Mathematics (Syllabus D) 4024 / 11 — October/November 2024

(b)

The temperature at midnight is $9\,^{\circ}\text{C}$ lower than the temperature at midday.

Find the temperature at midnight.

______ $^{\circ}\text{C}$

Q22MMedium-EasyNumber

Amber and Pablo share $280 in the ratio $2 : 5$ .

Work out Pablo's share.

$ ______

DifficultyMedium-Easy

Worked solution

Approach

To divide an amount in a given ratio, first calculate the total number of parts by adding the numbers in the ratio. Then, determine the value of a single part by dividing the total amount by the total number of parts. Finally, multiply the value of one part by the specific number of parts allocated to Pablo.

Working

The ratio of Amber's share to Pablo's share is $2 : 5$ .

First, find the total number of parts:

2 + 5 = 7 \text{ parts}

The total amount shared is $280. Calculate the value of one part:

\frac{280}{7} = 40 \text{ (dollars per part)}

Pablo's share corresponds to $5$ parts. Multiply the value of one part by $5$ :

5 \times 40 = 200

Alternatively, using the fraction method directly:
Pablo receives $\frac{5}{7}$ of the total amount.

\frac{5}{7} \times 280 = 200

Answer

200

Final answer

200

Detailed explanation

Walkthrough

The problem asks us to share a total sum of money ($280) between two people based on a specific ratio ( $2:5$ ).

Understand the Ratio: The ratio $2:5$ means that for every $2$ units Amber gets, Pablo gets $5$ units. Together, they share $2 + 5 = 7$ equal units (or parts) of the total amount.
Find the Value of One Unit: Since the total amount of $280 represents these $7$ equal units, we can find the value of just one unit by dividing the total amount by the total number of units:
$\text{Value of 1 part} = \frac{\text{Total Amount}}{\text{Total Parts}} = \frac{280}{7}$
Performing the division: $280 \div 7 = 40$ . So, each part is worth $40.
Calculate Pablo's Share: The problem specifies that Pablo's share corresponds to $5$ parts of the ratio. Therefore, we multiply the value of one part by $5$ :
$\text{Pablo's Share} = 5 \times 40 = 200$
Verification: We can check our work by calculating Amber's share as well. Amber gets $2$ parts:
$2 \times 40 = 80$
Adding both shares together should give the original total:
$80 + 200 = 280$
The calculation is correct.

Key Takeaways

Total Parts: Always add the numbers in the ratio together to find how many equal parts the total quantity is divided into.
Unitary Method: Divide the total quantity by the total number of parts to find the value of a single part. This is often the safest and most intuitive method.
Fractional Method: You can also solve this by multiplying the total by the fraction representing the person's share (e.g., $\frac{5}{7}$ for Pablo).

Common Mistakes

Multiplying instead of dividing: A common error is to multiply the total by the ratio numbers (e.g., $280 \times 5$ ) without dividing by the total parts first. Remember that the ratio describes parts of the whole, not a multiplier of the whole.
Ignoring the total parts: Simply taking $\frac{5}{280}$ or calculating $\frac{280}{5}$ directly misses the fact that the denominator must be the sum of the ratio terms ( $2+5=7$ ).
Giving the wrong share: Calculating Amber's share ( $80$ ) instead of Pablo's, or vice versa. Always double-check which number in the ratio belongs to whom.

Things to Be Careful About

Units: Ensure the final answer includes the currency symbol if required by the context, although the blank line often implies just the number. Here, the question provides "$ _____", so the numerical value $200$ is sufficient.
Exactness: In non-calculator papers, ensure your division is exact. If the result were a decimal, you might need to round to appropriate significant figures or decimal places, but with integers like $280$ and $7$ , the result is an integer.

Techniques used

calculate the total number of parts in a ratiofind the value of one partmultiply to find the required share

Q3Medium-EasyStatistics

Here are eight integers.

-1 \quad -5 \quad -1 \quad -3 \quad -3 \quad 2 \quad -1 \quad -7

Find

(a)

the mode

______

(b)

the median.

______

DifficultyMedium-Easy

Worked solution

Approach

To find the median, first arrange the integers in ascending order (from smallest to largest). Then, identify the middle value(s).

Working

The original list is:

-1, -5, -1, -3, -3, 2, -1, -7

There are $8$ numbers in total ( $n=8$ ). Since $8$ is an even number, the median will be the average of the $4^{th}$ and $5^{th}$ values in the ordered list.

Step 1: Order the data.
The smallest number is $-7$ . The next smallest is $-5$ , followed by $-3$ (twice), then $-1$ (three times), and finally $2$ .

Ordered list:

-7, -5, -3, -3, -1, -1, -1, 2

Step 2: Locate the middle values.
The position of the median values is given by $\frac{n}{2}$ and $\frac{n}{2} + 1$ .
For $n=8$ , we look for the $4^{th}$ and $5^{th}$ terms.

1st: $-7$
2nd: $-5$
3rd: $-3$
4th: $-3$
5th: $-1$
6th: $-1$
7th: $-1$
8th: $2$

Step 3: Calculate the average of the two middle values.

\text{Median} = \frac{-3 + (-1)}{2}

\text{Median} = \frac{-4}{2}

\text{Median} = -2

Answer

-2

Final answer

-2

Detailed explanation

Walkthrough

The median is the middle value of a data set when it is arranged in numerical order.

First, we sort the eight integers: $-1, -5, -1, -3, -3, 2, -1, -7$ .
Sorting them from lowest to highest gives: $-7, -5, -3, -3, -1, -1, -1, 2$ .

Because there is an even number of items ( $8$ ), there is no single middle number. Instead, we take the two numbers in the exact center. These are the $4^{th}$ and $5^{th}$ numbers in the sorted sequence.

The $4^{th}$ number is $-3$ .
The $5^{th}$ number is $-1$ .

The median is the arithmetic mean (average) of these two values:

\frac{-3 + (-1)}{2} = \frac{-4}{2} = -2

Key Takeaways

Always order the data before finding the median.
For an odd number of data points, the median is the single middle value.
For an even number of data points, the median is the average of the two middle values.

Common Mistakes

Failing to sort the data correctly, especially with negative numbers (e.g., thinking $-1$ is smaller than $-7$ ).
Picking just one of the middle values instead of averaging them.
Miscounting the positions of the middle values (e.g., picking the 3rd and 4th instead of the 4th and 5th).

Things to Be Careful About

The mark scheme awards a B1 mark for providing the correct ordered list. While the final answer requires calculation, showing the ordered list helps verify your method.

Techniques used

order the data set from least to greatestfind the middle value(s) of an ordered data set

Q4Medium-EasyGeometry

(a)

Shade one more small square so the diagram has one line of symmetry.

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

To create exactly one line of symmetry, we can use the main diagonal running from the top-left to the bottom-right of the 5 by 5 grid. We reflect each already-shaded square across this diagonal and shade any missing square that is required.

Working

The coordinates of the shaded squares are (row 1, col 4), (row 2, col 2), (row 3, col 5) and (row 5, col 3).

Reflecting across the main diagonal (where row number equals column number):

(row 1, col 4) reflects to (row 4, col 1).
(row 2, col 2) lies on the diagonal and is its own reflection.
(row 3, col 5) reflects to (row 5, col 3), which is already shaded.

To complete the line of symmetry along the main diagonal, we must shade the square at (row 4, col 1). Checking other potential lines of symmetry (horizontal, vertical, or the other diagonal) shows that none of them can be completed with just one additional square. Thus, the main diagonal is the unique line of symmetry.

Answer

Shade the square in row 4, column 1.

Final answer

Row 4, column 1

Detailed explanation

Walkthrough

The question asks to shade exactly one more square in a 5 by 5 grid so that the resulting diagram has exactly one line of symmetry. We test the possible lines of symmetry for a square grid: horizontal, vertical, and the two diagonals.

Horizontal line (middle row): The shaded squares are in rows 1, 2, 3, and 5. Row 4 is empty. Reflecting across the horizontal midline would require adding squares in rows 4, 3, 2, and 1, which is more than one square.
Vertical line (middle column): The shaded squares are in columns 2, 3, 4, and 5. Column 1 is empty. Reflecting across the vertical midline would require adding squares in columns 4, 3, 2, and 1, which is more than one square.
Anti-diagonal (top-right to bottom-left): Reflecting (1, 4) gives (2, 5), which is not shaded. This would require multiple additions.
Main diagonal (top-left to bottom-right): Reflection maps (row, col) to (col, row).
- (1, 4) maps to (4, 1). This square is unshaded.
- (2, 2) maps to (2, 2). It is on the line of symmetry.
- (3, 5) maps to (5, 3). This square is already shaded.
- (5, 3) maps to (3, 5). This square is already shaded.

Only one square, (4, 1), needs to be shaded to make the diagram symmetric about the main diagonal. No other lines of symmetry are present, satisfying the condition of having exactly one line of symmetry.

Key Takeaways

A square grid has four possible lines of symmetry: horizontal, vertical, and two diagonals.
To find the missing square for a given line of symmetry, reflect each existing shaded square across that line and identify which reflected position is not yet shaded.
Always verify that the completed diagram does not accidentally gain additional lines of symmetry.

Common Mistakes

Shading a square that creates a second line of symmetry (e.g., creating both diagonal symmetries).
Miscounting rows and columns, especially when reflecting across a diagonal (mixing up row and column indices).
Assuming the line of symmetry must be horizontal or vertical.

Things to Be Careful About

The question specifies "one line of symmetry". Ensure the final shaded pattern does not possess rotational symmetry or additional reflectional symmetry that would introduce a second line.
When shading on a grid, clearly define your coordinate system (e.g., row from top, column from left) to avoid off-by-one errors.

Techniques used

identify a line of symmetry on the gridreflect the given shaded squares across the lineshade the missing square to complete the symmetry

(b)

Here is a regular polygon.

Fig. 2

Complete the description of the rotational symmetry of this polygon.

The polygon has rotational symmetry of order ______

DifficultyEasy

Worked solution

Approach

The rotational symmetry of a patterned shape is the number of times the pattern coincides with itself during one full 360-degree rotation. We examine the shaded sectors of the regular hexagon to find the repeating unit.

Working

The regular hexagon is divided into 6 equal equilateral triangles. Looking at Fig. 2, the triangles are shaded in an alternating pattern: shaded, unshaded, shaded, unshaded, shaded, unshaded.

The pattern repeats every 2 triangles. Since there are 6 triangles in total, the pattern repeats $6 \div 2 = 3$ times in a full rotation.

Therefore, the polygon (with its shading) has rotational symmetry of order 3.

Answer

3

Final answer

3

Detailed explanation

Walkthrough

Rotational symmetry of order $n$ means that a shape looks exactly the same $n$ times during a full $360^\circ$ rotation. For a regular polygon with a patterned interior, we must consider both the boundary of the polygon and the shading inside.

The regular hexagon has 6 equal triangular sectors. In Fig. 2, the sectors are shaded alternately: one shaded, one unshaded, one shaded, and so on. This means the pattern repeats every 2 sectors.

To find the order of rotational symmetry, we divide the total number of sectors by the number of sectors in one repeating unit:

\text{Order} = \frac{6}{2} = 3

If we rotate the hexagon by $120^\circ$ (which is $360^\circ \div 3$ ), the shaded sectors map onto shaded sectors and the unshaded sectors map onto unshaded sectors. The shape coincides with itself 3 times in a full rotation.

Key Takeaways

The order of rotational symmetry is the number of times a shape coincides with itself in a $360^\circ$ rotation.
When a regular polygon has internal shading, the rotational symmetry is determined by the repeating pattern of the shading, not just the polygon's boundary.
For a regular $n$ -gon with alternating shaded sectors, the order of rotational symmetry is $n \div 2$ .

Common Mistakes

Answering 6 (the rotational symmetry of the regular hexagon boundary alone) without considering the shading pattern.
Answering 2 (the number of shaded sectors) instead of counting how many times the full pattern repeats.
Forgetting that the order must be an integer that divides evenly into 360.

Things to Be Careful About

Always consider the entire figure, including any internal lines or shading, when determining rotational symmetry.
The order of rotational symmetry must be a whole number.

Techniques used

identify the repeating unit in the patterncount the number of times the pattern repeats in a full 360-degree rotation

Q5EasyAlgebra and Graphs

(a)

Simplify.

2a - 3b + 4b - 5a

______

DifficultyEasy

Worked solution

Approach

To simplify the expression $2a - 3b + 4b - 5a$ , we need to collect all the 'like terms' together. Like terms are terms that have exactly the same variable part (in this case, terms with ' $a$ ' and terms with ' $b$ ').

Working

First, let's group the terms containing $a$ :

2a - 5a

Now, combine these coefficients:

2 - 5 = -3

So, the combined $a$ term is:

-3a

Next, let's group the terms containing $b$ :

-3b + 4b

Combine these coefficients:

-3 + 4 = 1

So, the combined $b$ term is:

1b \text{ or simply } b

Finally, put the two simplified parts together:

-3a + b

Answer

-3a + b

Final answer

-3a + b

Detailed explanation

Walkthrough

The goal is to make the expression shorter and simpler. We can only add or subtract terms if they are 'like terms', meaning they share the exact same variables raised to the exact same powers. Here, we have ' $a$ ' terms and ' $b$ ' terms.

Group the $a$ terms: The expression has $+2a$ and $-5a$ . Think of it as having 2 apples and taking away 5 apples, which leaves you with a debt of 3 apples ( $-3a$ ).
Group the $b$ terms: The expression has $-3b$ and $+4b$ . Think of it as owing 3 pears but then receiving 4 pears. You pay off the debt and have 1 pear left ( $+1b$ ).
Combine: Join the results: $-3a + b$ .

