Mathematics (Syllabus D) 4024 / 21 — May/June 2024

(ii)

The cost of one ticket at $67.60 is 4% more than the cost of one ticket last year.

Calculate the cost of one ticket last year.

$ ______

(b)

The number of seats in the basketball stadium is 20 545.
The number of seats sold for the first match of the season is 19 340.

Calculate the percentage of the seats in the stadium that are sold.

______ %

(c)

A team plays 41 matches.
For the 41 matches, the mean number of seats sold per match is 16 440.
The total number of seats sold for the first 21 matches is 329 000.

Calculate the mean number of seats sold per match for the last 20 matches.

______

DifficultyMedium-Easy

Worked solution

Approach

First find the total number of seats sold over all 41 matches by multiplying the mean by 41. Then subtract the total for the first 21 matches and divide by the remaining 20 matches.

Working

\text{Total for 41 matches} = 16440 \times 41

= 674040

\text{Total for last 20 matches} = 674040 - 329000 = 345040

\text{Mean for last 20 matches} = \frac{345040}{20}

= 17252

Answer

$17252$ seats per match

Final answer

17252

Detailed explanation

Walkthrough

Since the mean number of seats sold per match over 41 matches is $16440$ , the total number sold over those 41 matches is

16440 \times 41 = 674040

The total for the first 21 matches is already given as $329000$ . Therefore the total sold in the last 20 matches is

674040 - 329000 = 345040

Finally, divide this remaining total by the number of remaining matches, 20:

\frac{345040}{20} = 17252

So the mean seats sold per match over the last 20 matches is $17252$ .

Key Takeaways

The mean can be used in reverse to find the total: mean = total ÷ number of values.
When some part of the data is known, subtract the known total from the full total.

Common Mistakes

Finding the mean of the two given numbers $16440$ and $329000$ instead of working with totals.
Dividing the total for the last 20 matches by 41 or by 21 instead of 20.
Forgetting to subtract before dividing.

Things to Be Careful About

The total of the last 20 matches must be divided by 20, not by the number of matches remaining as given.
The final answer is an integer here, so no rounding is necessary.

Techniques used

find the total number of seats sold over all 41 matchessubtract the total for the first 21 matchesdivide by 20 to find the mean

(d)

The table shows the salaries of three basketball players.

Basketball player	Salary ($)
Stephen	$8.27 \times 10^6$
Joe	$4.29 \times 10^6$
Tristan	$3.64 \times 10^7$

3M

(i)

Find the difference between the salaries of Tristan and Stephen.

$ ______

DifficultyEasy

Worked solution

Approach

Write both salaries using the same power of 10, subtract, then express the answer in standard form.

Working

Tristan: $3.64 \times 10^7 = 36.4 \times 10^6$

Stephen: $8.27 \times 10^6$

36.4 \times 10^6 - 8.27 \times 10^6 = 28.13 \times 10^6

= 2.813 \times 10^7

Answer

$28130000$ or $2.813 \times 10^7$

Final answer

28130000

Detailed explanation

Walkthrough

The salary of Tristan is $3.64 \times 10^7$ , which is $36.4 \times 10^6$ . Stephen's salary is $8.27 \times 10^6$ . Now both values have the same power of 10, so their coefficients can be subtracted directly:

36.4 \times 10^6 - 8.27 \times 10^6 = 28.13 \times 10^6

Since $28.13 \times 10^6$ is not standard form, rewrite it as $2.813 \times 10^7$ , or as the ordinary number $28130000$ .

Key Takeaways

Standard form subtraction requires equal powers of 10.
A number in standard form must have its first part between 1 and 10 before multiplying by a power of 10.

Common Mistakes

Subtracting $8.27$ from $3.64$ directly without aligning the powers.
Writing $28.13\times 10^6$ as the final answer because it is not standard form.
Misreading the sign: Tristan's salary is greater than Stephen's, so the difference is positive.

Things to Be Careful About

Check to see whether the ordinary number form $28130000$ or the standard form $2.813\times 10^7$ is expected; the mark scheme accepts both.
Ensure the subtraction is done from Tristan's salary to Stephen's salary, not the reverse.

Techniques used

rewrite the numbers with the same power of 10subtract the coefficientsrewrite the result in standard form

(ii)

The total Joe earns is his salary plus a bonus of $x.
The total he earns is 102.5% of his salary.

Calculate the value of $x$ .

$x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Joe's total $102.5\%$ of his salary. Calculate the total, then subtract the actual salary to find the bonus $x$ .

Working

Total earnings:

\frac{102.5}{100} \times 4.29 \times 10^6

= 1.025 \times 4.29 \times 10^6

= 4.39725 \times 10^6

So the bonus is:

x = 4.39725 \times 10^6 - 4.29 \times 10^6

x = 0.10725 \times 10^6 = 107250

Answer

$107250$

Final answer

107250

Detailed explanation

Walkthrough

The total Joe earns is $102.5\%$ of his salary. Write this as a decimal multiplier:

$1.025$ .

Then the total is

1.025 \times 4.29 \times 10^6 = 4.39725 \times 10^6

The bonus $x$ is the total minus the salary:

4.39725 \times 10^6 - 4.29 \times 10^6 = 0.10725 \times 10^6

0.10725 \times 10^6 = 107250

So the value of $x$ is 107250.

Key Takeaways

A percentage above 100% means that the quantity is increased and the total is the salary plus the bonus.
Subtracting the original value from the increased value yields the increase itself.

Common Mistakes

Subtracting $100\%$ from $102.5\%$ incorrectly and applying $2.5\%$ directly without converting to decimal.
Forgetting to multiply by $10^6$ when working with standard form.
Answering with $4.39725\times10^6$ (the total) instead of the bonus.

Things to Be Careful About

The mark scheme accepts a method that uses the equation

\frac{4.29\times10^6+x}{4.29\times10^6}\times 100 = 102.5

Alternatively, the simplest route is to recognise that $x$ is $2.5\%$ of $4.29\times10^6$ and compute

x=0.025\times 4.29\times10^6=107250

The answer can also be given in standard form as $1.0725\times10^5$ .

Techniques used

express 102.5% as a decimal multiplierfind the total amount earnedsubtract the salary to find the bonus

Q2MediumStatistics

(a)

The temperature at midday was recorded at ten different heights on a mountain.
The results are shown in the table.

Height (m)	300	825	600	425	900	100	1250	1450	1125	1350
Temperature (°C)	3.0	$-0.8$	0.0	1.2	$-1.9$	3.5	$-4.6$	$-6.4$	$-4.0$	$-3.8$

5M

(i)

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

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

Five points have already been plotted. The remaining five points from the table are:

(100,\; 3.5),\quad (1250,\; -4.6),\quad (1450,\; -6.4),\quad (1125,\; -4.0),\quad (1350,\; -3.8)

Each point is plotted as a cross at the intersection of its height (x-axis) and temperature (y-axis) values.

Working

Answer

All five remaining points plotted correctly on the scatter diagram.

Final answer

Points plotted at (100, 3.5), (1125, -4.0), (1250, -4.6), (1350, -3.8), (1450, -6.4)

Detailed explanation

Walkthrough

The table gives ten (height, temperature) pairs. The first five — (300, 3.0), (825, -0.8), (600, 0.0), (425, 1.2), (900, -1.9) — are already shown as crosses on the grid. We must plot the remaining five.

Reading from the table:

Height 100 m, Temperature 3.5 °C → point (100, 3.5)
Height 1250 m, Temperature -4.6 °C → point (1250, -4.6)
Height 1450 m, Temperature -6.4 °C → point (1450, -6.4)
Height 1125 m, Temperature -4.0 °C → point (1125, -4.0)
Height 1350 m, Temperature -3.8 °C → point (1350, -3.8)

Each point is marked with a cross at the correct grid intersection. The x-axis has major lines every 250 m and minor subdivisions every 50 m; the y-axis has major lines every 1 °C and minor subdivisions every 0.2 °C. Points like (1125, -4.0) fall exactly on the midpoint between 1000 and 1250, and (1350, -3.8) is two minor subdivisions above -4.0.

Key Takeaways

Scatter diagrams plot one variable on each axis; the order matters (height on x, temperature on y here).
Every data pair must be plotted; missing even a few points loses marks.

Common Mistakes

Swapping the coordinates (plotting temperature on the x-axis and height on the y-axis).
Misreading negative temperatures as positive.
Plotting to the wrong minor subdivision on the axes.

Things to Be Careful About

The axes are labelled clearly: Height (m) on the horizontal axis, Temperature (°C) on the vertical axis.
Negative temperatures are below the x-axis; make sure points like -4.6 are plotted below 0.
Accuracy of plotting is assessed by whether 3 or 4 of the remaining points are correct (B1).

Techniques used

read coordinates from the data tableplot points on a scatter diagram grid

(ii)

Describe the type of correlation shown in the scatter diagram.

______

(iii)

Draw a line of best fit on the scatter diagram.

(iv)

Another reading is taken at a height of 1000 m.

Use your line of best fit to estimate the temperature at this height.

______ °C

DifficultyMedium-Easy

Worked solution

Approach

Locate 1000 m on the horizontal axis, move vertically up to the line of best fit, then move horizontally to read the temperature on the vertical axis.

Working

At height $x = 1000$ m, the line of best fit is at approximately $y = -2.8$ °C. (Any reading between -2.5 and -3.2 °C is acceptable, depending on the exact line drawn.)

Using the approximate line from (0, 4.5) to (1500, -6.5):

\text{Slope} = \frac{-6.5 - 4.5}{1500 - 0} = \frac{-11}{1500}

\text{Temperature at } 1000 \text{ m} \approx 4.5 + \left(\frac{-11}{1500}\right) \times 1000 = 4.5 - 7.33 \approx -2.8\;°\text{C}

Answer

-2.8\;°\text{C}

(Acceptable range: approximately $-2.5$ to $-3.2$ °C, depending on the candidate's line of best fit.)

Final answer

-2.8

Detailed explanation

Walkthrough

To estimate the temperature at 1000 m:

Find 1000 on the horizontal (height) axis.
Move vertically up from 1000 until you meet the line of best fit.
From that intersection, move horizontally to the left to read the temperature on the vertical axis.

On the line drawn in diagram-2, this gives approximately -2.8 °C. Since different candidates will draw slightly different lines of best fit, the mark scheme accepts any reasonable reading from the student's own line (typically between -2.5 and -3.2 °C).

Key Takeaways

Estimation from a line of best fit is an interpolation task.
The answer depends on the candidate's own line, so there is a range of acceptable answers.
Always read from the line, not from the data points (unless the line passes exactly through a point).

Common Mistakes

Reading the value from a data point near 1000 m instead of from the line of best fit. (There is no data point at exactly 1000 m.)
Reading the wrong axis.
Not using a ruler to read off the value accurately.

Things to Be Careful About

The mark scheme explicitly states follow-through from the ruled line of best fit. A candidate who drew a poor line will get a correspondingly different (but still accepted) answer.
The answer must be in °C and given to a reasonable number of decimal places (typically 1 or 2).

