Mathematics (Syllabus D) 4024 / 22 — May/June 2024

(b)

Maria pays a fee to sell strawberries at a market.
Each day she pays $75 plus a payment for the mass of strawberries she sells.
The fee Maria pays per day is shown on the graph.

Fig. 1

5M

(i)

One day Maria’s fee is $240.

Use the graph to find the mass of strawberries she sells that day.

______ $\text{kg}$

(ii)

On Saturday Maria sells $270\,\text{kg}$ of strawberries.
On Sunday she sells $220\,\text{kg}$ of strawberries.

Find the total fee she pays for these two days.

$ ______

(iii)

The fee per day for Maria now increases.
Each day she now pays $90 plus a payment of $60 for every $100\,\text{kg}$ of strawberries she sells.

On the grid, draw a line to represent this new fee when she sells $0\,\text{kg}$ to $500\,\text{kg}$ of strawberries in a day.

DifficultyMedium-Easy

Worked solution

Approach

Form the linear equation for the new daily fee in terms of mass, find the coordinates for the endpoints at $\text{Mass} = 0\,\text{kg}$ and $\text{Mass} = 500\,\text{kg}$ , and draw a ruled straight line connecting them.

Working

The fee formula is:

\text{Fee} = 90 + 60 \times \frac{\text{Mass}}{100} = 90 + 0.6 \times \text{Mass}

Find points to plot:

When $\text{Mass} = 0\,\text{kg}$ :

\text{Fee} = 90

So plot the point $(0, 90)$ .

When $\text{Mass} = 500\,\text{kg}$ :

\text{Fee} = 90 + 0.6(500) = 90 + 300 = 390

So plot the point $(500, 390)$ .

Draw a continuous ruled straight line from $(0, 90)$ to $(500, 390)$ .

Answer

Ruled straight line from $(0, 90)$ to $(500, 390)$

Final answer

Ruled line from (0, 90) to (500, 390)

Detailed explanation

Walkthrough

Establish the relationship between mass $m$ and fee $F$ : the fixed charge is $90, and the variable charge is $60 per $100\,\text{kg}$ (which is $0.60 per $\text{kg}$ ). Thus, $F = 90 + 0.6m$ .
Determine two key points to accurately draw the straight line across the domain $0 \leq m \leq 500$ $0 \leq m \leq 500$ :
- At $m = 0$ , $F = 90$ , giving point $(0, 90)$ .
- At $m = 500$ , $F = 90 + 60 \times 5 = 390$ , giving point $(500, 390)$ .
Using a ruler, draw a single straight line starting at $(0, 90)$ and terminating at $(500, 390)$ .

Key Takeaways

A linear model of the form $y = mx + c$ can be plotted accurately by calculating the coordinates of its endpoints and joining them with a ruled line.

Common Mistakes

Drawing a freehand or unruled line instead of using a ruler.
Stopping the line short before reaching $500\,\text{kg}$ .
Starting at $(0, 0)$ or $(0, 60)$ instead of the $y$ -intercept $(0, 90)$ .

Things to Be Careful About

Ensure the line is drawn cleanly from $0$ to $500$ on the horizontal axis.

Techniques used

form linear equation from contextplot points and draw a straight line

(c)

Write the ratio $1.6\,\text{kg} : 600\,\text{g} : 2.4\,\text{kg}$ in its simplest form.

______ : ______ : ______

Q2Medium-EasyProbability

Fig. 2

The diagram shows a fair spinner numbered from 1 to 5.
The score is the number the spinner lands on.

(a)

The spinner is spun once.

Find the probability that the score is

2M

(i)

3

______

(ii)

even.

______

(b)

The spinner is spun twice.
The two scores are added together.

5M

(i)

Complete the possibility diagram to show all the outcomes.

Fig. 3

DifficultyMedium-Easy

Worked solution

Approach

The spinner is spun twice. We create a possibility diagram (grid) where the columns represent the first spin and the rows represent the second spin. Each cell contains the sum of the two scores.

Working

The grid headers are:

Columns (First spin): 1, 2, 3, 4, 5
Rows (Second spin): 1, 2, 3, 4, 5

We calculate the sum for each cell (Row value + Column value):

Row 1 (Second spin = 1): $1+1=2, 1+2=3, 1+3=4, 1+4=5, 1+5=6$ (Already given)
Row 2 (Second spin = 2): $2+1=3, 2+2=4, 2+3=5, 2+4=6, 2+5=7$ (Already given)
Row 3 (Second spin = 3): $3+1=4, 3+2=5, 3+3=6$ (Given). Missing: $3+4=7, 3+5=8$ .
Row 4 (Second spin = 4): $4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9$ .
Row 5 (Second spin = 5): $5+1=6, 5+2=7, 5+3=8, 5+4=9, 5+5=10$ .

Answer

The completed grid values for the blank cells are:

Row 3: 7, 8
Row 4: 5, 6, 7, 8, 9
Row 5: 6, 7, 8, 9, 10

Final answer

Grid completed with sums: Row 3 ends with 7, 8; Row 4 is 5, 6, 7, 8, 9; Row 5 is 6, 7, 8, 9, 10

Detailed explanation

Walkthrough

We are adding two scores from a spinner numbered 1-5. We use a grid (possibility diagram) to list all 25 possible outcomes. The column headers are the first spin (1-5) and row headers are the second spin (1-5). Each cell is the sum of the row and column headers.

Given:
Row 1: 2, 3, 4, 5, 6
Row 2: 3, 4, 5, 6, 7
Row 3 starts with: 4, 5, 6. We need to add 3 to 4 and 5: $3+4=7, 3+5=8$ .
Row 4 needs: $4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9$ .
Row 5 needs: $5+1=6, 5+2=7, 5+3=8, 5+4=9, 5+5=10$ .

Key Takeaways

A possibility diagram (grid) is a useful tool for visualizing all outcomes of two independent events. For addition, the value in each cell is the sum of the corresponding row and column headers.

Common Mistakes

Adding the wrong numbers (e.g., row + row instead of row + column).
Arithmetic errors in addition.
Not filling in all cells (the mark scheme awards partial credit for at least 7 correct values).

Things to Be Careful About

The grid must be fully completed for subsequent parts to use follow-through marks. Ensure all 25 cells are filled.

Techniques used

construct an addition grid for two spinscalculate the sum of first and second spin scoresfill in the missing cells in the sample space

(ii)

Find the probability that the outcome is 4.

______

(iii)

Find the probability that the outcome is greater than 6.

______

DifficultyMedium-Easy

Worked solution

Approach

We need to find the probability that the sum is greater than 6. We count the number of cells in the grid with values 7, 8, 9, or 10 and divide by 25.

Working

From the grid, the values greater than 6 are:

Row 2: 7 (1 value)
Row 3: 7, 8 (2 values)
Row 4: 7, 8, 9 (3 values)
Row 5: 7, 8, 9, 10 (4 values)

Total number of outcomes $> 6$ :

1 + 2 + 3 + 4 = 10

Alternatively, listing them:

Sum = 7: $(2,5), (3,4), (4,3), (5,2)$ -> 4 outcomes
Sum = 8: $(3,5), (4,4), (5,3)$ -> 3 outcomes
Sum = 9: $(4,5), (5,4)$ -> 2 outcomes
Sum = 10: $(5,5)$ -> 1 outcome
Total = $4 + 3 + 2 + 1 = 10$ .

\text{P}(\text{sum} > 6) = \frac{10}{25}

This can be simplified to $\frac{2}{5}$ , but $\frac{10}{25}$ is accepted.

Answer

\frac{10}{25} \quad \text{or} \quad \frac{2}{5}

Final answer

10/25

Detailed explanation

Walkthrough

We look for sums strictly greater than 6 in the grid. These are 7, 8, 9, 10.

Row 1 max is 6. (0 outcomes)
Row 2: only 7 is > 6. (1 outcome: 2+5)
Row 3: 7, 8 are > 6. (2 outcomes: 3+4, 3+5)
Row 4: 7, 8, 9 are > 6. (3 outcomes: 4+3, 4+4, 4+5)
Row 5: 7, 8, 9, 10 are > 6. (4 outcomes: 5+2, 5+3, 5+4, 5+5)
Total count = $1 + 2 + 3 + 4 = 10$ .
Probability = $\frac{10}{25}$ .

Key Takeaways

When counting outcomes satisfying a condition like $> 6$ , it's often easier to count from the diagonal or rows where the values exceed the threshold.

Common Mistakes

Including 6 in the count (question says 'greater than 6', not '6 or more').
Miscounting the number of cells (e.g., getting 8 or 12 instead of 10).
Not simplifying the fraction (though $\frac{10}{25}$ is often accepted in 4024 unless simplification is asked for, the mark scheme says 'oe').

Things to Be Careful About

The mark scheme allows follow-through from an incorrect grid (if they counted correctly from their wrong grid). However, for the correct answer, ensure all cells $>6$ are counted. $\frac{10}{25}$ is acceptable.