Key Takeaways

Always look for variables first. Only combine numbers attached to the same variable.
Be careful with negative signs. Subtracting a larger number from a smaller one (like $2 - 5$ ) results in a negative value.
A coefficient of $1$ is usually written implicitly (e.g., $1b$ becomes $b$ ).

Common Mistakes

Combining different variables (e.g., writing $-3ab$ ). This is incorrect because $a$ and $b$ are different quantities.
Arithmetic errors with negatives (e.g., calculating $2 - 5 = 3$ instead of $-3$ ).
Forgetting to carry the sign of the first term when grouping.

Things to Be Careful About

The mark scheme awards marks for seeing either $-3a$ or $b$ correctly, so double-check each individual combination before writing the final answer.

Techniques used

group like termscombine coefficients of like terms

(b)

Expand.

5(3x - 2)

______

Q6Medium-EasyStatistics

The table shows the time spent on a homework task and the number of errors made for some students in a class.

Time (minutes)	79	92	91	85	82	95	60	65	63	70
Number of errors	5	1	3	3	5	0	9	7	8	7

(a)

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

Fig. 3

DifficultyMedium-Easy

Worked solution

Approach

The scatter diagram already has six points plotted. We need to plot the remaining four points from the table.

Working

The remaining data pairs (Time, Number of errors) are:

\begin{aligned} (60, 9) \\ (63, 8) \\ (65, 7) \\ (70, 7) \end{aligned}

Plot each of these four points on the scatter diagram as crosses at the corresponding coordinates.

Answer

(60, 9), \quad (63, 8), \quad (65, 7), \quad (70, 7)

Final answer

Points plotted at (60, 9), (63, 8), (65, 7), and (70, 7).

Detailed explanation

Walkthrough

The question provides a table of 10 data points and a scatter diagram with 6 points already plotted. The first 6 points correspond to the first 6 columns of the table: (79, 5), (92, 1), (91, 3), (85, 3), (82, 5), and (95, 0). The remaining 4 points must be plotted to complete the diagram. These are (60, 9), (63, 8), (65, 7), and (70, 7). To plot them, locate the time value on the horizontal axis and the number of errors on the vertical axis, then mark the intersection with a cross. The mark scheme awards 1 mark for 2 or 3 correct plots and 1 mark for all 4 correct plots.

Key Takeaways

A scatter diagram is used to show the relationship between two variables.
Each point represents a single observation, with its x-coordinate from one variable and its y-coordinate from the other.
All data points from the table must be plotted to fully complete the scatter diagram.

Common Mistakes

Plotting the coordinates in the wrong order (e.g., plotting (9, 60) instead of (60, 9)). Always remember the horizontal axis is the first variable (Time) and the vertical axis is the second (Number of errors).
Misreading the grid lines. The horizontal axis has major markings every 5 minutes and minor markings every 1 minute. The vertical axis has markings every 1 error.

Things to Be Careful About

Ensure points are plotted as crosses (×) as shown in the existing diagram, not dots.
Check that the x-axis is indeed Time (minutes) and not the other way around.
The mark scheme allows partial credit (B1 for 2 or 3 correct plots), so accuracy on all 4 points is not strictly required for full marks, but all 4 should be attempted.

Techniques used

extract coordinates from a data tableplot points on a scatter diagram

(b)

On the scatter diagram, draw a line of best fit.

DifficultyMedium-Easy

Worked solution

Approach

A line of best fit is a straight line that passes through the middle of the data points, balancing the number of points above and below it. The data shows a negative correlation (as time increases, errors decrease), so the line must slope downwards from left to right.

Working

Using a ruler, draw a straight line that passes as close as possible to all the plotted points. Ensure there are roughly equal numbers of points on either side of the line and that the line follows the general negative trend of the data.

Answer

\text{A straight line of best fit with a negative gradient is drawn.}

Final answer

A straight line with negative gradient passing through the middle of the points.

Detailed explanation

Walkthrough

The line of best fit summarises the relationship between the two variables. Because the number of errors decreases as time spent increases, there is a negative correlation. The line must therefore have a negative gradient (slope downwards from left to right). To draw it correctly, use a ruler to sketch a straight line that runs through the centre of the cloud of points. It should not necessarily pass through every point, nor does it need to pass through the origin. The goal is to have roughly equal numbers of points above and below the line, and to minimise the overall distance from the points to the line.

Key Takeaways

A line of best fit is always a straight line in this context (linear relationship).
It must reflect the direction of the correlation (negative gradient here).
It does not need to pass through every point or the origin (0, 0).

Common Mistakes

Drawing a curved line instead of a straight ruled line.
Drawing a line with the wrong gradient (positive instead of negative).
Forcing the line to pass through the origin (0, 0), which is not required unless the data clearly supports it.
Not using a ruler, resulting in a wobbly or curved line.

Things to Be Careful About

The mark scheme requires a 'ruled line', so use a ruler.
The line must have a negative gradient; a positive gradient will score no marks.
This part is a drawing task, so the exact position of the line will vary between candidates, which is acceptable as long as it is reasonable.

Techniques used

draw a line of best fit through the scatter plot

(c)

Another of the students in the class made 6 errors.

Use your line of best fit to estimate the time this student spent on the homework task.

______ $\text{minutes}$

DifficultyMedium-Easy

Worked solution

Approach

To estimate the time for a student who made 6 errors, locate 6 on the vertical axis (Number of errors), move horizontally to the line of best fit, and then move vertically down to read the corresponding time on the horizontal axis.

Working

Find $y = 6$ on the vertical axis (Number of errors).
Move horizontally to the right until you meet the line of best fit drawn in part (b).
From that intersection point, move vertically down to the horizontal axis to read the time.
Reading from a typical line of best fit for this data, the time is approximately $68$ minutes. (Acceptable range: $66$ to $72$ minutes depending on the exact line drawn in part (b).)

t \approx 68 \text{ minutes}

Answer

68

Final answer

68

Detailed explanation

Walkthrough

This part asks for an estimation using the line of best fit, not the exact data. Since no student in the table made exactly 6 errors (the closest are 5 errors at 79 and 82 minutes, and 7 errors at 65 and 70 minutes), we must interpolate using our line from part (b).

To do this:

Locate the value $6$ on the vertical axis (Number of errors).
Draw a horizontal line from $y = 6$ across to the right until it intersects your line of best fit.
From that intersection point, draw a vertical line down to the horizontal axis (Time in minutes).
Read the value on the horizontal axis. For a correctly drawn line of best fit, this value will be around $68$ minutes. Because the line is hand-drawn, answers in the range of $66$ to $72$ minutes are all acceptable, as long as they are consistent with the line drawn in part (b).

Key Takeaways

Interpolation on a scatter diagram uses the line of best fit, not the nearest data point.
The line of best fit smooths out variations in the data to provide a general estimate.
Readings from hand-drawn lines allow for a range of acceptable answers.

Common Mistakes

Reading the value from the nearest data point instead of the line of best fit (e.g., guessing 70 or 79 minutes based on the table).
Reading the axes in the wrong direction (e.g., reading the y-value when looking for the x-value).
Not aligning the horizontal and vertical reading lines properly with the grid.

Things to Be Careful About

The answer must be a reading from the candidate's own line of best fit. If their line is unreasonable, the reading will be wrong, but the mark is still awarded for the correct reading method.
The mark scheme states 'Dependent on negative gradient', meaning if the line in part (b) had a positive gradient, the reading would be incorrect and score no marks.
Give the answer to the nearest minute or as a reasonable whole number, as the horizontal axis is marked in minutes.

Techniques used

read a value from the line of best fit

Q72MEasyNumber

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

\frac{3.1 \times 26.7}{6.9 - 2.3}

______

DifficultyEasy

Worked solution

Approach

Round each number in the expression to 1 significant figure. Then substitute these rounded values into the expression and calculate the result.

Working

The numbers are $3.1$ , $26.7$ , $6.9$ , and $2.3$ .

Round $3.1$ to 1 significant figure:
The first digit is $3$ . The next digit is $1$ , so we round down (keep as is).
$3.1 \approx 3$
Round $26.7$ to 1 significant figure:
The first digit is $2$ . The next digit is $6$ , so we round up.
$26.7 \approx 30$
Round $6.9$ to 1 significant figure:
The first digit is $6$ . The next digit is $9$ , so we round up.
$6.9 \approx 7$
Round $2.3$ to 1 significant figure:
The first digit is $2$ . The next digit is $3$ , so we round down.
$2.3 \approx 2$

Substitute these values back into the original expression:
$\frac{3.1 \times 26.7}{6.9 - 2.3} \approx \frac{3 \times 30}{7 - 2}$

Calculate the numerator:
$3 \times 30 = 90$

Calculate the denominator:
$7 - 2 = 5$

Divide the numerator by the denominator:
$\frac{90}{5} = 18$

Answer

$18$

Final answer

18

Detailed explanation

Walkthrough

To estimate the value of the expression, we first simplify the numbers involved by rounding them to 1 significant figure. This makes mental calculation easier while keeping the answer reasonably close to the exact value.

For $3.1$ , the first significant figure is $3$ . Since the following digit ( $1$ ) is less than $5$ , it rounds to $3$ .
For $26.7$ , the first significant figure is $2$ . The following digit ( $6$ ) is $5$ or more, so we round up the $2$ to a $3$ and replace the rest with a zero, giving $30$ .
For $6.9$ , the first significant figure is $6$ . The following digit ( $9$ ) causes us to round up to $7$ .
For $2.3$ , the first significant figure is $2$ . The following digit ( $3$ ) means we round down to $2$ .

Now substitute these estimated values into the fraction:
$\frac{3 \times 30}{7 - 2}$

Perform the multiplication in the numerator: $3 \times 30 = 90$ .
Perform the subtraction in the denominator: $7 - 2 = 5$ .
Finally, divide: $90 \div 5 = 18$ .

Key Takeaways

Rounding to 1 significant figure looks at the first non-zero digit and the digit immediately following it.
If the following digit is 5 or greater, round up; otherwise, round down.
Estimation involves replacing complex numbers with simpler ones to make calculation manageable.

Common Mistakes

Rounding to the nearest whole number instead of 1 significant figure (e.g., rounding $26.7$ to $27$ instead of $30$ ). Note that for single-digit integers like $3.1$ , they might be the same, but for two-digit numbers starting with 1-4, 1 s.f. changes the magnitude significantly (e.g., $26.7 \to 30$ , not $27$ ).
Arithmetic errors in the final calculation.

Things to Be Careful About

Ensure you are rounding to significant figures, not decimal places. For $26.7$ , 1 s.f. is $30$ , whereas 1 d.p. would be $26.7$ (or 0 d.p. is $27$ ).
The mark scheme awards B1 for seeing three of the correct rounded values ( $3, 30, 7, 2$ ), so identifying these correctly is crucial even if the final calculation has a minor error.

Techniques used

round numbers to 1 significant figureestimate the value of a fraction

Q8Medium-EasyAlgebra and Graphs

These are the first four terms of a sequence.

2 \quad 8 \quad 14 \quad 20

(a)

Find the next number in the sequence.

______

(b)

Find an expression, in terms of $n$ , for the $n$ th term of this sequence.

______

DifficultyMedium-Easy

Worked solution

Approach

For an arithmetic sequence, the $n$ th term is of the form $dn + c$ , where $d$ is the common difference. Once $d$ is found, substitute a known term (e.g., $n=1$ ) to solve for $c$ .

Working

From part (a), we know the common difference $d = 6$ . Therefore, the $n$ th term starts with:

6n

Let the $n$ th term be $6n + c$ . We can find $c$ by comparing $6n$ to the actual terms.

For $n = 1$ (the first term):

6(1) = 6

The actual first term is $2$ . So:

6 + c = 2

Solving for $c$ :

c = 2 - 6 = -4

Thus, the expression for the $n$ th term is:

6n - 4

Check with the second term ( $n=2$ ):

6(2) - 4 = 12 - 4 = 8 \quad (\text{Correct})

Check with the fourth term ( $n=4$ ):

6(4) - 4 = 24 - 4 = 20 \quad (\text{Correct})

Answer

6n - 4

Final answer

6n - 4

Detailed explanation

Walkthrough

An arithmetic sequence generates terms by adding a fixed amount each time. In algebraic terms, this fixed amount becomes the coefficient of $n$ . Since the common difference is $6$ , the core of our formula is $6n$ .

However, $6n$ does not give the correct terms directly. For example, when $n=1$ , $6n = 6$ , but the first term is $2$ . We need to adjust $6n$ to match the sequence. The difference between what $6n$ gives and what we actually want is the constant $c$ .

Using the first term ( $n=1$ ):
Desired value: $2$
Value from $6n$ : $6$
Adjustment needed: $2 - 6 = -4$

So we subtract $4$ from $6n$ , giving the formula $6n - 4$ . It is good practice to check this formula against another term, such as the 3rd term ( $n=3$ ): $6(3) - 4 = 18 - 4 = 14$ , which matches the sequence.

Key Takeaways

The coefficient of $n$ in the $n$ th term formula is always equal to the common difference of the arithmetic sequence.
The constant term is determined by finding the "zeroth term" (the value before the first term) or by solving $dn + c = \text{term}_1$ .

Common Mistakes

Writing $6n + 4$ instead of $6n - 4$ (sign error when finding the adjustment).
Assuming the constant is related to the first term directly without subtraction (e.g., writing $6n + 2$ ).
Failing to simplify if the result was derived differently (though here it is simple).

Things to Be Careful About

Ensure you use the correct value for $n$ . The first term corresponds to $n=1$ , not $n=0$ . If you used $n=0$ to find the constant, you would find the zeroth term is $-4$ , leading to $6(n-0) - 4 = 6n - 4$ , which yields the same result but requires careful interpretation.