Techniques used

read a value from a line of best fit by interpolation

(b)

The table summarises the times taken by 80 adults to climb the mountain.

Time taken ( $h$ hours)	$5.5 < h \leq 6.5$	$6.5 < h \leq 7.5$	$7.5 < h \leq 8$	$8 < h \leq 8.5$	$8.5 < h \leq 10.5$
Frequency	8	15	20	23	14

4M

(i)

Calculate an estimate of the mean time.

______ hours

DifficultyMedium

Worked solution

Approach

For grouped data, estimate the mean by using the midpoint of each class interval as the representative value, multiplying by the frequency, summing these products, and dividing by the total frequency.

Working

The class intervals and their midpoints are:

Class	Midpoint $x$	Frequency $f$	$fx$
$5.5 < h \leq 6.5$	$\frac{5.5+6.5}{2} = 6.0$	8	$6.0 \times 8 = 48.0$
$6.5 < h \leq 7.5$	$\frac{6.5+7.5}{2} = 7.0$	15	$7.0 \times 15 = 105.0$
$7.5 < h \leq 8$	$\frac{7.5+8}{2} = 7.75$	20	$7.75 \times 20 = 155.0$
$8 < h \leq 8.5$	$\frac{8+8.5}{2} = 8.25$	23	$8.25 \times 23 = 189.75$
$8.5 < h \leq 10.5$	$\frac{8.5+10.5}{2} = 9.5$	14	$9.5 \times 14 = 133.0$

Total frequency:

\sum f = 8 + 15 + 20 + 23 + 14 = 80

Sum of $fx$ :

\sum fx = 48.0 + 105.0 + 155.0 + 189.75 + 133.0 = 630.75

Estimated mean:

\bar{x} = \frac{\sum fx}{\sum f} = \frac{630.75}{80} = 7.884375

Rounding to 3 significant figures:

\bar{x} \approx 7.88\;\text{hours}

Answer

7.88\;\text{hours}

Final answer

7.88

Detailed explanation

Walkthrough

The data is given in grouped form with unequal class widths. To estimate the mean:

Find the midpoint of each class. The midpoint is the average of the lower and upper class boundaries.
- For $5.5 < h \leq 6.5$ : midpoint = $(5.5 + 6.5)/2 = 6.0$
- For $6.5 < h \leq 7.5$ : midpoint = $(6.5 + 7.5)/2 = 7.0$
- For $7.5 < h \leq 8$ : midpoint = $(7.5 + 8)/2 = 7.75$ (note: this class has width 0.5, not 1.0)
- For $8 < h \leq 8.5$ : midpoint = $(8 + 8.5)/2 = 8.25$
- For $8.5 < h \leq 10.5$ : midpoint = $(8.5 + 10.5)/2 = 9.5$ (this class has width 2.0)
Multiply each midpoint by its frequency to get $fx$ .
Sum all $fx$ values and divide by the total frequency (80).

The key trap here is that the class widths are not all equal (0.5, 1.0, 1.0, 2.0), so you must be careful with the midpoints. The midpoint of $7.5 < h \leq 8$ is 7.75, not 7.5 or 8.0.

Key Takeaways

For grouped data, always use class midpoints to estimate the mean.
Unequal class widths do not affect the mean calculation — only the midpoint and frequency matter.
The estimated mean is not the exact mean; it is an approximation based on the assumption that all values in a class equal the midpoint.

Common Mistakes

Using the lower or upper class boundary instead of the midpoint.
Forgetting that $7.5 < h \leq 8$ has midpoint 7.75 (not 7.5 or 8).
Adding the frequencies incorrectly (total should be 80).
Arithmetic errors in multiplying midpoints by frequencies.
Rounding too early — keep full precision until the final division.

Things to Be Careful About

The mark scheme accepts 7.88 or 7.884... — give at least 3 significant figures.
The class $7.5 < h \leq 8$ has width 0.5, and $8.5 < h \leq 10.5$ has width 2.0. These unequal widths matter for the histogram in part (ii) but not for the mean calculation.

Techniques used

calculate class midpointsmultiply midpoints by frequenciessum and divide by total frequency

(ii)

A histogram is drawn to show this information.
The height of the bar representing $5.5 < h \leq 6.5$ is 8 mm.

Calculate the height of the bar representing $8 < h \leq 8.5$ .

______ mm

DifficultyMedium-Easy

Worked solution

Approach

In a histogram with unequal class widths, the bar height is proportional to the frequency density (frequency divided by class width), not the frequency itself.

Working

Step 1: Find the frequency density of the first bar.

For $5.5 < h \leq 6.5$ :

Frequency = 8
Class width = $6.5 - 5.5 = 1.0$ hour
Frequency density = $\frac{8}{1.0} = 8$

The bar height is 8 mm, so the scale is:

1\;\text{mm} = 1\;\text{unit of frequency density}

Step 2: Calculate the frequency density for $8 < h \leq 8.5$ .

Frequency = 23
Class width = $8.5 - 8 = 0.5$ hour
Frequency density = $\frac{23}{0.5} = 46$

Step 3: Convert to bar height.

\text{Bar height} = 46 \times 1\;\text{mm} = 46\;\text{mm}

Answer

46\;\text{mm}

Final answer

46

Detailed explanation

Walkthrough

In a histogram, when class widths are unequal, the bar height represents frequency density, not frequency. Frequency density is defined as:

\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}

From the given bar:

Class $5.5 < h \leq 6.5$ : frequency = 8, width = 1.0
Frequency density = $8 / 1.0 = 8$
Bar height = 8 mm
Therefore: 1 mm on the y-axis = 1 unit of frequency density

For the required bar:

Class $8 < h \leq 8.5$ : frequency = 23, width = 0.5
Frequency density = $23 / 0.5 = 46$
Bar height = $46 \times 1 = 46$ mm

Key Takeaways

In a histogram with unequal class widths, bar height ∝ frequency density.
Frequency density = frequency ÷ class width.
Always use the given bar to establish the scale before calculating other bar heights.

Common Mistakes

Using frequency directly instead of frequency density (would give 23 mm, which is wrong).
Using the wrong class width (e.g., using 1.0 instead of 0.5 for the class $8 < h \leq 8.5$ ).
Forgetting to establish the scale from the given bar first.

Things to Be Careful About

The class $8 < h \leq 8.5$ has width 0.5, which is half the width of most other classes. This makes its frequency density much larger than its frequency would suggest.
The answer is 46 mm — a whole number, which is a good sanity check.

Techniques used

calculate frequency density for the known baruse the scale to find the height of another bar

Q3Medium-EasyGeometry

(a)

Fig. 2

Triangle $PQR$ is isosceles with $PQ = QR$ .
The exterior angle of the triangle at $R$ is $142°$ .

Calculate angle $PQR$ .

$P\hat{Q}R$ = ______

DifficultyMedium-Easy

Worked solution

Approach

The exterior angle at $R$ and the interior angle $Q\hat{R}P$ lie on a straight line, so they add to $180°$ . Since $PQ = QR$ , the base angles $Q\hat{P}R$ and $Q\hat{R}P$ are equal. Finally, use the angle sum of a triangle to find $P\hat{Q}R$ .

Working

\angle QRP = 180° - 142° = 38°

Since $PQ = QR$ , triangle $PQR$ is isosceles, so:

\angle QPR = \angle QRP = 38°

The angles in triangle $PQR$ add to $180°$ :

P\hat{Q}R = 180° - (38° + 38°) = 180° - 76° = 104°

Answer

P\hat{Q}R = 104°

Final answer

104°

Detailed explanation

Walkthrough

First, we find the interior angle at $R$ . The exterior angle ( $142°$ ) and the interior angle $Q\hat{R}P$ are on a straight line, meaning they are supplementary and add up to $180°$ . Subtracting $142°$ from $180°$ gives $38°$ .

Next, we use the property of isosceles triangles. Because sides $PQ$ and $QR$ are equal, the angles opposite them — $Q\hat{P}R$ and $Q\hat{R}P$ — must be equal. Therefore, $Q\hat{P}R$ is also $38°$ .

Finally, we apply the triangle angle sum theorem. The three angles in triangle $PQR$ must total $180°$ . Subtracting the two base angles ( $38° + 38° = 76°$ ) from $180°$ leaves $104°$ for the apex angle $P\hat{Q}R$ .

Key Takeaways

Angles on a straight line add to $180°$ .
In an isosceles triangle, the base angles (opposite the equal sides) are equal.
The interior angles of a triangle always sum to $180°$ .

Common Mistakes

Assuming the exterior angle equals the sum of the two opposite interior angles without first finding the interior angle at $R$ . (Note: The exterior angle theorem states the exterior angle equals the sum of the two opposite interior angles, so $142° = P\hat{Q}R + Q\hat{P}R$ . Since $Q\hat{P}R = 38°$ , then $P\hat{Q}R = 142° - 38° = 104°$ . This is a valid alternative route, but the mark scheme specifically awards method marks for the $180 - 2 \times 38$ approach).
Forgetting that the base angles are the ones opposite the equal sides.
Adding the angles incorrectly or subtracting from the wrong total.

Things to Be Careful About

The exterior angle is at $R$ , not at $Q$ . Ensure you subtract from $180°$ to get the interior angle $Q\hat{R}P$ , not assume $Q\hat{R}P = 142°$ .
Always include the degree symbol in your final answer if the question uses it, though numerical answers are often accepted without it.

Techniques used

use a straight line to find the interior angleapply the isosceles triangle base angle propertyuse the triangle angle sum to find the apex angle

(b)

Fig. 3

The diagonals of trapezium $ABCD$ meet at $E$ .

Show that triangle $ABE$ is similar to triangle $CDE$ .
Give a reason for each statement you make.

DifficultyMedium-Easy

Worked solution

Approach

To prove triangle $ABE$ is similar to triangle $CDE$ , we need to show that all three corresponding angles are equal. We can find these using the properties of vertically opposite angles and alternate angles formed by the parallel sides $AB$ and $DC$ cut by the diagonals acting as transversals.

Working

In triangles $ABE$ and $CDE$ :

\angle AEB = \angle CED \quad \text{[vertically opposite angles]}

Since $AB \parallel DC$ and $AC$ is a transversal:

\angle EAB = \angle ECD \quad \text{[alternate angles]}

Since $AB \parallel DC$ and $BD$ is a transversal:

\angle EBA = \angle EDC \quad \text{[alternate angles]}

Since all three pairs of corresponding angles are equal:

\triangle ABE \sim \triangle CDE \quad \text{[AAA similarity]}

Answer

Triangle $ABE$ is similar to triangle $CDE$ by AAA similarity.

Final answer

Triangle ABE is similar to triangle CDE (AAA)

Detailed explanation

Walkthrough

We are asked to prove that triangle $ABE$ is similar to triangle $CDE$ . Similarity in triangles can be proven if all three corresponding angles are equal (AAA criterion).

First, look at the intersection point $E$ . The diagonals $AC$ and $BD$ cross at $E$ , forming two pairs of vertically opposite angles. Therefore, $\angle AEB = \angle CED$ .