Techniques used

identify outcomes greater than 6 from the gridcount the number of such outcomescalculate probability as a fraction

Q3MediumNumber

(a)

The exchange rate between dollars ($) and Malaysian Ringgits (MYR) is $1 = 4.19 MYR.
The exchange rate between dollars ($) and Pakistani Rupees (PKR) is $1 = 179.12 PKR.

Find the exchange rate between Malaysian Ringgits and Pakistani Rupees.

1 MYR = ______ PKR

DifficultyMedium-Easy

Worked solution

Approach

The dollar connects the two other currencies. Since the same number of Malaysian Ringgits is equal to the same number of Pakistani Rupees, divide the PKR rate by the MYR rate to get the exchange rate from MYR to PKR.

Working

The two exchange rates both describe one dollar, so

4.19\,\text{MYR} = 179.12\,\text{PKR}

Divide both sides by $4.19$ :

1\,\text{MYR} = \frac{179.12}{4.19}\,\text{PKR}

Evaluating the division gives

1\,\text{MYR} = 42.75\,\text{PKR}

Answer

1 MYR = 42.75 PKR

Final answer

42.75 PKR

Detailed explanation

Walkthrough

The two exchange rates tell us the value of one dollar in two different currencies. Therefore, the same dollar amount buys $4.19$ MYR and also buys $179.12$ PKR. This means those two amounts have equal value, so $4.19\,\text{MYR} = 179.12\,\text{PKR}$ . To find how many PKR are equal to one MYR, divide the number of PKR by the number of MYR. So the required rate is $\frac{179.12}{4.19}$ , which works out to $42.75$ PKR per MYR.

Key Takeaways

Currency conversions can be done by treating both currencies as linked through a common base currency.
The order of the division determines the units of the answer.
In this case, the answer must be in PKR per MYR, not MYR per PKR.

Common Mistakes

Dividing $4.19$ by $179.12$ instead. This gives about $0.0234$ MYR per PKR, which is not what the question asks for.
Multiplying the two exchange rates together. Since one MYR is less than one dollar, the result must be smaller than $179.12$ PKR.
Writing the answer with the units reversed, such as writing MYR instead of PKR.

Things to Be Careful About

Make sure the factor placed in the denominator is the currency you are converting from: here it is MYR.
On the calculator component, it is acceptable to evaluate the fraction on the calculator, but you should still write down the substituted expression first.
Both given rates are to two decimal places, so the expected answer is also given to two decimal places: $42.75$ .

Techniques used

use a common base currency to link two other currenciesdivide the number of one currency by the number of the other currencycheck the order of the division

(b)

Farhad invests $1500 in an account paying compound interest at a rate of 4% per year.
Gulsan invests $1500 in an account paying simple interest at a rate of $x$ % per year.

Farhad and Gulsan have the same amount of money in their accounts at the end of 2 years.

Calculate the value of $x$ .

$x$ = ______

DifficultyMedium

Worked solution

Approach

First find Gulsan's balance? No, first find Farhad's balance using compound interest, then use the simple interest formula for Gulsan, equate the two balances, and solve for $x$ .

Working

Farhad's balance after 2 years is

1500\left(1+\frac{4}{100}\right)^2 = 1500(1.04)^2 = 1500 \times 1.0816 = 1622.40

So the interest Farhad earns over the two years is

1622.40 - 1500 = 122.40

Gulsan receives simple interest for two years, so her simple interest is

\frac{1500 \times x \times 2}{100} = 30x

Her balance at the end of two years is therefore

1500 + 30x

Since the balances are equal:

1500 + 30x = 1622.40

Subtract $1500$ from both sides:

30x = 122.40

Divide by $30$ :

x = \frac{122.40}{30} = 4.08

This is the same as the mark scheme formula:

x = \frac{1500\left(1+\frac{4}{100}\right)^2 - 1500}{1500 \times 2} \times 100 = 4.08

Answer

$x = 4.08$

Final answer

4.08

Detailed explanation

Walkthrough

For Farhad, compound interest means the balance is multiplied by $1.04$ each year. After two years it is $1500(1.04)^2 = 1622.40$ .

The interest Farhad has earned is the balance minus the original amount: $1622.40 - 1500 = 122.40$ .

For Gulsan, simple interest means interest is added in equal instalments. Each year she earns

\frac{1500x}{100} = 15x

so over two years she earns $30x$ . Her total balance is the original amount plus the interest:

1500 + 30x

Since Farhad and Gulsan finish with equal balances, the equation is:

1500 + 30x = 1622.40

Solving this gives $x = 4.08$ . The single mark-scheme formula does the same thing: it takes the interest earned by compound interest, divides by the product of the principal and the number of years, and multiplies by 100 to express the result as a percentage.

Key Takeaways

Compound interest uses a growth factor of $1 + \frac{r}{100}$ for each year.
Simple interest is linear: the interest is the same each year.
To compare two different interest schemes, set their final balances equal and solve.
A compound interest rate can be converted to an equivalent simple interest rate for the same period.

Common Mistakes

Using $1500 \times (0.04)^2$ for Farhad's balance. This omits the principal and gives the wrong final amount.
Wrongly using simple interest for Farhad too, for example $1500\left(1 + 2 \times \frac{4}{100}\right)$ . That would give $1620$ , not $1622.40$ .
Forgetting to add the original $1500$ to Gulsan's interest before setting up the equation.
Omitting the factor of 100 in the final rate calculation, which would give $0.0408$ instead of $4.08$ .
Rounding intermediate values such as $1.04^2$ too early.

Things to Be Careful About

The unknown $x$ is the percentage rate, so the final answer must be $4.08$ , not a decimal $0.0408$ and not a percentage sign attached as if it were a multiplication.
The time period for both investments is exactly 2 years, so the simple interest formula must use $2$ years.
Keep all amounts in the same unit, here the same currency, so that they can be equated directly.
This is a calculator component, but the substituted expressions must still be written out so each method step can be followed.

Techniques used

apply the compound interest growth factor for two yearsapply the simple interest formulaequate the two account balancessolve a linear equation for the unknown rate

Q4MediumAlgebra and Graphs

(a)

The diagrams show the first four patterns in a sequence.

Fig. 4

6M

(i)

Draw Pattern 5 on the grid below.

Fig. 5

DifficultyMedium-Easy

Worked solution

Approach

Pattern $n$ is a large equilateral triangle made of $n^2$ small triangles arranged in $n$ rows. The grey (shaded) triangles are the downward-pointing ones. In row $k$ (from the top), there are $k-1$ grey triangles and $k$ white triangles. For Pattern 5, there are 5 rows, giving $5^2 = 25$ small triangles in total, with 10 grey and 15 white.

Working

Answer

A large equilateral triangle composed of 25 small triangles in 5 rows, with the 10 downward-pointing triangles shaded grey and the 15 upward-pointing triangles left white.

Final answer

Pattern 5 drawn: 5 rows, 25 small triangles, 10 grey (downward-pointing), 15 white (upward-pointing)

Detailed explanation

Walkthrough

Each pattern $n$ forms a large equilateral triangle with $n$ rows. Row 1 (top) has 1 triangle, row 2 has 3, row 3 has 5, and row $k$ has $2k-1$ triangles. The total is $1 + 3 + 5 + \dots + (2n-1) = n^2$ . The grey triangles are always the downward-pointing ones. In row $k$ , there are $k-1$ downward-pointing triangles and $k$ upward-pointing triangles. For Pattern 5, row 1 has 1 white, row 2 has 1 grey and 2 white, row 3 has 2 grey and 3 white, row 4 has 3 grey and 4 white, and row 5 has 4 grey and 5 white. Summing these gives 10 grey and 15 white triangles.

Key Takeaways

In a triangular grid pattern, row $k$ contains $2k-1$ small triangles. Downward-pointing triangles are grey, upward-pointing are white. The total number of triangles in pattern $n$ is $n^2$ .

Common Mistakes

Counting the total number of small triangles incorrectly (e.g. counting only the upward-pointing ones).
Shading the wrong triangles (shading upward-pointing instead of downward-pointing).
Forgetting that row 1 has 0 grey triangles.

Things to Be Careful About

The answer is a drawing. Ensure the large triangle has exactly 5 rows and the correct number of shaded and unshaded small triangles. The grid provided is isometric, so the triangles must be drawn aligned with the grid lines.

Techniques used

recognise the visual pattern in the sequenceextend the shading rule to the next rowdraw the completed pattern on the grid

(ii)

Complete the table.

Pattern ( $n$ )	1	2	3	4
Total number of triangles	1	4	9	16
Number of grey triangles	0	1	3
Number of white triangles	1	3	6

DifficultyMedium-Easy

Worked solution

Approach

The total number of triangles follows the sequence $1, 4, 9, 16, \ldots$ , which is $n^2$ . The grey and white triangle counts can be extended by finding the first differences of their respective sequences.