Techniques used

use the common difference as the coefficient of ncalculate the zeroth term to find the constant offsetcombine to form the nth term expression

Q9MediumTransformations and Vectors

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

Fig. 4

(a)

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

______

3M Similar

DifficultyMedium

Worked solution

Approach

Compare the sizes and orientations of triangles $A$ and $B$ . Since triangle $B$ is larger than triangle $A$ and they have the same orientation (the right angle is at the bottom-left in both), the transformation is an enlargement. Find the scale factor by comparing corresponding side lengths, then find the centre of enlargement by drawing lines through corresponding vertices and finding their intersection.

Working

Step 1: Identify corresponding vertices.

Triangle $A$ has vertices $(1, 1)$ , $(2, 1)$ , $(1, 3)$ . Triangle $B$ has vertices $(-3, 1)$ , $(-3, -5)$ , $(0, -5)$ .

The right-angle vertex of $A$ is $(1, 1)$ and the right-angle vertex of $B$ is $(-3, -5)$ . The vertex $(2, 1)$ (bottom-right of $A$ ) corresponds to $(0, -5)$ (bottom-right of $B$ ). The vertex $(1, 3)$ (top of $A$ ) corresponds to $(-3, 1)$ (top of $B$ ).

Step 2: Calculate the scale factor.

The vertical side of $A$ from $(1, 1)$ to $(1, 3)$ has length $2$ . The corresponding vertical side of $B$ from $(-3, -5)$ to $(-3, 1)$ has length $6$ .

\text{Scale factor} = \frac{6}{2} = 3

Step 3: Find the centre of enlargement.

Draw lines through corresponding vertices. The centre is where these lines meet.

Line through $(1, 1)$ and $(-3, -5)$ :

\text{slope} = \frac{-5 - 1}{-3 - 1} = \frac{-6}{-4} = \frac{3}{2}

y - 1 = \frac{3}{2}(x - 1) \implies y = \frac{3}{2}x - \frac{1}{2}

Line through $(2, 1)$ and $(0, -5)$ :

\text{slope} = \frac{-5 - 1}{0 - 2} = \frac{-6}{-2} = 3

y - 1 = 3(x - 2) \implies y = 3x - 5

Set equal to find intersection:

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

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

\frac{9}{2} = \frac{3}{2}x

x = 3

Substitute $x = 3$ into $y = 3x - 5$ :

y = 3(3) - 5 = 4

The centre of enlargement is $(3, 4)$ . Verify with the third pair: line through $(1, 3)$ and $(-3, 1)$ has slope $\frac{1 - 3}{-3 - 1} = \frac{1}{2}$ , giving $y = \frac{1}{2}x + \frac{5}{2}$ . At $x = 3$ , $y = \frac{3}{2} + \frac{5}{2} = 4$ . ✓

Answer

Enlargement with scale factor $3$ and centre $(3, 4)$

Final answer

Enlargement, scale factor 3, centre (3, 4)

Detailed explanation

Walkthrough

Step 1 — Identify the transformation type.
Triangle $B$ is clearly larger than triangle $A$ , so the transformation must involve a change in size. Both triangles have the same orientation: the right angle is at the bottom-left corner, the horizontal side runs to the right, and the vertical side runs upward. This rules out rotation or reflection (which would change orientation or flip the shape). A translation cannot change size. The only single transformation that changes size while preserving orientation is an enlargement.

Step 2 — Find the scale factor.
Pick any pair of corresponding sides and divide the length in $B$ by the length in $A$ . The vertical side of $A$ goes from $(1, 1)$ to $(1, 3)$ , length $2$ . The corresponding vertical side of $B$ goes from $(-3, -5)$ to $(-3, 1)$ , length $6$ . The scale factor is $6 \div 2 = 3$ .

Step 3 — Find the centre of enlargement.
For an enlargement, every point $P$ maps to an image $P'$ such that $P'$ , $P$ , and the centre $C$ are collinear, with $CP' = k \cdot CP$ . Therefore, drawing a line through any pair of corresponding vertices will pass through the centre. Draw lines through $(1, 1)$ and $(-3, -5)$ , and through $(2, 1)$ and $(0, -5)$ . Their intersection is the centre.

The line through $(1, 1)$ and $(-3, -5)$ has slope $\frac{3}{2}$ and equation $y = \frac{3}{2}x - \frac{1}{2}$ . The line through $(2, 1)$ and $(0, -5)$ has slope $3$ and equation $y = 3x - 5$ . Setting them equal gives $x = 3$ , $y = 4$ . So the centre is $(3, 4)$ .

Step 4 — Verify.
Check that the third pair of corresponding vertices also lies on a line through $(3, 4)$ . The line through $(1, 3)$ and $(-3, 1)$ has equation $y = \frac{1}{2}x + \frac{5}{2}$ , and at $x = 3$ gives $y = 4$ . ✓

Key Takeaways

An enlargement preserves orientation and changes size by a scale factor $k$ .
The scale factor is found by dividing a length in the image by the corresponding length in the original.
The centre of enlargement lies on the line joining every point to its image; intersecting two such lines gives the centre.
Always verify the centre using a third pair of corresponding points.

Common Mistakes

Stating "enlargement" without giving both the scale factor and the centre — all three pieces are required for full marks (B1 each).
Getting the scale factor wrong by dividing in the wrong order (e.g. $2 \div 6 = \frac{1}{3}$ instead of $3$ ). Remember: image $\div$ original.
Reading the centre coordinates from the grid incorrectly — always verify by substituting back into the line equations.
Forgetting that the centre of enlargement can lie outside the shape; $(3, 4)$ is not inside either triangle.

Things to Be Careful About

The answer must include all three elements: the type of transformation (enlargement), the scale factor ( $3$ ), and the centre coordinates ( $(3, 4)$ ). Missing any one costs a mark.
Coordinates must be given as an ordered pair $(x, y)$ , not as separate numbers.
The scale factor is a pure number (no units).
In the non-calculator component, show the line equations and the algebra that gives the intersection point — do not just state the centre.

Techniques used

compare corresponding side lengths to find scale factordraw lines through corresponding vertices to locate centre of enlargementsolve simultaneous linear equations to find intersection point

(b)

Triangle $A$ is mapped onto triangle $C$ by the translation $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ .

Draw triangle $C$ .

DifficultyEasy

Worked solution

Approach

A translation by vector $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ moves every point 3 units left and 2 units up. Add the vector components to each vertex of triangle $A$ to find the vertices of triangle $C$ , then draw the triangle.

Working

Triangle $A$ has vertices $(1, 1)$ , $(2, 1)$ , $(1, 3)$ .

Apply the translation $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ to each vertex:

(1, 1) \to (1 - 3,\; 1 + 2) = (-2, 3)

(2, 1) \to (2 - 3,\; 1 + 2) = (-1, 3)

(1, 3) \to (1 - 3,\; 3 + 2) = (-2, 5)

Triangle $C$ has vertices $(-2, 3)$ , $(-1, 3)$ , and $(-2, 5)$ .

Answer

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

Final answer

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

Detailed explanation

Walkthrough

A translation moves every point by the same amount in the same direction. The vector $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ means: move 3 units in the negative $x$ -direction (left) and 2 units in the positive $y$ -direction (up). To find the image of each vertex, simply add the vector components to the coordinates.

$(1, 1) + (-3, 2) = (-2, 3)$
$(2, 1) + (-3, 2) = (-1, 3)$
$(1, 3) + (-3, 2) = (-2, 5)$

Plot these three points and join them to draw triangle $C$ .

Key Takeaways

A translation by $\begin{pmatrix} a \\ b \end{pmatrix}$ adds $a$ to the $x$ -coordinate and $b$ to the $y$ -coordinate of every point.
The shape, size, and orientation are all preserved under translation.
Always apply the translation to each vertex individually.

Common Mistakes

Subtracting instead of adding (or vice versa) — remember the vector components are added to the original coordinates.
Mixing up the order: the top number is the $x$ -change, the bottom is the $y$ -change.
Drawing the triangle from triangle $B$ instead of triangle $A$ — the question says triangle $A$ is mapped onto triangle $C$ .

Things to Be Careful About

The answer is a drawing, so accuracy matters. Plot the vertices precisely on the grid and join them with straight lines.
The mark scheme awards B1 for correct vertices and B1 for the triangle being drawn. If the triangle is wrong, check the SC1 alternative: translating shape $B$ by $\begin{pmatrix} -3 \\ 2 \end{pmatrix}$ gives vertices $(-6, 3)$ , $(-6, -3)$ , $(-3, -3)$ , which would score 1 mark.
Coordinates should be read correctly from the grid; double-check each vertex before drawing.

Techniques used

apply translation vector to each vertexadd the vector components to the coordinates

Q103MMedium-EasyAlgebra and Graphs

Solve the simultaneous equations.
Show your working.

\begin{aligned} 3a + b &= -4 \\ 2a + 3b &= 9 \end{aligned}

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Approach

We are given two linear equations in two variables, $a$ and $b$ . To solve them simultaneously, we will use the elimination method: scale one equation so that the coefficient of $b$ matches the other, then subtract to remove $b$ and solve for $a$ . Finally, substitute $a$ back into an original equation to find $b$ .

Working

Label the equations:
(1) $3a + b = -4$
(2) $2a + 3b = 9$

Multiply equation (1) by 3 so the $b$ terms match:

9a + 3b = -12 \quad \text{(3)}

Subtract equation (2) from equation (3):

(9a + 3b) - (2a + 3b) = -12 - 9

7a = -21

Divide by 7:

a = -3

Substitute $a = -3$ into equation (1):

3(-3) + b = -4

-9 + b = -4

Add 9 to both sides:

b = 5

Check with equation (2): $2(-3) + 3(5) = -6 + 15 = 9$ . The values satisfy both equations.

Answer

a = -3, \quad b = 5

Final answer

a = -3, b = 5

Detailed explanation

Walkthrough

We start with two equations that share the same unknowns, $a$ and $b$ . Because both equations must be true at the same time, we can combine them to cancel out one variable. Multiplying the first equation by 3 gives us $3b$ , which exactly matches the $3b$ in the second equation. When we subtract the second equation from this scaled version, the $3b$ terms disappear, leaving a simple one-variable equation for $a$ . Solving it gives $a = -3$ . With $a$ known, we plug it back into the first original equation to isolate $b$ , yielding $b = 5$ . Substituting both values into the second equation confirms they work.

Key Takeaways

Elimination works best when you can easily make the coefficients of one variable identical (or opposite) across the two equations. Always show the scaling step explicitly to earn the method mark. After finding one variable, substitute immediately into an original equation rather than the scaled one to avoid carrying forward arithmetic errors.

Common Mistakes

Failing to multiply every term in an equation when scaling it up (e.g., writing $9a + 3b = -4$ instead of $-12$ ).
Sign errors during subtraction, particularly $-12 - 9$ being calculated as $-3$ instead of $-21$ .
Providing only one variable's value, which limits the score to A1 according to the mark scheme.
Skipping working entirely; without visible elimination steps, the M1 mark is lost (though SC1 may still apply if the answers are correct).

Things to Be Careful About

Show every algebraic step clearly. The mark scheme awards M1 specifically for a correct elimination method, and A2 requires both correct values.
Watch negative number arithmetic carefully throughout the process.
Verify your final pair by substituting into the unused original equation before finishing.
Write the final answer clearly with both variables stated, as requested by the question format.

Techniques used

eliminate one variable by scaling equationssolve the resulting linear equationsubstitute back to find the second variable

Q11Medium-EasyCoordinate Geometry

Point $A$ $(2, 4)$ is joined to point $B$ $(5, -2)$ by a straight line.

(a)

Find the coordinates of the midpoint of $AB$ .

( ______ , ______ )

DifficultyEasy

Worked solution

Approach

The midpoint $M$ of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by the average of the coordinates:

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Working

Given points $A(2, 4)$ and $B(5, -2)$ , we substitute these values into the formula.

For the $x$ -coordinate of the midpoint:

x_m = \frac{2 + 5}{2}

x_m = \frac{7}{2} = 3.5

For the $y$ -coordinate of the midpoint:

y_m = \frac{4 + (-2)}{2}

y_m = \frac{2}{2} = 1

Thus, the midpoint is $(3.5, 1)$ . The mark scheme accepts $3.5$ or its equivalent fraction $\frac{7}{2}$ .

Answer

(3.5, 1)

Final answer

(3.5, 1)

Detailed explanation

Walkthrough

To find the midpoint of a straight line joining two points, we need to find the point exactly halfway between them on both the horizontal ( $x$ ) and vertical ( $y$ ) axes.

Identify the coordinates: Point $A$ is at $(2, 4)$ and Point $B$ is at $(5, -2)$ . So, $x_1 = 2$ , $y_1 = 4$ , $x_2 = 5$ , and $y_2 = -2$ .
Apply the midpoint formula: The midpoint is the average of the $x$ $x$ -coordinates and the average of the $y$ $y$ -coordinates separately.
- For the $x$ -coordinate: Add the $x$ -values ( $2 + 5 = 7$ ) and divide by 2. This gives $3.5$ (or $\frac{7}{2}$ ).
- For the $y$ -coordinate: Add the $y$ -values ( $4 + (-2) = 2$ ) and divide by 2. This gives $1$ .
Combine: The final coordinate pair is $(3.5, 1)$ .

Key Takeaways

The midpoint formula is simply the arithmetic mean of the coordinates: $\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$ .
You can use decimals or fractions for the answer unless specified otherwise (here, $3.5$ is accepted).

Common Mistakes

Adding the coordinates but forgetting to divide by 2 (e.g., answering $(7, 2)$ ).
Subtracting the coordinates instead of adding (e.g., $5 - 2 = 3$ ).
Mixing up $x$ and $y$ values when substituting.

Things to Be Careful About

Ensure you handle negative signs correctly in the addition (e.g., $4 + (-2)$ is not $4 - 2$ in terms of sign handling logic, though numerically it results in subtraction; keep track of the negative value clearly).
The question asks for exact coordinates; do not round if the result is an exact decimal like 3.5.

