Mathematics (Syllabus D) 4024 / 21 — October/November 2024

(b)

Basma employs 5 sales assistants and 2 supervisors.
On one particular week, the 5 sales assistants each work for 30 hours and the 2 supervisors each work for 38 hours.
For that week, the total amount Basma pays these 7 employees is $3324.70 .
Basma pays each sales assistant $13.45 per hour.

Calculate the amount Basma pays each supervisor per hour.

$ ______ $\text{per hour}$

DifficultyMedium-Easy

Worked solution

Approach

Find the total pay of the 5 sales assistants. Subtract that from the total paid to all 7 employees to find the total supervisor pay, then divide by the total number of supervisor hours.

Working

Total sales-assistant pay:

5 \times 30 \times 13.45 = 150 \times 13.45 = 2017.50

Total supervisor pay:

3324.70 - 2017.50 = 1307.20

Total supervisor hours:

2 \times 38 = 76

Hourly supervisor rate:

\frac{1307.20}{76} = 17.20

Answer

Each supervisor is paid 17.20 dollars per hour ($17.20 per hour).

Final answer

17.20 dollars per hour

Detailed explanation

Walkthrough

The total payroll of 3324.70 dollars includes both the sales assistants and the supervisors.

First calculate how much the 5 sales assistants receive. Each works 30 hours, so they work $5 \times 30 = 150$ hours altogether. At 13.45 dollars per hour, their total is:

150 \times 13.45 = 2017.50

So the total supervisor pay is:

3324.70 - 2017.50 = 1307.20

The two supervisors each work 38 hours, so their combined hours are:

2 \times 38 = 76

Because rate is total pay divided by total hours:

\frac{1307.20}{76} = 17.20

Key Takeaways

Total pay is hours multiplied by the hourly rate.
When one part of a total is known, subtract it to find the other part.
An hourly rate is the total amount paid divided by the number of hours.

Common Mistakes

Using 3324.70 as if it were the supervisors' total without first subtracting the sales assistants' pay.
Forgetting that there are 2 supervisors, each working 38 hours, so the total is 76 hours.
Multiplying 5 by 13.45 but not by the 30 hours.

Things to Be Careful About

The final answer is a rate in dollars per hour. The mark scheme gives the answer as 17.2[0], so both 17.2 and 17.20 are accepted. Show the subtract and division steps because the marks are awarded for the method as well as the final answer.

Techniques used

calculate total sales assistant paysubtract from the total payrolldivide by total supervisor hours

(c)

The exchange rate between dollars ($) and pounds (£) is $1 = £0.77 .
Basma buys 50 identical games for a total of £245.
She makes a profit of 39% when she sells each game.

Calculate the selling price of one game in dollars.
Give your answer correct to the nearest cent.

$ ______

DifficultyMedium

Worked solution

Approach

Find the cost of one game in pounds. Add 39% profit by multiplying by 1.39 to get the selling price in pounds. Then convert pounds to dollars using the exchange rate $1 = £0.77.

Working

Cost of one game in pounds:

\frac{245}{50} = 4.90

Add the 39% profit:

4.90 \times 1.39 = 6.811

Convert to dollars:

\frac{6.811}{0.77} = 8.84545\ldots

Rounded to the nearest cent:

8.84545\ldots \approx 8.85

Answer

The selling price is 8.85 dollars ($8.85).

Final answer

8.85 dollars

Detailed explanation

Walkthrough

Basma bought 50 games for 245 pounds, so the cost of each game is:

\frac{245}{50} = 4.90

She makes 39% profit when she sells a game. That means the selling price is the cost plus 39% of the cost, i.e. 139% of the cost. Using a percentage multiplier:

4.90 \times 1.39 = 6.811

Now convert that selling price from pounds to dollars. The rate says $1 = £0.77, so every 0.77 of a pound equals 1 dollar. Therefore divide the price in pounds by 0.77:

\frac{6.811}{0.77} = 8.84545\ldots

Finally, to the nearest cent, this is 8.85 dollars.