Next, use the fact that $AB$ is parallel to $DC$ . The diagonal $AC$ acts as a transversal cutting these parallel lines. The angles $\angle EAB$ and $\angle ECD$ are alternate angles (forming a Z-shape), so they are equal.

Similarly, the diagonal $BD$ acts as a second transversal. The angles $\angle EBA$ and $\angle EDC$ are alternate angles, so they are equal.

With three pairs of equal angles established, the triangles are similar by the AAA (Angle-Angle-Angle) criterion.

Key Takeaways

Vertically opposite angles are equal when two lines intersect.
Alternate angles are equal when a transversal cuts two parallel lines.
Triangles are similar if all three corresponding angles are equal (AAA similarity).

Common Mistakes

Stating "corresponding angles" instead of "alternate angles" for the angles formed by the parallel lines and transversals. Corresponding angles are in the same relative position at each intersection, which is not the case here.
Forgetting to provide a reason for each angle equality. The mark scheme explicitly requires a reason for each statement.
Concluding similarity based on only one or two pairs of angles without stating the AAA criterion.

Things to Be Careful About

Always name the angles correctly using three letters where the vertex is in the middle (e.g., $\angle EAB$ , not $\angle A$ ).
Ensure the reasons match the angle relationships: "vertically opposite angles" for the angles at $E$ , and "alternate angles" for the angles between the parallel lines $AB$ and $DC$ .
The mark scheme awards marks for two correct pairs with reasons, or one correct pair with a reason, but stating all three is the most robust approach.

Techniques used

identify vertically opposite angles at the intersection of diagonalsidentify alternate angles formed by parallel lines and a transversalstate the AAA similarity criterion

Q4Medium-EasyNumber

(a)

Two of the factors of 50 are square numbers.
One of these square numbers is 1.

Find the other square number that is a factor of 50.

______

(b)

\begin{aligned} A &= 2^{x-1} \times 3^{2y} \times 7 \\ B &= 2^{x+3} \times 3^y \times 5 \end{aligned}

The numbers $A$ and $B$ are written as the product of their prime factors, where $x$ and $y$ are positive integers.

4M

(i)

Find the highest common factor (HCF) of $A$ and $B$ in terms of $x$ and $y$ .

______

DifficultyMedium-Easy

Worked solution

Approach

For two numbers written in prime factor form, the HCF is the product of the primes common to both numbers, each raised to the smaller index.

Working

A = 2^{x-1} \times 3^{2y} \times 7, \qquad B = 2^{x+3} \times 3^y \times 5

The primes common to $A$ and $B$ are $2$ and $3$ .

For the prime $2$ : compare $x-1$ and $x+3$ . Since $x+3 > x-1$ , the smaller index is $x-1$ , giving $2^{x-1}$ .

For the prime $3$ : compare $2y$ and $y$ . Since $y$ is positive, $2y > y$ , so the smaller index is $y$ , giving $3^y$ .

The primes $5$ and $7$ are not in both numbers, so they are not part of the HCF.

Therefore,

\text{HCF}(A,B) = 2^{x-1} \times 3^y

Answer

2^{x-1} \times 3^y

Final answer

2^(x-1) * 3^y

Detailed explanation

Walkthrough

The HCF must divide both $A$ and $B$ . A prime can appear in the HCF only if it is a factor of both numbers. Here $5$ appears only in $B$ and $7$ appears only in $A$ , so neither is included. The common primes are $2$ and $3$ .

For the prime $2$ , the exponents are $x-1$ and $x+3$ . Since $x$ is positive, $x+3 > x-1$ , so take the smaller exponent $x-1$ .

For the prime $3$ , the exponents are $2y$ and $y$ . Since $2y > y$ , take the smaller exponent $y$ .

Multiplying the selected prime powers gives the HCF.

Key Takeaways

HCF uses the primes common to every number and takes the smallest index for each common prime.
Comparing variable indices is the same idea as comparing numerical indices.

Common Mistakes

Including $5$ or $7$ , because a prime must appear in both factorisations to be in the HCF.
Choosing the larger index $x+3$ or $2y$ ; these indices are used for the LCM, not the HCF.

Things to Be Careful About

Since $x$ and $y$ are positive integers, the inequalities $x+3 > x-1$ and $2y > y$ are correct.
The answer should remain as a product of prime powers in terms of $x$ and $y$ .

Techniques used

identify the prime factors common to both numberscompare the indices of the shared primesselect the smaller index for the HCF

(ii)

Find the lowest common multiple (LCM) of $A$ and $B$ in terms of $x$ and $y$ .

______

DifficultyMedium-Easy

Worked solution

Approach

The LCM of two numbers in prime factor form includes every prime that appears in either number, raised to the larger index.

Working

A = 2^{x-1} \times 3^{2y} \times 7, \qquad B = 2^{x+3} \times 3^y \times 5

For each prime, choose the larger index:

Prime $2$ : $x-1$ and $x+3$ ; the larger is $x+3$ , so include $2^{x+3}$ .
Prime $3$ : $2y$ and $y$ ; the larger is $2y$ , so include $3^{2y}$ .
Prime $5$ : appears only in $B$ , so include $5$ .
Prime $7$ : appears only in $A$ , so include $7$ .

Therefore,

\text{LCM}(A,B) = 2^{x+3} \times 3^{2y} \times 5 \times 7

Answer

2^{x+3} \times 3^{2y} \times 5 \times 7

Final answer

2^(x+3) * 3^(2y) * 5 * 7

Detailed explanation

Walkthrough

The LCM must be divisible by both $A$ and $B$ . It therefore needs every prime factor that appears in either number, and each prime must appear with the largest index required by either factorisation.

For prime $2$ , the indices are $x-1$ and $x+3$ , so take $x+3$ because it is larger.

For prime $3$ , the indices are $2y$ and $y$ , so take $2y$ .

The prime $5$ occurs only in $B$ , so it is included once. The prime $7$ occurs only in $A$ , so it is included once.

Multiplying these prime powers gives the LCM.

Key Takeaways

The LCM uses every distinct prime factor from the numbers, each with the largest index.
The HCF and LCM differ: HCF uses only common primes and smaller indices; LCM uses all primes and larger indices.

Common Mistakes

Using the smaller indices for the LCM; that gives the HCF.
Omitting $5$ or $7$ because they do not appear in both numbers. The LCM includes primes that appear in any of the numbers.
Writing $3^y$ instead of $3^{2y}$ ; the larger index for $3$ is $2y$ .

Things to Be Careful About

The final expression must contain all four prime factors: $2$ , $3$ , $5$ and $7$ .
The answer is expected in terms of $x$ and $y$ , so it is left as a product of prime powers, not evaluated as a number.

Techniques used

list every distinct prime factor in either numberchoose the larger index for each primemultiply the selected prime powers

Q5MediumAlgebra and GraphsNumberMensuration

(a)

Two companies move boxes.
Company $A$ charges $0.50 for each box plus a fixed fee of $125.
Company $B$ charges only a fixed fee of $350.

Find the number of boxes moved when Company $A$ charges the same as Company $B$ .

______

DifficultyMedium-Easy

Worked solution

Approach

We need to find the number of boxes where the total cost for Company $A$ equals the total cost for Company $B$ . We will define a variable for the number of boxes, write an expression for each company's charge, set them equal to form an equation, and solve for the variable.

Working

Let $x$ be the number of boxes moved.

The cost for Company $A$ is $0.50 per box plus a fixed fee of $125:

\text{Cost}_A = 0.50x + 125

The cost for Company $B$ is a fixed fee of $350:

\text{Cost}_B = 350

Set the costs equal to each other:

0.50x + 125 = 350

Subtract 125 from both sides:

0.50x = 225

Divide by 0.50 (which is equivalent to multiplying by 2):

x = 450

Answer

450 \text{ boxes}

Final answer

450

Detailed explanation

Walkthrough

First, we translate the word problem into mathematical expressions.

Company A has a variable cost ($0.50 per box) and a fixed cost ($125). If $x$ is the number of boxes, the total is $0.50x + 125$ .
Company B has only a fixed cost ($350).

To find when they charge the same amount, we equate the two expressions: $0.50x + 125 = 350$ . This is a standard linear equation. To isolate $x$ , we first remove the constant term (+125) by subtracting 125 from the right side ( $350 - 125 = 225$ ). Then, we divide by the coefficient of $x$ (0.50). Dividing by 0.50 is the same as multiplying by 2, giving $x = 450$ .

Key Takeaways

Variable vs Fixed Costs: Understand that 'per item' costs depend on quantity (variable), while flat fees do not (fixed).
Equating Quantities: When asked for the point where two different schemes result in the same value, setting their expressions equal is the correct method.

Common Mistakes

Forgetting to include the fixed fee in Company A's expression (writing just $0.50x$ ).
Adding instead of subtracting when moving the 125 across the equals sign.
Arithmetic errors with decimals (e.g., calculating $350 - 125$ incorrectly).

Things to Be Careful About

Ensure you use the correct currency symbol if required, though the numerical answer is the primary requirement here.
Check your answer by plugging $x=450$ back into both original expressions: $0.50(450) + 125 = 225 + 125 = 350$ . It matches.

Techniques used

form a linear equation from a worded cost contextsolve the equation by isolating the variable

(b)

The maximum mass a van can carry is exactly $770\,\text{kg}$ .
The van carries boxes each of mass $4\,\text{kg}$ , correct to the nearest kilogram.

Find the upper bound for the number of boxes this van can carry.

______

DifficultyMedium

Worked solution

Approach

To find the maximum number of boxes the van can carry, we must consider the scenario that allows the most boxes to fit. Since the total mass limit is fixed, the boxes must be as light as possible to allow more of them to be carried. Therefore, we use the lower bound of the mass of a single box.

Working

The mass of a box is given as $4\,\text{kg}$ , correct to the nearest kilogram.
This means the actual mass $m$ satisfies:

3.5 \leq m < 4.5

The lower bound of the mass of one box is $3.5\,\text{kg}$ .

The maximum mass the van can carry is exactly $770\,\text{kg}$ . Let $n$ be the number of boxes. The total mass of $n$ boxes must be less than or equal to $770\,\text{kg}$ :

n \times (\text{mass of one box}) \leq 770

To find the largest possible integer $n$ , we divide the total capacity by the smallest possible mass per box (the lower bound):

n \leq \frac{770}{3.5}

Calculate the division:

\frac{770}{3.5} = \frac{7700}{35} = 220

Since the calculation results in an exact integer, and any box heavier than the lower bound would reduce the count, the maximum number of boxes is 220.

Answer

220

Final answer

220

Detailed explanation

Walkthrough

This is a bounds problem involving discrete items (boxes).

Identify Bounds: The mass is $4\,\text{kg}$ to the nearest kg. The upper bound is $4.5$ and the lower bound is $3.5$ .
Logical Reasoning: We want the maximum number of boxes. If every box were heavy (near $4.5\,\text{kg}$ ), fewer would fit. If every box were light (near $3.5\,\text{kg}$ ), more would fit. To find the theoretical maximum capacity, we assume every box weighs the minimum possible amount, which is the lower bound ( $3.5\,\text{kg}$ ).
Calculation: Divide the total weight limit ( $770\,\text{kg}$ ) by the minimum weight per box ( $3.5\,\text{kg}$ ).
$\frac{770}{3.5} = 220$
Discrete Constraint: Since 220 is an integer, exactly 220 boxes of mass $3.5\,\text{kg}$ weigh exactly $770\,\text{kg}$ . Any real box weighing slightly more than $3.5\,\text{kg}$ would mean we could carry fewer than 220. Thus, 220 is the strict upper bound for the count.