Techniques used

calculate the midpoint of a line segment

(b)

Find the gradient of the line $AB$ .

______

DifficultyMedium-Easy

Worked solution

Approach

The gradient $m$ of a straight line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated as the change in $y$ divided by the change in $x$ :

m = \frac{y_2 - y_1}{x_2 - x_1}

This is often described as "rise over run".

Working

Using the points $A(2, 4)$ and $B(5, -2)$ :

Let $(x_1, y_1) = (2, 4)$ and $(x_2, y_2) = (5, -2)$ .

Substitute these values into the gradient formula:

m = \frac{-2 - 4}{5 - 2}

Calculate the numerator (change in $y$ ):

-2 - 4 = -6

Calculate the denominator (change in $x$ ):

5 - 2 = 3

Divide the numerator by the denominator:

m = \frac{-6}{3}

m = -2

Note: The order of subtraction must be consistent. Using $\frac{4 - (-2)}{2 - 5}$ yields $\frac{6}{-3} = -2$ , which is also correct.

Answer

-2

Final answer

-2

Detailed explanation

Walkthrough

Gradient measures the steepness and direction of a line. It is defined as the vertical change divided by the horizontal change between any two points on the line.

Formula: Use $m = \frac{y_2 - y_1}{x_2 - x_1}$ . It is crucial that you subtract the $y$ -coordinates together and the $x$ -coordinates together.
Substitution:
- Change in $y$ : Take the $y$ -value of the second point ( $-2$ ) and subtract the $y$ -value of the first point ( $4$ ). Result: $-6$ .
- Change in $x$ : Take the $x$ -value of the second point ( $5$ ) and subtract the $x$ -value of the first point ( $2$ ). Result: $3$ .
Calculation: Divide the change in $y$ by the change in $x$ : $\frac{-6}{3} = -2$ .
Sign Check: Since the line goes down as you move from left to right (from $y=4$ to $y=-2$ ), the gradient must be negative. Our result $-2$ confirms this.

Key Takeaways

Gradient is always $\frac{\text{change in } y}{\text{change in } x}$ .
The order of points does not matter as long as you are consistent (i.e., if you do $y_2 - y_1$ , you must do $x_2 - x_1$ ).
A negative gradient indicates a downward slope.

Common Mistakes

Swapping the numerator and denominator (calculating $\frac{x_2-x_1}{y_2-y_1}$ ).
Inconsistent subtraction order (e.g., $y_2 - y_1$ but $x_1 - x_2$ ), which flips the sign of the answer.
Arithmetic errors with negative numbers (e.g., thinking $-2 - 4 = -2$ or $6$ ).

Things to Be Careful About

The mark scheme shows the method mark (M1) is awarded for the correct substitution expression like $\frac{4 - (-2)}{2 - 5}$ . Even if the final calculation is wrong, showing this step correctly earns partial credit.
Do not simplify the fraction incorrectly. Here $\frac{-6}{3}$ simplifies cleanly to $-2$ .

Techniques used

calculate the gradient of a straight line

Q12Medium-EasyNumber

(a)

Write $0.000257$ in standard form.

______

(b)

Work out

\frac{2 \times 10^4}{4 \times 10^{-5}}

Give your answer in standard form.

______

DifficultyMedium-Easy

Worked solution

Approach

Divide the coefficients and subtract the power of ten in the denominator from the power in the numerator. Then adjust the result so the coefficient lies between 1 and 10.

Working

\frac{2 \times 10^4}{4 \times 10^{-5}} = \frac{2}{4} \times 10^{4 - (-5)}

= 0.5 \times 10^{9}

Since $0.5 = 5 \times 10^{-1}$ , multiply the powers of ten:

0.5 \times 10^{9} = 5 \times 10^{-1} \times 10^{9} = 5 \times 10^{8}

Answer

5 \times 10^{8}

Final answer

5 × 10^8

Detailed explanation

Walkthrough

Split the calculation into coefficients and powers of ten: divide $2$ by $4$ to get $0.5$ , and divide $10^4$ by $10^{-5}$ by subtracting the exponents: $4 - (-5) = 9$ . This gives $0.5 \times 10^9$ , which is not standard form because $0.5 < 1$ . Rewrite $0.5$ as $5 \times 10^{-1}$ , so $0.5 \times 10^9 = 5 \times 10^{-1} \times 10^9 = 5 \times 10^8$ .

Key Takeaways

When dividing numbers in standard form, divide the coefficients and subtract the powers of ten. The final answer must have a coefficient between 1 and 10.

Common Mistakes

Leaving the answer as $0.5 \times 10^9$ loses the final mark because $0.5$ is not between 1 and 10. Another common error is subtracting the powers incorrectly: $4 - (-5) = 9$ , not $-1$ or $1$ . Also, forgetting to divide the coefficients and only dividing the powers.

Things to Be Careful About

The mark scheme gives B1 for $0.5 \times 10^9$ oe seen, or for an answer $A \times 10^8$ with $1 \leq A < 10$ . The final answer must be $5 \times 10^8$ cao. On the non-calculator paper, show the exponent subtraction and the adjustment from $0.5 \times 10^9$ to $5 \times 10^8$ to make the method clear.

Techniques used

divide the coefficientssubtract the powers of ten when dividingrewrite a coefficient less than 1 in standard form

Q132MMedium-EasyNumber

Work out $2\frac{1}{5} \div \frac{3}{4}$ .

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

______

DifficultyMedium-Easy

Worked solution

Approach

To divide by a fraction, we multiply by its reciprocal. First, convert the mixed number $2\frac{1}{5}$ into an improper fraction. Then perform the multiplication.

Working

Convert the mixed number $2\frac{1}{5}$ to an improper fraction:

2\frac{1}{5} = \frac{(2 \times 5) + 1}{5} = \frac{11}{5}

Rewrite the division as multiplication by the reciprocal of $\frac{3}{4}$ :

\frac{11}{5} \div \frac{3}{4} = \frac{11}{5} \times \frac{4}{3}

Multiply the numerators together and the denominators together:

\frac{11 \times 4}{5 \times 3} = \frac{44}{15}

Convert the improper fraction $\frac{44}{15}$ back to a mixed number. We know that $15 \times 2 = 30$ , so:

\frac{44}{15} = \frac{30 + 14}{15} = 2\frac{14}{15}

Check if $\frac{14}{15}$ can be simplified further. The factors of $14$ are $1, 2, 7, 14$ . The factors of $15$ are $1, 3, 5, 15$ . They share no common factors other than $1$ , so the fraction is in its simplest form.

Answer

2\frac{14}{15}

Final answer

2 14/15

Detailed explanation

Walkthrough

The problem asks us to divide a mixed number by a proper fraction: $2\frac{1}{5} \div \frac{3}{4}$ .

Step 1: Convert the mixed number.
Mixed numbers are difficult to work with directly in multiplication or division. It is much easier to convert them into improper fractions. To convert $2\frac{1}{5}$ , we multiply the whole number part ( $2$ ) by the denominator ( $5$ ) and add the numerator ( $1$ ). This gives $11$ , which becomes the new numerator over the original denominator $5$ . So, $2\frac{1}{5} = \frac{11}{5}$ .

Step 2: Change division to multiplication.
The rule for dividing by a fraction is to multiply by its reciprocal (flip the second fraction upside down). The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$ . Therefore, the operation becomes $\frac{11}{5} \times \frac{4}{3}$ .

Step 3: Multiply the fractions.
To multiply two fractions, we multiply the top numbers (numerators) together and the bottom numbers (denominators) together.
Numerator: $11 \times 4 = 44$
Denominator: $5 \times 3 = 15$
This gives us the improper fraction $\frac{44}{15}$ .

Step 4: Convert back to a mixed number.
The question requires the answer as a mixed number. We need to find how many times $15$ goes into $44$ .
$15 \times 1 = 15$
$15 \times 2 = 30$
$15 \times 3 = 45$ (this is too big)
So, $15$ goes into $44$ exactly $2$ times.
Now we find the remainder: $44 - 30 = 14$ .
The remainder ( $14$ ) becomes the new numerator, while the denominator stays $15$ .
The result is $2\frac{14}{15}$ .

Step 5: Check for simplification.
The fraction part is $\frac{14}{15}$ . We check if they share any common factors. $14 = 2 \times 7$ and $15 = 3 \times 5$ . There are no common factors, so the fraction cannot be simplified further.

Key Takeaways

Always convert mixed numbers to improper fractions before performing multiplication or division.
Dividing by a fraction is equivalent to multiplying by its reciprocal.
When the final answer must be a mixed number, convert the improper fraction back using integer division to find the whole number part and the remainder.

Common Mistakes

Forgetting to invert the second fraction (e.g., multiplying straight across to get $\frac{33}{20}$ ).
Incorrectly converting mixed numbers (e.g., calculating $2\frac{1}{5}$ as $\frac{21}{5}$ instead of $\frac{11}{5}$ ).
Leaving the answer as an improper fraction when a mixed number is requested.
Failing to simplify the fractional part of the mixed number if it is possible (though not applicable here, it is a frequent error).

Things to Be Careful About

Ensure the final answer is in its simplest form. In this case, $\frac{14}{15}$ is irreducible. If the answer had been $2\frac{16}{15}$ , it would be invalid because the fraction part must be less than $1$ .
On the non-calculator component, all steps must be shown clearly to earn method marks.

Techniques used

convert mixed numbers to improper fractionsinvert the divisor for divisionmultiply the numerators and denominatorssimplify the resulting fractionconvert an improper fraction back to a mixed number

Q14MediumNumber

(a)

Write $360$ as a product of its prime factors.

______

DifficultyMedium-Easy

Worked solution

Approach

We decompose 360 by dividing by the smallest prime numbers successively until we reach 1. We then group identical factors to write the answer in index form.

Working

First, divide 360 by 2:

360 \div 2 = 180

Divide 180 by 2:

180 \div 2 = 90

Divide 90 by 2:

90 \div 2 = 45

Now 45 is odd, so we try the next smallest prime, which is 3:

45 \div 3 = 15

Divide 15 by 3:

15 \div 3 = 5

Finally, 5 is prime:

5 \div 5 = 1

The prime factors are $2, 2, 2, 3, 3,$ and $5$ . Grouping them gives:

360 = 2^3 \times 3^2 \times 5

Answer

2^3 \times 3^2 \times 5

Final answer

2^3 x 3^2 x 5

Detailed explanation

Walkthrough

To find the prime factorization of 360, we use repeated division. We start with the smallest prime number, 2. Since 360 is even, it is divisible by 2. We keep dividing by 2 as long as the result remains even: $360 \to 180 \to 90 \to 45$ . At this point, we have used the factor 2 three times ( $2^3$ ). The remaining number is 45, which is not divisible by 2. We move to the next prime number, 3. Since the sum of digits of 45 (4+5=9) is divisible by 3, 45 is divisible by 3. Dividing gives 15, and dividing again gives 5. This accounts for two factors of 3 ( $3^2$ ). The final number is 5, which is itself a prime number. Thus, the complete list of prime factors is $2, 2, 2, 3, 3, 5$ . Writing these in index notation gives $2^3 \times 3^2 \times 5$ .

Key Takeaways

Prime factorization breaks any integer down into a unique product of prime numbers.
Index notation provides a compact way to write repeated multiplication of the same factor.
Always check divisibility by small primes (2, 3, 5, etc.) in order.

Common Mistakes

Forgetting that 1 is not a prime number.
Stopping the factorization too early (e.g., stopping at 9 instead of breaking it down to $3 \times 3$ ).
Incorrectly counting the powers when writing the final answer.

Things to Be Careful About

Ensure you write the final answer in the requested format (index notation vs. product of primes). Both $2^3 \times 3^2 \times 5$ and $2 \times 2 \times 2 \times 3 \times 3 \times 5$ are generally accepted unless specified otherwise.

Techniques used

decompose a composite number into prime factorsexpress the result in index notation

(b)

Find the smallest positive integer $n$ such that $360n$ is a cube number.

$n$ = ______

DifficultyMedium

Worked solution

Approach

A number is a perfect cube if all the exponents in its prime factorization are multiples of 3. We examine the factors found in part (a) and determine what additional factors are needed to raise each exponent to the next multiple of 3.

Working

From part (a), the prime factorization of 360 is:

360 = 2^3 \times 3^2 \times 5^1

For $360n$ to be a cube number, every prime factor's exponent must be a multiple of 3.

For the base 2: The exponent is already 3, which is a multiple of 3. No additional 2s are needed.
For the base 3: The exponent is 2. The next multiple of 3 is 3. We need one more factor of 3 ( $3^{3-2} = 3^1$ ).
For the base 5: The exponent is 1. The next multiple of 3 is 3. We need two more factors of 5 ( $5^{3-1} = 5^2$ ).

Therefore, $n$ must provide these missing factors:

n = 3^1 \times 5^2

Calculate the value of $n$ :

n = 3 \times 25 = 75

Check: $360 \times 75 = 27000$ . Since $30^3 = 27000$ , the result is correct.

Answer

75

Final answer

75

Detailed explanation

Walkthrough

We established in part (a) that $360 = 2^3 \times 3^2 \times 5$ . A perfect cube has prime factors where every exponent is divisible by 3.

Looking at the term $2^3$ : The exponent 3 is already a multiple of 3. So, $n$ does not need to contain any factor of 2.

Looking at the term $3^2$ : The exponent 2 is not a multiple of 3. To make it a cube, we need to reach an exponent of 3 (the next multiple of 3). This means we need to multiply by $3^1$ .

Looking at the term $5^1$ : The exponent 1 is not a multiple of 3. To make it a cube, we need to reach an exponent of 3. This means we need to multiply by $5^2$ .

So, the smallest positive integer $n$ is the product of these required missing factors: $n = 3^1 \times 5^2 = 3 \times 25 = 75$ .