Key Takeaways

A 39% profit is obtained by multiplying the cost by 1.39, not by 0.39.
When the exchange rate is $1 = £0.77, pounds are converted to dollars by dividing by 0.77.
Worth while to keep the intermediate unrounded value before the nearest-cent rounding.

Common Mistakes

Multiplying 6.811 by 0.77 instead of dividing by 0.77, which gives the wrong currency direction.
Using 0.39 instead of 1.39, forgetting to add the profit.
Not dividing 245 by 50 first, so the profit is applied to the whole batch instead of one game.
Rounding 8.845 prematurely to 8.8 rather than 8.85.

Things to Be Careful

When rounding to the nearest cent, the answer requires two decimal places. The mark scheme also shows the combined form:

\frac{245 \times 1.39}{50 \times 0.77}

But showing each intermediate step clearly makes it much easier to secure the method marks.

Techniques used

divide total cost by the number of gamesapply a percentage profit multiplierconvert pounds to dollarsround to the nearest cent

(d)

Basma invests $12000 in an account paying compound interest at a rate of 1.5% per year.
At the end of year 1, she invests another $12000 in the same account.
At the end of year 4, she takes $20000 out of the account.

Calculate the amount of money remaining in the account at the end of year 4.
Give your answer correct to the nearest cent.

$ ______

DifficultyMedium

Worked solution

Approach

Deposit 1 earns interest for the whole 4 years. Deposit 2 is added at the beginning of year 1, so it added nothing for year 1; it receives interest for 3 years. Use a compound interest in both, add the two final balances, then subtract the $20000 withdrawal.

Working

First deposit after 4 years:

12000\left(1+\frac{1.5}{100}\right)^4 = 12000 \times 1.061363550625 = 12736.36

Second deposit after 3 years:

12000\left(1+\frac{1.5}{100}\right)^3 = 12000 \times 1.045678375 = 12548.14

Total before withdrawal:

12736.36 + 12548.14 = 25284.50

After withdrawing $20000:

25284.50 - 20000 = 5284.50

Answer

The amount remaining at the end of year 4 is 5284.50 dollars ($5284.50).

Final answer

5284.50 dollars

Detailed explanation

Walkthrough

The two deposits start at different times, so they earn interest over different numbers of years.

The first $12000 is invested at the start of year 1. It earns interest in years 1, 2, 3 and 4, so it is compounded 4 times.
At the end of year 1, a second $12000 is added. It earns interest only in years 2, 3 and 4, so it is compounded 3 times.

The annual multiplier for 1.5% is $1 + \frac{1.5}{100} = 1.015$ .

So the balance is:

12000 \times 1.015^4 + 12000 \times 1.015^3

Using calculator value:

12000 \times 1.061363550625 = 12736.36

12000 \times 1.045678375 = 12548.14

Total before withdrawal:

12736.36 + 12548.14 = 25284.50

Finally subtract the withdrawal of $20000:

25284.50 - 20000 = 5284.50

Key Takeaways

Compound interest uses an exponent, because each year interest is added to the balance before the next year.
Every new deposit must be compounded for the number of complete years it remains in the account.
Withdrawals are subtracted after all interest for that part of the problem has been applied.

Common Mistakes

Forgetting that the second deposit starts one year later and therefore only has 3 compound periods.
Using $0.015$ in the multiplier instead of $1.015$ , which treats interest as going down by 1.5%.
Adding the second deposit at the start of year 3 or giving it a full 4 years.
Rounding the factor $1.015^4$ before finishing the calculation.

Things to Be Careful About

The question asks for the answer to the nearest cent, so the final balance is 5284.50 dollars. The compound interest formula must be used here, not simple interest. The mark scheme also accepts the combined form:

12000 \times 1.015^3 + 12000 \times 1.015^3 + 20000? = ?

Actually the standard accepted calculation is:

12000 \times 1.015^4 + 12000 \times 1.015^3 = 25284.50

Keep the full unrounded values shown in the Working before making the final 5284.50.

Techniques used

apply the compound interest formula to each deposit separatelyuse different numbers of compounding periods for each depositsum the two balances and subtract the withdrawal

Q2Medium-EasyNumberStatistics

In a traffic survey, information about the vehicles passing a checkpoint is recorded.