Working

Total number of triangles:

\begin{aligned} \text{For } n=5: \quad 5^2 &= 25 \\ \text{For } n=6: \quad 6^2 &= 36 \end{aligned}

Number of grey triangles:
The sequence is $0, 1, 3, 6$ . The first differences are $1, 2, 3$ . The next differences are $4$ and $5$ .

\begin{aligned} \text{For } n=5: \quad 6 + 4 &= 10 \\ \text{For } n=6: \quad 10 + 5 &= 15 \end{aligned}

Number of white triangles:
The sequence is $1, 3, 6, 10$ . The first differences are $2, 3, 4$ . The next differences are $5$ and $6$ .

\begin{aligned} \text{For } n=5: \quad 10 + 5 &= 15 \\ \text{For } n=6: \quad 15 + 6 &= 21 \end{aligned}

Answer

Pattern ( $n$ )	1	2	3	4	5	6
Total number of triangles	1	4	9	16	25	36
Number of grey triangles	0	1	3	6	10	15
Number of white triangles	1	3	6	10	15	21

Final answer

Total: 25, 36; Grey: 10, 15; White: 15, 21

Detailed explanation

Walkthrough

The total number of triangles is given by $n^2$ , so for $n=5$ it is $25$ and for $n=6$ it is $36$ . For the grey triangles, the sequence is $0, 1, 3, 6$ . The differences between consecutive terms are $1, 2, 3$ , which increase by $1$ each time. The next differences are $4$ and $5$ , giving $6+4=10$ and $10+5=15$ . For the white triangles, the sequence is $1, 3, 6, 10$ . The differences are $2, 3, 4$ , increasing by $1$ . The next differences are $5$ and $6$ , giving $10+5=15$ and $15+6=21$ .

Key Takeaways

When a sequence does not have a constant first difference, look at the second differences. If the second differences are constant, the sequence is quadratic, and the first differences increase linearly.

Common Mistakes

Assuming the first differences are constant when they are not.
Forgetting to add the difference to the correct previous term.
Mixing up the grey and white triangle counts.

Things to Be Careful About

Ensure the table is completed for both $n=5$ and $n=6$ in all three rows. The marks are awarded for at least one correct row or column, but all should be filled.

Techniques used

calculate the total number of triangles for n=5 and n=6find the differences in the grey and white triangle countsextend the sequences to the 5th and 6th terms

(iii)

Write an expression, in terms of $n$ , for the total number of triangles in Pattern $n$ .

______

(iv)

Write an expression, in terms of $n$ , for the number of white triangles in Pattern $n$ .

______

DifficultyMedium

Worked solution

Approach

The number of white triangles is $1, 3, 6, 10, 15, 21$ . The first differences are $2, 3, 4, 5, 6$ and the second difference is constant at $1$ . This indicates a quadratic sequence of the form $an^2 + bn + c$ .

Working

The second difference is $2a = 1$ , so $a = \frac{1}{2}$ .
The expression is $\frac{1}{2}n^2 + bn + c$ .
For $n=1$ : $\frac{1}{2}(1)^2 + b(1) + c = 1 \implies b + c = \frac{1}{2}$
For $n=2$ : $\frac{1}{2}(2)^2 + b(2) + c = 3 \implies 2 + 2b + c = 3 \implies 2b + c = 1$

Subtracting the first equation from the second:

(2b + c) - (b + c) = 1 - \frac{1}{2}

b = \frac{1}{2}

Substituting $b = \frac{1}{2}$ into $b + c = \frac{1}{2}$ :

\frac{1}{2} + c = \frac{1}{2} \implies c = 0

So the expression is:

\frac{1}{2}n^2 + \frac{1}{2}n

Alternatively, this is the sum of the first $n$ natural numbers: $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$ .

Answer

\frac{1}{2}n^2 + \frac{1}{2}n

Final answer

1/2 n^2 + 1/2 n

Detailed explanation

Walkthrough

The sequence of white triangles is $1, 3, 6, 10$ . The first differences are $2, 3, 4$ and the second difference is $1$ . For a quadratic sequence $an^2 + bn + c$ , the second difference is $2a$ , so $2a = 1 \implies a = \frac{1}{2}$ . Substituting $n=1$ and $n=2$ into $\frac{1}{2}n^2 + bn + c$ gives two equations: $b + c = \frac{1}{2}$ and $2b + c = 1$ . Solving these yields $b = \frac{1}{2}$ and $c = 0$ . The expression is $\frac{1}{2}n^2 + \frac{1}{2}n$ , which can also be written as $\frac{n(n+1)}{2}$ .

Key Takeaways

For a quadratic sequence, the coefficient of $n^2$ is half the second difference. The sequence of triangular numbers $1, 3, 6, 10, \ldots$ has the formula $\frac{n(n+1)}{2}$ .

Common Mistakes

Forgetting to halve the second difference to find $a$ .
Making arithmetic errors when solving the simultaneous equations for $b$ and $c$ .
Writing $\frac{1}{2}n^2 + \frac{1}{2}n + 1$ or other incorrect constant terms.

Things to Be Careful About

The mark scheme accepts $\frac{1}{2}n^2 + \frac{1}{2}n$ or equivalent forms like $\frac{n(n+1)}{2}$ . Ensure the expression is in terms of $n$ .

Techniques used

find the first and second differences of the white triangle sequenceuse the second difference to determine the quadratic coefficientsolve for the linear and constant terms

(b)

The 3rd term of a linear sequence is 34.
The 8th term of the same linear sequence is 14.

3M

(i)

Find the value of the first term of this sequence.

______

(ii)

Find the value of the first negative term of this sequence.

______

Q5MediumTransformations and VectorsMensuration

Fig. 6

(a)

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

______

DifficultyMedium-Easy

Worked solution

Approach

Compare the coordinates of corresponding vertices of shapes $A$ and $B$ to determine the transformation. A reflection in the line $y = -x$ maps any point $(x, y)$ to $(-y, -x)$ .

Working

Map the vertices of shape $A$ using the rule $(x, y) \to (-y, -x)$ :

\begin{aligned} (-2, 4) &\to (-4, 2) \\ (-1, 4) &\to (-4, 1) \\ (-1, 6) &\to (-6, 1) \\ (-2, 5) &\to (-5, 2) \end{aligned}

These image coordinates match the vertices of shape $B$ (allowing for the standard grid interpretation of the trapezoid's vertices).

Answer

Reflection in the line $y = -x$

Final answer

Reflection in the line y = -x

Detailed explanation

Walkthrough

To find the single transformation mapping shape $A$ onto shape $B$ , we compare their vertices. Shape $A$ has vertices at $(-2, 4)$ , $(-1, 4)$ , $(-1, 6)$ , and $(-2, 5)$ . Shape $B$ has vertices at $(-4, 2)$ , $(-4, 1)$ , $(-6, 1)$ , and $(-5, 2)$ . Notice that each $x$ -coordinate of $A$ becomes the negative of the $y$ -coordinate of $B$ , and each $y$ -coordinate of $A$ becomes the negative of the $x$ -coordinate of $B$ . This is exactly the rule for a reflection in the line $y = -x$ , which maps $(x, y)$ to $(-y, -x)$ . Since the shapes are congruent and oriented in a way that requires a flip across this diagonal, the transformation is a reflection.

Key Takeaways

A reflection in the line $y = -x$ swaps the coordinates and negates both: $(x, y) \to (-y, -x)$ . Always verify by mapping at least two vertices to confirm the transformation.

Common Mistakes

Stating only "reflection" without giving the equation of the mirror line; the mark scheme requires both.
Confusing the line $y = -x$ with $y = x$ (which maps $(x, y) \to (y, x)$ ).
Forgetting to negate both coordinates when reflecting in $y = -x$ .

Things to Be Careful About

A full description of a transformation must include the type of transformation and the specific line, point, or vector involved. For reflections, this means stating the equation of the mirror line. Here, $y = -x$ is the mirror line.

Techniques used

identify the mirror line by comparing corresponding verticesverify the transformation using coordinate rules

(b)

Shape $A$ is mapped onto shape $C$ by an enlargement of scale factor 3.
Two of the vertices of shape $C$ are $(2, 5)$ and $(5, 2)$ .

4M

(i)

Find the coordinates of the centre of the enlargement.

( ______ , ______ )

DifficultyMedium

Worked solution

Approach

Let the centre of enlargement be $(h, k)$ . The formula for an enlargement with scale factor $k_s = 3$ maps a point $(x, y)$ to $(h + 3(x - h), k + 3(y - k))$ . We use two known vertices of shape $A$ and their corresponding images in shape $C$ to find $h$ and $k$ .

Working

Shape $A$ has vertices including $(-2, 5)$ and $(-1, 4)$ . These map to $(2, 5)$ and $(5, 2)$ in shape $C$ respectively.

For the vertex $(-2, 5) \to (2, 5)$ :

\begin{aligned} h + 3(-2 - h) &= 2 \\ k + 3(5 - k) &= 5 \end{aligned}

Solving the $x$ -equation:

\begin{aligned} h - 6 - 3h &= 2 \\ -2h &= 8 \\ h &= -4 \end{aligned}

Solving the $y$ -equation:

\begin{aligned} k + 15 - 3k &= 5 \\ -2k &= -10 \\ k &= 5 \end{aligned}

The centre of enlargement is $(-4, 5)$ .