Key Takeaways

Maximizing Count: To maximize the number of items within a fixed total limit, use the lower bound of the individual item size/mass.
Minimizing Count: Conversely, to find the minimum number of items needed to exceed a limit, use the upper bound of the individual item.
Exact Integers: If the division yields an exact integer, that is the answer. If it yielded a decimal (e.g., 220.9), we would take the floor (220) because you can't have a fraction of a box.

Common Mistakes

Using the upper bound of the mass ( $4.5\,\text{kg}$ ): This calculates the minimum number of boxes required to reach a certain weight, or the maximum number if you assumed they were heaviest, which minimizes the count.
Rounding up the final answer: You cannot carry 220.1 boxes. You must round down (floor) for discrete objects, unless the question asks for the next whole unit to cover a distance/weight.
Ignoring the 'nearest kg' implication and using exactly 4 kg.

Things to Be Careful About

'Correct to the nearest...': Always check the precision. Nearest kg $ightarrow$ 0.5 margin. Nearest 10g $ightarrow$ 5g margin.
'Exactly': The van's limit is exact, so no bounds apply to the 770 figure itself.

Techniques used

determine the upper bound of a measured quantitydivide a maximum capacity by the minimum unit massinterpret the floor function for discrete items

(c)

A lorry contains boxes of three sizes $S$ , $M$ and $L$ .
The ratio of the number of boxes $S : M = 2 : 7$ .
The ratio of the number of boxes $S : L = 5 : 4$ .
The lorry contains 72 boxes of size $L$ .

Find the total number of boxes in the lorry.

______

DifficultyMedium

Worked solution

Approach

We are given two ratios involving the same category (Size $S$ ): $S:M$ and $S:L$ . We need to combine these into a single ratio $S:M:L$ . Once combined, we can use the known number of $L$ boxes to find the value of one 'part' of the ratio, and then calculate the total number of boxes.

Working

Given:
$S : M = 2 : 7$
$S : L = 5 : 4$

To combine these, the term for $S$ must be the same in both ratios. The least common multiple of 2 and 5 is 10.

Scale the first ratio ( $S:M = 2:7$ ) by multiplying by 5:
$S : M = 10 : 35$

Scale the second ratio ( $S:L = 5:4$ ) by multiplying by 2:
$S : L = 10 : 8$

Now we have a consistent ratio for $S:M:L$ :
$S : M : L = 10 : 35 : 8$

We are told there are 72 boxes of size $L$ . In our ratio, $L$ corresponds to 8 parts.
Let $k$ be the multiplier for the ratio parts. Then:
$8k = 72$
$k = \frac{72}{8} = 9$

Now find the total number of boxes. The total number of parts is:
$10 + 35 + 8 = 53 \text{ parts}$

Total boxes = $53 \times k$ :
$53 \times 9 = 477$

Alternatively, calculate each size individually:

$S = 10 \times 9 = 90$
$M = 35 \times 9 = 315$
$L = 8 \times 9 = 72$
Total = $90 + 315 + 72 = 477$

Answer

477

Final answer

477

Detailed explanation

Walkthrough

Combine Ratios: We have $S:M = 2:7$ and $S:L = 5:4$ . The link is $S$ . In the first ratio, $S$ is 2 units. In the second, $S$ is 5 units. To make them compatible, we multiply the first ratio by 5 and the second by 2, so $S$ becomes 10 in both.
- $S:M$ becomes $10:35$ .
- $S:L$ becomes $10:8$ .
- Combined: $S:M:L = 10:35:8$ .
Find the Multiplier: We know the actual number of $L$ boxes is 72. In the ratio, $L$ is represented by 8 parts. So, 8 parts = 72 boxes. One part = $72 / 8 = 9$ boxes.
Calculate Total: The total number of boxes corresponds to the sum of all ratio parts: $10 (S) + 35 (M) + 8 (L) = 53$ parts.
Total boxes = $53 \text{ parts} \times 9 \text{ boxes/part} = 477$ .

Key Takeaways

Common Term Method: When combining two ratios $A:B$ and $A:C$ , always equalize the $A$ terms using the LCM.
Parts Method: Converting the specific known quantity to a 'value per part' is often faster than solving for individual variables first.

Common Mistakes

Adding the ratios directly ( $2+5 : 7 : 4$ ), which is incorrect because the scales are different.
Forgetting to sum all three parts ( $S, M, L$ ) to get the total, stopping at just finding $S$ or $M$ .
Arithmetic errors in multiplication ( $53 \times 9$ ).

Things to Be Careful About

Ensure the order of the letters in the final combined ratio matches the question ( $S:M:L$ ).
Double-check that the sum of the calculated individual quantities ( $90+315+72$ ) equals the calculated total ( $477$ ).

Techniques used

combine two ratios sharing a common termuse a known quantity to find the scale factorsum the parts of the combined ratio

Q6MediumTransformations and Vectors

Fig. 4

(a)

Triangle $A$ is mapped onto triangle $P$ by a translation of $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ .

Draw triangle $P$ .

DifficultyEasy

Worked solution

Approach

Apply the translation vector to each vertex of triangle $A$ by adding the vector components to the coordinates.

Working

Triangle $A$ has vertices at $(4, 1)$ , $(6, 1)$ and $(4, -3)$ . The translation is $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ , so we add $1$ to each $x$ -coordinate and $-3$ to each $y$ -coordinate:

\begin{aligned} (4, 1) &\to (4+1,\; 1-3) = (5, -2) \\ (6, 1) &\to (6+1,\; 1-3) = (7, -2) \\ (4, -3) &\to (4+1,\; -3-3) = (5, -6) \end{aligned}

Answer

Triangle $P$ has vertices at $(5, -2)$ , $(7, -2)$ and $(5, -6)$ .

Final answer

Triangle P with vertices (5, -2), (7, -2) and (5, -6)

Detailed explanation

Walkthrough

A translation moves every point by the same vector. The vector $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ means move 1 unit right and 3 units down. We apply this to each vertex of triangle $A$ by adding $1$ to the $x$ -coordinate and subtracting $3$ from the $y$ -coordinate. This gives the three vertices of triangle $P$ , which we then plot on the grid and join to form the triangle.

Key Takeaways

A translation by $\begin{pmatrix} a \\ b \end{pmatrix}$ adds $a$ to every $x$ -coordinate and $b$ to every $y$ -coordinate.
The shape, size and orientation of the figure do not change under a translation.

Common Mistakes

Adding the vector components to only some vertices instead of all three.
Subtracting instead of adding (or vice versa) when the vector component is negative.
Forgetting to plot all three vertices and join them to form a closed triangle.

Things to Be Careful About

The question asks you to draw triangle $P$ , so the answer is the drawing itself — mark the three vertices and join them.
Check that the shape is congruent to $A$ and has the same orientation; a wrong sign in the vector will flip or shift the triangle incorrectly.
Verify at least one vertex by eye against the grid to catch arithmetic errors.

Techniques used

add translation vector components to each vertex coordinateplot the image vertices on the grid

(b)

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

______

DifficultyMedium

Worked solution

Approach

Compare corresponding sides of triangles $A$ and $B$ to find the scale factor, then locate the centre of enlargement by extending lines through corresponding vertices.

Working

Triangle $A$ has vertices $(4, 1)$ , $(6, 1)$ and $(4, -3)$ . Triangle $B$ has vertices $(-3, 1)$ , $(-2, 1)$ and $(-2, 3)$ .

Step 1: Find the scale factor.

The horizontal side of $A$ from $(4, 1)$ to $(6, 1)$ has length $2$ . The corresponding horizontal side of $B$ from $(-3, 1)$ to $(-2, 1)$ has length $1$ . So the scale factor in magnitude is $\frac{1}{2}$ .

The vertical side of $A$ goes down from $(4, 1)$ to $(4, -3)$ , while the corresponding vertical side of $B$ goes up from $(-2, 1)$ to $(-2, 3)$ . Since the orientation is reversed, the scale factor is negative:

k = -\frac{1}{2}

Step 2: Find the centre of enlargement.

Draw lines through corresponding vertices: $(4, 1)$ to $(-2, 1)$ , $(6, 1)$ to $(-3, 1)$ and $(4, -3)$ to $(-2, 3)$ . All three lines pass through the point $(0, 1)$ .

We can verify: with centre $(0, 1)$ and scale factor $-\frac{1}{2}$ :

\begin{aligned} (4, 1) &\to \left(0 + \left(-\tfrac{1}{2}\right)(4 - 0),\; 1 + \left(-\tfrac{1}{2}\right)(1 - 1)\right) = (-2, 1) \\ (6, 1) &\to \left(0 + \left(-\tfrac{1}{2}\right)(6 - 0),\; 1 + \left(-\tfrac{1}{2}\right)(1 - 1)\right) = (-3, 1) \\ (4, -3) &\to \left(0 + \left(-\tfrac{1}{2}\right)(4 - 0),\; 1 + \left(-\tfrac{1}{2}\right)(-3 - 1)\right) = (-2, 3) \end{aligned}

All three map correctly to the vertices of $B$ .

Answer

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

Final answer

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

Detailed explanation

Walkthrough

Step 1 — Identify the transformation type. Triangles $A$ and $B$ are similar (same shape, different size), so the transformation is an enlargement. We need to find the scale factor and the centre.

Step 2 — Find the scale factor. Measure a side of $A$ and the corresponding side of $B$ . The horizontal side of $A$ from $(4, 1)$ to $(6, 1)$ has length $2$ ; the corresponding horizontal side of $B$ from $(-3, 1)$ to $(-2, 1)$ has length $1$ . The ratio is $\frac{1}{2}$ .

Now check orientation. In $A$ , the vertical side goes downward from $(4, 1)$ to $(4, -3)$ . In $B$ , the corresponding vertical side goes upward from $(-2, 1)$ to $(-2, 3)$ . The reversal of direction means the scale factor is negative: $k = -\frac{1}{2}$ .

Step 3 — Find the centre. For an enlargement with centre $(c_x, c_y)$ and scale factor $k$ , each point satisfies $\text{image} - \text{centre} = k(\text{original} - \text{centre})$ . Equivalently, lines drawn through corresponding vertices all pass through the centre. Drawing lines from $(4, 1)$ to $(-2, 1)$ , from $(6, 1)$ to $(-3, 1)$ and from $(4, -3)$ to $(-2, 3)$ , we see they all intersect at $(0, 1)$ .

Step 4 — Verify. Substitute $(0, 1)$ and $k = -\frac{1}{2}$ into the enlargement formula and confirm all three vertices of $A$ map to the vertices of $B$ .

