthrough

A bearing is an angle measured clockwise from the North line.

First, consider village $A$ . The North arrow is already provided. A bearing of $060^\circ$ means we measure $60^\circ$ clockwise from the North line at $A$ and draw a ray extending from $A$ in that direction.

Next, consider village $B$ . The North arrow is also provided here. A bearing of $320^\circ$ means we measure $320^\circ$ clockwise from the North line at $B$ . Since a full circle is $360^\circ$ , this is equivalent to measuring $40^\circ$ counter-clockwise (or to the left) from the North line at $B$ . Draw a ray extending from $B$ in this direction.

The position of village $C$ must satisfy both conditions simultaneously, so it must lie on both rays. Mark the point where these two rays intersect and label it $C$ .

Key Takeaways

Bearings are always measured clockwise from the North line.
When constructing a point from two bearings, draw rays from both reference points and find their intersection.
North lines at different points on a scale drawing are parallel to each other.

Common Mistakes

Measuring the bearing counter-clockwise instead of clockwise.
Measuring from the wrong reference line (e.g., from the line $AB$ instead of the North line).
Forgetting that North lines at different points are parallel and should be drawn vertically upwards at each point.

Things to Be Careful About

Ensure the protractor is aligned correctly with the North line at each village.
Bearings must be given as three figures (e.g., $060^\circ$ , not $60^\circ$ ).
The intersection point must be clearly marked and labeled $C$ .

Techniques used

draw a bearing from a given pointfind the intersection of two bearing lines

Q7Medium-EasyNumber

Evaluate.

(a)

$\sqrt[3]{125}$

______

(b)

$4^{-2}$

______

Q8Medium-EasyStatistics

Asha records the distance she walks and the time she takes for each of 10 walks.
The table shows her results.

Distance (km)	4.5	4.6	7.2	8.4	5.5	7.5	4.2	9.0	3.8	5.6
Time (minutes)	52	60	93	105	65	100	52	116	49	62

(a)

Complete the scatter diagram.
The first 6 points have been plotted for you.

Fig. 4

DifficultyMedium-Easy

Worked solution

Approach

Identify the four data pairs that have not yet been plotted on the scatter diagram and mark them as crosses at the correct coordinates.

Working

The complete list of (Distance, Time) pairs is:
(4.5, 52), (4.6, 60), (7.2, 93), (8.4, 105), (5.5, 65), (7.5, 100), (4.2, 52), (9.0, 116), (3.8, 49) and (5.6, 62).

The first six are already plotted. The four remaining points to plot are:

$x = 4.2, y = 52$
$x = 9.0, y = 116$
$x = 3.8, y = 49$
$x = 5.6, y = 62$

Locate each of these on the grid and mark with a cross.

Answer

Four points correctly plotted at (4.2, 52), (9.0, 116), (3.8, 49) and (5.6, 62).

Final answer

Four points plotted at (4.2, 52), (9.0, 116), (3.8, 49) and (5.6, 62)

Detailed explanation

Walkthrough

The table provides 10 (Distance, Time) pairs. The scatter diagram already shows six of these points. To complete the diagram, we simply read off the remaining four pairs from the table and plot them at the corresponding coordinates on the grid.

The missing pairs are:

Distance $4.2\,\text{km}$ , Time $52\,\text{minutes}$
Distance $9.0\,\text{km}$ , Time $116\,\text{minutes}$
Distance $3.8\,\text{km}$ , Time $49\,\text{minutes}$
Distance $5.6\,\text{km}$ , Time $62\,\text{minutes}$

For each pair, move along the horizontal axis to the distance value, then move up to the time value, and place a cross. For example, for $(3.8, 49)$ , go to $3.8$ on the Distance axis and $49$ on the Time axis.

Key Takeaways

A scatter diagram is completed by plotting every data pair from the table. The horizontal axis represents the independent variable (Distance) and the vertical axis represents the dependent variable (Time).

Common Mistakes

Reading the axes in the wrong order (plotting Time on the horizontal axis and Distance on the vertical axis).
Misreading the minor gridlines: the horizontal axis has major lines at every $1\,\text{km}$ and minor lines at every $0.2\,\text{km}$ ; the vertical axis has major lines at every $10\,\text{minutes}$ and minor lines at every $2\,\text{minutes}$ .
Forgetting to plot a point entirely.

Things to Be Careful About

Ensure each point is plotted to the correct minor gridline. For instance, $x = 3.8$ is four minor gridlines to the right of $3$ , and $y = 49$ is four and a half minor gridlines above $40$ . Accuracy here is essential for part (b).

Techniques used

extract un-plotted data pairs from the tableplot points on a scatter diagram with given axis scales

(b)

Draw a line of best fit.

DifficultyMedium-Easy

Worked solution

Approach

Draw a straight ruled line that passes through the centre of the data pattern, with roughly equal numbers of points above and below the line.

Working

A straight line of best fit is drawn through the plotted points. The line should have a positive gradient, reflecting the positive correlation between distance and time. It should not necessarily pass through every point, but should balance the points on either side. For example, a line passing near $(4, 50)$ and $(9, 115)$ is appropriate.

Answer

A straight ruled line of best fit with a positive gradient drawn through the data points.

Final answer

A straight ruled line of best fit with a positive gradient drawn through the data points

Detailed explanation

Walkthrough

A line of best fit summarises the trend in the scatter diagram. Since time increases as distance increases, the line must have a positive gradient.

Draw a straight ruler line through the middle of the cloud of points. The key criteria are:

The line must be straight and ruled (not freehand).
It must show the positive trend (going up from left to right).
There should be roughly equal numbers of points above and below the line.
The line should extend across the range of the data.

Key Takeaways

The line of best fit is a visual summary of the relationship between two variables. It is used to estimate values within the range of the data (interpolation).

Common Mistakes

Drawing a curved line instead of a straight ruled line.
Forcing the line to pass through the origin $(0, 0)$ , which is not appropriate here.
Not using a ruler, resulting in a wobbly line.

Things to Be Careful About

The line must be straight and ruled. In part (c), any estimate must be read from this specific line, so the candidate's line determines the acceptable answer range. The line must have a positive gradient.

Techniques used

draw a ruled straight line of best fitbalance points above and below the line

(c)

Asha goes for another walk.
She walks a distance of $6.8\,\text{km}$ .

Use your line of best fit to estimate the time Asha takes for this walk.

______ $\text{minutes}$

DifficultyMedium-Easy

Worked solution

Approach

Locate $6.8\,\text{km}$ on the horizontal axis, move vertically up to the line of best fit, then move horizontally to read the corresponding time on the vertical axis.

Working

Find $6.8$ on the Distance (horizontal) axis. This is four minor gridlines to the left of $7$ .
Move vertically upwards from $x = 6.8$ until you meet the line of best fit.
From that intersection point, move horizontally to the left to read the value on the Time (vertical) axis.

Using a typical line of best fit passing near $(4, 50)$ and $(9, 115)$ :

\text{Gradient} = \frac{115 - 50}{9 - 4} = \frac{65}{5} = 13

At $x = 6.8$ :

y = 50 + 13(6.8 - 4) = 50 + 13(2.8) = 50 + 36.4 = 86.4

Reading from the graph, the time is approximately $86$ to $88\,\text{minutes}$ . A value such as $87\,\text{minutes}$ is acceptable.

Answer

$87\,\text{minutes}$

Final answer

87

Detailed explanation

Walkthrough

We are asked to estimate the time for a walk of $6.8\,\text{km}$ using the line of best fit drawn in part (b).

Locate $6.8$ on the horizontal Distance axis. Since the major gridlines are at every $1\,\text{km}$ and minor at every $0.2\,\text{km}$ , $6.8$ is four minor gridlines to the left of $7$ .
Draw a vertical line up from $x = 6.8$ until it intersects the line of best fit.
From that intersection, draw a horizontal line to the left to read the value on the vertical Time axis.

The exact reading depends on the specific line of best fit the candidate drew, but it should be around $86$ to $88\,\text{minutes}$ . The mark scheme accepts any reasonable reading from a valid line of best fit with a positive gradient.

Key Takeaways

A line of best fit can be used to interpolate (estimate) values within the range of the data. Always read from the line of best fit, not from the raw data points.

Common Mistakes

Reading the value from a raw data point instead of the line of best fit.
Reading the axes in the wrong order (reading distance from the vertical axis).
Not drawing the vertical and horizontal guide lines clearly, leading to an inaccurate reading.

Things to Be Careful About

The answer is dependent on the candidate's line of best fit. Any reading between $85$ and $90$ is likely to be accepted, provided it comes from a straight line with a positive gradient. Give the answer as a whole number of minutes.

Techniques used

locate a value on the horizontal axisread the corresponding value from the line of best fit

Q92MEasyNumber

The diagram shows a rectangle.

Fig. 5

By writing each number correct to 1 significant figure, find an estimate for the area of the rectangle.

______ $\text{mm}^2$

Q10Medium-EasyNumber

(a)

Write 228 as a product of its prime factors.