Answer

$(-4, 5)$

Final answer

(-4, 5)

Detailed explanation

Walkthrough

We are given that shape $A$ is enlarged by a scale factor of $3$ to produce shape $C$ , and we know two corresponding vertices: $(-2, 5)$ maps to $(2, 5)$ , and $(-1, 4)$ maps to $(5, 2)$ . Let the centre of enlargement be $(h, k)$ . The general formula for the image of a point $(x, y)$ under an enlargement with centre $(h, k)$ and scale factor $3$ is $(h + 3(x - h), k + 3(y - k))$ . Substituting the known point $(-2, 5)$ and its image $(2, 5)$ gives two equations: $h + 3(-2 - h) = 2$ and $k + 3(5 - k) = 5$ . Solving these yields $h = -4$ and $k = 5$ . We can verify this with the second pair: $(-4 + 3(-1 - (-4)), 5 + 3(4 - 5)) = (-4 + 9, 5 - 3) = (5, 2)$ , which matches.

Key Takeaways

The enlargement formula $(x', y') = (h + k_s(x - h), k + k_s(y - k))$ allows you to find the centre of enlargement if you know the scale factor and at least one pair of corresponding points.

Common Mistakes

Assuming the centre is the origin or misreading the coordinates from the grid.
Forgetting to subtract the centre coordinates before multiplying by the scale factor.
Not verifying the answer with the second pair of vertices.

Things to Be Careful About

The scale factor is $3$ , not $\frac{1}{3}$ . Ensure you are solving for $(h, k)$ correctly and not confusing it with the image coordinates. The mark scheme accepts partial credit for identifying that the $x$ -coordinate is less than $-2$ or the $y$ -coordinate is $5$ .

Techniques used

use the enlargement formula to set up equations for the centre coordinatessolve simultaneous equations to find the centre

(ii)

Find the area of shape $C$ .

______ $\text{units}^2$

DifficultyMedium-Easy

Worked solution

Approach

First, find the area of shape $A$ by counting grid squares or decomposing it into simpler shapes. Then, use the area scale factor, which is the square of the linear scale factor ( $3^2 = 9$ ), to find the area of shape $C$ .

Working

Shape $A$ has vertices at $(-2, 4)$ , $(-1, 4)$ , $(-1, 6)$ , and $(-2, 5)$ . It can be seen as a $1 \times 2$ rectangle (area $= 2$ ) with a right-angled triangle of base $1$ and height $1$ (area $= 0.5$ ) removed from the top-left corner.

\text{Area of } A = 2 - 0.5 = 1.5 \text{ units}^2

The linear scale factor is $3$ , so the area scale factor is $3^2 = 9$ .

\text{Area of } C = 9 \times 1.5 = 13.5 \text{ units}^2

Alternatively, using the trapezium formula on shape $C$ with parallel sides $3$ and $6$ and height $3$ :

\text{Area} = \frac{1}{2}(3 + 6) \times 3 = 13.5

Answer

$13.5$

Final answer

13.5

Detailed explanation

Walkthrough

To find the area of shape $C$ , we first calculate the area of shape $A$ . Shape $A$ is a quadrilateral that can be viewed as a $1 \times 2$ rectangle (from $x = -2$ to $-1$ and $y = 4$ to $6$ ) minus a right-angled triangle in the top-left corner (vertices $(-2, 5)$ , $(-2, 6)$ , $(-1, 6)$ ). The rectangle has area $2$ , and the triangle has area $\frac{1}{2} \times 1 \times 1 = 0.5$ . Thus, the area of $A$ is $1.5$ . Since shape $C$ is an enlargement of $A$ with a linear scale factor of $3$ , its area is multiplied by $3^2 = 9$ . Therefore, the area of $C$ is $1.5 \times 9 = 13.5$ .

Key Takeaways

When a shape is enlarged by a linear scale factor $k$ , its area is multiplied by $k^2$ . Always calculate the original area first if it is not given.

Common Mistakes

Forgetting to square the scale factor when finding the area scale factor (using $3$ instead of $9$ ).
Miscalculating the area of the original shape by miscounting grid squares or misidentifying the decomposition.
Rounding the answer unnecessarily; $13.5$ is exact.

Things to Be Careful About

The question asks for the area in $\text{units}^2$ . Ensure your final answer is in the correct form. The mark scheme accepts $13\frac{1}{2}$ or $13.5$ . Working must be shown to earn method marks for the area calculation or the application of the scale factor.

Techniques used

calculate the area of the original shape using grid squares or decompositionapply the area scale factor (square of the linear scale factor)

(c)

Transformation T is represented by the matrix

\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

Transformation T maps shape $A$ onto shape $D$ .

5M

(i)

On the grid, draw shape $D$ .

DifficultyMedium-Easy

Worked solution

Approach

The transformation matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ maps a point $\begin{pmatrix} x \\ y \end{pmatrix}$ to $\begin{pmatrix} -y \\ x \end{pmatrix}$ . Apply this to each vertex of shape $A$ to find the vertices of shape $D$ , then draw it on the grid.

Working

Vertices of shape $A$ : $(-2, 4)$ , $(-1, 4)$ , $(-1, 6)$ , $(-2, 5)$ .

\begin{aligned} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -2 \\ 4 \end{pmatrix} &= \begin{pmatrix} -4 \\ -2 \end{pmatrix} \\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 \\ 4 \end{pmatrix} &= \begin{pmatrix} -4 \\ -1 \end{pmatrix} \\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 \\ 6 \end{pmatrix} &= \begin{pmatrix} -6 \\ -1 \end{pmatrix} \\ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -2 \\ 5 \end{pmatrix} &= \begin{pmatrix} -5 \\ -2 \end{pmatrix} \end{aligned}

The vertices of shape $D$ are $(-4, -2)$ , $(-4, -1)$ , $(-6, -1)$ , and $(-5, -2)$ .

Answer

Shape $D$ is drawn with vertices at $(-6, -1)$ , $(-4, -1)$ , $(-5, -2)$ , and $(-4, -2)$ .

Final answer

Shape D with vertices at (-6, -1), (-4, -1), (-5, -2), (-4, -2)

Detailed explanation

Walkthrough

The matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ represents a standard rotation. To find the image of shape $A$ , we multiply this matrix by the column vector of each vertex. For a point $(x, y)$ , the product is $\begin{pmatrix} 0 \cdot x + (-1) \cdot y \\ 1 \cdot x + 0 \cdot y \end{pmatrix} = \begin{pmatrix} -y \\ x \end{pmatrix}$ . Applying this to the vertices of $A$ : $(-2, 4) \to (-4, -2)$ , $(-1, 4) \to (-4, -1)$ , $(-1, 6) \to (-6, -1)$ , and $(-2, 5) \to (-5, -2)$ . Plotting these points and joining them in order gives shape $D$ .

Key Takeaways

Matrix multiplication $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$ is used to find image coordinates. The matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is a standard rotation matrix for $90^\circ$ anticlockwise about the origin.

Common Mistakes

Making sign errors when multiplying the matrix (e.g., forgetting the negative sign on the $y$ -coordinate).
Plotting the points incorrectly on the grid, especially in the third quadrant.
Not joining the vertices in the correct order to form the shape.

Things to Be Careful About

The question asks to draw shape $D$ on the grid. Ensure the vertices are plotted accurately to within half a grid square. The mark scheme awards marks for at least three correct vertices or three correct pairs of coordinates.

Techniques used

apply the transformation matrix to each vertex of shape Aplot the resulting vertices to draw shape D

(ii)

Describe fully the single transformation represented by the matrix

\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

______

DifficultyMedium-Easy

Worked solution

Approach

The matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is a standard rotation matrix. Recall that a rotation of $\theta$ anticlockwise about the origin is given by $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ . Match the entries to find $\theta$ .

Working

\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

This gives $\cos\theta = 0$ and $\sin\theta = 1$ , so $\theta = 90^\circ$ .

Since the matrix is about the origin, the centre is $(0, 0)$ .

Answer

Rotation of $90^\circ$ anticlockwise about $(0, 0)$

Final answer

Rotation 90 degrees anticlockwise about (0, 0)

Detailed explanation

Walkthrough

The given matrix is $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ . We compare this to the general rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ . From the top-left and bottom-right entries, $\cos\theta = 0$ . From the bottom-left entry, $\sin\theta = 1$ . The angle $\theta$ that satisfies both is $90^\circ$ . Since the sine is positive and cosine is zero, the rotation is anticlockwise. The absence of translation terms in the matrix means the centre of rotation is the origin $(0, 0)$ .

Key Takeaways

Standard rotation matrices are: $90^\circ$ anticlockwise: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ ; $180^\circ$ : $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ ; $90^\circ$ clockwise: $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ . Memorizing these saves time and reduces errors.

Common Mistakes

Confusing $90^\circ$ anticlockwise with $90^\circ$ clockwise (the signs of the off-diagonal entries are swapped).
Forgetting to state the centre of rotation; a full description requires the type, angle, direction, and centre.
Stating "rotation about the origin" without specifying the angle and direction.

Things to Be Careful About

The mark scheme requires three specific pieces of information: "Rotation", " $90^\circ$ anticlockwise", and " $(0, 0)$ ". Missing any of these will cost a mark. Ensure you write "anticlockwise" and not just "counterclockwise" if the mark scheme is strict, though both are generally accepted. The angle must include the degree symbol.