Key Takeaways

A number is a perfect cube if and only if the exponent of every prime in its prime factorization is a multiple of 3.
To find the smallest multiplier, identify how many more of each prime factor are needed to reach the next multiple of 3.

Common Mistakes

Assuming $n$ must be a prime number.
Multiplying the existing factors together instead of finding the missing ones.
Misidentifying the target exponent (e.g., aiming for 4 or 5 instead of 3 or 6).

Things to Be Careful About

Remember that $5$ is effectively $5^1$ . It is easy to forget that it needs two more factors of 5 to become $5^3$ .

Techniques used

identify missing factors to complete cube exponentsmultiply known prime factors to find the multiplier

Q154MMedium-EasyMensuration

A sector of a circle with angle $60^{\circ}$ has arc length $4\pi\,\text{cm}$ .

Find the area of the sector.
Give your answer, as simply as possible, in terms of $\pi$ .

______ $\text{cm}^2$

DifficultyMedium-Easy

Worked solution

Approach

The sector angle is $60^\circ$ , which is $\frac{60}{360} = \frac{1}{6}$ of a full circle. Use the arc length formula to find the radius $r$ , then use the sector area formula.

Working

Arc length of a sector with angle $\theta$ is

\frac{\theta}{360} \times 2\pi r

Here $\theta = 60^\circ$ and arc length $= 4\pi$ , so

\frac{60}{360} \times 2\pi r = 4\pi

Simplify $\frac{60}{360}$ :

\frac{1}{6} \times 2\pi r = 4\pi

\frac{2\pi r}{6} = 4\pi

\frac{\pi r}{3} = 4\pi

Divide both sides by $\pi$ :

\frac{r}{3} = 4

r = 12

Now the area of a sector is

\frac{60}{360} \times \pi r^2

Substitute $r = 12$ :

\frac{1}{6} \times \pi \times 12^2

= \frac{144\pi}{6}

= 24\pi

Answer

24\pi\ \text{cm}^2

Final answer

24π cm^2

Detailed explanation

Walkthrough

The sector angle $60^\circ$ is $\frac{60}{360} = \frac{1}{6}$ of a full circle. Therefore both the arc length and the area of this sector are $\frac{1}{6}$ of the corresponding full-circle values.

The arc length of the whole circle is its circumference, $2\pi r$ . So the arc length of the sector is

\frac{1}{6} \times 2\pi r

We are told this equals $4\pi$ , giving

\frac{1}{6} \times 2\pi r = 4\pi

Simplify the left-hand side:

\frac{2\pi r}{6} = 4\pi \quad \Rightarrow \quad \frac{\pi r}{3} = 4\pi

Dividing both sides by $\pi$ gives $\frac{r}{3} = 4$ , so $r = 12$ . This radius is needed before the area can be found.

The area of a full circle is $\pi r^2$ , so the sector area is

\frac{1}{6} \times \pi \times 12^2 = \frac{144\pi}{6} = 24\pi

Key Takeaways

A sector angle written as a fraction of $360^\circ$ gives the fraction of the circle used for both arc length and area.
Arc length of a sector: $\frac{\theta}{360} \times 2\pi r$ .
Area of a sector: $\frac{\theta}{360} \times \pi r^2$ .
When a question asks for an answer in terms of $\pi$ , leave $\pi$ in the answer and do not convert to a decimal.

Common Mistakes

Forgetting to multiply by the sector fraction $\frac{60}{360}$ .
Using the arc length as if it were the radius.
Finding the area of the full circle $\pi r^2$ instead of the sector area.
Writing a decimal approximation such as $75.4$ instead of $24\pi$ .

Things to Be Careful About

The mark scheme awards B2 for finding $r = 12$ ; if you only set up the equation, you may earn M1.
Working must be shown (nfww) to gain full marks; an unsupported final answer may not be accepted.
Keep the answer exactly as $24\pi\,\text{cm}^2$ ; do not round.
On the non-calculator paper, all simplification must be done by hand, as shown in the working.

Techniques used

write the arc length formula for a sectorsolve the resulting linear equation for the radiussubstitute into the sector area formulasimplify the exact expression in terms of pi

Q16Medium-Easy[Legacy] Matrices

(a)

Express as a single matrix.

2\begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}

______

DifficultyMedium-Easy

Worked solution

Approach

Multiply the first matrix by the scalar $2$ , then subtract the second matrix element by element.

Working

2 \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 6 & -2 \\ 4 & 8 \end{pmatrix}

Now subtract:

\begin{pmatrix} 6 & -2 \\ 4 & 8 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 6-1 & -2-3 \\ 4-(-2) & 8-5 \end{pmatrix}

= \begin{pmatrix} 5 & -5 \\ 6 & 3 \end{pmatrix}

Answer

\begin{pmatrix} 5 & -5 \\ 6 & 3 \end{pmatrix}

Final answer

[[5, -5], [6, 3]]

Detailed explanation

Walkthrough

The expression contains a scalar multiplication followed by a subtraction. First multiply every entry of the first matrix by $2$ , giving entries $6$ , $-2$ , $4$ and $8$ . Then subtract the second matrix entry by entry: $6-1=5$ , $-2-3=-5$ , $4-(-2)=6$ and $8-5=3$ . This produces the final matrix with entries $5$ , $-5$ , $6$ and $3$ .

Key Takeaways

Matrix multiplication by a scalar is performed on every element. Matrix addition and subtraction are performed element by element, so the two matrices must have the same size. Since both matrices are $2 \times 2$ , this subtraction is valid.

Common Mistakes

Multiplying only some entries by the scalar.
Subtracting incorrectly when the second matrix has a negative entry; $4-(-2)$ is $6$ , not $2$ .
Combining the matrices before carrying out the scalar multiplication.
Writing only the intermediate scaled matrix as the final answer; this can earn partial credit but not full marks.

Things to Be Careful About

Work through the four entries systematically. The final matrix must be written as one $2 \times 2$ matrix. Since the answer is exact, no rounding is needed.

Techniques used

multiply every element of the matrix by the scalarsubtract corresponding elements of the matriceswrite the result as a single matrix

(b)

$\mathbf{A} = \begin{pmatrix} 4 & k \\ 2 & 1 \end{pmatrix}$

The determinant of $\mathbf{A}$ is $10$ .

2M

(i)

Find the value of $k$ .

$k$ = ______

(ii)

Find $\mathbf{A}^{-1}$ .

$\mathbf{A}^{-1}$ = ______

DifficultyEasy

Worked solution

Approach

Use the inverse formula for a $2 \times 2$ matrix:

\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

From part (i), $\det(\mathbf{A}) = 10$ and $k = -3$ . Substitute $a=4$ , $b=k=-3$ , $c=2$ and $d=1$ .

Working

\mathbf{A}^{-1} = \frac{1}{10} \begin{pmatrix} 1 & -k \\ -2 & 4 \end{pmatrix}

Since $k = -3$ , we have $-k = 3$ :

\mathbf{A}^{-1} = \frac{1}{10} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}

This can also be written as

\begin{pmatrix} \frac{1}{10} & \frac{3}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}

Answer

\mathbf{A}^{-1} = \frac{1}{10} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}

Final answer

1/10 [[1, 3], [-2, 4]]

Detailed explanation

Walkthrough

For a $2 \times 2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ , the inverse is $\frac{1}{ad-bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$ . From part (i), the determinant is $10$ and $k=-3$ . Substituting $a=4$ , $b=k=-3$ , $c=2$ , $d=1$ gives the inverse as $\frac{1}{10}\begin{pmatrix}1 & -k \\ -2 & 4\end{pmatrix}$ . Since $-k=3$ , this becomes $\frac{1}{10}\begin{pmatrix}1 & 3 \\ -2 & 4\end{pmatrix}$ .

Key Takeaways

The inverse of a $2 \times 2$ matrix is the reciprocal of the determinant multiplied by a rearranged matrix. This question also shows follow-through: once $k$ is found, it is substituted directly into the inverse formula.

Common Mistakes

Forgetting to divide by the determinant.
Using the wrong sign pattern: the top-right entry should be $-k$ , and the bottom-left entry should be $-2$ .
Using $k=3$ instead of $k=-3$ , or failing to simplify $-k$ to $3$ .
Leaving the scalar $\frac{1}{10}$ outside but then not applying it to every entry if the answer is written as individual fractions.

Things to Be Careful About

The answer may be left as $\frac{1}{10}\begin{pmatrix}1 & 3 \\ -2 & 4\end{pmatrix}$ or written with each entry divided by $10$ . The mark scheme allows follow-through from the value of $k$ found in part (i), so if a different $k$ was obtained, that value should be used in this part.

Techniques used

substitute entries and determinant into the inverse formulareplace k with the value found in part (i)write the inverse as a scalar multiple of a matrix

Q17Medium-EasyAlgebra and Graphs

The diagram shows the lines $y = 2x + 1$ and $x + y = 2$ .

Fig. 5

(a)

The region $R$ is defined by these three inequalities.

y \leq 2x + 1 \quad x + y \leq 2 \quad y \geq -2

On the diagram, shade and label the region $R$ .

DifficultyMedium-Easy

Worked solution

Approach

The region $R$ is bounded by three lines:

$y = 2x + 1$ (from $y \leq 2x + 1$ )
$x + y = 2$ , i.e. $y = -x + 2$ (from $x + y \leq 2$ )
$y = -2$ (from $y \geq -2$ )

For each inequality, we shade the side that satisfies it, then the region $R$ is where all three shaded areas overlap.

Working

Boundary line 1: $y = 2x + 1$

The inequality $y \leq 2x + 1$ means we shade the region below (or on) this line. The line passes through $(0, 1)$ and $(1, 3)$ as shown in the diagram.

Boundary line 2: $x + y = 2$ , or $y = -x + 2$

The inequality $x + y \leq 2$ means we shade the region below (or on) this line. The line passes through $(0, 2)$ and $(2, 0)$ as shown in the diagram.

Boundary line 3: $y = -2$

The inequality $y \geq -2$ means we shade the region above (or on) this horizontal line.

Finding the vertices of region $R$ :

Intersection of $y = 2x + 1$ and $y = -x + 2$ :

\begin{aligned} 2x + 1 &= -x + 2 \\ 3x &= 1 \\ x &= \frac{1}{3}, \quad y = \frac{5}{3} \end{aligned}

Vertex: $\left(\frac{1}{3}, \frac{5}{3}\right)$

Intersection of $y = 2x + 1$ and $y = -2$ :

\begin{aligned} 2x + 1 &= -2 \\ 2x &= -3 \\ x &= -\frac{3}{2} \end{aligned}

Vertex: $\left(-\frac{3}{2}, -2\right)$

Intersection of $y = -x + 2$ and $y = -2$ :

\begin{aligned} -x + 2 &= -2 \\ x &= 4 \end{aligned}

Vertex: $(4, -2)$

The region $R$ is the triangle with vertices at $\left(\frac{1}{3}, \frac{5}{3}\right)$ , $\left(-\frac{3}{2}, -2\right)$ , and $(4, -2)$ , shaded dark grey.

Answer

The region $R$ is the triangular area bounded by $y = 2x + 1$ , $y = -x + 2$ , and $y = -2$ , with vertices at $\left(\frac{1}{3}, \frac{5}{3}\right)$ , $\left(-\frac{3}{2}, -2\right)$ , and $(4, -2)$ .

Final answer

Shaded triangular region with vertices at (1/3, 5/3), (-3/2, -2), and (4, -2)

Detailed explanation

Walkthrough

The question asks us to shade the region satisfying three simultaneous inequalities. Each inequality defines a half-plane, and the region $R$ is the intersection of all three half-planes.

Step 1: Identify the three boundary lines.

From $y \leq 2x + 1$ , the boundary is $y = 2x + 1$ . Since the inequality is $\leq$ , we shade below this line.
From $x + y \leq 2$ , the boundary is $y = -x + 2$ . Since the inequality is $\leq$ , we shade below this line.
From $y \geq -2$ , the boundary is $y = -2$ . Since the inequality is $\geq$ , we shade above this line.

Step 2: Find the vertices of the region.
The vertices occur where pairs of boundary lines intersect. We solve each pair of equations simultaneously:

For $y = 2x + 1$ and $y = -x + 2$ : setting $2x + 1 = -x + 2$ gives $3x = 1$ , so $x = \frac{1}{3}$ and $y = \frac{5}{3}$ .

For $y = 2x + 1$ and $y = -2$ : setting $2x + 1 = -2$ gives $2x = -3$ , so $x = -\frac{3}{2}$ .

For $y = -x + 2$ and $y = -2$ : setting $-x + 2 = -2$ gives $x = 4$ .

Step 3: Shade the region.
The region $R$ is the triangle enclosed by all three lines, shaded where all three conditions are met simultaneously.

Key Takeaways

Each linear inequality $y \leq mx + c$ or $y \geq mx + c$ defines a half-plane below or above the boundary line.
The solution to a system of inequalities is the region where all individual shaded areas overlap.
Finding vertices of the solution region involves solving pairs of boundary equations simultaneously.

Common Mistakes

Shading the wrong side of a boundary line (e.g., shading above $y = 2x + 1$ instead of below).
Forgetting to include the boundary line itself (the inequality is $\leq$ or $\geq$ , not $<$ or $>$ ).
Drawing $y = -2$ incorrectly or forgetting to shade above it.
Not labelling the region $R$ on the diagram.

Things to Be Careful About

The inequalities use $\leq$ and $\geq$ , so the boundary lines are included in the region (solid lines, not dashed).
The region must satisfy ALL three inequalities simultaneously — shade carefully and find the overlap.
Always label the shaded region with the letter given in the question ( $R$ ).
The vertices may not be at integer coordinates, so check your intersection calculations carefully.