______

(b)

$228^2 = 51984$

Write 51984 as a product of its prime factors.

______

Q11MediumAlgebra and Graphs

The mass of a small box is $x\,\text{kg}$ .
The mass of a large box is $y\,\text{kg}$ .

(a)

The total mass of 4 small boxes and 6 large boxes is $30\,\text{kg}$ .

Show that $2x + 3y = 15$ .

(b)

The total mass of 6 small boxes and 1 large box is $13\,\text{kg}$ .

Use this information to write down an equation in terms of $x$ and $y$ .

______

(c)

Solve the simultaneous equations to find the mass of a small box and the mass of a large box.
You must show all your working.

Small box = ______ $\text{kg}$
Large box = ______ $\text{kg}$

DifficultyMedium

Worked solution

Approach

We now have a system of two simultaneous equations:

$2x + 3y = 15$
$6x + y = 13$
We will use the elimination method. We will multiply the second equation by 3 so that the $y$ terms match, then subtract to eliminate $y$ .

Working

Equation (1): $2x + 3y = 15$
Equation (2): $6x + y = 13$

Multiply Equation (2) by 3:

3(6x + y) = 3(13)

18x + 3y = 39 \quad \text{--- (3)}

Subtract Equation (1) from Equation (3):

(18x + 3y) - (2x + 3y) = 39 - 15

16x = 24

Solve for $x$ :

x = \frac{24}{16} = \frac{3}{2} = 1.5

Substitute $x = 1.5$ into Equation (2) to find $y$ :

6(1.5) + y = 13

9 + y = 13

y = 13 - 9

y = 4

Check with Equation (1): $2(1.5) + 3(4) = 3 + 12 = 15$ . This is correct.

Answer

Small box = $1.5$ kg
Large box = $4$ kg

Final answer

Small box = 1.5 kg, Large box = 4 kg

Detailed explanation

Walkthrough

We start with the two equations found in the previous parts. To solve them, we need to remove one variable. Looking at $2x + 3y = 15$ and $6x + y = 13$ , it is easy to make the $y$ terms match by multiplying the second equation by 3. This gives $18x + 3y = 39$ . Now both equations have $+3y$ . Subtracting the first equation from this new one removes the $y$ 's, leaving an equation with only $x$ : $16x = 24$ . Solving this gives $x = 1.5$ . Finally, plug $x = 1.5$ back into one of the original equations (the simpler one, $6x + y = 13$ ) to find $y$ . $6 \times 1.5$ is 9, so $9 + y = 13$ , meaning $y = 4$ .

Key Takeaways

The elimination method works best when coefficients are multiples of each other.
Always substitute your answer back into an equation to verify it works.

Common Mistakes

Subtracting the constants incorrectly (e.g., $39 - 15$ calculated wrongly).
Substituting the value of $x$ into the wrong equation or making an arithmetic error during substitution.
Failing to show the working for the elimination step (M1 mark requires visible scaling/subtraction).

Things to Be Careful About

The mark scheme awards marks for the method (M1) and the answers (A2/A1). Even if the final answer is wrong, showing the correct elimination process earns partial credit.
Ensure units (kg) are included in the final answer if required by the format.

Techniques used

form two equations from previous partseliminate one variable by scaling and subtractingsubstitute back to find the second variable

Q12Medium-EasyTransformations and Vectors

Triangle $A$ and triangle $B$ are drawn on the grid.

Fig. 6

(a)

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

______

DifficultyMedium-Easy

Worked solution

Approach

To describe the transformation fully, we need the type of transformation, the angle (if rotation), and the centre (if rotation or enlargement). We test mapping the vertices of triangle $A$ to triangle $B$ and find the fixed point (centre) and the angle.

Working

Triangle $A$ has vertices at $(1, 1)$ , $(4, 1)$ and $(3, 2)$ . Triangle $B$ has vertices at $(-2, -2)$ , $(-2, -5)$ and $(-1, -4)$ .

Join corresponding vertices: $(1, 1)$ to $(-2, -2)$ , and $(4, 1)$ to $(-2, -5)$ . The perpendicular bisectors of these join lines intersect at the centre of rotation.

Midpoint of $(1, 1)$ and $(-2, -2)$ is $(-0.5, -0.5)$ . The slope of the join line is $\frac{-2 - 1}{-2 - 1} = 1$ , so the perpendicular bisector has slope $-1$ :

y - (-0.5) = -1(x - (-0.5)) \\ y = -x - 1

Midpoint of $(4, 1)$ and $(-2, -5)$ is $(1, -2)$ . The slope of the join line is $\frac{-5 - 1}{-2 - 4} = 1$ , so the perpendicular bisector also has slope $-1$ :

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

Both bisectors lie on the line $y = -x - 1$ . Testing the point $(-2, 1)$ :

1 = -(-2) - 1 = 1 \quad \checkmark

So the centre is $(-2, 1)$ . Checking the angle: the vector from $(-2, 1)$ to $(1, 1)$ is $(3, 0)$ . The vector from $(-2, 1)$ to $(-2, -2)$ is $(0, -3)$ . This is a $90^\circ$ clockwise rotation.

Answer

\text{Rotation } 90^\circ \text{ clockwise about } (-2, 1)

Final answer

Rotation 90° clockwise about (-2, 1)

Detailed explanation

Walkthrough

To find the single transformation mapping triangle $A$ onto triangle $B$ , we first observe that the orientation has changed in a way consistent with a rotation. We join corresponding vertices and construct their perpendicular bisectors to find the centre of rotation. The join line from $(1, 1)$ to $(-2, -2)$ has a midpoint of $(-0.5, -0.5)$ and a gradient of $1$ , giving a perpendicular bisector of $y = -x - 1$ . The join line from $(4, 1)$ to $(-2, -5)$ has a midpoint of $(1, -2)$ and the same gradient of $1$ , giving the same perpendicular bisector $y = -x - 1$ . Any point on this line is a candidate for the centre. Testing $(-2, 1)$ , we see it lies on the line. To confirm it is the centre, we check the vectors from $(-2, 1)$ to the original vertex $(1, 1)$ which is $(3, 0)$ , and to the image vertex $(-2, -2)$ which is $(0, -3)$ . The vector $(3, 0)$ rotated $90^\circ$ clockwise gives $(0, -3)$ , confirming the transformation is a rotation of $90^\circ$ clockwise about $(-2, 1)$ .

Key Takeaways

A single rotation is fully described by its centre, angle, and direction. The centre can be found by intersecting the perpendicular bisectors of the lines joining corresponding vertices of the object and its image.

Common Mistakes

Forgetting to state the direction of rotation (clockwise vs anticlockwise).
Giving the centre incorrectly, often by misreading the grid coordinates.
Describing the transformation as a reflection or translation when the orientation clearly indicates rotation.

Things to Be Careful About

A full description of a rotation requires three pieces of information: the type (rotation), the angle ( $90^\circ$ ), and the centre $(-2, 1)$ . Omitting any one of these will cost marks. Always verify the centre by checking the distance from the centre to corresponding vertices is equal.

Techniques used

identify the centre of rotation by intersecting perpendicular bisectors of joining linesdetermine the angle and direction of rotation

(b)

Draw the image of triangle $A$ after a reflection in the line $x = -1$ .

DifficultyMedium-Easy

Worked solution

Approach

Reflect each vertex of triangle $A$ in the line $x = -1$ . For a vertical line $x = k$ , the $y$ -coordinate is unchanged and the new $x$ -coordinate is $k - (x - k) = 2k - x$ .

Working

The line of reflection is $x = -1$ .

Vertex $(1, 1)$ : new $x = 2(-1) - 1 = -3$ . Image is $(-3, 1)$ .
Vertex $(4, 1)$ : new $x = 2(-1) - 4 = -6$ . Image is $(-6, 1)$ .
Vertex $(3, 2)$ : new $x = 2(-1) - 3 = -5$ . Image is $(-5, 2)$ .

Draw the triangle with vertices at $(-6, 1)$ , $(-3, 1)$ and $(-5, 2)$ on the grid.

Answer

Triangle with vertices at $(-6, 1)$ , $(-3, 1)$ and $(-5, 2)$ .

Final answer

Triangle with vertices at (-6, 1), (-3, 1) and (-5, 2)

Detailed explanation

Walkthrough

To reflect triangle $A$ in the line $x = -1$ , we keep the $y$ -coordinates the same and calculate the new $x$ -coordinates. The distance from each vertex to the line $x = -1$ is measured horizontally, and the image is placed the same distance on the opposite side.

For $(1, 1)$ , the distance to $x = -1$ is $2$ units to the right. The image is $2$ units to the left: $x = -1 - 2 = -3$ . So $(-3, 1)$ .
For $(4, 1)$ , the distance is $5$ units to the right. The image is $5$ units to the left: $x = -1 - 5 = -6$ . So $(-6, 1)$ .
For $(3, 2)$ , the distance is $4$ units to the right. The image is $4$ units to the left: $x = -1 - 4 = -5$ . So $(-5, 2)$ .