Techniques used

recognize the standard rotation matrixidentify the angle and direction of rotationstate the centre of rotation

Q6Medium-EasyNumberProbability

(a)

In the Venn diagram, shade the region represented by $P \cup Q$ .

Fig. 7

(b)

Use set notation to describe the shaded region in the Venn diagram.

Fig. 8

______

(c)

$\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
$A = \{x : x \text{ is a factor of } 40\}$
$B = \{x : x \text{ is an odd number\}}$

4M

(i)

Complete the Venn diagram.

Fig. 9

DifficultyMedium-Easy

Worked solution

Approach

Determine the elements of sets $A$ and $B$ from the universal set $\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , and place each element in the correct region of the Venn diagram.

Working

List the elements belonging to each definition within $\mathscr{E}$ :

Universal set: $\mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
Factors of $40$ in $\mathscr{E}$ : $A = \{1, 2, 4, 5, 8, 10\}$
Odd numbers in $\mathscr{E}$ : $B = \{1, 3, 5, 7, 9\}$

Now distribute the elements into the four regions:

Intersection $A \cap B$ (factors of 40 that are odd):

A \cap B = \{1, 5\}

$A$ only ( $A \cap B'$ ) (factors of 40 that are even):

\{2, 4, 8, 10\}

$B$ only ( $A' \cap B$ ) (odd numbers that are not factors of 40):

\{3, 7, 9\}

Outside both circles ( $(A \cup B)'$ ) (numbers that are neither factors of 40 nor odd, i.e., even non-factors: $6$ ):
Note that checking $6$ gives outside both circles: $\{6\}$ .

Checking all 10 elements:

Circle $A$ only: $10, 2, 8, 4$
Intersection: $1, 5$
Circle $B$ only: $3, 7, 9$
Outside both: $6$

Answer

Final answer

Venn diagram completed with: A only: 2, 4, 8, 10; Intersection: 1, 5; B only: 3, 7, 9; Outside: 6

Detailed explanation

Walkthrough

Identify the universal set: integers from $1$ to $10$ .
Identify elements of $A$ (factors of 40 in the range): $1, 2, 4, 5, 8, 10$ .
Identify elements of $B$ (odd numbers in the range): $1, 3, 5, 7, 9$ .
Find the overlap $A \cap B$ : elements appearing in both lists are $1$ and $5$ .
Place the remaining elements of $A$ ( $2, 4, 8, 10$ ) in the ' $A$ only' region (since $10$ is already printed, add $2, 4, 8$ ).
Place the remaining elements of $B$ ( $7, 9$ ) in the ' $B$ only' region (since $3$ is already printed, add $7, 9$ ).
Place the remaining elements of $\mathscr{E}$ that are in neither set ( $6$ ) outside both circles.

Key Takeaways

Every element of the universal set must appear exactly once in the Venn diagram.
Identifying the intersection first makes placing the remaining elements straightforward.

Common Mistakes

Forgetting numbers that belong outside both circles (e.g. omitting $6$ ).
Misidentifying factors of $40$ (e.g. thinking $4$ is not a factor or including multiples instead).

Things to Be Careful About

Ensure no elements from $\mathscr{E}$ are omitted or repeated.

Techniques used

list set elements from descriptionsplace elements into correct Venn diagram regions

(ii)

List the elements of $A' \cap B$ .

______

(iii)

One element of $\mathscr{E}$ is chosen at random.

Find the probability that this element is in $A \cap B$ .

______

Q7MediumAlgebra and Graphs

(a)

Solve $\frac{x}{3} = 7$ .

$x$ = ______

(b)

Solve $6x - 5 = 2(x + 3)$ .

$x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

First, expand the bracket on the right-hand side. Then, collect all $x$ terms on one side and constant terms on the other. Finally, divide to find $x$ .

Working

Given:

6x - 5 = 2(x + 3)

Expand the bracket:

6x - 5 = 2x + 6

Subtract $2x$ from both sides to group $x$ terms:

6x - 2x - 5 = 6

4x - 5 = 6

Add $5$ to both sides to group constants:

4x = 6 + 5

4x = 11

Divide by $4$ :

x = \frac{11}{4}

Convert to a mixed number:

x = 2\frac{3}{4}

Answer

x = 2\frac{3}{4}

Final answer

2 3/4

Detailed explanation

Walkthrough

The equation is $6x - 5 = 2(x + 3)$ .

Expand: Multiply the term outside the bracket ( $2$ ) by each term inside ( $x$ and $3$ ). This gives $2x + 6$ . The equation becomes $6x - 5 = 2x + 6$ .
Collect x terms: Subtract $2x$ from both sides. $6x - 2x$ leaves $4x$ on the left. The equation is now $4x - 5 = 6$ .
Collect constants: Add $5$ to both sides. $-5$ cancels out on the left, and $6 + 5$ becomes $11$ on the right. The equation is $4x = 11$ .
Solve: Divide by $4$ . $x = \frac{11}{4}$ .
Format: The mark scheme accepts improper fractions or mixed numbers. $\frac{11}{4}$ is $2$ with a remainder of $3$ , so $2\frac{3}{4}$ .

Key Takeaways

When solving linear equations with brackets: expand first, then move variables to one side and numbers to the other. Always check signs when moving terms across the equals sign.

Common Mistakes

Forgetting to distribute the multiplier to both terms inside the bracket (e.g., writing $2x + 3$ ). Incorrectly combining negative and positive numbers (e.g., $6 - 5 = 1$ instead of adding $5$ to $6$ ).

Things to Be Careful About

The question asks for $x$ . Ensure the final answer is simplified. $2\frac{3}{4}$ is preferred over $2.75$ if the mark scheme specifies 'oe' (otherwise equivalent), but $\frac{11}{4}$ is also correct. Note that $M1$ is awarded for setting up $6x - 2x = 5 + 6$ .

Techniques used

expand bracketscollect like termssolve for the variable

(c)

Factorise $3x^2 - 2x - 8$ .

______

DifficultyMedium

Worked solution

Approach

We need to factorise $3x^2 - 2x - 8$ . Since the coefficient of $x^2$ is $3$ (not $1$ ), we look for two binomials $(3x + m)(x + n)$ such that $m \times n = -8$ and the cross-terms sum to $-2x$ . Alternatively, use the 'splitting the middle term' method.

Working

Method: Splitting the middle term.
Find two numbers that multiply to $a \times c = 3 \times (-8) = -24$ and add to $b = -2$ .

Factors of $-24$ :
$1, -24$ (sum $-23$ )
$2, -12$ (sum $-10$ )
$3, -8$ (sum $-5$ )
$4, -6$ (sum $-2$ ) -> These are the numbers.

Rewrite $-2x$ as $+4x - 6x$ :

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

Group the terms:

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

Factor out the common bracket $(3x + 4)$ :

(3x + 4)(x - 2)

Answer

(3x + 4)(x - 2)

Final answer

(3x + 4)(x - 2)

Detailed explanation

Walkthrough

To factorise $3x^2 - 2x - 8$ , we look for factors of the form $(3x + A)(x + B)$ . Expanding this gives $3x^2 + 3Bx + Ax + AB = 3x^2 + (3B+A)x + AB$ .
We need $AB = -8$ and $3B + A = -2$ .
Alternatively, using the product-sum method: multiply $a \cdot c = 3 \cdot -8 = -24$ . Find factors of $-24$ that add to $-2$ . These are $4$ and $-6$ .
Split the middle term: $3x^2 + 4x - 6x - 8$ .
Factor by grouping: $x(3x + 4) - 2(3x + 4)$ .
Result: $(3x + 4)(x - 2)$ .

Key Takeaways

For quadratics $ax^2 + bx + c$ , find two numbers that multiply to $ac$ and add to $b$ . Use these to split the middle term, then factor by grouping.

Common Mistakes

Choosing factors of $-8$ directly without considering the coefficient $3$ . For example, trying $(x-2)(x+4)$ which expands to $x^2+2x-8$ , missing the $3x^2$ term. Sign errors when splitting the middle term.

Things to Be Careful About

Check your answer by expanding it back out. $(3x+4)(x-2) = 3x^2 - 6x + 4x - 8 = 3x^2 - 2x - 8$ . Correct. The mark scheme awards M1 for seeing the correct structure or partial factorisation.

Techniques used

factorise a quadratic expression with a leading coefficient greater than 1

(d)

$(ax + b)^2 = 4x^2 - 12x + c$

Given that $a > 0$ , find the value of each of $a$ , $b$ and $c$ .

$a$ = ______
$b$ = ______
$c$ = ______

DifficultyMedium

Worked solution

Approach

Expand the left-hand side $(ax + b)^2$ and compare the coefficients of the resulting terms with the right-hand side $4x^2 - 12x + c$ .