Techniques used

identify the three boundary linesdetermine which side of each line satisfies the inequalityfind the intersection points of the boundary linesshade the region satisfying all three inequalities

(b)

The point $(k, k - 2)$ lies in the region $R$ where $k$ is an integer.

List the possible values of $k$ .

______

DifficultyMedium-Easy

Worked solution

Approach

The point $(k, k-2)$ lies in region $R$ , so it must satisfy all three inequalities defining $R$ . Substitute $x = k$ and $y = k - 2$ into each inequality and solve for $k$ .

Working

Inequality 1: $y \leq 2x + 1$

k - 2 \leq 2k + 1

-2 - 1 \leq 2k - k

-3 \leq k

So $k \geq -3$ .

Inequality 2: $x + y \leq 2$

k + (k - 2) \leq 2

2k - 2 \leq 2

2k \leq 4

k \leq 2

Inequality 3: $y \geq -2$

k - 2 \geq -2

k \geq 0

Combining all three conditions:

k \geq -3, \quad k \leq 2, \quad k \geq 0

The most restrictive conditions are $k \geq 0$ and $k \leq 2$ , giving:

0 \leq k \leq 2

Since $k$ is an integer, the possible values are:

k = 0, \quad k = 1, \quad k = 2

Verification:

$k = 0$ : point $(0, -2)$ . Check: $-2 \leq 1$ ✓, $0 + (-2) = -2 \leq 2$ ✓, $-2 \geq -2$ ✓
$k = 1$ : point $(1, -1)$ . Check: $-1 \leq 3$ ✓, $1 + (-1) = 0 \leq 2$ ✓, $-1 \geq -2$ ✓
$k = 2$ : point $(2, 0)$ . Check: $0 \leq 5$ ✓, $2 + 0 = 2 \leq 2$ ✓, $0 \geq -2$ ✓

Answer

k = 0, 1, 2

Final answer

0, 1, 2

Detailed explanation

Walkthrough

The point $(k, k-2)$ lies in region $R$ , meaning it must satisfy all three inequalities that define $R$ . We substitute $x = k$ and $y = k - 2$ into each inequality and solve for $k$ .

Step 1: Apply $y \leq 2x + 1$ .
Substituting gives $k - 2 \leq 2k + 1$ . Rearranging: $-3 \leq k$ , so $k \geq -3$ .

Step 2: Apply $x + y \leq 2$ .
Substituting gives $k + (k - 2) \leq 2$ , which simplifies to $2k \leq 4$ , so $k \leq 2$ .

Step 3: Apply $y \geq -2$ .
Substituting gives $k - 2 \geq -2$ , which simplifies to $k \geq 0$ .

Step 4: Combine the conditions.
We need $k \geq -3$ AND $k \leq 2$ AND $k \geq 0$ . The most restrictive lower bound is $k \geq 0$ (since $0 > -3$ ), and the upper bound is $k \leq 2$ . So $0 \leq k \leq 2$ .

Step 5: List integer values.
Since $k$ must be an integer, the possible values are $k = 0, 1, 2$ .

Key Takeaways

When a point with a parameter lies in a region defined by inequalities, substitute the coordinates into each inequality.
Each inequality gives a constraint on the parameter; the valid range is the intersection of all constraints.
Always check that the final answer satisfies the original conditions, especially when combining inequalities.

Common Mistakes

Making sign errors when rearranging inequalities (e.g., forgetting to flip the sign when dividing by a negative — though not needed here).
Forgetting that $k$ must be an integer and giving a continuous range like $0 \leq k \leq 2$ .
Missing one of the three inequalities when substituting.
Not verifying the answer by checking each value in the original inequalities.

Things to Be Careful About

The question states $k$ is an integer, so the answer must be a list of discrete values, not an inequality.
All three inequalities must be satisfied simultaneously — check each one.
The point $(k, k-2)$ lies on the line $y = x - 2$ , which is different from any of the boundary lines of region $R$ . Verify the point is actually inside the shaded region.

Techniques used

substitute the point coordinates into each inequalitysolve the resulting inequalities for kfind the integer values of k in the valid range

Q18Medium-EasyStatistics

50 adults each take part in a quiz.
The cumulative frequency diagram shows their scores.

Fig. 6

(a)

Use the diagram to find an estimate of the interquartile range.

______

DifficultyMedium-Easy

Worked solution

Approach

The interquartile range (IQR) is the difference between the upper quartile (UQ) and the lower quartile (LQ). For 50 data points, the LQ is at the $\frac{1}{4} \times 50 = 12.5$ th position and the UQ is at the $\frac{3}{4} \times 50 = 37.5$ th position. We read the corresponding scores from the cumulative frequency diagram.

Working

Total number of adults = $50$ .

\text{Position of LQ} = \frac{1}{4} \times 50 = 12.5

Locate $12.5$ on the cumulative frequency axis, move horizontally to the curve, and then vertically down to the score axis. The score is approximately $17$ .

\text{LQ} \approx 17

\text{Position of UQ} = \frac{3}{4} \times 50 = 37.5

Locate $37.5$ on the cumulative frequency axis, move horizontally to the curve, and then vertically down to the score axis. The score is approximately $68$ .

\text{UQ} \approx 68

\text{IQR} = \text{UQ} - \text{LQ} = 68 - 17 = 51

(Any answer between $49$ and $53$ is accepted based on reading the graph.)

Answer

51

Final answer

51

Detailed explanation

Walkthrough

The interquartile range is a measure of spread calculated as the upper quartile minus the lower quartile. With a total frequency of $50$ , the lower quartile is found at the $12.5$ th value and the upper quartile at the $37.5$ th value. On the cumulative frequency diagram, we find $12.5$ on the vertical axis, trace across to the curve, and read down to the horizontal axis to get the lower quartile score (around $17$ ). We repeat the process for $37.5$ to find the upper quartile score (around $68$ ). Subtracting the lower quartile from the upper quartile gives the interquartile range.

Key Takeaways

The lower quartile is at $\frac{1}{4}$ of the total frequency, and the upper quartile is at $\frac{3}{4}$ .
To read a value from a cumulative frequency diagram, always start on the vertical (cumulative frequency) axis, move horizontally to the curve, then vertically down to the horizontal axis.
The interquartile range is simply $\text{UQ} - \text{LQ}$ .

Common Mistakes

Starting on the horizontal axis instead of the vertical axis when reading the diagram.
Using $\frac{1}{4}$ and $\frac{3}{4}$ of the horizontal axis range instead of the cumulative frequency.
Forgetting to subtract the LQ from the UQ, and just giving one of the quartile values.

Things to Be Careful About

Reading from a graph always involves estimation, so a range of acceptable answers is given. Ensure you read horizontally from the cumulative frequency axis and vertically down to the score axis.
The question asks for an estimate, so small reading errors (e.g., reading $16$ or $18$ for the LQ) are accepted as long as the final IQR is consistent with the readings.

Techniques used

find the lower quartile from the cumulative frequency diagramfind the upper quartile from the cumulative frequency diagramsubtract the lower quartile from the upper quartile

(b)

20% of the adults win a prize for getting a high score in the quiz.

Use the diagram to work out the minimum score needed to win a prize.

______

DifficultyMedium-Easy

Worked solution

Approach

The top $20\%$ of adults win a prize. We first calculate how many adults this represents, then find the cumulative frequency that separates the top $20\%$ from the rest. Finally, we read the corresponding score from the diagram.

Working

Total number of adults = $50$ .

20\% \text{ of } 50 = 0.20 \times 50 = 10

The top $10$ adults have the highest scores. Since there are $50$ adults in total, the top $10$ are those with scores above the $40$ th value ( $50 - 10 = 40$ ).

Locate $40$ on the cumulative frequency axis, move horizontally to the curve, and then vertically down to the score axis.

\text{Score at CF} = 40 \approx 75

(Any answer between $74$ and $76$ is accepted.)

Answer

75

Final answer

75

Detailed explanation

Walkthrough

The question asks for the minimum score to be in the top $20\%$ . First, find $20\%$ of the total number of adults: $0.20 \times 50 = 10$ . This means the top $10$ people win. On a cumulative frequency diagram, the highest values are at the top right. To find the cutoff for the top $10$ , we subtract $10$ from the total frequency: $50 - 10 = 40$ . This means anyone with a cumulative frequency of $40$ or more is in the top $10$ . We find $40$ on the vertical axis, trace to the curve, and read the score on the horizontal axis, which is approximately $75$ .

Key Takeaways

"Top $x\%$ " means you need to find the cumulative frequency that leaves $x\%$ at the upper end of the distribution.
Subtract the top percentage count from the total frequency to find the correct cumulative frequency value to read from the graph.

Common Mistakes

Reading the score corresponding to $20\%$ of the maximum score ( $20$ ) instead of $20\%$ of the total frequency.
Finding the score at CF = $10$ instead of CF = $40$ , which would give the bottom $20\%$ rather than the top $20\%$ .

Things to Be Careful About

Always verify which end of the distribution the question is asking about. "High score" means the right side of the graph, so we work backwards from the maximum cumulative frequency ( $50$ ).

Techniques used

calculate 20% of the total frequencyfind the corresponding score on the cumulative frequency diagram

(c)

Use the diagram to complete the frequency table.

Score ( $s$ )	$0 < s \leq 20$	$20 < s \leq 40$	$40 < s \leq 60$	$60 < s \leq 80$	$80 < s \leq 100$
Frequency					8

DifficultyEasy

Worked solution

Approach

The frequency for each class is the difference between the cumulative frequency at the upper boundary of the class and the cumulative frequency at the lower boundary. We read the cumulative frequencies from the diagram at $s = 20, 40, 60, 80$ and subtract consecutive values.

Working

From the diagram:

At $s = 20$ , cumulative frequency = $14$
At $s = 40$ , cumulative frequency = $27$
At $s = 60$ , cumulative frequency = $35$
At $s = 80$ , cumulative frequency = $42$
At $s = 100$ , cumulative frequency = $50$ (given)

Calculate the frequency for each class:

\text{For } 0 < s \leq 20: \quad 14 - 0 = 14

\text{For } 20 < s \leq 40: \quad 27 - 14 = 13

\text{For } 40 < s \leq 60: \quad 35 - 27 = 8

\text{For } 60 < s \leq 80: \quad 42 - 35 = 7

\text{For } 80 < s \leq 100: \quad 50 - 42 = 8 \quad (\text{given})

Answer

Score ( $s$ )	$0 < s \leq 20$	$20 < s \leq 40$	$40 < s \leq 60$	$60 < s \leq 80$	$80 < s \leq 100$
Frequency	$14$	$13$	$8$	$7$	$8$

The missing frequencies are $14, 13, 8, 7$ .

Final answer

14, 13, 8, 7

Detailed explanation

Walkthrough

A cumulative frequency diagram shows the running total of frequencies. To find the frequency for a specific class interval, we find the cumulative frequency at the upper end of the interval and subtract the cumulative frequency at the lower end. For the first class ( $0 < s \leq 20$ ), the cumulative frequency at $20$ is $14$ , and at $0$ it is $0$ , so the frequency is $14 - 0 = 14$ . For the next class ( $20 < s \leq 40$ ), the cumulative frequency at $40$ is $27$ , so the frequency is $27 - 14 = 13$ . We repeat this for the remaining classes: $35 - 27 = 8$ and $42 - 35 = 7$ .

Key Takeaways

The frequency of a class is the difference between consecutive cumulative frequencies.
Always read the cumulative frequency at the upper boundary of each class interval.

Common Mistakes

Copying the cumulative frequency value directly as the frequency.
Subtracting in the wrong order (e.g., lower minus upper).

Things to Be Careful About

Ensure the class boundaries match the values on the horizontal axis of the diagram. The diagram gives cumulative frequencies at $s = 20, 40, 60, 80, 100$ , which align perfectly with the table's class boundaries.

Techniques used

read cumulative frequencies at class boundariessubtract consecutive cumulative frequencies to find class frequencies

Q192MMedium-EasyAlgebra and Graphs

$x$ is inversely proportional to the square root of $y$ .
When $x = 2$ , $y = 16$ .

Find $y$ when $x = 32$ .

$y$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Since $x$ is inversely proportional to the square root of $y$ , we write $x = \frac{k}{\sqrt{y}}$ (or equivalently $x\sqrt{y} = k$ ). We substitute the given pair $(x=2, y=16)$ to find the constant $k$ . Then we use this $k$ and the new value $x=32$ to find the corresponding $y$ .

Working

The relationship is:

x = \frac{k}{\sqrt{y}}

Substitute $x = 2$ and $y = 16$ :

2 = \frac{k}{\sqrt{16}}

Calculate $\sqrt{16}$ :

\sqrt{16} = 4

So the equation becomes:

2 = \frac{k}{4}

Solve for $k$ by multiplying both sides by 4:

k = 2 \times 4 = 8

Now use $k = 8$ and the new value $x = 32$ to find $y$ :

32 = \frac{8}{\sqrt{y}}

Rearrange to isolate $\sqrt{y}$ :

\sqrt{y} = \frac{8}{32}

Simplify the fraction:

\sqrt{y} = \frac{1}{4}

Square both sides to find $y$ :

y = \left(\frac{1}{4}\right)^2

y = \frac{1}{16}

Answer

y = \frac{1}{16}

Final answer

1/16

Detailed explanation

Walkthrough

Formulate the equation: The phrase " $x$ is inversely proportional to the square root of $y$ " translates directly into the algebraic formula $x = \frac{k}{\sqrt{y}}$ , where $k$ is a constant. Alternatively, this can be written as $x\sqrt{y} = k$ .
Find the constant ( $k$ ): Use the specific values provided in the question to determine $k$ . We are told that when $x = 2$ , $y = 16$ . Substitute these into the equation:
$2 = \frac{k}{\sqrt{16}}$
Since $\sqrt{16} = 4$ , this simplifies to $2 = \frac{k}{4}$ . Multiplying by 4 gives $k = 8$ .
Solve for the unknown: Now that we have the constant $k=8$ , the specific relationship for this problem is $x = \frac{8}{\sqrt{y}}$ . We need to find $y$ when $x = 32$ . Substitute $x = 32$ :
$32 = \frac{8}{\sqrt{y}}$
Rearranging to make $\sqrt{y}$ the subject:
$\sqrt{y} = \frac{8}{32}$
Simplifying the fraction $\frac{8}{32}$ gives $\frac{1}{4}$ . So:
$\sqrt{y} = \frac{1}{4}$
Finally, square both sides to find $y$ :
$y = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$

Key Takeaways

Inverse Proportion Language: "Inversely proportional to the square root of $y$ " always means dividing the constant by $\sqrt{y}$ , not squaring it or anything else.
Finding Constants: You always need one complete set of matching values ( $x$ and $y$ ) to find the constant $k$ before you can solve for any other unknown.
Order of Operations: When solving $x = \frac{k}{\sqrt{y}}$ , remember to isolate $\sqrt{y}$ first, then square to get $y$ . Don't forget the final squaring step!