Draw and label the new triangle on the grid.

Key Takeaways

Reflection in a vertical line $x = k$ leaves $y$ unchanged and maps $x$ to $2k - x$ . The shape and size remain the same, but the orientation is reversed.

Common Mistakes

Forgetting to change the sign of the distance from the line of reflection.
Drawing the triangle in the wrong position or with incorrect dimensions.

Things to Be Careful About

The answer must be drawn on the grid. Ensure the vertices are plotted exactly at the calculated coordinates and the triangle is closed and labelled clearly.

Techniques used

reflect each vertex across the given vertical linedraw and label the reflected triangle

(c)

Triangle $A$ is enlarged with scale factor $k$ and centre $(0, 0)$ .
Triangle $C$ is the image of triangle $A$ after the enlargement.
The coordinates of one vertex of triangle $C$ are $(12, 3)$ .

3M

(i)

Find the value of $k$ .

$k$ = ______

DifficultyEasy

Worked solution

Approach

Under an enlargement with centre $(0, 0)$ and scale factor $k$ , a point $(x, y)$ maps to $(kx, ky)$ . We use the given vertex of $A$ and its corresponding vertex of $C$ to find $k$ .

Working

The vertices of triangle $A$ are $(1, 1)$ , $(4, 1)$ and $(3, 2)$ . One vertex of triangle $C$ is $(12, 3)$ .

Testing each vertex of $A$ multiplied by a scale factor $k$ :

If $(1, 1) \to (k, k)$ , this cannot be $(12, 3)$ since $12 \neq 3$ .
If $(4, 1) \to (4k, k)$ , setting $4k = 12$ gives $k = 3$ , and $k = 3$ gives $(12, 3)$ . This matches.
If $(3, 2) \to (3k, 2k)$ , this would be $(9, 6)$ for $k = 3$ .

So the scale factor is $k = 3$ .

Answer

k = 3

Final answer

3

Detailed explanation

Walkthrough

An enlargement with centre $(0, 0)$ and scale factor $k$ maps any point $(x, y)$ to $(kx, ky)$ . We are given that one vertex of the image triangle $C$ is $(12, 3)$ . We test the vertices of triangle $A$ to see which one maps to $(12, 3)$ when multiplied by $k$ .

$(1, 1) \times k = (k, k)$ . For this to be $(12, 3)$ , we would need $k = 12$ and $k = 3$ , which is impossible.
$(4, 1) \times k = (4k, k)$ . Setting $4k = 12$ gives $k = 3$ . Checking the $y$ -coordinate: $1 \times 3 = 3$ , which matches the $y$ -coordinate of $(12, 3)$ .
$(3, 2) \times k = (3k, 2k)$ . For $k = 3$ , this is $(9, 6)$ , which is not $(12, 3)$ .

Thus, the vertex $(4, 1)$ of $A$ maps to $(12, 3)$ of $C$ , and the scale factor is $k = 3$ .

Key Takeaways

For an enlargement from the origin, the scale factor is the ratio of any coordinate of the image to the corresponding coordinate of the object: $k = \frac{x'}{x} = \frac{y'}{y}$ .

Common Mistakes

Trying to match the wrong vertices (e.g., assuming $(1, 1)$ maps to $(12, 3)$ ).
Calculating $k$ incorrectly by dividing in the wrong order (object divided by image instead of image divided by object).

Things to Be Careful About

The scale factor must be consistent for both $x$ and $y$ coordinates. Always verify by checking both coordinates of the matched vertex pair.

Techniques used

use the given vertex and its image to calculate the scale factordivide image coordinates by original coordinates

(ii)

Find the coordinates of the other two vertices of triangle $C$ .

( ______ , ______ ) and ( ______ , ______ )

Q13Medium-EasyNumber

In a sale, a shop reduces all prices by 20%.

(a)

A coat costs $85 before the sale.

Work out the sale price of the coat.

$ ______

(b)

The sale price of a shirt is $40.

Work out the cost of the shirt before the sale.

$ ______

DifficultyMedium-Easy

Worked solution

Approach

The sale price is after a 20% reduction, so it represents 80% of the original price. Let the original price be $x$ . Then 80% of $x$ is $40. Solve for $x$ .

Working

Let the original price be $x$ . Since the sale price is 80% of the original price:

\frac{80}{100}x = 40

Simplify the fraction:

\frac{4}{5}x = 40

Multiply both sides by $\frac{5}{4}$ :

x = 40 \times \frac{5}{4} = 50

Answer

The cost of the shirt before the sale was $50.

Final answer

50 dollars

Detailed explanation

Walkthrough

The shirt's sale price is $40 after a 20% reduction. That means the $40 is 80% of the original price, not 100%. Let the original price be $x$ . Write 80% as $\frac{80}{100}$ , so the equation is $\frac{80}{100}x = 40$ . Simplify $\frac{80}{100}$ to $\frac{4}{5}$ . Then multiply both sides by $\frac{5}{4}$ to get $x = 40 \times \frac{5}{4} = 50$ . So the original price was $50. This matches the mark scheme: M1 for forming the equation (or equivalent, seen or implied).

Key Takeaways

In a reverse percentage problem, the given amount is the result after the change, so divide by the remaining percentage.
A 20% reduction leaves 80%, so original = sale price $\div 0.8$ = sale price $\times \frac{5}{4}$ .
Always identify whether the given amount is before or after the percentage change.

Common Mistakes

Multiplying 40 by 1.2 to get 48; this is a 20% increase, not the reverse of a 20% decrease.
Multiplying 40 by 0.2 and subtracting (40 - 8 = 32); this treats 40 as the original price, but 40 is the sale price.
Forgetting that 40 is 80% of the original, not 100%.
Not showing the equation; the mark scheme requires M1 for the equation, seen or implied.

Things to Be Careful About

The original price is larger than the sale price, so the answer must be greater than 40. If you get less than 40, check your method.
On the non-calculator paper, simplify $\frac{80}{100}$ to $\frac{4}{5}$ and multiply by hand: $40 \times \frac{5}{4} = 50$ .
The final answer should be in dollars: $50.

Techniques used

recognise the sale price as a percentage of the originalform a linear equationsolve by multiplying by the reciprocal

Q14MediumTransformations and VectorsCoordinate GeometryNumber

Fig. 7

$A$ is the point $(-4, 5)$ and $B$ is the point $(6, 1)$ .

\overrightarrow{BC} = \begin{pmatrix}-3 \\ -4\end{pmatrix}

(a)

Find the column vector $\overrightarrow{AB}$ .

$\overrightarrow{AB}$ = ______

DifficultyEasy

Worked solution

Approach

To find the column vector $\overrightarrow{AB}$ , subtract the position vector of $A$ from the position vector of $B$ :

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}

Working

The position vectors of $A(-4, 5)$ and $B(6, 1)$ are:

\overrightarrow{OA} = \begin{pmatrix} -4 \\ 5 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}

Calculate $\overrightarrow{AB}$ :

\overrightarrow{AB} = \begin{pmatrix} 6 \\ 1 \end{pmatrix} - \begin{pmatrix} -4 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 - (-4) \\ 1 - 5 \end{pmatrix} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}

Answer

\begin{pmatrix} 10 \\ -4 \end{pmatrix}

Final answer

(10, -4)

Detailed explanation

Walkthrough

To go from point $A$ to point $B$ , we determine the horizontal and vertical displacements:

The horizontal movement ( $x$ -component) is the final $x$ -coordinate minus the initial $x$ -coordinate:

6 - (-4) = 10

The vertical movement ( $y$ -component) is the final $y$ -coordinate minus the initial $y$ -coordinate:

1 - 5 = -4

Writing these as a column vector gives $\begin{pmatrix} 10 \\ -4 \end{pmatrix}$ .

Key Takeaways

For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ , the column vector $\overrightarrow{AB}$ is given by $\begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$ .

Common Mistakes

Subtracting in the wrong direction (calculating $\overrightarrow{BA}$ instead of $\overrightarrow{AB}$ ), which leads to $\begin{pmatrix} -10 \\ 4 \end{pmatrix}$ .
Sign errors when subtracting negative numbers, e.g., writing $6 - (-4) = 2$ instead of $10$ .

Things to Be Careful About

Ensure the answer is written as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$ with top and bottom components, not as a coordinate or fraction.

Techniques used

calculate a displacement vector from two pointssubtract position vectors

(b)

Find the coordinates of point $C$ .

( ______ , ______ )

(c)

$ABCD$ is a trapezium.
$AB$ is parallel to $DC$ and $AB = 2DC$ .

5M

(i)

Find the coordinates of point $D$ .