Working

Expand $(ax + b)^2$ :

(ax + b)^2 = a^2x^2 + 2abx + b^2

Equate this to $4x^2 - 12x + c$ :

a^2x^2 + 2abx + b^2 = 4x^2 - 12x + c

Compare coefficients of corresponding powers of $x$ :

Coefficient of $x^2$ : $a^2 = 4$
Since $a > 0$ , we have $a = 2$ .
Coefficient of $x$ : $2ab = -12$
Substitute $a = 2$ :
$2(2)b = -12$ $4b = -12$ $b = -3$
Constant term: $b^2 = c$
Substitute $b = -3$ :
$c = (-3)^2$ $c = 9$

Answer

a = 2, \quad b = -3, \quad c = 9

Final answer

a=2, b=-3, c=9

Detailed explanation

Walkthrough

The identity is $(ax + b)^2 = 4x^2 - 12x + c$ .
First, expand the left side: $(ax+b)(ax+b) = a^2x^2 + 2abx + b^2$ .
Now match the coefficients with the right side $4x^2 - 12x + c$ .

The $x^2$ term: $a^2 = 4$ . Since $a>0$ , $a=2$ .
The $x$ term: $2ab = -12$ . With $a=2$ , $4b = -12$ , so $b=-3$ .
The constant term: $b^2 = c$ . With $b=-3$ , $c = (-3)^2 = 9$ .

Key Takeaways

When two polynomials are equal for all values of $x$ , their corresponding coefficients must be equal. Expanding a perfect square $(A+B)^2 = A^2 + 2AB + B^2$ is a key skill.

Common Mistakes

Taking $a = -2$ despite the condition $a > 0$ . Calculating $b$ incorrectly from $2ab = -12$ . Forgetting to square $b$ to find $c$ (e.g., saying $c=-3$ ). Not squaring the negative sign for $b$ when finding $c$ .

Things to Be Careful About

Pay close attention to the condition $a > 0$ . If $a$ were allowed to be negative, $b$ would change sign accordingly, but $c$ would remain the same. Ensure you follow the follow-through marks if an earlier value was wrong, but here we calculate exact values.

Techniques used

expand a squared binomialcompare coefficients

Q8MediumMensurationTrigonometryNumber

(a)

[Volume of a cone = $\frac{1}{3}\pi r^2 h$ ]
[Volume of a sphere = $\frac{4}{3}\pi r^3$ ]
[Curved surface area of cone = $\pi r l$ ]
[Surface area of a sphere = $4\pi r^2$ ]

Fig. 10

A solid is formed by placing a cone on top of a hemisphere.
The cone and the hemisphere each have radius $r\,\text{cm}$ .
The height of the solid is $10\,\text{cm}$ .
The curved surface area of the hemisphere is $145\,\text{cm}^2$ .

11M

(i)

Show that $r = 4.80$ , correct to 2 decimal places.

DifficultyMedium-Easy

Worked solution

Approach

The curved surface area of a full sphere is $4\pi r^2$ , so the curved surface area of a hemisphere is $2\pi r^2$ . Set this equal to $145\,\text{cm}^2$ and solve for $r$ , showing intermediate decimal places before rounding.

Working

2\pi r^2 = 145

r^2 = \frac{145}{2\pi} \approx 23.07746...

r = \sqrt{\frac{145}{2\pi}} \approx 4.8039...

Rounding to $2$ decimal places:

r = 4.80\,\text{cm}

Answer

r = 4.80

Final answer

4.80

Detailed explanation

Walkthrough

The formula for the surface area of a sphere is $4\pi r^2$ . Since a hemisphere is half of a sphere, its curved surface area is $\frac{1}{2} \times 4\pi r^2 = 2\pi r^2$ .
Set up the equation $2\pi r^2 = 145$ .
Divide both sides by $2\pi$ to find $r^2 = \frac{145}{2\pi} \approx 23.0775$ .
Take the positive square root to find $r = \sqrt{23.0775} \approx 4.8039\,\text{cm}$ .
Rounding to 2 decimal places yields $4.80$ , matching the required value.

Key Takeaways

For "show that" questions, you must show unrounded intermediate values (at least 3 or 4 significant figures or decimal places) before stating the final rounded answer to prove you arrived at the result legitimately.

Common Mistakes

Using $4\pi r^2 = 145$ (forgetting to divide the sphere's surface area by 2 for a hemisphere).
Not showing the unrounded value (e.g. $4.803...$ or $4.804...$ ) before writing $4.80$ .

Things to Be Careful About

Ensure you divide by the product $(2\pi)$ on your calculator rather than dividing by $2$ and multiplying by $\pi$ .

Techniques used

equate hemisphere surface area formula to given valuesolve for radiusround to required decimal places

(ii)

Calculate the volume of the solid.

______ $\text{cm}^3$

DifficultyMedium

Worked solution

Approach

The solid is composed of a cone on top of a hemisphere.

Find the height of the cone, $h$ , by subtracting the radius of the hemisphere from the total height of $10\,\text{cm}$ .
Calculate the volume of the cone using $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$ .
Calculate the volume of the hemisphere using $V_{\text{hemisphere}} = \frac{2}{3}\pi r^3$ .
Add the two volumes together to find the total volume.

Working

Height of the cone:

h = 10 - 4.80 = 5.20\,\text{cm}

Volume of the cone:

V_{\text{cone}} = \frac{1}{3}\pi (4.80)^2(5.20) \approx 125.453\,\text{cm}^3

Volume of the hemisphere:

V_{\text{hemisphere}} = \frac{2}{3}\pi (4.80)^3 \approx 231.606\,\text{cm}^3

Total volume:

V = V_{\text{cone}} + V_{\text{hemisphere}} = 125.453 + 231.606 = 357.059\ldots\,\text{cm}^3

Rounding to 3 significant figures gives $357\,\text{cm}^3$ (or $358\,\text{cm}^3$ if using more precise unrounded values of $r$ ).

Answer

357\,\text{cm}^3

Final answer

357

Detailed explanation

Walkthrough

The total height is $10\,\text{cm}$ . The hemisphere extends downward from the circular boundary by its radius, $r = 4.80\,\text{cm}$ . Therefore, the vertical height of the cone is $h = 10 - 4.80 = 5.20\,\text{cm}$ .
The volume of a cone is $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times (4.80)^2 \times 5.20 \approx 125.45\,\text{cm}^3$ .
The volume of a hemisphere is half the volume of a sphere: $\frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \pi \times (4.80)^3 \approx 231.61\,\text{cm}^3$ .
Summing both components gives $125.45 + 231.61 = 357.06\,\text{cm}^3 \approx 357\,\text{cm}^3$ .

Key Takeaways

The height of a hemisphere from its flat circular base to its lowest point equals its radius $r$ .
The total volume of a compound solid is found by summing the individual standard solid volumes.

Common Mistakes

Using $h = 10\,\text{cm}$ as the height of the cone instead of subtracting $r$ .
Using the formula for a full sphere ( $\frac{4}{3}\pi r^3$ ) instead of a hemisphere ( $\frac{2}{3}\pi r^3$ ).

Things to Be Careful About

Ensure accurate intermediate values are carried through calculation to prevent premature rounding errors.

Techniques used

find cone height by subtractioncalculate cone volumecalculate hemisphere volumesum individual volumes

(iii)

Calculate the curved surface area of the cone.

______ $\text{cm}^2$

DifficultyMedium

Worked solution

Approach

Find the slant height $l$ of the cone using Pythagoras' theorem: $l = \sqrt{r^2 + h^2}$ .
Compute the curved surface area of the cone using the formula $A = \pi r l$ .

Working

Slant height $l$ :

l = \sqrt{4.80^2 + 5.20^2} = \sqrt{23.04 + 27.04} = \sqrt{50.08} \approx 7.07672\,\text{cm}

Curved surface area of the cone:

A = \pi r l = \pi \times 4.80 \times 7.07672 \approx 106.719\ldots\,\text{cm}^2

Rounding to 3 significant figures gives $107\,\text{cm}^2$ .

Answer

107\,\text{cm}^2

Final answer

107

Detailed explanation

Walkthrough

The formula for the curved surface area of a cone is $\pi r l$ , where $l$ is the slant height.
The vertical height of the cone is $h = 5.20\,\text{cm}$ and the base radius is $r = 4.80\,\text{cm}$ .
Using Pythagoras' theorem on the right-angled cross-section of the cone:

l = \sqrt{r^2 + h^2} = \sqrt{4.80^2 + 5.20^2} = \sqrt{50.08} \approx 7.0767\,\text{cm}

Substituting $r$ and $l$ into the curved surface area formula:

\text{Area} = \pi \times 4.80 \times 7.0767 \approx 106.72\,\text{cm}^2

Rounded to 3 significant figures, this gives $107\,\text{cm}^2$ (or $106.7\,\text{cm}^2$ to 4 s.f.).

Key Takeaways

The formula $\pi r l$ uses the slant height $l$ , not the perpendicular height $h$ .
Slant height is found via $l = \sqrt{r^2 + h^2}$ .

Common Mistakes

Using the perpendicular height $h = 5.20\,\text{cm}$ directly in place of $l$ .
Using the total solid height $10\,\text{cm}$ in Pythagoras' theorem instead of the cone's height $5.20\,\text{cm}$ .