Key Takeaways

When the image is smaller and inverted relative to the original, the scale factor is negative and between $-1$ and $0$ .
The centre of enlargement is the intersection of lines joining corresponding vertices.
A full description requires both the scale factor (with sign) and the centre coordinates.

Common Mistakes

Giving the scale factor as $+\frac{1}{2}$ and missing the negative sign. This is the most common error — the reversed orientation is the key clue.
Giving only the scale factor without the centre, or vice versa. Both are required for full marks.
Misidentifying corresponding vertices. The right-angle vertex of $A$ at $(4, 1)$ corresponds to the right-angle vertex of $B$ at $(-2, 1)$ , not to $(-3, 1)$ .
Writing the centre as $(0, 0)$ by misreading the grid.

Things to Be Careful About

The mark scheme awards B1 for each of: enlargement, scale factor $-\frac{1}{2}$ (or equivalent), and centre $(0, 1)$ . All three must be stated.
"oe" (or equivalent) is allowed for the scale factor — $-0.5$ or $-\frac{1}{2}$ are both acceptable.
When describing the transformation, state it as "enlargement with scale factor $-\frac{1}{2}$ about centre $(0, 1)$ " — the order and wording matter for clarity.

Techniques used

compare side lengths to find scale factordetect negative scale factor from reversed orientationfind centre of enlargement by intersecting lines through corresponding vertices

(c)

Transformation $\mathbf{M}$ is a reflection in the line $y = -1$ .
Transformation $\mathbf{R}$ is a rotation $90°$ clockwise about $(1, 1)$ .
$\mathbf{RM}(B) = Q$ .

Draw triangle $Q$ .

DifficultyMedium

Worked solution

Approach

$\mathbf{RM}(B) = Q$ means apply $\mathbf{M}$ first, then $\mathbf{R}$ . Find the image of each vertex of $B$ under $\mathbf{M}$ , then apply $\mathbf{R}$ to those intermediate points.

Working

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

Step 1: Apply $\mathbf{M}$ — reflection in the line $y = -1$ .

A reflection in $y = -1$ maps $(x, y)$ to $(x, -2 - y)$ because the line $y = -1$ is the midpoint between $y$ and its image.

\begin{aligned} (-3, 1) &\to (-3,\; -2 - 1) = (-3, -3) \\ (-2, 1) &\to (-2,\; -2 - 1) = (-2, -3) \\ (-2, 3) &\to (-2,\; -2 - 3) = (-2, -5) \end{aligned}

Let this intermediate triangle be $B'$ with vertices $(-3, -3)$ , $(-2, -3)$ and $(-2, -5)$ .

Step 2: Apply $\mathbf{R}$ — rotation $90°$ clockwise about $(1, 1)$ .

For a $90°$ clockwise rotation about $(a, b)$ , the mapping is:

(x, y) \to (a + (y - b),\; b - (x - a))

With $(a, b) = (1, 1)$ this becomes $(x, y) \to (y,\; 2 - x)$ .

\begin{aligned} (-3, -3) &\to (-3,\; 2 - (-3)) = (-3, 5) \\ (-2, -3) &\to (-3,\; 2 - (-2)) = (-3, 4) \\ (-2, -5) &\to (-5,\; 2 - (-2)) = (-5, 4) \end{aligned}

Answer

Triangle $Q$ has vertices at $(-5, 4)$ , $(-3, 4)$ and $(-3, 5)$ .

Final answer

Triangle Q with vertices (-5, 4), (-3, 4) and (-3, 5)

Detailed explanation

Walkthrough

The notation $\mathbf{RM}(B) = Q$ means the transformation $\mathbf{M}$ is applied first, then $\mathbf{R}$ . We track each vertex of $B$ through both steps.

Step 1 — Reflection $\mathbf{M}$ in $y = -1$ . The line $y = -1$ is horizontal. A point at height $y$ is at distance $y - (-1) = y + 1$ above this line. Its image is the same distance below, at height $-1 - (y + 1) = -2 - y$ . The $x$ -coordinate is unchanged. Applying this to each vertex of $B$ : $(-3, 1) \to (-3, -3)$ , $(-2, 1) \to (-2, -3)$ , $(-2, 3) \to (-2, -5)$ . Call this intermediate triangle $B'$ .

Step 2 — Rotation $\mathbf{R}$ by $90°$ clockwise about $(1, 1)$ . To rotate about a point other than the origin, translate so the centre becomes the origin, rotate, then translate back. For $90°$ clockwise about the origin, $(x, y) \to (y, -x)$ . So about $(1, 1)$ : $(x, y) \to (1 + (y - 1),\; 1 - (x - 1)) = (y,\; 2 - x)$ . Applying to each vertex of $B'$ : $(-3, -3) \to (-3, 5)$ , $(-2, -3) \to (-3, 4)$ , $(-2, -5) \to (-5, 4)$ .

The final triangle $Q$ has vertices $(-5, 4)$ , $(-3, 4)$ and $(-3, 5)$ .

Key Takeaways

$\mathbf{RM}$ means apply $\mathbf{M}$ first, then $\mathbf{R}$ — the order is right-to-left in function notation.
For reflection in a horizontal line $y = c$ , the formula is $(x, y) \to (x, 2c - y)$ .
For $90°$ clockwise rotation about $(a, b)$ , use $(x, y) \to (a + y - b,\; b - x + a)$ .
Always find the intermediate image before applying the second transformation.

Common Mistakes

Applying $\mathbf{R}$ first and then $\mathbf{M}$ , which gives the wrong answer. $\mathbf{RM}(B)$ means $\mathbf{M}$ first.
Using the wrong rotation formula. $90°$ clockwise about the origin is $(x, y) \to (y, -x)$ , not $(x, y) \to (-y, x)$ (which is $90°$ anticlockwise).
Forgetting to translate back after rotating about a non-origin centre.
Sign errors when computing $2c - y$ or $2 - x$ .
The mark scheme awards partial credit for a correctly sized and oriented triangle in the wrong position (B1), or a triangle at incorrect positions entirely (SC1 or SC2), so showing the intermediate step $B'$ is important for method marks.

Things to Be Careful About

The mark scheme allows partial credit: B2 for the correct vertices $(-5, 4)$ , $(-3, 4)$ , $(-3, 5)$ ; B1 for correct size and orientation but wrong position; special credit (SC) for specific wrong positions at $(3, -6)$ , $(1, -6)$ , $(1, -7)$ or $(3, 2)$ , $(1, 2)$ , $(1, 1)$ or $(7, -2)$ , $(5, -2)$ , $(5, -3)$ . These SC positions correspond to common errors such as applying $\mathbf{R}$ first or using the wrong rotation direction.
Always verify by checking that the final triangle is congruent to $B$ (same size and shape) — a rotation and reflection preserve size, so $Q$ must be congruent to $B$ .
When drawing, mark all three vertices clearly and join them to form a closed triangle.

Techniques used

apply reflection formula to each vertex of Bapply rotation formula to each intermediate vertexplot the final image triangle Q

Q7Medium-HardMensurationTrigonometryGeometry

(a)

A cuboid has dimensions $5\,\text{cm}$ by $12\,\text{cm}$ by $h\,\text{cm}$ .
The volume of the cuboid is $480\,\text{cm}^3$ .

Calculate the value of $h$ .

$h$ = ______

(b)

Fig. 5

$ABCDE$ is a pentagon.
$AE$ is parallel to $BD$ .
$AE = 21\,\text{cm}$ , $BD = 16\,\text{cm}$ and $DE = 8\,\text{cm}$ .
Angle $DEA = 90°$ and angle $CBD = 65°$ .

8M

(i)

Calculate angle $BAE$ .

Angle $BAE$ = ______

DifficultyMedium

Worked solution

Approach

To find angle $BAE$ , construct a perpendicular from $B$ to $AE$ to form a right-angled triangle. Use the properties of parallel lines and the given lengths to find the opposite and adjacent sides of this triangle, then apply the tangent ratio.

Working

Draw a perpendicular from $B$ to $AE$ , meeting $AE$ at $F$ . Since $AE \parallel BD$ and $DE \perp AE$ , the shape $BFDE$ is a rectangle.

Therefore:

$BF = DE = 8\,\text{cm}$
$FE = BD = 16\,\text{cm}$

The length $AF$ is:

AF = AE - FE = 21 - 16 = 5\,\text{cm}

In the right-angled triangle $ABF$ , we use the tangent ratio for angle $BAF$ (which is the same as angle $BAE$ ):

\tan(\angle BAE) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BF}{AF} = \frac{8}{5} = 1.6

Calculate the angle:

\angle BAE = \tan^{-1}(1.6) \approx 57.9946^\circ

Rounding to one decimal place:

\angle BAE = 58.0^\circ

Answer

\angle BAE = 58.0^\circ

Final answer

58.0

Detailed explanation

Walkthrough

We need to find angle $BAE$ . The pentagon has parallel sides $AE$ and $BD$ , with $DE$ perpendicular to $AE$ . By dropping a perpendicular from $B$ to $AE$ at point $F$ , we create a rectangle $BFDE$ and a right-angled triangle $ABF$ . The height $BF$ equals $DE$ , which is $8\,\text{cm}$ . The segment $FE$ equals $BD$ , which is $16\,\text{cm}$ . Since $AE$ is $21\,\text{cm}$ , the remaining segment $AF$ is $21 - 16 = 5\,\text{cm}$ . In triangle $ABF$ , the tangent of angle $BAE$ is the ratio of the opposite side ( $BF = 8$ ) to the adjacent side ( $AF = 5$ ). Taking the inverse tangent of $1.6$ gives approximately $58.0^\circ$ .

Key Takeaways

When dealing with shapes that have parallel sides and right angles, constructing auxiliary perpendiculars can create right-angled triangles where trigonometric ratios can be applied. Always identify the correct opposite and adjacent sides relative to the angle you are finding.

Common Mistakes

Using the wrong side lengths for the tangent ratio (e.g., using $21$ instead of $5$ as the adjacent side).
Forgetting to subtract the parallel segment length ( $16$ ) from the total base length ( $21$ ) to find the adjacent side.
Rounding the angle to the nearest whole number ( $58^\circ$ ) instead of the required one decimal place ( $58.0^\circ$ ).

Things to Be Careful About

The question does not specify accuracy, but $58.0^\circ$ is the standard expected form. Ensure you use the inverse tangent function correctly on your calculator. The horizontal distance is $21 - 16 = 5$ , not $21$ or $16$ .

Techniques used

construct an auxiliary perpendicular to form a right-angled triangleuse the tangent ratio to find an anglecalculate the inverse tangent

(ii)

The area of pentagon $ABCDE$ is $200\,\text{cm}^2$ .

Calculate the length of $BC$ .

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

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Split the pentagon $ABCDE$ into a trapezium $ABDE$ and a triangle $BCD$ . Calculate the area of the trapezium, subtract it from the total area to find the area of the triangle, then use the sine area formula to find the length $BC$ .