( ______ , ______ )

DifficultyMedium

Worked solution

Approach

Since $ABCD$ is a trapezium with $AB$ parallel to $DC$ and $AB = 2DC$ , the vector $\overrightarrow{DC}$ is in the same direction as $\overrightarrow{AB}$ and has half its magnitude:

\overrightarrow{DC} = \frac{1}{2} \overrightarrow{AB}

To find the position vector of $D$ , use:

\overrightarrow{OD} = \overrightarrow{OC} - \overrightarrow{DC}

Working

From part (a), $\overrightarrow{AB} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}$ . Thus:

\overrightarrow{DC} = \frac{1}{2}\begin{pmatrix} 10 \\ -4 \end{pmatrix} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}

From part (b), $\overrightarrow{OC} = \begin{pmatrix} 3 \\ -3 \end{pmatrix}$ .

Now, calculate $\overrightarrow{OD}$ :

\overrightarrow{OD} = \begin{pmatrix} 3 \\ -3 \end{pmatrix} - \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 - 5 \\ -3 - (-2) \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \end{pmatrix}

Therefore, the coordinates of point $D$ are $(-2, -1)$ .

Answer

(-2, -1)

Final answer

(-2, -1)

Detailed explanation

Walkthrough

In the trapezium $ABCD$ , the sides are traversed in the order $A \to B \to C \to D$ . Thus, the side parallel to vector $\overrightarrow{AB}$ is vector $\overrightarrow{DC}$ .
Since $AB$ is parallel to $DC$ and $AB = 2DC$ , the vector $\overrightarrow{DC} = \frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ .
Since point $C$ is at $(3, -3)$ , going from $D$ to $C$ adds $\begin{pmatrix} 5 \\ -2 \end{pmatrix}$ . To go backwards from $C$ to $D$ , we subtract this vector:

x_D = 3 - 5 = -2

y_D = -3 - (-2) = -1

So $D$ has coordinates $(-2, -1)$ .

Key Takeaways

In a polygon named $ABCD$ , the order of vertices means side $AB$ corresponds vectorially to $\overrightarrow{AB}$ , and the opposite side directed the same way is $\overrightarrow{DC}$ (not $\overrightarrow{CD}$ ).
Parallel vectors with a known ratio of lengths are scalar multiples of each other.

Common Mistakes

Taking $\overrightarrow{CD} = \frac{1}{2}\overrightarrow{AB}$ , which reverses the direction and gives the wrong coordinates for $D$ .
Adding $\overrightarrow{DC}$ to $\overrightarrow{OC}$ instead of subtracting it.

Things to Be Careful About

Be mindful of negative signs when subtracting components, e.g., $-3 - (-2) = -1$ .

Techniques used

use parallel vector propertiesscale a vectorfind a position vector by subtracting a vector

(ii)

Find the length of line $AD$ .
Give your answer as a surd in its simplest form.

______

DifficultyMedium-Easy

Worked solution

Approach

Use the distance formula between points $A(-4, 5)$ and $D(-2, -1)$ :

AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Then simplify the resulting surd into the form $k\sqrt{n}$ .

Working

Substitute the coordinates of $A(-4, 5)$ and $D(-2, -1)$ :

\begin{aligned} AD &= \sqrt{((-2) - (-4))^2 + ((-1) - 5)^2} \\ &= \sqrt{(2)^2 + (-6)^2} \\ &= \sqrt{4 + 36} \\ &= \sqrt{40} \end{aligned}

Simplify the surd $\sqrt{40}$ :

\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}

Answer

2\sqrt{10}

Final answer

$2\sqrt{10}$

Detailed explanation

Walkthrough

Find the horizontal and vertical distances between $A(-4, 5)$ $A (- 4, 5)$ and $D(-2, -1)$ $D (- 2, - 1)$ :
- Change in $x$ : $-2 - (-4) = 2$
- Change in $y$ : $-1 - 5 = -6$
Use Pythagoras' theorem to find the length of line segment $AD$ :

AD^2 = 2^2 + (-6)^2 = 4 + 36 = 40

Take the square root: $AD = \sqrt{40}$ .
Simplify the surd by identifying the largest square factor of $40$ , which is $4$ :

\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}

Key Takeaways

The length of a line segment is found using $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ .
A surd $\sqrt{N}$ is in its simplest form when no square factors remain inside the square root.

Common Mistakes

Forgetting to square the negative value properly, e.g., evaluating $(-6)^2$ as $-36$ instead of $+36$ .
Leaving the answer unsimplified as $\sqrt{40}$ , which loses the final mark.
Converting to a decimal (e.g. $6.32$ ) instead of keeping the exact surd form requested.

Things to Be Careful About

The question specifically states: "Give your answer as a surd in its simplest form." Exact simplified surd form is required.

Techniques used

apply distance formula between two pointssimplify a surd

Q15Medium-EasyNumber

(a)

Simplify.

\sqrt{175} - \sqrt{28}

______

DifficultyMedium-Easy

Worked solution

Approach

Simplify each surd by extracting the largest square factor from the radicand, then subtract the resulting like surds.

Working

\sqrt{175} = \sqrt{25 \times 7} = \sqrt{25} \times \sqrt{7} = 5\sqrt{7}

\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}

Now subtract:

5\sqrt{7} - 2\sqrt{7} = 3\sqrt{7}

Answer

3\sqrt{7}

Final answer

3√7

Detailed explanation

Walkthrough

We need to simplify $\sqrt{175} - \sqrt{28}$ . Neither 175 nor 28 is a perfect square, but each has a square factor. $175 = 25 \times 7$ and 25 is a perfect square, so $\sqrt{175} = \sqrt{25 \times 7} = \sqrt{25} \times \sqrt{7} = 5\sqrt{7}$ . Similarly $28 = 4 \times 7$ and 4 is a perfect square, so $\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}$ . Now both terms are multiples of $\sqrt{7}$ , so we can combine them like terms: $5\sqrt{7} - 2\sqrt{7} = 3\sqrt{7}$ . The answer is $3\sqrt{7}$ .

Key Takeaways

The core skill is recognising the largest square factor of a number under a square root and extracting it. Once both surds share the same irrational part, they combine exactly like algebraic terms.

Common Mistakes

Not extracting the largest square factor (e.g. writing $\sqrt{175} = \sqrt{5 \times 35}$ and getting stuck).
Forgetting that $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ only when both $a$ and $b$ are non-negative.
Mixing up the signs when subtracting.

Things to Be Careful About

The mark scheme awards B1 for either $5\sqrt{7}$ or $2\sqrt{7}$ , and the final answer must be exactly $3\sqrt{7}$ (cao). Ensure both surds are fully simplified before combining. This is a non-calculator component, so all working must be by hand.

Techniques used

simplify surds by extracting square factorscombine like surds

(b)

Rationalise the denominator.

\frac{1}{\sqrt{5}}

______

Q16Medium-EasyStatistics

A group of 80 people each record their journey time from home to work one day.
The cumulative frequency diagram shows the results.

Fig. 8

(a)

Use the cumulative frequency diagram to find an estimate of

5M

(i)

the median

______ $\text{minutes}$

(ii)

the interquartile range

______ $\text{minutes}$

DifficultyMedium-Easy

Worked solution

Approach

The lower quartile (LQ) is at cumulative frequency $\frac{1}{4} \times 80 = 20$ , and the upper quartile (UQ) is at cumulative frequency $\frac{3}{4} \times 80 = 60$ . Read these values from the graph and subtract LQ from UQ to find the interquartile range.

Working

Lower quartile cumulative frequency = $80 \div 4 = 20$ .
On the diagram, CF = 20 corresponds to a journey time of 25 minutes. So LQ = 25 minutes.

Upper quartile cumulative frequency = $80 \times \frac{3}{4} = 60$ .
On the diagram, CF = 60 corresponds to a journey time of 35 minutes. So UQ = 35 minutes.

Interquartile range = UQ - LQ = $35 - 25 = 10$ minutes.

Answer

10

Final answer

10

Detailed explanation

Walkthrough

The interquartile range (IQR) measures the spread of the middle 50% of the data. It is calculated as Upper Quartile (UQ) minus Lower Quartile (LQ).

The LQ is at $\frac{1}{4}$ of the total frequency: $\frac{80}{4} = 20$ . Reading across from 20 on the vertical axis to the curve and down to the horizontal axis gives 25 minutes.
The UQ is at $\frac{3}{4}$ of the total frequency: $\frac{3 \times 80}{4} = 60$ . Reading across from 60 on the vertical axis to the curve and down to the horizontal axis gives 35 minutes.
IQR = $35 - 25 = 10$ minutes.

Key Takeaways

LQ and UQ on a cumulative frequency graph are found at $\frac{1}{4}$ and $\frac{3}{4}$ of the total frequency, respectively. The IQR is the difference between these two values.

Common Mistakes

Using the wrong fraction for the quartiles (e.g., using $\frac{1}{2}$ for LQ).
Reading the frequency value (20 or 60) as the answer instead of the corresponding time value on the horizontal axis.
Subtracting in the wrong order (LQ - UQ gives a negative number).

Things to Be Careful About

Ensure you read the time values from the horizontal axis, not the cumulative frequency values from the vertical axis. The scale on the horizontal axis has 5 small squares between each 10-minute mark, meaning each small square represents 2 minutes.