Things to Be Careful About

Retain sufficient decimal places for $l$ before multiplying by $\pi \times r$ .

Techniques used

use Pythagoras' theorem to find slant heightcalculate curved surface area of cone

(b)

Fig. 11

The diagram shows two sectors of circles.
Sector $A$ has angle $x$ and radius $y$ .

The angle of sector $B$ is 20% smaller than the angle of sector $A$ .
The radius of sector $B$ is 20% longer than the radius of sector $A$ .

Calculate the area of sector $B$ as a percentage of the area of sector $A$ .

______ %

DifficultyMedium

Worked solution

Approach

Express the angle and radius of sector $B$ in terms of $x$ and $y$ .
Write down the area formulas for sector $A$ and sector $B$ .
Compute the ratio $\frac{\text{Area of Sector } B}{\text{Area of Sector } A} \times 100\%$ .

Working

For sector $A$ :

Angle $= x$
Radius $= y$

\text{Area}_A = \frac{x}{360} \times \pi y^2

For sector $B$ :

Angle is $20\%$ smaller: $\theta_B = (1 - 0.20)x = 0.8x$
Radius is $20\%$ longer: $r_B = (1 + 0.20)y = 1.2y$

\text{Area}_B = \frac{0.8x}{360} \times \pi (1.2y)^2 = \frac{0.8x}{360} \times \pi \times 1.44y^2 = 0.8 \times 1.44 \times \left(\frac{x}{360}\pi y^2\right)

\text{Area}_B = 1.152 \times \text{Area}_A

Calculating $\text{Area}_B$ as a percentage of $\text{Area}_A$ :

\frac{\text{Area}_B}{\text{Area}_A} \times 100\% = 1.152 \times 100\% = 115.2\%

Answer

115.2\%

Final answer

115.2

Detailed explanation

Walkthrough

The area of any sector with angle $\theta$ and radius $r$ is given by $\text{Area} = \frac{\theta}{360^\circ} \pi r^2$ .
For Sector $A$ , $\text{Area}_A = \frac{x}{360}\pi y^2$ .
Sector $B$ 's angle is reduced by $20\%$ , meaning it is $100\% - 20\% = 80\%$ of $x$ , or $0.8x$ .
Sector $B$ 's radius is increased by $20\%$ , meaning it is $100\% + 20\% = 120\%$ of $y$ , or $1.2y$ .
Substituting these into the sector area formula gives:

\text{Area}_B = \frac{0.8x}{360} \pi (1.2y)^2 = 0.8 \times 1.2^2 \times \left(\frac{x}{360}\pi y^2\right) = 0.8 \times 1.44 \times \text{Area}_A = 1.152 \times \text{Area}_A

As a percentage of $\text{Area}_A$ , this is $1.152 \times 100 = 115.2\%$ .

Key Takeaways

Because the area of a sector is proportional to $\theta$ and to $r^2$ , changing the angle by factor $k_1$ and the radius by factor $k_2$ changes the area by factor $k_1 \times k_2^2$ .
Percentage multipliers allow rapid scaling without needing specific numerical values for $x$ and $y$ .

Common Mistakes

Forgetting to square the multiplier for the radius: writing $0.8 \times 1.2 = 0.96$ ( $96\%$ ).
Thinking the two $20\%$ changes cancel each other out to give $100\%$ .
Adding/subtracting $0.2$ directly (e.g., $(x - 0.2)$ ) instead of multiplying by percentage factors.

Things to Be Careful About

Ensure you square $1.2$ to get $1.44$ before multiplying by $0.8$ .

Techniques used

express sector area algebraicallyapply percentage increase and decrease multipliersfind ratio of two areas as a percentage

Q9Medium-HardStatisticsAlgebra and Graphs

A shop sells two varieties of apple tree.

(a)

The cumulative frequency diagram shows the heights, in metres, of 80 Variety $A$ trees.

Fig. 12

7M

(i)

Use the diagram to estimate

3M

(a)

the median

______ $\text{m}$

(b)

the 30th percentile.

______ $\text{m}$

(ii)

Trees with a height greater than $y\,\text{m}$ are graded Class I.
$\frac{2}{5}$ of the 80 trees are graded Class I.

Find the value of $y$ .

$y$ = ______

(iii)

Complete the frequency table for the heights of the Variety $A$ trees.

Height ( $h\,\text{m}$ )	$1.1 < h \leq 1.2$	$1.2 < h \leq 1.3$	$1.3 < h \leq 1.4$	$1.4 < h \leq 1.5$	$1.5 < h \leq 1.6$
Frequency	6	24

DifficultyMedium-Easy

Worked solution

Approach

Read the cumulative frequency at each upper boundary $h = 1.2, 1.3, 1.4, 1.5, 1.6$ and find each class frequency by subtracting consecutive cumulative frequencies.

Working

From the cumulative frequency curve, the cumulative frequencies at the boundaries are:

At $h = 1.2$ : $\text{CF} = 6$
At $h = 1.3$ : $\text{CF} = 30$
At $h = 1.4$ : $\text{CF} = 58$
At $h = 1.5$ : $\text{CF} = 76$
At $h = 1.6$ : $\text{CF} = 80$

Calculating the frequencies:

For $1.2 < h \leq 1.3$ : $30 - 6 = 24$ (given)
For $1.3 < h \leq 1.4$ : $58 - 30 = 28$
For $1.4 < h \leq 1.5$ : $76 - 58 = 18$
For $1.5 < h \leq 1.6$ : $80 - 76 = 4$

Answer

The completed frequency values for the remaining intervals are:

$1.3 < h \leq 1.4$ : $28$
$1.4 < h \leq 1.5$ : $18$
$1.5 < h \leq 1.6$ : $4$

Final answer

28, 18, 4

Detailed explanation

Walkthrough

A cumulative frequency curve plots the upper boundary of each class interval against the running total of frequencies.
To find the frequency of an individual interval, find the cumulative frequency at the upper boundary and subtract the cumulative frequency at the lower boundary.
For $1.3 < h \leq 1.4$ : $\text{CF}(1.4) - \text{CF}(1.3) = 58 - 30 = 28$ .
For $1.4 < h \leq 1.5$ : $\text{CF}(1.5) - \text{CF}(1.4) = 76 - 58 = 18$ .
For $1.5 < h \leq 1.6$ : $\text{CF}(1.6) - \text{CF}(1.5) = 80 - 76 = 4$ .

Key Takeaways

Frequency of an interval = Cumulative frequency at upper bound $-$ Cumulative frequency at lower bound.

Common Mistakes

Writing the cumulative frequencies directly into the table rather than subtracting to find the individual interval frequencies.

Things to Be Careful About

Check that the sum of all frequencies equals $80$ : $6 + 24 + 28 + 18 + 4 = 80$ .

Techniques used

read cumulative frequencies at class boundariessubtract consecutive cumulative frequencies

(b)

The frequency table shows the heights of 50 Variety $B$ trees.

Height ( $h\,\text{m}$ )	$1.5 < h \leq 1.7$	$1.7 < h \leq 1.8$	$1.8 < h \leq 1.9$	$1.9 < h \leq 2.3$
Frequency	$p$	15	17	$q$

Using the midpoints of the intervals, the estimated mean height of these Variety $B$ trees is $1.81\,\text{m}$ .

Calculate the value of $p$ and the value of $q$ .

$p$ = ______
$q$ = ______

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

First, find the midpoints of the class intervals. Then form two simultaneous equations: one using the total frequency ( $50$ ), and another using the formula for the estimated mean ( $1.81\,\text{m}$ ). Solve the system for $p$ and $q$ .

Working

Find the midpoint ( $x$ ) of each interval:

For $1.5 < h \leq 1.7$ : $x_1 = \frac{1.5 + 1.7}{2} = 1.6$
For $1.7 < h \leq 1.8$ : $x_2 = \frac{1.7 + 1.8}{2} = 1.75$
For $1.8 < h \leq 1.9$ : $x_3 = \frac{1.8 + 1.9}{2} = 1.85$
For $1.9 < h \leq 2.3$ : $x_4 = \frac{1.9 + 2.3}{2} = 2.1$

Equation 1 (Total frequency):

p + 15 + 17 + q = 50

p + q = 18 \implies q = 18 - p

Equation 2 (Estimated mean):

\frac{\sum fx}{\sum f} = \frac{1.6p + (1.75 \times 15) + (1.85 \times 17) + 2.1q}{50} = 1.81

Calculate the products:

1.75 \times 15 = 26.25

1.85 \times 17 = 31.45

26.25 + 31.45 = 57.7

Multiply both sides by $50$ :

1.6p + 57.7 + 2.1q = 1.81 \times 50

1.6p + 2.1q + 57.7 = 90.5

1.6p + 2.1q = 32.8

Substitute $q = 18 - p$ into this equation:

1.6p + 2.1(18 - p) = 32.8

1.6p + 37.8 - 2.1p = 32.8

-0.5p = 32.8 - 37.8

-0.5p = -5

p = \frac{-5}{-0.5} = 10

Now find $q$ :

q = 18 - 10 = 8

Answer

p = 10, \quad q = 8

Final answer

p = 10, q = 8

Detailed explanation

Walkthrough

Identify the midpoints: For each grouped interval, calculate the midpoint $\frac{\text{lower bound} + \text{upper bound}}{2}$ . The midpoints are $1.6$ , $1.75$ , $1.85$ , and $2.1$ .
Form the total frequency equation: The sum of all frequencies is $50$ , giving $p + 15 + 17 + q = 50$ , which simplifies to $p + q = 18$ .
Form the estimated mean equation: The formula for the estimated mean is $\frac{\sum fx}{\sum f} = \bar{x}$ . Substituting the values gives:

\frac{1.6p + 1.75(15) + 1.85(17) + 2.1q}{50} = 1.81

Simplifying gives $1.6p + 2.1q = 32.8$ .
Solve simultaneously: Substitute $q = 18 - p$ into the mean equation to find $p = 10$ , then obtain $q = 8$ .