Working

The area of trapezium $ABDE$ (with parallel sides $AE = 21$ and $BD = 16$ , and height $DE = 8$ ):

\text{Area of } ABDE = \frac{1}{2}(AE + BD) \times DE = \frac{1}{2}(21 + 16) \times 8

\text{Area of } ABDE = \frac{1}{2} \times 37 \times 8 = 148\,\text{cm}^2

The total area of pentagon $ABCDE$ is $200\,\text{cm}^2$ . The area of triangle $BCD$ is:

\text{Area of } \triangle BCD = 200 - 148 = 52\,\text{cm}^2

The area of triangle $BCD$ can also be expressed using the sine formula with the included angle $\angle CBD = 65^\circ$ :

\text{Area of } \triangle BCD = \frac{1}{2} \times BC \times BD \times \sin(\angle CBD)

Substitute the known values:

52 = \frac{1}{2} \times BC \times 16 \times \sin(65^\circ)

52 = 8 \times BC \times \sin(65^\circ)

Solve for $BC$ :

BC = \frac{52}{8 \times \sin(65^\circ)}

BC = \frac{52}{8 \times 0.906307787}

BC = \frac{52}{7.250462298} \approx 7.1717\,\text{cm}

Rounding to three significant figures:

BC = 7.17\,\text{cm}

Answer

BC = 7.17\,\text{cm}

Final answer

7.17

Detailed explanation

Walkthrough

The pentagon is split into two simpler shapes: trapezium $ABDE$ and triangle $BCD$ . The trapezium has parallel sides $AE = 21\,\text{cm}$ and $BD = 16\,\text{cm}$ , with perpendicular height $DE = 8\,\text{cm}$ . Its area is $\frac{1}{2}(21 + 16) \times 8 = 148\,\text{cm}^2$ . Subtracting this from the total pentagon area ( $200\,\text{cm}^2$ ) gives the area of triangle $BCD$ as $52\,\text{cm}^2$ . Using the sine area formula for triangle $BCD$ with the known side $BD = 16\,\text{cm}$ and included angle $\angle CBD = 65^\circ$ , we set up $52 = \frac{1}{2} \times BC \times 16 \times \sin(65^\circ)$ . Solving for $BC$ gives approximately $7.17\,\text{cm}$ .

Key Takeaways

Compound shapes can often be split into simpler geometric figures (trapeziums, triangles, rectangles) whose areas are easier to calculate. When finding a missing side in a triangle where the area and an included angle are known, the sine area formula $\text{Area} = \frac{1}{2}ab\sin C$ is the most direct tool.

Common Mistakes

Calculating the area of the trapezium incorrectly (e.g., forgetting to divide by 2, or using the wrong parallel sides).
Using the wrong formula for the area of triangle $BCD$ (e.g., $\frac{1}{2} \times \text{base} \times \text{height}$ without correctly identifying the height).
Algebraic errors when rearranging the sine area formula to solve for $BC$ .
Rounding too early in the calculation, which can lead to an incorrect final answer.

Things to Be Careful About

The mark scheme accepts answers from $7.171$ to $7.172$ , so $7.17$ (to 3 significant figures) is correct. Ensure you use the sine area formula correctly: $\frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{included angle})$ . The angle $65^\circ$ must be the included angle between the two sides, which it is ( $\angle CBD$ is between $BC$ and $BD$ ).

Techniques used

split compound shape into simpler shapescalculate area of trapeziumuse sine area formula to find side length

Q8MediumNumberAlgebra and Graphs

(a)

8M

(i)

Complete the table for $y = \frac{2^x}{5}$ .

$x$	0	1	2	3	4	5
$y$	0.2	0.4	0.8	1.6	3.2

(ii)

On the grid, draw the graph of $y = \frac{2^x}{5}$ for $0 \leq x \leq 5$ .

Fig. 6

DifficultyMedium-Easy

Worked solution

Approach

Use the completed table of values from part (a)(i) to plot the points on the given grid, then draw a smooth curve through them.

Working

The coordinates to plot are:

(0, 0.2), \quad (1, 0.4), \quad (2, 0.8), \quad (3, 1.6), \quad (4, 3.2), \quad (5, 6.4)

Plot these six points on the grid. Then draw a smooth, continuous curve that passes through all the plotted points. The curve should start near the x-axis at $x=0$ and rise increasingly steeply as $x$ increases, characteristic of an exponential growth function.

Answer

A smooth curve passing through $(0, 0.2)$ , $(1, 0.4)$ , $(2, 0.8)$ , $(3, 1.6)$ , $(4, 3.2)$ , and $(5, 6.4)$ .

Final answer

Smooth curve through (0, 0.2), (1, 0.4), (2, 0.8), (3, 1.6), (4, 3.2), (5, 6.4)

Detailed explanation

Walkthrough

The question asks to draw the graph of $y = \frac{2^x}{5}$ for $0 \leq x \leq 5$ . We already have the table of values from part (a)(i). We plot each $(x, y)$ pair as a point on the grid. After plotting all six points, we connect them with a smooth curve. Because the function is exponential ( $2^x$ ), the curve will not be a straight line; it will start relatively flat and curve upwards with increasing steepness.

Key Takeaways

How to plot points from a table of values and recognize the shape of an exponential graph (smooth, continuously increasing curve).

Common Mistakes

Drawing straight line segments between the points instead of a smooth curve. Forgetting to plot all points, which reduces the marks awarded. Plotting points at incorrect coordinates.

Things to Be Careful About

The mark scheme awards follow-through marks (B2FT or B1FT) based on the number of correctly plotted points (5 or 6 for full marks, 3 or 4 for partial). Ensure the curve is smooth and not jagged or made of straight lines. The grid has subdivisions, so read the y-values carefully (e.g., $0.2$ is two small squares above the x-axis if each major square is 1 unit and there are 5 subdivisions per unit).

Techniques used

plot points from a table of valuesdraw a smooth curve through plotted points

(iii)

$2^{x+3} = 100$

4M

(a)

Show that $\frac{2^x}{5} = \frac{5}{2}$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the index law $a^{m+n} = a^m \times a^n$ to expand $2^{x+3}$ , then rearrange the equation to show that $\frac{2^x}{5}$ equals $\frac{5}{2}$ .

Working

Start with the given equation:

2^{x+3} = 100

Apply the index law $a^{m+n} = a^m \times a^n$ :

2^x \times 2^3 = 100

Evaluate $2^3$ :

2^x \times 8 = 100

Divide both sides by 8:

2^x = \frac{100}{8} = 12.5

We need to show that $\frac{2^x}{5} = \frac{5}{2}$ . Divide both sides of $2^x = 12.5$ by 5:

\frac{2^x}{5} = \frac{12.5}{5}

Simplify the right side:

\frac{12.5}{5} = 2.5 = \frac{5}{2}

Therefore:

\frac{2^x}{5} = \frac{5}{2}

Answer

Shown as required.

Final answer

Shown

Detailed explanation

Walkthrough

We are given $2^{x+3} = 100$ and need to show that $\frac{2^x}{5} = \frac{5}{2}$ . First, we use the index law for multiplying powers with the same base: $2^{x+3} = 2^x \times 2^3$ . Since $2^3 = 8$ , the equation becomes $8 \times 2^x = 100$ . Dividing by 8 gives $2^x = 12.5$ . The target expression is $\frac{2^x}{5}$ , so we divide our result for $2^x$ by 5: $\frac{12.5}{5} = 2.5$ . Finally, we convert $2.5$ to the fraction $\frac{5}{2}$ , which completes the proof.

Key Takeaways

The product law of indices ( $a^{m+n} = a^m \times a^n$ ) and how to manipulate equations to isolate a specific algebraic expression.

Common Mistakes

Forgetting to expand $2^{x+3}$ as $2^x \times 2^3$ and instead trying to solve for $x$ directly using logarithms (which is not allowed or expected at this stage). Arithmetic errors when dividing 100 by 8 or 12.5 by 5.

Things to Be Careful About

The mark scheme explicitly allows doing the division by 2 three times as an alternative route (i.e., $100 \div 2 = 50$ , $50 \div 2 = 25$ , $25 \div 2 = 12.5$ ). Both methods are valid. Ensure the final form matches $\frac{5}{2}$ exactly.

Techniques used

apply index laws to expand a powerrearrange an equation to isolate a fraction

(b)

By drawing a suitable line on the grid, solve $2^{x+3} = 100$ .

$x$ = ______

DifficultyMedium

Worked solution

Approach

From part (a)(iii)(a), we established that solving $2^{x+3} = 100$ is equivalent to finding $x$ when $\frac{2^x}{5} = \frac{5}{2} = 2.5$ . Draw the horizontal line $y = 2.5$ on the graph from part (a)(ii) and read the x-coordinate where it intersects the curve.

Working

We want to solve $2^{x+3} = 100$ . From part (a)(iii)(a), this is equivalent to:

\frac{2^x}{5} = 2.5

On the graph of $y = \frac{2^x}{5}$ , this corresponds to finding the x-value where $y = 2.5$ . Draw a horizontal line at $y = 2.5$ across the grid.

The line $y = 2.5$ intersects the curve at a point between $x = 3$ and $x = 4$ . Reading the graph carefully:

At $x = 3$ , $y = 1.6$
At $x = 4$ , $y = 3.2$

The intersection occurs closer to $x = 4$ . By estimating or reading from the grid, the x-coordinate is approximately $3.6$ to $3.7$ . The mark scheme accepts any value from $3.5$ to $3.8$ .

Answer

x \approx 3.6

Final answer

3.6

Detailed explanation

Walkthrough

In part (a)(iii)(a), we showed that $2^{x+3} = 100$ implies $\frac{2^x}{5} = 2.5$ . The graph we drew in part (a)(ii) is exactly $y = \frac{2^x}{5}$ . Therefore, to solve $2^{x+3} = 100$ , we simply need to find the x-value on our graph where the y-value is $2.5$ . We draw a horizontal line at $y = 2.5$ and see where it crosses the curve. Looking at the plotted points, at $x=3$ , $y=1.6$ , and at $x=4$ , $y=3.2$ . Since $2.5$ is between $1.6$ and $3.2$ , the x-value must be between 3 and 4. Reading the grid, the intersection is at approximately $x = 3.6$ .

Key Takeaways

How to use an existing graph to solve a related algebraic equation by drawing a suitable line (in this case, a horizontal line $y = k$ ).

Common Mistakes

Drawing the wrong line (e.g., $y = 100$ or $x = 2.5$ ). Failing to link the equation $2^{x+3} = 100$ to the result of part (a)(iii)(a). Reading the intersection point inaccurately from the grid.

Things to Be Careful About

The mark scheme accepts a range of answers from $3.5$ to $3.8$ because this is a graphical solution and reading precision is limited. Ensure you draw the line $y = 2.5$ , not $y = 2.5x$ or anything else. The line must be clearly drawn to earn the method mark.

Techniques used

draw a horizontal line to represent a constant valueread the x-coordinate of the intersection with the curve

(b)

Fig. 7

This is a sketch of the graph $y = a + bx - x^2$ .
The graph crosses the $x$ -axis at integer values of $x$ .

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

$a$ = ______
$b$ = ______

DifficultyMedium

Worked solution

Approach

The equation is $y = a + bx - x^2$ . The y-intercept gives $a$ directly. The x-intercepts are integer values, and their product is related to $a$ . Use these to find the roots, then expand to find $b$ .