Techniques used

calculate lower and upper quartile cumulative frequenciesread quartile values from the cumulative frequency graphsubtract lower quartile from upper quartile

(iii)

the number of people who had a journey time of 40 minutes or more.

______

(b)

Each of the 80 people also record their journey time from work to home that day.
The table shows the results.

Median	35 minutes
Interquartile range	15 minutes

Jay says:

"The journey times from home to work are more consistent than the journey times from work to home."

Is Jay correct?
Explain how you decide.

______ because ______

DifficultyMedium-Easy

Worked solution

Approach

Consistency of data is measured by the spread, such as the interquartile range (IQR). A smaller IQR indicates more consistent data. Compare the IQR of the journey from home to work with the IQR of the journey from work to home.

Working

From part (a)(ii), the IQR for the journey from home to work is 10 minutes.
The IQR for the journey from work to home is given as 15 minutes.

Since $10 < 15$ , the journey times from home to work have a smaller spread and are therefore more consistent.

Jay is correct.

Answer

Yes, because the IQR for the journey from home to work (10 minutes) is lower than the IQR for the journey from work to home (15 minutes).

Final answer

Yes, because the IQR for the journey from home to work is lower (10 minutes compared to 15 minutes).

Detailed explanation

Walkthrough

Jay claims that the journey times from home to work are more consistent than from work to home. Consistency in statistics is indicated by a smaller spread of data, which can be measured using the interquartile range (IQR).

From part (a)(ii), the IQR for the journey from home to work is 10 minutes.
The table gives the IQR for the journey from work to home as 15 minutes.
Since 10 minutes is less than 15 minutes, the home-to-work journey times are less spread out and therefore more consistent.
Jay's statement is correct.

Key Takeaways

A lower interquartile range (IQR) indicates that the middle 50% of the data is more tightly clustered, meaning the data is more consistent.

Common Mistakes

Comparing the medians instead of the IQRs to determine consistency. The median measures central tendency, not spread.
Failing to explicitly state the numerical values being compared in the explanation.

Things to Be Careful About

When comparing consistency, always use a measure of spread such as the IQR or range, not the median or mean. The explanation must clearly state which value is lower and conclude accordingly.

Techniques used

compare interquartile ranges to assess consistency

Q17MediumAlgebra and GraphsCoordinate Geometry

The equation of line $L$ is $5y + 3x = 10$ .

(a)

Rearrange $5y + 3x = 10$ to make $y$ the subject.

$y$ = ______

DifficultyMedium-Easy

Worked solution

Approach

To make $y$ the subject of the equation $5y + 3x = 10$ , we need to isolate $y$ on one side of the equation. This involves moving the $x$ term to the other side and then dividing by the coefficient of $y$ .

Working

Start with the given equation:

5y + 3x = 10

Subtract $3x$ from both sides to move the $x$ term:

5y = 10 - 3x

Divide every term by $5$ to isolate $y$ :

y = \frac{10}{5} - \frac{3x}{5}

Simplify the fraction $\frac{10}{5}$ :

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

This can also be written as:

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

Answer

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

Final answer

y = 2 - 3/5 x

Detailed explanation

Walkthrough

The goal is to write the equation in the form $y = mx + c$ .

We start with $5y + 3x = 10$ . The term containing $y$ is $5y$ . To get $y$ by itself, we first remove the $+ 3x$ term. We do this by subtracting $3x$ from both sides, giving $5y = 10 - 3x$ . This corresponds to mark M1 for isolating $5y$ or an equivalent step like dividing by 5 immediately ( $y + \frac{3}{5}x = 2$ ).
Next, we divide the entire equation by the coefficient of $y$ , which is $5$ . This gives $y = \frac{10 - 3x}{5}$ .
Simplifying each term separately results in $y = 2 - \frac{3}{5}x$ . This is the required form.

Key Takeaways

When rearranging equations, perform inverse operations in reverse order of operations (BIDMAS/BODMAS). Here, subtraction (of $3x$ ) comes before division (by $5$ ).
Always check that the subject variable has a coefficient of $1$ .

Common Mistakes

Forgetting to divide the $x$ term by $5$ when isolating $y$ . A common error is writing $y = 10 - \frac{3}{5}x$ .
Sign errors: forgetting that subtracting $3x$ makes it negative on the right-hand side.
Not simplifying the constant term ( $\frac{10}{5} = 2$ ).

Things to Be Careful About

The question asks for $y$ as the subject. Do not stop at $5y = 10 - 3x$ .
The final answer should be simplified. $2 - \frac{3}{5}x$ is preferred over $\frac{10 - 3x}{5}$ unless specified otherwise, though both are often accepted as 'oe' (other equivalent). However, separating the terms makes the gradient and y-intercept clear.

Techniques used

rearrange a linear equation to make y the subjectdivide by the coefficient of ysimplify the constant term

(b)

Line $P$ is perpendicular to line $L$ .
Line $P$ passes through the point $(6, 7)$ .

Find the equation of line $P$ .

______

DifficultyMedium

Worked solution

Approach

Line $P$ is perpendicular to line $L$ . First, identify the gradient of line $L$ from the answer to part (a). Then, use the property that the product of gradients of perpendicular lines is $-1$ to find the gradient of line $P$ . Finally, substitute the coordinates of the given point $(6, 7)$ into the equation $y = m_P x + c$ to find the y-intercept $c$ .

Working

From part (a), the equation of line $L$ is:

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

The gradient of line $L$ , $m_L$ , is $-\frac{3}{5}$ .

Let $m_P$ be the gradient of line $P$ . Since line $P$ is perpendicular to line $L$ :

m_L \times m_P = -1

-\frac{3}{5} \times m_P = -1

m_P = \frac{-1}{-\frac{3}{5}} = \frac{5}{3}

So, the equation of line $P$ is of the form:

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

Line $P$ passes through the point $(6, 7)$ . Substitute $x = 6$ and $y = 7$ into the equation to find $c$ :

7 = \frac{5}{3}(6) + c

Calculate $\frac{5}{3} \times 6$ :

\frac{5}{3} \times 6 = 5 \times 2 = 10

So:

7 = 10 + c

Solve for $c$ :

c = 7 - 10 = -3

Substitute $c = -3$ back into the equation for line $P$ :

y = \frac{5}{3}x - 3

Answer

y = \frac{5}{3}x - 3

Final answer

y = 5/3 x - 3

Detailed explanation

Walkthrough

Find the gradient of L: From part (a), we established that line $L$ has the equation $y = 2 - \frac{3}{5}x$ . The coefficient of $x$ is the gradient, so $m_L = -\frac{3}{5}$ .
Find the gradient of P: Perpendicular lines have gradients that are negative reciprocals of each other. If $m_L = -\frac{3}{5}$ , then $m_P = -\frac{1}{m_L} = -\frac{1}{-3/5} = \frac{5}{3}$ . This earns B1 if stated correctly or derived.
Use the point to find the intercept: Line $P$ has the form $y = \frac{5}{3}x + c$ . It passes through $(6, 7)$ . Substituting these values gives $7 = \frac{5}{3}(6) + c$ . Calculating $\frac{5}{3} \times 6$ yields $10$ . So, $7 = 10 + c$ . Solving for $c$ gives $c = -3$ . This substitution step earns M1.
Write the final equation: Combining the gradient and intercept gives $y = \frac{5}{3}x - 3$ .

Key Takeaways

The condition for perpendicularity is $m_1 m_2 = -1$ . This means you flip the fraction and change the sign.
Once you have the gradient, any single point on the line is sufficient to find the exact equation.

Common Mistakes

Taking the wrong negative reciprocal (e.g., keeping the sign same or flipping without changing sign).
Arithmetic errors when calculating $\frac{5}{3} \times 6$ . Remember to simplify before multiplying.
Errors in solving for $c$ : $7 - 10 = -3$ , not $3$ .
Writing the equation in a different form (e.g., $3y = 5x - 9$ ) without converting to $y = mx + c$ if that format is expected, although $3y = 5x - 9$ is mathematically equivalent. The mark scheme accepts 'oe' but standard form $y=...$ is safest.

Things to Be Careful About

Ensure you use the gradient from your answer in part (a). If you made an error in (a), you might still get marks for method in (b) if you use your own incorrect gradient consistently ('follow-through'), but the final answer would be based on that error. In this case, using the correct gradient $-3/5$ leads to $5/3$ .
The question does not specify the form of the final answer, but $y = mx + c$ is standard for such questions.

Techniques used

identify the gradient of the original linecalculate the negative reciprocal gradient for a perpendicular linesubstitute a point to find the y-interceptform the equation of the new line

Q18MediumAlgebra and Graphs

The diagram shows the speed–time graph for a cyclist's journey.

Fig. 9

(a)

Describe the motion of the cyclist between $t = 40$ and $t = 100$ .

______

(b)

The acceleration of the cyclist between $t = 0$ and $t = 40$ is $0.25\,\text{m/s}^2$ .