Common Mistakes

Incorrect Formula: Writing $x = \frac{k^2}{y}$ or $x = \frac{k}{y}$ instead of $x = \frac{k}{\sqrt{y}}$ .
Calculation Errors: Forgetting that $\sqrt{16} = 4$ or making arithmetic mistakes when calculating $k$ .
Premature Squaring: Squaring $x$ and $k$ incorrectly or forgetting to square the result at the end to find $y$ .
Fraction Simplification: Failing to simplify $\frac{8}{32}$ to $\frac{1}{4}$ , which can lead to calculation errors if squaring unsimplified fractions.

Things to Be Careful About

Exact Form: The answer should be left as an exact fraction ( $\frac{1}{16}$ ). Do not convert to a decimal unless asked.
Working Shown: Ensure you show the substitution steps clearly to earn method marks, especially for finding $k$ and isolating $y$ .

Techniques used

state the general formula for inverse proportionsubstitute known values to find the constant of proportionalitysolve for the unknown variable using the constant

Q202MMedium-EasyNumber

$(ax^n)^{\frac{2}{3}} = 4x^{10}$

Work out the value of $a$ and the value of $n$ .

$a$ = ______
$n$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Expand the left-hand side using the laws of indices: $(xy)^m = x^my^m$ and $(x^a)^b = x^{ab}$ . Then compare the numerical coefficient and the power of $x$ with the right-hand side to find $a$ and $n$ .

Working

Given:

(ax^n)^{\frac{2}{3}} = 4x^{10}

First, apply the outer exponent $\frac{2}{3}$ to each factor inside the bracket:

a^{\frac{2}{3}} \cdot (x^n)^{\frac{2}{3}} = 4x^{10}

Next, simplify the $x$ term using the rule $(x^n)^m = x^{nm}$ :

a^{\frac{2}{3}} \cdot x^{\frac{2n}{3}} = 4x^{10}

Now equate the coefficients (the numbers) on both sides:

a^{\frac{2}{3}} = 4

To solve for $a$ , raise both sides to the reciprocal power $\frac{3}{2}$ :

(a^{\frac{2}{3}})^{\frac{3}{2}} = 4^{\frac{3}{2}}

a = (\sqrt{4})^3

a = 2^3 = 8

Then equate the exponents of $x$ on both sides:

\frac{2n}{3} = 10

Multiply both sides by 3:

2n = 30

Divide by 2:

n = 15

Answer

a = 8, \quad n = 15

Final answer

a = 8, n = 15

Detailed explanation

Walkthrough

The problem gives an equation involving variables raised to fractional powers. To solve for the unknown constants $a$ and $n$ , we must first simplify the expression on the left-hand side so that it looks like the one on the right-hand side.

Expand the bracket: The expression is $(ax^n)^{\frac{2}{3}}$ . We distribute the exponent $\frac{2}{3}$ to both $a$ and $x^n$ . This uses the law $(xy)^m = x^m y^m$ . So we get $a^{\frac{2}{3}}(x^n)^{\frac{2}{3}}$ .
Simplify the variable part: For the $x$ term, we have a power raised to another power: $(x^n)^{\frac{2}{3}}$ . The rule here is to multiply the exponents: $n \times \frac{2}{3} = \frac{2n}{3}$ . The equation becomes $a^{\frac{2}{3}}x^{\frac{2n}{3}} = 4x^{10}$ .
Compare coefficients: Since the equation holds for all values of $x$ , the constant parts must be equal. Therefore, $a^{\frac{2}{3}} = 4$ . To isolate $a$ , we raise both sides to the power of $\frac{3}{2}$ (the reciprocal of $\frac{2}{3}$ ). Note that $4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$ . Thus, $a=8$ .
Compare exponents: Similarly, the exponents of the base $x$ must be equal. So, $\frac{2n}{3} = 10$ . Solving this linear equation involves multiplying by 3 to get $2n=30$ , then dividing by 2 to get $n=15$ .

Key Takeaways

Power of a Product Rule: $(AB)^m = A^m B^m$ . Every factor inside the brackets gets raised to the power outside.
Power of a Power Rule: $(A^m)^n = A^{mn}$ . Multiply the exponents together.
Equating Components: In an identity like $C_1 x^{p_1} = C_2 x^{p_2}$ , you can set $C_1 = C_2$ and $p_1 = p_2$ separately to solve for unknowns.

Common Mistakes

Forgetting to distribute the exponent to the coefficient $a$ . Writing $a x^{\dots}$ instead of $a^{\frac{2}{3}} x^{\dots}$ .
Adding exponents instead of multiplying them when simplifying $(x^n)^{\frac{2}{3}}$ . Writing $x^{n + \frac{2}{3}}$ .
Incorrectly solving $a^{\frac{2}{3}} = 4$ . Trying to divide by $\frac{2}{3}$ instead of raising to the reciprocal power.

Things to Be Careful About

When calculating $4^{\frac{3}{2}}$ , it is often easier to take the square root first ( $\sqrt{4}=2$ ) and then cube the result ( $2^3=8$ ), rather than cubing 4 first ( $64$ ) and then taking the square root.
Ensure working is shown clearly, as marks are awarded for the method of equating coefficients and powers.

Techniques used

apply the power of a product ruleapply the power of a power ruleequating coefficients to solve for unknowns

Q212MMedium-EasyAlgebra and Graphs

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

Fig. 7

Calculate the total distance travelled.

______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Approach

The distance travelled is equal to the area under the speed–time graph. We can calculate this area using the formula for a trapezium or by splitting the shape into two triangles and a rectangle.

Working

Method 1: Trapezium formula

The region under the graph is a trapezium. The two parallel sides lie along the time axis and have lengths $60$ (from $t = 0$ to $t = 60$ ) and $20$ (from $t = 10$ to $t = 30$ ). The perpendicular height is $10$ m/s.

\text{Distance} = \left( \frac{60 + 20}{2} \right) \times 10

\text{Distance} = \frac{80}{2} \times 10 = 40 \times 10 = 400

Method 2: Splitting into simpler shapes

Split the area under the graph into two triangles and a rectangle:

First triangle (from $t = 0$ to $t = 10$ ):

\text{Area} = \frac{10 \times 10}{2} = 50

Rectangle (from $t = 10$ to $t = 30$ ):

\text{Area} = (30 - 10) \times 10 = 20 \times 10 = 200

Second triangle (from $t = 30$ to $t = 60$ ):

\text{Area} = \frac{(60 - 30) \times 10}{2} = \frac{30 \times 10}{2} = 150

Total distance:

50 + 200 + 150 = 400

Answer

400

Final answer

400

Detailed explanation

Walkthrough

The problem asks for the total distance travelled given a speed–time graph. A fundamental principle in kinematics is that the distance travelled is equal to the area under the speed–time graph.

The graph forms a trapezium with vertices at $(0,0)$ , $(10,10)$ , $(30,10)$ , and $(60,0)$ . We can find the area of this trapezium directly using the formula $\frac{a+b}{2} \times h$ , where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular height. Here, the parallel sides are along the time axis: the bottom side has length $60 - 0 = 60$ , and the top side has length $30 - 10 = 20$ . The height is the maximum speed, $10$ . Substituting these values gives $\frac{60 + 20}{2} \times 10 = 400$ .

Alternatively, the area can be split into three simpler shapes: a right-angled triangle from $t=0$ to $t=10$ , a rectangle from $t=10$ to $t=30$ , and another right-angled triangle from $t=30$ to $t=60$ . Calculating the area of each and adding them together also gives $50 + 200 + 150 = 400$ .

Key Takeaways

The distance travelled is always equal to the area under a speed–time graph.
Areas under such graphs can be calculated using standard geometric formulas for trapeziums, rectangles, and triangles.
Units of speed (m/s) multiplied by units of time (s) yield units of distance (m).

Common Mistakes

Calculating the area using the wrong dimensions, such as using the time values on the vertical axis or speed values on the horizontal axis.
Forgetting to subtract the start time from the end time when finding the length of the parallel sides (e.g., using $30$ instead of $30 - 10 = 20$ for the top parallel side).
Adding the speeds or times directly instead of finding the area under the curve.

Things to Be Careful About

Ensure the final answer is in the correct units (metres, as requested by the question).
The diagram is marked 'NOT TO SCALE', so do not attempt to measure lengths with a ruler; always use the numerical values given on the axes.
Both the trapezium method and the composite shape method are acceptable, but all working must be shown to earn the method mark (M1).

Techniques used

recognise that distance is the area under a speed-time graphcalculate the area of the trapezium (or sum of triangles and rectangle)substitute values into the area formula

Q22MediumAlgebra and Graphs

$f(x) = 3x^2 + 5$

(a)

Work out $f(-1)$ .

______

(b)

Solve $f(2x) = 17$ .

$x$ = ______ or $x$ = ______

3M Similar

DifficultyMedium

Worked solution

Approach

We are given $f(2x) = 17$ . First, we determine the expression for $f(2x)$ by substituting $(2x)$ for $x$ in the original function definition $f(x) = 3x^2 + 5$ . Then we set this equal to 17 and solve the resulting quadratic equation for $x$ .

Working

Substitute $2x$ into $f(x)$ :

f(2x) = 3(2x)^2 + 5

Expand $(2x)^2$ :

(2x)^2 = 4x^2

So,

f(2x) = 3(4x^2) + 5 = 12x^2 + 5

Set this expression equal to 17:

12x^2 + 5 = 17

Subtract 5 from both sides:

12x^2 = 12

Divide by 12:

x^2 = 1

Take the square root of both sides. Remember to consider both positive and negative roots:

x = \pm\sqrt{1}

x = 1 \quad \text{or} \quad x = -1

Answer

x = 1 \quad \text{or} \quad x = -1

Final answer

x = 1 or x = -1

Detailed explanation

Walkthrough

The question gives us the output of the function when the input is $2x$ . We start by finding what $f(2x)$ actually is. Since $f(x) = 3x^2 + 5$ , replacing $x$ with $2x$ gives $f(2x) = 3(2x)^2 + 5$ . It is crucial to expand $(2x)^2$ correctly as $4x^2$ , not $2x^2$ . Multiplying by 3 gives $12x^2$ . So the expression is $12x^2 + 5$ . We are told this equals 17, so we set up the equation $12x^2 + 5 = 17$ . Subtracting 5 gives $12x^2 = 12$ . Dividing by 12 isolates $x^2$ , giving $x^2 = 1$ . Taking the square root yields two solutions because both $1^2$ and $(-1)^2$ equal 1.

Key Takeaways

Substitution into functions: If the input is a compound term like $2x$ , treat it as a single unit inside brackets: $f(2x)$ .
Expanding squared binomials/terms: $(ax)^2 = a^2x^2$ . Here $(2x)^2 = 4x^2$ .
Solving quadratics: When solving $x^2 = k$ , there are always two solutions ( $+\sqrt{k}$ and $-\sqrt{k}$ ) unless $k=0$ .

Common Mistakes

Expanding $(2x)^2$ as $2x^2$ instead of $4x^2$ . The coefficient must also be squared.
Forgetting the negative root when solving $x^2 = 1$ . Students often write only $x=1$ .
Arithmetic errors in rearranging the equation (e.g., adding 5 instead of subtracting).

Things to Be Careful About

Always check if the mark scheme requires specific working marks (M1). Showing the expansion $12x^2 + 5 = 17$ is typically required for method marks.
Ensure the final answer lists both values clearly.

Techniques used

form an equation from a function definitionexpand algebraic termssolve a simple quadratic equation

Q23Medium-EasyNumber

A rectangle has length $32\,\text{cm}$ and width $15\,\text{cm}$ .
Each measurement is given correct to the nearest centimetre.

(a)

Write down the upper bound for the length.

______ $\text{cm}$

(b)

Calculate the upper bound for the difference between the length and the width.

______ $\text{cm}$

DifficultyMedium-Easy

Worked solution

Approach

To find the upper bound of the difference between two quantities ( $L - W$ ), we must maximize the result. This happens when the first quantity (Length) is at its largest possible value (upper bound) and the second quantity (Width) is at its smallest possible value (lower bound).

\text{Upper Bound}(L - W) = \text{Upper Bound}(L) - \text{Lower Bound}(W)

Working

From part (a), the upper bound for the length ( $L$ ) is $32.5\,\text{cm}$ .

The width ( $W$ ) is $15\,\text{cm}$ correct to the nearest cm. Its lower bound is:

\text{Lower Bound}(W) = 15 - 0.5 = 14.5\,\text{cm}

Now calculate the upper bound of the difference:

\text{Upper Bound}(L - W) = 32.5 - 14.5 = 18

Answer

18

Final answer

18

Detailed explanation

Walkthrough

We want to find the maximum possible difference between the length and the width. To make a subtraction result as large as possible, you need the biggest possible number being subtracted FROM and the smallest possible number doing the SUBTRACTING.