Working

Find $a$ :
The graph crosses the y-axis at $y = 7$ . Substitute $x = 0$ into the equation:

y = a + b(0) - (0)^2 = a

Therefore:

a = 7

Find $b$ :
The equation is now $y = 7 + bx - x^2$ . The graph crosses the x-axis at integer values, so $y = 0$ has integer roots. Let the roots be $c$ and $d$ . By Vieta's formulas (or by factoring), the product of the roots is:

c \times d = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{7}{-1} = -7

Since $c$ and $d$ are integers and their product is $-7$ , the possible pairs are $(1, -7)$ or $(-1, 7)$ . The sketch shows one negative root and one positive root, with the positive root further from the origin. Thus, the roots are $x = -1$ and $x = 7$ .

We can write the equation in factored form:

y = -(x + 1)(x - 7)

Expand this expression:

y = -(x^2 - 7x + x - 7) = -(x^2 - 6x - 7)

y = -x^2 + 6x + 7 = 7 + 6x - x^2

Comparing this with $y = 7 + bx - x^2$ , we get:

b = 6

Answer

a = 7, \quad b = 6

Final answer

a = 7, b = 6

Detailed explanation

Walkthrough

We are given the quadratic $y = a + bx - x^2$ and a sketch. First, find $a$ using the y-intercept. When $x = 0$ , $y = a$ . The sketch shows the curve crossing the y-axis at $y = 7$ , so $a = 7$ . Next, find $b$ using the x-intercepts. The curve crosses the x-axis at integer values, meaning $7 + bx - x^2 = 0$ has integer roots. Rearranging gives $x^2 - bx - 7 = 0$ . The product of the roots is $-7$ . The only integer pairs with product $-7$ are $(1, -7)$ and $(-1, 7)$ . The sketch shows the positive root is further from the origin than the negative root, so the roots are $-1$ and $7$ . The sum of the roots is $-1 + 7 = 6$ . Since the sum of the roots for $x^2 - bx - 7 = 0$ is $b$ , we have $b = 6$ . Alternatively, substitute $x = -1, y = 0$ into $y = 7 + bx - x^2$ : $0 = 7 - b - 1 \implies b = 6$ .

Key Takeaways

How to use intercepts (y-intercept for the constant term, x-intercepts for the linear coefficient) to determine the equation of a quadratic curve.

Common Mistakes

Assuming the roots are $1$ and $-7$ without checking the sketch (the sketch shows the positive root is larger in magnitude). Forgetting that the coefficient of $x^2$ is negative, which affects the product and sum of roots. Arithmetic errors when expanding $(x+1)(x-7)$ .

Things to Be Careful About

The mark scheme accepts $a = 7$ as a direct mark. For $b$ , it accepts finding the roots $(-1, 0)$ and $(7, 0)$ or using the property that the product of the x-intercepts $c \times d = -a$ . Ensure you read the y-intercept correctly from the sketch (it is clearly marked as 7). The diagram is NOT TO SCALE, but the relative positions of the intercepts (one negative, one positive, positive further out) are reliable indicators.

Techniques used

identify the y-intercept to find a constantuse the x-intercepts to form a factored quadraticexpand and compare coefficients

Q9MediumGeometryTrigonometryNumber

Fig. 8

The diagram shows the positions of three ports $A$ , $B$ and $C$ .
The bearing of port $B$ from port $A$ is $107°$ .
The bearing of port $C$ from port $A$ is $192°$ .
$AB = 176\,\text{km}$ and $AC = 132\,\text{km}$ .

(a)

Find the bearing of $A$ from $B$ .

______

(b)

Calculate $BC$ .

$BC$ = ______ $\text{km}$

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

First, find the angle $CAB$ by subtracting the bearing of $B$ from the bearing of $C$ . Then, use the cosine rule in triangle $ABC$ to calculate the length $BC$ .

Working

\angle CAB = 192^\circ - 107^\circ = 85^\circ

Apply the cosine rule:

BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle CAB)

Substitute the known values:

BC^2 = 176^2 + 132^2 - 2 \times 176 \times 132 \times \cos(85^\circ)

BC^2 = 30976 + 17424 - 46464 \times 0.0871557\ldots

BC^2 = 48400 - 4049.65\ldots = 44350.35\ldots

BC = \sqrt{44350.35\ldots} \approx 210.595\ldots

Rounding to 3 significant figures:

BC \approx 211

Answer

211

Final answer

211

Detailed explanation

Walkthrough

The bearing of $B$ from $A$ is 107 degrees and the bearing of $C$ from $A$ is 192 degrees. The angle between them, $\angle CAB$ , is the difference: $192 - 107 = 85$ degrees. With two sides and the included angle known, the cosine rule is the correct tool to find the third side $BC$ . Substituting $AB = 176$ , $AC = 132$ , and $\angle CAB = 85^\circ$ into $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle CAB)$ gives $BC^2 \approx 44350.35$ . Taking the square root yields $BC \approx 210.6$ km, which rounds to 211 km to 3 significant figures.

Key Takeaways

The angle between two bearings measured from the same point is simply the difference between the two bearing values. The cosine rule $c^2 = a^2 + b^2 - 2ab\cos(C)$ is used when two sides and the included angle of a triangle are known.

Common Mistakes

Using the wrong angle in the cosine rule (e.g., using 107 or 192 instead of the difference 85).
Rounding intermediate values too early, which can lead to an incorrect final answer.
Forgetting to take the square root at the end.

Things to Be Careful About

Ensure the calculator is in degree mode. The mark scheme accepts 210.5 to 210.6 or 211. Always show the substituted cosine rule expression to earn method marks.

Techniques used

find the included angle between two bearingsapply the cosine rule to find the third side

(c)

Boat $B$ leaves port $B$ at 10.00 am.
It sails directly to port $A$ at an average speed of $48\,\text{km/h}$ .

Boat $C$ leaves port $C$ at 10.15 am.
It sails directly to port $A$ and arrives there 7 minutes before boat $B$ .

Find the average speed of boat $C$ in $\text{km/h}$ .

______ $\text{km/h}$

5M Similar

DifficultyMedium

Worked solution

Approach

Calculate the time taken for boat $B$ to travel from $B$ to $A$ . Use this to find the arrival time of boat $B$ . Determine the arrival time of boat $C$ (7 minutes earlier), then find the duration of boat $C$ 's journey. Finally, use the distance $AC$ and this duration to find the average speed of boat $C$ .

Working

Time taken by boat $B$ :

\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{176}{48} = 3\frac{2}{3} \text{ hours} = 3 \text{ hours } 40 \text{ minutes}

Arrival time of boat $B$ :
Boat $B$ leaves at 10:00 am, so it arrives at 1:40 pm.

Arrival time of boat $C$ :
Boat $C$ arrives 7 minutes before boat $B$ :

1:40 \text{ pm} - 7 \text{ minutes} = 1:33 \text{ pm}

Time taken by boat $C$ :
Boat $C$ leaves at 10:15 am and arrives at 1:33 pm.

\text{Duration} = 3 \text{ hours } 18 \text{ minutes} = 3\frac{18}{60} \text{ hours} = 3.3 \text{ hours}

Average speed of boat $C$ :

\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{132}{3.3} = 40 \text{ km/h}

Answer

40

Final answer

40

Detailed explanation

Walkthrough

Boat $B$ travels 176 km at 48 km/h. Time = distance / speed = 176 / 48 = 3.666... hours, which is 3 hours and 40 minutes. Leaving at 10:00 am, boat $B$ arrives at 1:40 pm. Boat $C$ arrives 7 minutes earlier, at 1:33 pm. Boat $C$ leaves at 10:15 am, so its journey takes from 10:15 am to 1:33 pm, which is 3 hours and 18 minutes. Converting 3 hours 18 minutes to hours gives 3.3 hours. The distance $AC$ is given as 132 km. Speed of boat $C$ = 132 / 3.3 = 40 km/h.

Key Takeaways

When dealing with travel problems involving clock times, convert all durations to a consistent unit (hours or minutes) before calculating speed or time. Remember that 18 minutes is $18/60 = 0.3$ hours.

Common Mistakes

Adding or subtracting minutes incorrectly when changing clock times.
Forgetting to convert the 18 minutes into hours (0.3 hours) before dividing distance by time.
Using the wrong distance for boat $C$ (must use $AC = 132$ km, not $AB = 176$ km).

Things to Be Careful About

Ensure time differences are calculated correctly across the 12-hour mark. The mark scheme allows working in hours and minutes or converting to fractions/decimals of hours early. Show the substituted expression $\frac{132}{3.3}$ or equivalent to earn method marks.

Techniques used

calculate the time taken by boat Bdetermine the arrival time of boat Bfind the travel time for boat Ccalculate the speed of boat C

Q10Medium-HardAlgebra and GraphsGeometry

(a)

$r = \frac{4p + 3t}{2}$

Find the value of $p$ when $r = 10$ and $t = -2$ .

$p$ = ______

(b)

Fig. 9

The diagram shows a quadrilateral.

Form an equation in $w$ and solve it to find the size of the largest angle in the quadrilateral.

Largest angle = ______

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

The sum of the interior angles in any quadrilateral is $360^\circ$ . Set up an equation by summing the four given expressions to $360^\circ$ , solve for $w$ , and then evaluate the expressions to find the largest angle.

Working

Sum of the interior angles of the quadrilateral:

w + 2w + (w + 10) + (w - 15) = 360

Combine like terms:

5w - 5 = 360

Add $5$ to both sides:

5w = 365

Divide by $5$ :

w = 73

Now calculate the size of each angle:

$w^\circ = 73^\circ$
$2w^\circ = 2 \times 73^\circ = 146^\circ$
$(w + 10)^\circ = 73^\circ + 10^\circ = 83^\circ$
$(w - 15)^\circ = 73^\circ - 15^\circ = 58^\circ$

The largest angle is $146^\circ$ .

Answer

146^\circ

Final answer

146

Detailed explanation

Walkthrough

Recall that the sum of the interior angles of any four-sided polygon (quadrilateral) is always $360^\circ$ .
Add the four angle expressions: $w + 2w + (w + 10) + (w - 15) = 360$ .
Collect like terms: the terms in $w$ sum to $w + 2w + w + w = 5w$ , and the constant terms sum to $10 - 15 = -5$ . This gives $5w - 5 = 360$ .
Solve for $w$ : add $5$ to get $5w = 365$ , then divide by $5$ to obtain $w = 73$ .
The question asks for the size of the largest angle, not just $w$ $w$ . Substitute $w = 73$ $w = 73$ into each angle expression:
- $w = 73^\circ$
- $2w = 146^\circ$
- $w + 10 = 83^\circ$
- $w - 15 = 58^\circ$
The largest angle is $146^\circ$ .

Key Takeaways

Always remember the polygon interior angle sum property: $(n - 2) \times 180^\circ$ , which gives $360^\circ$ for $n = 4$ .
Read the final demand of the question carefully to ensure you answer what is asked (here, the largest angle, not just $w$ ).