Show that $v = 10$ .

DifficultyEasy

Worked solution

Approach

Acceleration is defined as the rate of change of speed. On a speed-time graph, this is the gradient of the line. For a straight line, gradient $= \frac{\text{change in speed}}{\text{change in time}}$ .

Working

The cyclist accelerates from $t = 0$ to $t = 40$ . The speed increases from $0$ to $v$ .

\text{Acceleration} = \frac{v - 0}{40 - 0}

Given that the acceleration is $0.25\,\text{m/s}^2$ :

0.25 = \frac{v}{40}

Multiply both sides by $40$ :

v = 0.25 \times 40 = 10

Answer

$v = 10$ is shown.

Final answer

v = 10

Detailed explanation

Walkthrough

Acceleration is the gradient of the speed-time graph. Between $t = 0$ and $t = 40$ , the graph is a straight line from $(0, 0)$ to $(40, v)$ . The gradient is $\frac{v - 0}{40 - 0} = \frac{v}{40}$ . Setting this equal to the given acceleration of $0.25$ and solving for $v$ gives $v = 10$ .

Key Takeaways

Acceleration is the gradient of a speed-time graph.
For a straight line, gradient $= \frac{\Delta y}{\Delta x} = \frac{\text{change in speed}}{\text{change in time}}$ .

Common Mistakes

Using the wrong formula for acceleration, such as $\text{speed} \times \text{time}$ .
Forgetting that the initial speed is $0$ at $t = 0$ .

Things to Be Careful About

The mark scheme requires showing the working $0.25 \times 40 = 10$ . Simply writing $v = 10$ without method will not score the mark.
Units: acceleration is in $\text{m/s}^2$ , time is in seconds, so speed is in $\text{m/s}$ . No unit conversion is needed here.

Techniques used

use acceleration as the gradient of a speed-time graph

(c)

The total distance travelled by the cyclist between $t = 0$ and $t = T$ is $1.4\,\text{km}$ .

Find the value of $T$ .

$T$ = ______

DifficultyMedium

Worked solution

Approach

The total distance travelled is the area under the speed-time graph. The graph forms a trapezium with parallel sides along the time axis. We must first convert the total distance to metres to match the speed units, then equate the area of the trapezium to this distance and solve for $T$ .

Working

Convert the total distance to metres:

1.4\,\text{km} = 1400\,\text{m}

The area under the graph is a trapezium with:

Parallel sides: $T$ (the total time) and $(100 - 40) = 60$ (the time at constant speed).
Height: $v = 10\,\text{m/s}$ (from part (b)).

The formula for the area of a trapezium is:

\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

Substitute the known values:

1400 = \frac{1}{2} \times (T + 60) \times 10

Simplify the right side:

1400 = 5(T + 60)

Divide both sides by $5$ :

280 = T + 60

Solve for $T$ :

T = 280 - 60 = 220

Answer

T = 220

Final answer

220

Detailed explanation

Walkthrough

Distance is the area under a speed-time graph. The total distance is given as $1.4\,\text{km}$ , which must be converted to $1400\,\text{m}$ to be consistent with the speed in $\text{m/s}$ . The shape under the graph is a trapezium. The parallel sides are the total duration $T$ and the duration of constant speed $(100 - 40 = 60)$ . The height is the constant speed $v = 10\,\text{m/s}$ . Using the trapezium area formula $\frac{1}{2}(a+b)h$ , we set $\frac{1}{2}(T + 60) \times 10 = 1400$ . Solving this linear equation gives $T = 220$ .

Key Takeaways

Distance travelled $=$ area under a speed-time graph.
The area can be split into simpler shapes (triangle, rectangle, trapezium) or calculated directly using the trapezium formula if the graph forms one.
Always check unit consistency (km vs m, hours vs seconds) before equating areas to distances.

Common Mistakes

Forgetting to convert $1.4\,\text{km}$ to $1400\,\text{m}$ , leading to an incorrect equation like $1.4 = 5(T + 60)$ .
Misidentifying the parallel sides of the trapezium. The parallel sides are horizontal: the total time $T$ and the constant-speed time $(100 - 40 = 60)$ . The height is the vertical speed $v = 10$ .
Calculating the area as a rectangle plus two triangles and making an arithmetic error.

Things to Be Careful About

The mark scheme awards method marks for a correct method to find a relevant area under the graph, or specifically for the equation $\frac{1}{2} \times 10 \times (T + 60) = 1400$ .
Final answer must be a number; $T = 220$ is correct. If units were required, it would be seconds, but the question asks for the value of $T$ .
Ensure the equation is solved correctly: $5(T + 60) = 1400 \Rightarrow T + 60 = 280 \Rightarrow T = 220$ . Do not stop at $T + 60 = 280$ .

Techniques used

convert distance from kilometres to metrescalculate the area of a trapezium under a speed-time graphsolve a linear equation for an unknown time

Q194MMedium-EasyAlgebra and Graphs

Simplify.

\frac{3x^2 - 12}{2x^2 + 11x + 14}

______

DifficultyMedium-Easy

Worked solution

Approach

To simplify the rational expression, we must factorise both the numerator and the denominator completely. Once in factored form, any common factors can be cancelled.

Working

Step 1: Factorise the numerator.
The numerator is $3x^2 - 12$ . First, identify a common numerical factor of 3:

3x^2 - 12 = 3(x^2 - 4)

The expression inside the bracket, $x^2 - 4$ , is a difference of two squares ( $a^2 - b^2 = (a-b)(a+b)$ ):

x^2 - 4 = (x + 2)(x - 2)

So, the fully factored numerator is:

3(x + 2)(x - 2)

Step 2: Factorise the denominator.
The denominator is $2x^2 + 11x + 14$ . We look for two numbers that multiply to give $a \times c = 2 \times 14 = 28$ and add to give $b = 11$ . These numbers are 7 and 4.
Split the middle term using these numbers:

2x^2 + 4x + 7x + 14

Group the terms:

(2x^2 + 4x) + (7x + 14)

Factorise each group:

2x(x + 2) + 7(x + 2)

Extract the common bracket $(x + 2)$ :

(2x + 7)(x + 2)

Step 3: Cancel common factors.
Substitute the factored forms back into the fraction:

\frac{3(x + 2)(x - 2)}{(2x + 7)(x + 2)}

The term $(x + 2)$ appears in both the numerator and the denominator, so it cancels out (assuming $x \neq -2$ ):

\frac{3(x - 2)}{2x + 7}

This can also be written as $\frac{3x - 6}{2x + 7}$ .

Answer

\frac{3(x - 2)}{2x + 7}

Final answer

3(x-2)/(2x+7)

Detailed explanation

Walkthrough

The question asks us to simplify a fraction containing polynomials. The key strategy here is factorisation. If we can write the top and bottom as products of simpler brackets, we can remove anything that appears on both sides.

First, look at the top part (numerator): $3x^2 - 12$ . Both terms share a factor of 3. Pulling that out leaves $x^2 - 4$ . This is a classic 'difference of two squares' because $x^2$ is a square and 4 is a square. It splits into $(x+2)(x-2)$ .

Next, look at the bottom part (denominator): $2x^2 + 11x + 14$ . This is a quadratic with a leading coefficient greater than 1. A reliable method is to find two numbers that multiply to $2 \times 14 = 28$ and add to 11. Those numbers are 7 and 4. We rewrite $11x$ as $4x + 7x$ , then group terms to find the common factor $(x+2)$ . This results in $(2x+7)(x+2)$ .

Now the fraction is:

\frac{3(x + 2)(x - 2)}{(2x + 7)(x + 2)}

We see $(x+2)$ is on the top and bottom. We cancel it. What remains is $\frac{3(x-2)}{2x+7}$ .

Key Takeaways

Always check for a common numerical factor before attempting complex quadratic factorisation.
Recognise the 'difference of two squares' pattern ( $a^2 - b^2$ ).
For quadratics where the first term has a coefficient (like $2x^2$ ), finding factor pairs of $a \times c$ helps split the middle term correctly.

Common Mistakes

Incorrect factorisation of the numerator: Students might write $3(x^2 - 4)$ but forget to expand the difference of squares further, losing marks for incomplete simplification.
Wrong sign in difference of squares: Writing $(x-2)^2$ or $(x+2)^2$ instead of $(x+2)(x-2)$ .
Factoring errors in the denominator: Choosing number pairs that multiply to 28 but don't sum to 11 (e.g., 2 and 14). Or failing to distribute the leading coefficient correctly during grouping (e.g., writing $(x+2)(x+7)$ which expands to $x^2+8x+14$ ).
Leaving uncancelled factors: Failing to spot that $(x+2)$ is present in both parts.
Expanding the final answer: While $\frac{3x-6}{2x+7}$ is correct, usually the factored form is preferred unless specified otherwise, though both are accepted per the mark scheme.

Things to Be Careful About

Ensure you have found the greatest common factor in the numerator. Missing the 3 would lead to an incorrect denominator match later.
When splitting the middle term in the denominator, verify your result by expanding $(2x+7)(x+2)$ to ensure it equals $2x^2 + 11x + 14$ .