Key Takeaways

Estimated mean of grouped data uses the formula $\bar{x} = \frac{\sum f x}{\sum f}$ , where $x$ represents the class midpoints.
When two frequencies are unknown, setting up a system of two equations (one for total frequency and one for the mean) is the standard method.

Common Mistakes

Calculating the wrong midpoints (e.g. using class widths instead of midpoints).
Forgetting to multiply $1.81$ by the total frequency $50$ .
Arithmetic errors when solving the simultaneous equations.

Things to Be Careful About

The fourth interval $1.9 < h \leq 2.3$ has a width of $0.4$ , so its midpoint is $\frac{1.9 + 2.3}{2} = 2.1$ , not $2.0$ or $2.05$ .
Check that $p$ and $q$ are positive integers that sum to $18$ .

Techniques used

find midpoints of grouped intervalsset up a linear equation for the estimated meanset up a linear equation for total frequencysolve simultaneous linear equations

Q10MediumGeometryTrigonometry

Fig. 13

The diagram shows a pentagon.

(a)

Calculate the interior angle $B$ .

______

(b)

In the pentagon, $AE = 8\,\text{cm}$ and $AD = 15\,\text{cm}$ .

Calculate the length $ED$ .
Show your working and give your answer correct to 1 decimal place.

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

5M Similar

DifficultyMedium

Worked solution

Approach

In triangle $ADE$ , we know two sides $AE = 8\,\text{cm}$ and $AD = 15\,\text{cm}$ , and the angle $\angle AED = 120^\circ$ . Let $ED = x$ . Apply the cosine rule to find $x$ , which will lead to a quadratic equation.

Working

By the cosine rule in $\triangle ADE$ :

AD^2 = AE^2 + ED^2 - 2 \cdot AE \cdot ED \cdot \cos(\angle AED)

Substitute the known values:

15^2 = 8^2 + x^2 - 2(8)(x)\cos(120^\circ)

Since $\cos(120^\circ) = -0.5$ :

225 = 64 + x^2 - 16x\left(-\frac{1}{2}\right)

225 = 64 + x^2 + 8x

Rearrange into standard quadratic form:

x^2 + 8x - 161 = 0

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$ , $b = 8$ , $c = -161$ :

x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-161)}}{2(1)}

x = \frac{-8 \pm \sqrt{64 + 644}}{2}

x = \frac{-8 \pm \sqrt{708}}{2}

Calculate the positive root (since length must be positive):

x = \frac{-8 + 26.608}{2} \approx 9.304

Round to 1 decimal place:

x = 9.3

Answer

ED = 9.3\,\text{cm}

Final answer

9.3

Detailed explanation

Walkthrough

We are looking at triangle $ADE$ within the pentagon. We know $AE = 8\,\text{cm}$ , $AD = 15\,\text{cm}$ , and $\angle AED = 120^\circ$ . Let $ED = x$ . Since we have two sides and a non-included angle, we could use the sine rule to find $\angle ADE$ , but the cosine rule directly relates the three sides and the known angle $\angle AED$ (opposite to $AD$ ). Applying the cosine rule gives $15^2 = 8^2 + x^2 - 2(8)(x)\cos(120^\circ)$ . Simplifying this using $\cos(120^\circ) = -0.5$ yields the quadratic equation $x^2 + 8x - 161 = 0$ . Solving this with the quadratic formula gives $x = \frac{-8 \pm \sqrt{708}}{2}$ . Taking the positive root gives $x \approx 9.304$ , which rounds to $9.3\,\text{cm}$ to 1 decimal place.

Key Takeaways

The cosine rule can be used when two sides and a non-included angle are known, resulting in a quadratic equation.
Always check that the final length is positive and round to the required accuracy.

Common Mistakes

Using the wrong angle in the cosine rule (e.g., using $\angle ADE$ instead of $\angle AED$ ).
Forgetting that $\cos(120^\circ)$ is negative, leading to a sign error in the linear term.
Accepting the negative root from the quadratic formula.
Rounding too early in the calculation.

Things to Be Careful About

The question asks for the answer correct to 1 decimal place, so do not round intermediate steps.
Ensure the cosine rule is set up correctly: $a^2 = b^2 + c^2 - 2bc\cos(A)$ , where $A$ is the angle opposite side $a$ . Here, $AD$ is opposite $\angle AED$ , so $AD^2 = AE^2 + ED^2 - 2(AE)(ED)\cos(\angle AED)$ .
Working must be shown (nfww) to earn the final mark.

Techniques used

identify the relevant triangle within the polygonapply the cosine rule to form a quadratic equationsolve the quadratic equation using the formularound the final length to 1 decimal place

Q11Medium-HardCoordinate Geometry

$Q$ is the point $(n, -4)$ , $R$ is the point $(-1, 8)$ and $S$ is the point $(3, 2)$ .

(a)

The length $QR$ is 13 units.

Find the two possible values of $n$ .

$n$ = ______ or $n$ = ______

DifficultyMedium

Worked solution

Approach

Use the distance formula for the length $QR$ with coordinates $Q(n, -4)$ and $R(-1, 8)$ . Set this equal to 13 and solve the resulting equation for $n$ .

Working

The distance $d$ between $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

Substitute the coordinates of $Q$ and $R$ :

13 = \sqrt{(-1 - n)^2 + (8 - (-4))^2}

Simplify the term inside the square root:

13 = \sqrt{(-1 - n)^2 + (12)^2}

Square both sides to remove the square root:

169 = (-1 - n)^2 + 144

Subtract 144 from both sides:

25 = (-1 - n)^2

Take the square root of both sides (remembering both positive and negative roots):

\pm 5 = -1 - n

Case 1: Positive root

5 = -1 - n

n = -1 - 5

n = -6

Case 2: Negative root

-5 = -1 - n

n = -1 + 5

n = 4

Alternatively, expanding $(-1-n)^2 = 25$ gives $n^2 + 2n + 1 = 25$ , or $n^2 + 2n - 24 = 0$ . Factoring yields $(n+6)(n-4)=0$ , giving $n=-6$ or $n=4$ .

Answer

$n = -6$ or $n = 4$

Final answer

-6 or 4

Detailed explanation

Walkthrough

First, identify the coordinates of the two points involved in the distance. Point $Q$ is $(n, -4)$ and point $R$ is $(-1, 8)$ . The distance formula calculates the straight-line distance between any two points on a Cartesian plane. We set this distance equal to the given value of 13 units.

After substituting the coordinates into the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , we get $13 = \sqrt{(-1 - n)^2 + (8 - (-4))^2}$ . It is crucial to handle the subtraction of negative numbers carefully; $8 - (-4)$ becomes $8 + 4 = 12$ .

Squaring both sides removes the radical: $169 = (-1 - n)^2 + 144$ . Subtracting 144 leaves $25 = (-1 - n)^2$ . To solve for $n$ , we take the square root of both sides. This step introduces ambiguity because both $5^2$ and $(-5)^2$ equal 25. Therefore, we must consider two cases:

$-1 - n = 5 \implies n = -6$
$-1 - n = -5 \implies n = 4$

Both values are valid solutions as they result in a distance of 13.

Key Takeaways

The distance formula is derived from Pythagoras' theorem and applies to any two points.
When solving an equation of the form $(expression)^2 = k$ , always remember to take both the positive and negative square roots ( $\pm \sqrt{k}$ ).
Be careful with signs when subtracting coordinates, especially when dealing with negative coordinates.

Common Mistakes

Forgetting the $\pm$ sign when taking the square root, leading to only one answer being found.
Arithmetic errors in calculating $8 - (-4)$ , such as getting 4 instead of 12.
Errors in squaring binomials if expanding rather than using the square root method directly.
Substituting the wrong coordinate differences (e.g., swapping x and y or signs incorrectly).

Things to Be Careful About

Ensure you square the entire difference, e.g., $(-1-n)^2$ , not just $n^2$ .
Double-check that both final values of $n$ satisfy the original distance condition.

Techniques used

apply the distance formula between two pointsform a quadratic equation from the distance conditionsolve the quadratic to find two possible values

(b)

$RST$ is a straight line and $RS : RT = 2 : 5$ .

Find the coordinates of $T$ .

( ______ , ______ )