(a)

160 vehicles pass the checkpoint in the morning.
The table shows the number of people in each of these vehicles.

Number of people	Frequency	Pie chart angle
1	72
2	48
3 or more	40	$90^\circ$

3M

(i)

Complete the table.

DifficultyMedium-Easy

Worked solution

Approach

The total angle in a pie chart is $360^\circ$ . The angle for each category is proportional to its frequency. We use the total frequency of $160$ and the total angle to find the missing angles.

Working

For 1 person (frequency $72$ ):

\frac{72}{160} \times 360 = 162^\circ

For 2 people (frequency $48$ ):

\frac{48}{160} \times 360 = 108^\circ

Alternatively, using the given angle for "3 or more" ( $90^\circ$ ):

360^\circ - 90^\circ - 162^\circ = 108^\circ

Answer

The missing angles are $162^\circ$ and $108^\circ$ .

Final answer

162, 108

Detailed explanation

Walkthrough

A pie chart represents a whole as $360^\circ$ . To find the angle for a category, we take the fraction of the total frequency that the category represents and multiply by $360^\circ$ . For the "1 person" category, the fraction is $\frac{72}{160}$ . Multiplying by $360$ gives $162^\circ$ . For the "2 people" category, the fraction is $\frac{48}{160}$ . Multiplying by $360$ gives $108^\circ$ . We can verify this by adding all three angles: $162 + 108 + 90 = 360^\circ$ .

Key Takeaways

Pie chart angles are calculated using the formula: $\text{Angle} = \frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ$ .

Common Mistakes

Forgetting to multiply by $360^\circ$ and just writing the fraction.
Using the wrong total frequency.
Not ensuring the angles add up to $360^\circ$ .

Things to Be Careful About

Always check that the angles sum to $360^\circ$ .
The question asks to complete the table, so both numerical answers must be provided.

Techniques used

calculate pie chart angle from frequency and totaluse proportion of total angle

(ii)

Complete the pie chart to show the results.

Fig. 1

(b)

The histogram shows the speeds of vehicles passing the checkpoint in the afternoon.

Fig. 2

4M

(i)

Sanjay says the histogram shows that the range of the speeds is $50\,\text{km/h}$ .
Explain why he may not be correct.

______ because ______

(ii)

Complete the frequency table.

Speed ( $s$ km/h)	$20 < s \leq 30$	$30 < s \leq 40$	$40 < s \leq 45$	$45 < s \leq 50$	$50 < s \leq 70$
Frequency	24

DifficultyMedium-Easy

Worked solution

Approach

The frequency in a histogram is calculated using the formula: $\text{Frequency} = \text{Frequency Density} \times \text{Class Width}$ . We calculate this for each interval.

Working

For $30 < s \leq 40$ :
Class width $= 40 - 30 = 10$ . Frequency density $= 4.6$ .

\text{Frequency} = 4.6 \times 10 = 46

For $40 < s \leq 45$ :
Class width $= 45 - 40 = 5$ . Frequency density $= 8.8$ .

\text{Frequency} = 8.8 \times 5 = 44

For $45 < s \leq 50$ :
Class width $= 50 - 45 = 5$ . Frequency density $= 7.6$ .

\text{Frequency} = 7.6 \times 5 = 38

For $50 < s \leq 70$ :
Class width $= 70 - 50 = 20$ . Frequency density $= 1.4$ .

\text{Frequency} = 1.4 \times 20 = 28

Answer

Speed ( $s$ km/h)	$20 < s \leq 30$	$30 < s \leq 40$	$40 < s \leq 45$	$45 < s \leq 50$	$50 < s \leq 70$
Frequency	24	46	44	38	28

Final answer

46, 44, 38, 28

Detailed explanation

Walkthrough

In a histogram with unequal class widths, the area of each bar represents the frequency. Since $\text{Area} = \text{width} \times \text{height}$ , and $\text{height}$ is the frequency density, we have $\text{Frequency} = \text{Class Width} \times \text{Frequency Density}$ . We apply this formula to each interval to find the missing frequencies.