Techniques used

factorise the numerator by taking out a common factorfactorise the quadratic denominator by splitting the middle termcancel common binomial factors

Q204MMediumNumber

Work out.

0.1\dot{7} + \frac{5}{9}

Give your answer as a fraction in its simplest form.

______

DifficultyMedium

Worked solution

Approach

Write the recurring decimal as a fraction, then add $\frac{5}{9}$ using a common denominator.

Working

Let

x = 0.\dot{1}\dot{7} = 0.171717\ldots

Since the repeating block has 2 digits, multiply by $100$ :

100x = 17.171717\ldots

Subtract $x$ from $100x$ :

99x = 17

x = \frac{17}{99}

Now add $\frac{5}{9}$ :

\frac{17}{99} + \frac{5}{9} = \frac{17}{99} + \frac{55}{99} = \frac{72}{99}

Simplify by dividing the numerator and denominator by $9$ :

\frac{72}{99} = \frac{8}{11}

Answer

\frac{8}{11}

Final answer

8/11

Detailed explanation

Walkthrough

The decimal $0.\dot{1}\dot{7}$ means $0.171717\ldots$ , where the block $17$ repeats forever. To change this into a fraction, set $x = 0.171717\ldots$ and multiply by $100$ because the repeating block has two digits:

100x = 17.171717\ldots

Subtracting the original equation removes the recurring tail:

99x = 17

so

x = \frac{17}{99}.

Next, write $\frac{5}{9}$ with denominator $99$ :

\frac{5}{9} = \frac{55}{99}.

Add the two fractions:

\frac{17}{99} + \frac{55}{99} = \frac{72}{99}.

Finally simplify by dividing top and bottom by $9$ :

\frac{72}{99} = \frac{8}{11}.

Key Takeaways

A recurring decimal can be converted to a fraction by choosing the power of $10$ that matches the length of the repeating block, multiplying, and subtracting. Then use a common denominator to add fractions, and always reduce the final result.

Common Mistakes

Multiplying by $10$ instead of $100$ when the repeating block has two digits; this does not line up the recurring tails.
Leaving the answer as $\frac{72}{99}$ because the question asks for simplest form.
Misreading the recurring notation and using the wrong decimal; here the repeating block is $17$ .
Adding without a common denominator, which would give an invalid fraction.

Things to Be Careful About

Show your working: seeing $\frac{17}{99}$ earns partial credit, and the common-denominator step earns a method mark. On this non-calculator paper, present all fraction arithmetic by hand. The answer must be exact and in simplest form: $\frac{8}{11}$ .

Techniques used

convert a recurring decimal to a fraction by multiplying by a suitable power of tensubtract equations to eliminate the recurring partrewrite fractions with a common denominatorsimplify a fraction by dividing numerator and denominator by their HCF

Q21MediumTransformations and Vectors

Fig. 10

$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .
$X$ is a point on $AB$ where $AX : XB = 3 : 2$ .
$OXC$ is a straight line.
$AC$ is parallel to $OB$ .

(a)

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

$\overrightarrow{AX}$ = ______

DifficultyMedium-Easy

Worked solution

Approach

To find $\overrightarrow{AX}$ , first express $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ using the triangle law of vector addition. Then, use the given ratio $AX : XB = 3 : 2$ to find $\overrightarrow{AX}$ as a fraction of $\overrightarrow{AB}$ .

Working

By the triangle law of vector addition:

\begin{aligned} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ \overrightarrow{AB} &= \mathbf{b} - \mathbf{a} \end{aligned}

Since $X$ lies on $AB$ and $AX : XB = 3 : 2$ , the point $X$ divides $AB$ in the ratio $3 : 2$ . This means $\overrightarrow{AX}$ is $\frac{3}{3+2} = \frac{3}{5}$ of $\overrightarrow{AB}$ .

\begin{aligned} \overrightarrow{AX} &= \frac{3}{5} \overrightarrow{AB} \\ \overrightarrow{AX} &= \frac{3}{5}(\mathbf{b} - \mathbf{a}) \\ \overrightarrow{AX} &= \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a} \end{aligned}

Answer

\overrightarrow{AX} = \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}

Final answer

3/5 b - 3/5 a

Detailed explanation

Walkthrough

We are given the position vectors $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . To find $\overrightarrow{AX}$ , we first need $\overrightarrow{AB}$ . Using the triangle law, $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$ . The point $X$ is on $AB$ such that $AX : XB = 3 : 2$ . This means the total length $AB$ is split into $3 + 2 = 5$ equal parts, and $AX$ takes up $3$ of those parts. Therefore, $\overrightarrow{AX} = \frac{3}{5} \overrightarrow{AB}$ . Substituting the expression for $\overrightarrow{AB}$ gives $\frac{3}{5}(\mathbf{b} - \mathbf{a})$ , which simplifies to $\frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}$ .

Key Takeaways

The vector between two points can be found by subtracting their position vectors: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ .
If a point divides a line segment in a ratio $m : n$ , the vector from the first point to the dividing point is $\frac{m}{m+n}$ of the total vector.

Common Mistakes

Writing $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$ (reversing the order).
Forgetting to add the ratio parts together: using $\frac{3}{2}$ instead of $\frac{3}{3+2} = \frac{3}{5}$ .
Not expanding the bracket to give the answer in the simplest form $\frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}$ .

Things to Be Careful About

Ensure the final answer is written in terms of $\mathbf{a}$ and $\mathbf{b}$ as requested.
The ratio $AX : XB = 3 : 2$ means $AX$ is the $3$ -part share out of $5$ total parts, not $\frac{3}{2}$ of $XB$ .
Keep vectors bold or use arrow notation consistently as required by the question.

Techniques used

find vector AB as difference of position vectorsuse ratio to find AX as fraction of AB

(b)

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

$\overrightarrow{XC}$ = ______

DifficultyMedium

Worked solution

Approach

To find $\overrightarrow{XC}$ , we can use the fact that $OXC$ is a straight line and $AC$ is parallel to $OB$ . This creates two similar triangles, $\triangle AXC$ and $\triangle BXO$ . We first find $\overrightarrow{OX}$ using vector addition, then use the similarity ratio to find $\overrightarrow{XC}$ .

Working

First, find $\overrightarrow{OX}$ :

\begin{aligned} \overrightarrow{OX} &= \overrightarrow{OA} + \overrightarrow{AX} \\ \overrightarrow{OX} &= \mathbf{a} + \left(\frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}\right) \\ \overrightarrow{OX} &= \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \end{aligned}

Since $AC$ is parallel to $OB$ and $OXC$ is a straight line, $\triangle AXC$ and $\triangle BXO$ are similar. We can verify this:

$\angle AXC = \angle BXO$ (vertically opposite angles)
$\angle XAC = \angle XBO$ (alternate angles, since $AC \parallel OB$ and $AB$ is a transversal)

The ratio of similarity is determined by the known sides $AX$ and $XB$ :

\frac{AX}{XB} = \frac{3}{2}

Therefore, the corresponding sides $XC$ and $XO$ are in the same ratio:

\frac{XC}{XO} = \frac{3}{2} \implies \overrightarrow{XC} = \frac{3}{2} \overrightarrow{OX}

Substitute $\overrightarrow{OX}$ :

\begin{aligned} \overrightarrow{XC} &= \frac{3}{2} \left( \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} \right) \\ \overrightarrow{XC} &= \frac{3}{5}\mathbf{a} + \frac{9}{10}\mathbf{b} \end{aligned}

Answer

\overrightarrow{XC} = \frac{3}{5}\mathbf{a} + \frac{9}{10}\mathbf{b}

Final answer

3/5 a + 9/10 b

Detailed explanation

Walkthrough

We need $\overrightarrow{XC}$ . Since $OXC$ is a straight line, $\overrightarrow{XC}$ is a scalar multiple of $\overrightarrow{OX}$ . To find this scalar, we use the parallel condition: $AC \parallel OB$ . This makes $\triangle AXC$ and $\triangle BXO$ similar (AA similarity: vertically opposite angles at $X$ , and alternate angles at $A$ and $B$ ). The ratio of similarity is $AX : XB = 3 : 2$ . Thus, $XC : XO = 3 : 2$ , meaning $\overrightarrow{XC} = \frac{3}{2} \overrightarrow{OX}$ . We already know $\overrightarrow{OA} = \mathbf{a}$ and from part (a), $\overrightarrow{AX} = \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}$ . Adding these gives $\overrightarrow{OX} = \mathbf{a} + \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}$ . Multiplying by $\frac{3}{2}$ yields $\overrightarrow{XC} = \frac{3}{5}\mathbf{a} + \frac{9}{10}\mathbf{b}$ .

Key Takeaways

Parallel lines in a vector diagram often create similar triangles, allowing you to find scalar relationships between vectors.
Vector addition along a path ( $\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX}$ ) is a fundamental tool for finding position vectors of intermediate points.
Vertically opposite angles and alternate angles are key to proving triangle similarity in these problems.