Identify the Upper Bound of the Length: We already found this in part (a) to be $32.5\,\text{cm}$ .
Identify the Lower Bound of the Width: The width is $15\,\text{cm}$ . The lowest value that rounds to $15$ is $14.5\,\text{cm}$ .
Subtract the Lower Bound of the Width from the Upper Bound of the Length: $32.5 - 14.5 = 18$ .

Key Takeaways

For sums and differences, bounds behave differently:
- Max Sum = Upper Bound(A) + Upper Bound(B)
- Max Difference = Upper Bound(A) - Lower Bound(B)
- Min Difference = Lower Bound(A) - Upper Bound(B)
Always check if the question asks for the upper or lower bound of the final calculated value.

Common Mistakes

Subtracting the upper bound of the width from the upper bound of the length ( $32.5 - 15.5 = 17$ ). This gives an intermediate value, not the maximum possible difference.
Subtracting the lower bound of the length from the upper bound of the width ( $15.5 - 14.5 = 1$ ). This calculates the minimum difference.
Rounding intermediate bounds too early or incorrectly.

Things to Be Careful About

The phrase "difference between" usually implies a positive magnitude, but in bounds questions asking for the "upper bound of the difference", it specifically refers to maximizing the result of $A - B$ . If $B > A$ , the difference might be negative, but here $L > W$ so the max difference is positive.
Units must be consistent (both in cm).

Techniques used

calculate the upper bound of a difference between two measurements

Q243MMediumAlgebra and Graphs

Simplify.

\frac{2x^2 + 5x - 3}{2x^2 + 6x}

______

DifficultyMedium

Worked solution

Approach

To simplify the algebraic fraction $\frac{2x^2 + 5x - 3}{2x^2 + 6x}$ , we must factorise both the numerator and the denominator completely. Once factorised, any common factors can be cancelled.

Working

Step 1: Factorise the numerator $2x^2 + 5x - 3$
We look for two numbers that multiply to give $a \times c = 2 \times (-3) = -6$ and add to give $b = 5$ . These numbers are $+6$ and $-1$ .

Split the middle term:

2x^2 + 6x - x - 3

Group terms in pairs:

(2x^2 + 6x) - (x + 3)

Factorise each group:

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

Extract the common bracket:

(2x - 1)(x + 3)

Step 2: Factorise the denominator $2x^2 + 6x$
Identify the highest common factor of the two terms $2x^2$ and $6x$ , which is $2x$ .

Extract the common factor:

2x(x + 3)

Step 3: Simplify the fraction
Substitute the factorised forms back into the fraction:

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

Cancel the common factor $(x + 3)$ :

\frac{2x - 1}{2x}

Answer

The simplified expression is:

\frac{2x - 1}{2x}

Final answer

(2x - 1)/(2x)

Detailed explanation

Walkthrough

This question asks us to simplify an algebraic fraction. The key rule here is that we can only cancel factors (terms being multiplied), not individual parts of sums or differences. Therefore, the first step is always to try to write the top and bottom as products of brackets.

For the numerator ( $2x^2 + 5x - 3$ ), this is a quadratic expression where the coefficient of $x^2$ is not 1. A reliable method is 'splitting the middle term'. We need two numbers that multiply to make the product of the first and last coefficients ( $2 \times -3 = -6$ ) and add to make the middle coefficient ( $5$ ). The numbers $6$ and $-1$ fit this perfectly because $6 \times -1 = -6$ and $6 + (-1) = 5$ . We rewrite $5x$ as $6x - 1x$ , group the terms, and factorise to get $(2x - 1)(x + 3)$ .

For the denominator ( $2x^2 + 6x$ ), there is no constant term, so we simply look for the Highest Common Factor (HCF) between $2x^2$ and $6x$ . Both share a coefficient of $2$ and a variable of $x$ , so the HCF is $2x$ . Factoring this out leaves $(x + 3)$ inside the bracket.

Now the fraction looks like $\frac{(2x - 1)(x + 3)}{2x(x + 3)}$ . Since $(x + 3)$ appears in both the top and bottom as a factor, it cancels out, leaving the final answer $\frac{2x - 1}{2x}$ .

Key Takeaways

Factorise before simplifying: Never attempt to cancel terms in an algebraic fraction until you have fully factorised the numerator and denominator into products.
Quadratic factorisation: When the leading coefficient is not 1, use the 'product-sum' method (find factors of $ac$ that sum to $b$ ) or trial-and-error with bracket structures $(ax+b)(cx+d)$ .
Common factors: Always check for a simple numerical/variable HCF in all terms before attempting complex factorisation methods.

Common Mistakes

Cancellation errors: Cancelling $x$ from $2x^2$ and $5x$ individually without factoring (e.g., thinking the answer is just $x+...$ ). This is invalid because cancellation is only allowed for factors.
Incorrect quadratic splitting: Choosing numbers that multiply to $-3$ (the constant) rather than $-6$ (the product of coefficients).
Sign errors: Forgetting the negative sign when splitting the middle term or extracting factors.
Incomplete factorisation: Leaving the denominator as $2x^2 + 6x$ instead of $2x(x+3)$ .

Things to Be Careful About

Final Form: Ensure the final answer has no common factors remaining. The mark scheme accepts $\frac{2x - 1}{2x}$ . Do not expand the denominator back out (i.e., do not write $1 - \frac{1}{2x}$ unless specifically asked or if that is your preferred form, but the standard single-fraction form is safest).
Working Shown: The mark scheme awards B1 marks for seeing the correct factorisations. Writing down these intermediate steps is crucial for partial credit even if the final cancellation is missed.

Techniques used

factorise the numerator into two linear factorsfactorise the denominator by extracting a common factorcancel the common binomial factor

Q253MMedium-EasyTrigonometry

Fig. 8

The diagram shows a cuboid.
$EH = 6\,\text{cm}$ , $HG = 2\,\text{cm}$ and $EC = 7\,\text{cm}$ .

Calculate $CG$ .

$CG$ = ______ $\text{cm}$

DifficultyMedium-Easy

Worked solution

Approach

Identify the right-angled triangles in the cuboid. First, find the diagonal of the base rectangle $EFGH$ using Pythagoras' theorem. Then, use this diagonal and the space diagonal $EC$ to find the vertical edge $CG$ .

Working

In the base rectangle $EFGH$ , triangle $EHG$ is right-angled at $H$ .

\begin{aligned} EG^2 &= EH^2 + HG^2 \\ EG^2 &= 6^2 + 2^2 \\ EG^2 &= 36 + 4 = 40 \end{aligned}

In triangle $EGC$ , the edge $CG$ is perpendicular to the base plane, so it is perpendicular to $EG$ . Thus, triangle $EGC$ is right-angled at $G$ .

\begin{aligned} EC^2 &= EG^2 + CG^2 \\ 7^2 &= 40 + CG^2 \\ 49 &= 40 + CG^2 \\ CG^2 &= 49 - 40 = 9 \\ CG &= 3 \end{aligned}

Alternatively, using the 3D Pythagoras formula directly:

\begin{aligned} EC^2 &= EH^2 + HG^2 + CG^2 \\ 7^2 &= 6^2 + 2^2 + CG^2 \\ 49 &= 36 + 4 + CG^2 \\ CG^2 &= 49 - 40 = 9 \\ CG &= 3 \end{aligned}

Answer

CG = 3

Final answer

3

Detailed explanation

Walkthrough

The problem asks for the length of the vertical edge $CG$ of a cuboid, given two edges of the base ( $EH = 6$ cm, $HG = 2$ cm) and the space diagonal ( $EC = 7$ cm).

To solve this, we use Pythagoras' theorem. First, consider the base rectangle $EFGH$ . The diagonal $EG$ of this base forms a right-angled triangle $EHG$ with sides $EH$ and $HG$ . We calculate $EG^2 = EH^2 + HG^2 = 6^2 + 2^2 = 40$ .

Next, consider the triangle $EGC$ . Since the edge $CG$ is vertical and perpendicular to the base, it is perpendicular to every line in the base passing through $G$ , including $EG$ . Therefore, triangle $EGC$ is right-angled at $G$ . We apply Pythagoras' theorem again: $EC^2 = EG^2 + CG^2$ , which gives $49 = 40 + CG^2$ , so $CG^2 = 9$ and $CG = 3$ .

Alternatively, one can use the direct 3D Pythagoras formula $EC^2 = EH^2 + HG^2 + CG^2$ , which yields the same result in a single equation.

Key Takeaways

In a cuboid, the space diagonal $d$ relates to the three edge lengths $l, w, h$ by $d^2 = l^2 + w^2 + h^2$ .
Any vertical edge of a cuboid is perpendicular to the diagonal of the base at its foot, creating a right-angled triangle with the space diagonal.
Pythagoras' theorem can be applied in two stages (base diagonal then space diagonal) or in one step using the 3D formula.

Common Mistakes

Forgetting that the space diagonal is not the hypotenuse of a triangle formed by two edges of the base; it requires all three dimensions.
Adding the squares instead of subtracting: $CG^2 = 7^2 + 6^2 + 2^2$ instead of $7^2 - 6^2 - 2^2$ .
Taking the square root of the wrong value or failing to simplify $\sqrt{9}$ to $3$ .
Confusing the face diagonal with the space diagonal (e.g., using $6^2 + 2^2$ as the square of $EC$ ).

Things to Be Careful About

Ensure the answer is given to the correct form; here, an exact integer is expected.
Verify which diagonal is given: $EC$ passes through the interior (space diagonal), not along a face.
Remember that in a cuboid, all vertical edges are perpendicular to the base, which justifies the right angle at $G$ in triangle $EGC$ .
The diagram is marked 'NOT TO SCALE', so do not estimate lengths visually; rely entirely on the given numerical values and geometric properties.

Techniques used

apply Pythagoras' theorem in a 3D cuboidcalculate the square of the space diagonalsolve for the unknown edge length

Q264MMediumAlgebra and Graphs

Solve.

\frac{x}{x - 1} - \frac{3}{2x - 1} = 1

$x$ = ______

DifficultyMedium

Worked solution

Approach

To solve the equation involving algebraic fractions, we first eliminate the denominators by multiplying every term by the least common multiple (LCM) of $(x - 1)$ and $(2x - 1)$ . This transforms the fractional equation into a polynomial equation (specifically, a quadratic). We then expand, simplify, and solve for $x$ .

Working

The given equation is:

\frac{x}{x - 1} - \frac{3}{2x - 1} = 1

Multiply both sides by $(x - 1)(2x - 1)$ to clear the denominators:

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

Expand the brackets on both sides.
LHS: $2x^2 - x - 3x + 3$
RHS: $(x - 1)(2x - 1) = 2x^2 - x - 2x + 1$

So the equation becomes:

2x^2 - 4x + 3 = 2x^2 - 3x + 1

Subtract $2x^2$ from both sides (the quadratic terms cancel out):

-4x + 3 = -3x + 1

Add $4x$ to both sides and subtract $1$ from both sides to isolate $x$ :

3 - 1 = 4x - 3x

2 = x

Check the solution in the original denominators: if $x = 2$ , then $x - 1 = 1 \neq 0$ and $2x - 1 = 3 \neq 0$ . The solution is valid.

Answer

x = 2

Final answer

2

Detailed explanation

Walkthrough

The problem asks us to find the value of $x$ that satisfies an equation containing algebraic fractions. The most efficient method is to remove the fractions entirely. We do this by multiplying every term in the equation by the product of the denominators, which is $(x - 1)(2x - 1)$ .

On the left-hand side, the first term $\frac{x}{x - 1}$ multiplied by $(x - 1)(2x - 1)$ leaves $x(2x - 1)$ because the $(x - 1)$ cancels out. Similarly, the second term $\frac{3}{2x - 1}$ becomes $3(x - 1)$ . On the right-hand side, the number $1$ multiplied by $(x - 1)(2x - 1)$ simply becomes the expanded product $(x - 1)(2x - 1)$ .

Next, we expand all the brackets carefully. It is crucial not to miss any signs when expanding $-3(x - 1)$ or the product on the right. After expansion, we notice that the $2x^2$ term appears on both sides. Subtracting $2x^2$ simplifies the quadratic equation down to a linear one, making it straightforward to solve. Finally, we verify that our answer does not make any original denominator zero (which would be undefined).

Key Takeaways

When solving rational equations, multiply by the LCD (Least Common Denominator) to clear fractions early.
Be extremely careful with signs when expanding brackets, especially when there is a negative coefficient outside (e.g., $-3(x-1)$ becomes $-3x + 3$ ).
Always check if the highest degree terms cancel out; a "quadratic" might just be linear in disguise.

Common Mistakes

Sign Errors: Expanding $-3(x - 1)$ as $-3x - 3$ instead of $-3x + 3$ . Or expanding $(x - 1)(2x - 1)$ incorrectly as $2x^2 - 1$ .
Incomplete Multiplication: Forgetting to multiply the RHS '1' by the full denominator product.
Undefined Values: Not checking if the solution makes the denominator zero (though here $x=2$ is safe, $x=1$ or $x=1/2$ would be invalid).
Algebraic Simplification: Failing to cancel the $2x^2$ terms, leading to unnecessary complexity.

Things to Be Careful About

Ensure you show working for the expansion (nfww - no marks for wrong answer without working). The mark scheme explicitly looks for the expanded forms like $2x^2 - x - 3x + 3$ .
Remember that cancelling terms is only valid if the term is non-zero. Here we assume $x \neq 1$ and $x \neq 1/2$ .

Techniques used

eliminate fractions by multiplying through by the common denominatorexpand algebraic bracketscollect like terms to form a quadratic equationfactorise the quadratic equationsolve for x