Common Mistakes

Stopping after calculating $w = 73$ and not computing the largest angle.
Adding the terms incorrectly (e.g., getting $5w + 25 = 360$ ).

Things to Be Careful About

Double-check which expression produces the largest angle (for $w = 73$ , $2w = 146$ is larger than $w + 10 = 83$ ).

Techniques used

form an equation using the angle sum of a quadrilateralsolve a linear equationevaluate each angle expression to determine the maximum

(c)

Simplify.

\frac{2k^2 - 5k - 3}{k^2 - 9}

______

DifficultyMedium-Easy

Worked solution

Approach

Factorise both the numerator and denominator completely, then cancel any common factor.

Working

Factorise the numerator $2k^2 - 5k - 3$ :

2k^2 - 5k - 3 = (2k + 1)(k - 3)

Factorise the denominator $k^2 - 9$ as a difference of two squares:

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

Write the fraction in factorised form and cancel the common factor $(k - 3)$ :

\frac{2k^2 - 5k - 3}{k^2 - 9} = \frac{(2k + 1)(k - 3)}{(k + 3)(k - 3)} = \frac{2k + 1}{k + 3}

Answer

\frac{2k + 1}{k + 3}

Final answer

(2k + 1)/(k + 3)

Detailed explanation

Walkthrough

Factorise the numerator $2k^2 - 5k - 3$ $2 k^{2} - 5 k - 3$ :
- We need two numbers that multiply to $2 \times (-3) = -6$ and add up to $-5$ . These numbers are $-6$ and $1$ .
- Split the middle term: $2k^2 - 6k + k - 3 = 2k(k - 3) + 1(k - 3) = (2k + 1)(k - 3)$ .
Factorise the denominator $k^2 - 9$ $k^{2} - 9$ :
- Recognize this as a difference of two squares: $a^2 - b^2 = (a + b)(a - b)$ .
- Here, $k^2 - 3^2 = (k + 3)(k - 3)$ .
Cancel the common factor:
- Both numerator and denominator have the factor $(k - 3)$ .
- Dividing both by $(k - 3)$ leaves $\frac{2k + 1}{k + 3}$ .

Key Takeaways

To simplify algebraic fractions, always factorise the numerator and denominator fully first. Never cancel individual terms directly from unfactorised expressions.
Recognize standard forms such as the difference of two squares ( $a^2 - b^2$ ).

Common Mistakes

Attempting to cancel terms like $k^2$ or constants directly from the numerator and denominator without factorising.
Incorrectly factorising $k^2 - 9$ as $(k - 3)^2$ .

Things to Be Careful About

Ensure the final answer is left in its simplest form.

Techniques used

factorise a quadratic trinomial with leading coefficient greater than 1factorise a difference of two squarescancel common algebraic factors

(d)

Solve.

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

Show all your working and give your answers correct to 2 decimal places.

$x$ = ______ or $x$ = ______

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Multiply the entire equation by the common denominator $(x + 3)(x - 2)$ to eliminate the fractions, expand and rearrange the terms into a standard quadratic equation $ax^2 + bx + c = 0$ , and then solve using the quadratic formula.

Working

Given equation:

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

Multiply through by $(x + 3)(x - 2)$ :

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

Expand the brackets on both sides:

2x - 4 + 5x + 15 = x^2 - 2x + 3x - 6

Simplify each side:

7x + 11 = x^2 + x - 6

Rearrange to form a quadratic equation equal to zero:

x^2 + x - 7x - 6 - 11 = 0

x^2 - 6x - 17 = 0

Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$ , $b = -6$ , $c = -17$ :

x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-17)}}{2(1)}

x = \frac{6 \pm \sqrt{36 + 68}}{2}

x = \frac{6 \pm \sqrt{104}}{2}

Calculate the two values:

x = \frac{6 + \sqrt{104}}{2} \approx \frac{6 + 10.198039}{2} = \frac{16.198039}{2} \approx 8.10

x = \frac{6 - \sqrt{104}}{2} \approx \frac{6 - 10.198039}{2} = \frac{-4.198039}{2} \approx -2.10

Answer

x = 8.10 \quad \text{or} \quad x = -2.10

Final answer

x = 8.10 or x = -2.10

Detailed explanation

Walkthrough

Eliminate fractions: Multiply every term in the equation by the common denominator $(x + 3)(x - 2)$ . This eliminates both denominators, giving $2(x - 2) + 5(x + 3) = 1 \times (x + 3)(x - 2)$ .
Expand the expressions:
- Left side: $2x - 4 + 5x + 15 = 7x + 11$ .
- Right side: $(x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6$ .
Rearrange into standard form $ax^2 + bx + c = 0$ $a x^{2} + b x + c = 0$ :
- Subtract $(7x + 11)$ from both sides:
- $x^2 + x - 7x - 6 - 11 = 0 \implies x^2 - 6x - 17 = 0$ .
Apply the quadratic formula:
- Identify coefficients: $a = 1$ , $b = -6$ , $c = -17$ .
- Substitute into $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :
- $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-17)}}{2(1)} = \frac{6 \pm \sqrt{36 + 68}}{2} = \frac{6 \pm \sqrt{104}}{2}$ .
Evaluate and round:
- $\sqrt{104} \approx 10.198039$
- $x_1 = \frac{6 + 10.198039}{2} \approx 8.0990... \to 8.10$
- $x_2 = \frac{6 - 10.198039}{2} \approx -2.0990... \to -2.10$

Key Takeaways

When clearing denominators from an algebraic fraction equation, ensure every term (including the constant $1$ on the right-hand side) is multiplied by the common denominator.
When instructed to give answers to 2 decimal places, write trailing zeros if necessary (e.g., $8.10$ and $-2.10$ , not $8.1$ and $-2.1$ ).

Common Mistakes

Forgetting to multiply the right-hand side ( $1$ ) by the common denominator $(x + 3)(x - 2)$ .
Sign errors when expanding $-4(1)(-17)$ , incorrectly getting $36 - 68$ instead of $36 + 68$ .
Rounding $8.099...$ to $8.1$ instead of keeping the required two decimal places as $8.10$ .

Things to Be Careful About

Double check the signs when rearranging all terms to one side.
Ensure you clearly show the substitution step in the quadratic formula as method marks are awarded for it.

Techniques used

clear denominators in an algebraic fractional equationexpand and simplify to form a quadratic equationapply the quadratic formularound solutions to 2 decimal places

Q11MediumProbability

(a)

On any day in January, the probability the temperature at a weather station is above $14°C$ is 0.35 .

6M

(i)

There are 31 days in January.

Find the number of days in January when you would expect the temperature to be above $14°C$ .

______

(ii)

The temperature on two consecutive days in January is recorded.

5M

(a)

Complete the tree diagram.

Fig. 10

DifficultyMedium-Easy

Worked solution

Approach

The sum of probabilities on any branch from a single node must equal $1$ . Since the temperature on consecutive days is treated as independent, the probabilities on the second day's branches are the same as the first day's.

Working

First day, lower branch:

P(14°C \text{ or below}) = 1 - 0.35 = 0.65

Second day, upper branches (from 'Above 14°C'):

P(14°C \text{ or below}) = 1 - 0.35 = 0.65

Second day, lower branches (from '14°C or below'):

P(\text{Above } 14°C) = 0.35

P(14°C \text{ or below}) = 1 - 0.35 = 0.65

Answer

The completed probabilities on the blank lines are $0.65$ , $0.65$ , $0.35$ , and $0.65$ .

\begin{aligned} &\text{First day lower branch: } 0.65 \\ &\text{Second day from Above: lower branch } 0.65 \\ &\text{Second day from Below: upper branch } 0.35 \\ &\text{Second day from Below: lower branch } 0.65 \end{aligned}

Final answer

0.65, 0.65, 0.35, 0.65

Detailed explanation

Walkthrough

A probability tree diagram requires that the probabilities on all branches emerging from a single point sum to $1$ .

First day, lower branch: The probability of 'Above 14°C' is $0.35$ , so the probability of '14°C or below' is $1 - 0.35 = 0.65$ .
Second day, upper branches: The problem implies independence between days, meaning the probabilities are the same as day one. The upper branch is $0.35$ , so the lower branch ('14°C or below') is $1 - 0.35 = 0.65$ .
Second day, lower branches: From the '14°C or below' node on day one, the probability of 'Above 14°C' on day two is still $0.35$ (independence). The probability of '14°C or below' is $1 - 0.35 = 0.65$ .

Key Takeaways

Branches from the same node must sum to $1$ .
In independent events, the probabilities on subsequent branches remain constant.

Common Mistakes

Forgetting that the second day's probabilities are independent and trying to adjust them based on the first day's outcome.
Arithmetic errors when calculating $1 - 0.35$ .

Things to Be Careful About

Ensure all four blank lines are filled. The mark scheme awards partial credit (M1) for 2 or 3 correct probabilities.

Techniques used

complete probability tree using complementary probabilitiesapply independence assumption for consecutive days

(b)

Find the probability that the temperature is above $14°C$ on both days.

______

(c)

Find the probability that the temperature is above $14°C$ on only one of the two days.

______

DifficultyMedium-Easy

Worked solution

Approach

"Above 14°C on only one day" means either (Above then Below) or (Below then Above). These are two mutually exclusive paths. We calculate the probability for each and add them.

Working

Path 1: Above on Day 1, Below on Day 2

P(\text{Above, Below}) = 0.35 \times 0.65 = 0.2275

Path 2: Below on Day 1, Above on Day 2

P(\text{Below, Above}) = 0.65 \times 0.35 = 0.2275

Total probability:

P(\text{only one day above}) = 0.2275 + 0.2275 = 0.455

Alternatively, using symmetry:

2 \times (0.35 \times 0.65) = 0.455

Answer

0.455

Final answer

0.455

Detailed explanation

Walkthrough

The event 'temperature is above 14°C on only one of the two days' can happen in two ways:

Above 14°C on Day 1 AND 14°C or below on Day 2.
14°C or below on Day 1 AND Above 14°C on Day 2.

We calculate the probability of each path by multiplying the branch probabilities:

Path 1: $0.35 \times 0.65 = 0.2275$
Path 2: $0.65 \times 0.35 = 0.2275$

Since these paths cannot happen at the same time (they are mutually exclusive), we add the probabilities: $0.2275 + 0.2275 = 0.455$ .

Key Takeaways

When an event can occur via multiple mutually exclusive paths, calculate the probability of each path and sum them.

Common Mistakes

Only calculating one of the two paths (e.g., $0.35 \times 0.65 = 0.2275$ ) and forgetting to add the other.
Adding the probabilities of the branches instead of multiplying them along the path.

Things to Be Careful About

The mark scheme explicitly looks for the method $2 \times 0.35 \times 0.65$ or the sum of the two products. Accepts fractions like $\frac{91}{200}$ .

Techniques used

identify the two mutually exclusive paths for exactly one eventmultiply and add the probabilities of these paths

(b)

In a group of 14 children:

8 wear red T-shirts
1 wears a green T-shirt
5 wear blue T-shirts.

Two children are chosen from the group at random.

Find the probability that they wear different coloured T-shirts.

______