Common Mistakes

Assuming $\overrightarrow{XC} = \frac{3}{2} \overrightarrow{OB}$ directly without using the similar triangles correctly.
Forgetting that $\overrightarrow{XC}$ and $\overrightarrow{OX}$ are in the same direction (since $O-X-C$ is a straight line), so the scalar is positive.
Arithmetic errors when multiplying fractions, e.g., $\frac{3}{2} \times \frac{3}{5} = \frac{9}{10}$ , not $\frac{9}{7}$ .

Things to Be Careful About

The mark scheme accepts alternative routes, such as finding $\overrightarrow{AC} = \frac{3}{2}\mathbf{b}$ first (from similarity) and then using $\overrightarrow{XC} = \overrightarrow{XA} + \overrightarrow{AC} = -\overrightarrow{AX} + \frac{3}{2}\mathbf{b}$ . Both routes are valid.
Ensure the final answer is simplified and written in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Check that the direction of $\overrightarrow{XC}$ matches $\overrightarrow{OX}$ ; if you get a negative scalar, you may have the order of similarity reversed.

Techniques used

find OX as sum of OA and AXuse similar triangles to find ratio XC to OXscale OX to find XC

Q22MediumAlgebra and Graphs

(a)

Write $x^2 + 4x - 12$ in the form $(x + a)^2 + b$ .

______

(b)

Use your answer to part (a) to find the coordinates of the turning point of the graph of $y = x^2 + 4x - 12$ .

( ______ , ______ )

(c)

Sketch the graph of $y = x^2 + 4x - 12$ indicating the values where the graph crosses the axes.

Fig. 11

4M Similar

DifficultyMedium

Worked solution

Approach

To sketch the graph, find the intercepts with the axes. The y-intercept is found by setting $x = 0$ . The x-intercepts are found by setting $y = 0$ and solving the quadratic equation. Then draw a U-shaped curve (parabola) passing through these points with the turning point at $(-2, -16)$ .

Working

y-intercept:
Set $x = 0$ :

y = 0^2 + 4(0) - 12 = -12

The graph crosses the y-axis at $(0, -12)$ .

x-intercepts:
Set $y = 0$ :

x^2 + 4x - 12 = 0

Factorise the quadratic:

(x + 6)(x - 2) = 0

So $x = -6$ or $x = 2$ . The graph crosses the x-axis at $(-6, 0)$ and $(2, 0)$ .

Alternatively, using the answer from part (a):

(x + 2)^2 - 16 = 0 \\ (x + 2)^2 = 16 \\ x + 2 = \pm 4 \\ x = 2 \text{ or } x = -6

Sketch:
Draw a U-shaped curve opening upwards. Mark the x-intercepts at $-6$ and $2$ , and the y-intercept at $-12$ . The minimum point is at $(-2, -16)$ , which is in the third quadrant.

Answer

The graph crosses the axes at $(-6, 0)$ , $(2, 0)$ , and $(0, -12)$ .

Final answer

Graph crossing axes at (-6, 0), (2, 0), and (0, -12)

Detailed explanation

Walkthrough

First, find where the graph crosses the y-axis by substituting $x = 0$ into the equation, giving $y = -12$ . Next, find where it crosses the x-axis by setting $y = 0$ and solving $x^2 + 4x - 12 = 0$ . This factors to $(x + 6)(x - 2) = 0$ , giving $x = -6$ and $x = 2$ . With the intercepts and the turning point $(-2, -16)$ known, draw a smooth U-shaped curve passing through all these points.

Key Takeaways

Sketching a quadratic graph requires identifying the y-intercept, x-intercepts, and the turning point. The completed square form is especially useful for finding the turning point and x-intercepts.

Common Mistakes

Forgetting to label the intercepts on the sketch. Drawing the curve opening downwards instead of upwards (since the coefficient of $x^2$ is positive). Placing the turning point in the wrong quadrant.

Things to Be Careful About

The sketch must be a smooth curve, not straight lines. The turning point must be in the correct position relative to the intercepts. If the sketch is incorrect, a maximum of 3 marks can be awarded even if the intercepts are calculated correctly.

Techniques used

find the y-intercept by substituting x = 0find the x-intercepts by factorising or using the completed square formsketch the parabola with correct intercepts and turning point

Q235MMediumMensuration

Fig. 12

$OAB$ is a minor sector of a circle, centre $O$ .
$OCD$ is a major sector of a different circle, centre $O$ .

$OCA$ and $ODB$ are straight lines.
$OC = 6\,\text{cm}$ and $OA = 9\,\text{cm}$ .
The length of the minor arc $AB$ is $5\pi\,\text{cm}$ .

Work out the area of the major sector $OCD$ .
Give your answer in terms of $\pi$ .

______ $\text{cm}^2$

DifficultyMedium

Worked solution

Approach

First, use the arc length formula for minor sector $OAB$ to determine the sector angle $A\hat{O}B$ . Then, calculate the reflex angle for the major sector $OCD$ by subtracting $A\hat{O}B$ from $360^\circ$ . Finally, use the sector area formula with radius $OC = 6\,\text{cm}$ and the reflex angle to find the area in terms of $\pi$ .

Working

Let $\theta = A\hat{O}B$ be the angle of the minor sector $OAB$ .

The arc length of minor sector $OAB$ with radius $r = 9\,\text{cm}$ is given by:

\text{Arc length } AB = \frac{\theta}{360^\circ} \times 2\pi r

Substitute the known values:

\frac{\theta}{360^\circ} \times 2\pi \times 9 = 5\pi

Simplify the left-hand side:

\frac{\theta}{360^\circ} \times 18\pi = 5\pi

Divide both sides by $\pi$ :

\frac{18\theta}{360} = 5

\frac{\theta}{20} = 5

\theta = 5 \times 20 = 100^\circ

The sector $OCD$ is a major sector with radius $r = 6\,\text{cm}$ . Its angle is the reflex angle at $O$ :

\text{Reflex angle } COD = 360^\circ - 100^\circ = 260^\circ

Now, calculate the area of the major sector $OCD$ :

\begin{aligned} \text{Area of major sector } OCD &= \frac{260^\circ}{360^\circ} \times \pi \times 6^2 \\ &= \frac{26}{36} \times 36\pi \\ &= 26\pi \end{aligned}

Answer

26\pi

Final answer

26pi

Detailed explanation

Walkthrough

Find the angle of the minor sector ( $A\hat{O}B$ ):
We are given the radius of the larger circle $OA = 9\,\text{cm}$ and the length of the minor arc $AB = 5\pi\,\text{cm}$ . The formula for arc length is:
$\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r$
Setting up the equation:
$\frac{\theta}{360^\circ} \times 2\pi \times 9 = 5\pi$
Simplifying $2 \times 9 = 18$ gives $\frac{18\theta}{360} = \frac{\theta}{20}$ . Thus, $\frac{\theta}{20} = 5$ , which gives $\theta = 100^\circ$ .
Find the angle of the major sector $OCD$ :
Since $OCA$ and $ODB$ are straight lines, the minor angle $C\hat{O}D$ is equal to $A\hat{O}B = 100^\circ$ . The sector $OCD$ is described as the major sector, so its angle is the reflex angle around centre $O$ :
$\text{Angle} = 360^\circ - 100^\circ = 260^\circ$
Calculate the area of the major sector:
The radius of sector $OCD$ is $OC = 6\,\text{cm}$ . Using the area of a sector formula:
$\text{Area} = \frac{260^\circ}{360^\circ} \times \pi \times 6^2$
Since $6^2 = 36$ , the $36$ in the numerator cancels neatly with the denominator of $\frac{260}{360} = \frac{26}{36}$ , leaving:
$\text{Area} = 26\pi\,\text{cm}^2$

Key Takeaways

The formula for arc length is $\frac{\theta}{360^\circ} \times 2\pi r$ .
The formula for the area of a sector is $\frac{\theta}{360^\circ} \times \pi r^2$ .
A major sector corresponds to the reflex angle subtended at the centre, which is $360^\circ - \text{minor angle}$ .

Common Mistakes

Finding the area of the minor sector $OCD$ (using $100^\circ$ instead of $260^\circ$ ), which would give $10\pi$ .
Mixing up the radii: using $r = 9$ for the area of $OCD$ instead of $r = 6$ , or $r = 6$ for the arc length of $AB$ .
Using the area formula $\pi r^2$ when working with arc length instead of the circumference formula $2\pi r$ .
Evaluating $\pi$ as a decimal (e.g. $3.142$ ) instead of keeping the answer in exact form in terms of $\pi$ as explicitly instructed.

Things to Be Careful About

Ensure all calculations are done without a calculator, looking for common factors to simplify arithmetic by hand.
The final answer must be left in terms of $\pi$ .

Techniques used

calculate the angle subtended by an arcfind the reflex angle of a major sectorcalculate the area of a sector