For $30 < s \leq 40$ : width is $10$ , FD is $4.6$ , so frequency is $46$ .
For $40 < s \leq 45$ : width is $5$ , FD is $8.8$ , so frequency is $44$ .
For $45 < s \leq 50$ : width is $5$ , FD is $7.6$ , so frequency is $38$ .
For $50 < s \leq 70$ : width is $20$ , FD is $1.4$ , so frequency is $28$ .

Key Takeaways

Always remember that frequency equals frequency density multiplied by the class width in a histogram.

Common Mistakes

Forgetting to multiply by the class width and just using the frequency density as the frequency.
Calculating the class width incorrectly, especially for the last interval ( $50$ to $70$ has width $20$ , not $10$ or $20$ ).

Things to Be Careful About

Pay attention to the class widths, which are not all equal in this histogram.
Ensure the calculations are accurate.

Techniques used

calculate frequency from frequency density and class width

Q3Medium-EasyAlgebra and Graphs

(a)

Complete the table for $y = 4 + 2x - \frac{x^2}{2}$ .

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$		1.5	4	5.5	6	5.5	4	1.5

DifficultyEasy

Worked solution

Approach

Substitute the missing $x$ -values ( $x = -2$ and $x = 6$ ) into the formula $y = 4 + 2x - \frac{x^2}{2}$ to find the corresponding $y$ -values.

Working

For $x = -2$ :

\begin{aligned} y &= 4 + 2(-2) - \frac{(-2)^2}{2} \\ y &= 4 - 4 - \frac{4}{2} \\ y &= 0 - 2 \\ y &= -2 \end{aligned}

For $x = 6$ :

\begin{aligned} y &= 4 + 2(6) - \frac{6^2}{2} \\ y &= 4 + 12 - \frac{36}{2} \\ y &= 16 - 18 \\ y &= -2 \end{aligned}

Answer

The completed table has $y = -2$ for both $x = -2$ and $x = 6$ .

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$	$-2$	$1.5$	$4$	$5.5$	$6$	$5.5$	$4$	$1.5$	$-2$

Final answer

y = -2 for x = -2 and y = -2 for x = 6

Detailed explanation

Walkthrough

The question provides a table of $x$ and $y$ values for the quadratic function $y = 4 + 2x - \frac{x^2}{2}$ and asks to fill in the missing entries. The missing $x$ -values are $-2$ and $6$ .

To find $y$ when $x = -2$ , substitute $-2$ into the formula:
$y = 4 + 2(-2) - \frac{(-2)^2}{2}$
Calculate each term: $2(-2) = -4$ , and $(-2)^2 = 4$ , so $\frac{4}{2} = 2$ . Adding these together gives $4 - 4 - 2 = -2$ .

To find $y$ when $x = 6$ , substitute $6$ into the formula:
$y = 4 + 2(6) - \frac{6^2}{2}$
Calculate each term: $2(6) = 12$ , and $6^2 = 36$ , so $\frac{36}{2} = 18$ . Adding these together gives $4 + 12 - 18 = -2$ .

Both missing values are $-2$ , which also confirms the symmetry of the parabola around its axis.

Key Takeaways

Substitution into algebraic expressions is a fundamental skill for completing tables of values.
Careful attention to signs and order of operations (especially squaring negative numbers) is required.

Common Mistakes

Forgetting that $(-2)^2 = 4$ , not $-4$ , which would incorrectly give $y = 4 - 4 - (-2) = 2$ .
Arithmetic errors when combining positive and negative integers, such as $4 + 12 - 18$ .

Things to Be Careful About

Always square the entire negative number: $(-2)^2 = 4$ . If you write $-2^2$ , it means $-(2^2) = -4$ , which is incorrect here.
Follow the order of operations: exponents first, then multiplication, then addition and subtraction.

Techniques used

substitute x = -2 into the quadratic expressionsubstitute x = 6 into the quadratic expressionsimplify the arithmetic to find the missing y-values

(b)

Draw the graph of $y = 4 + 2x - \frac{x^2}{2}$ for $-2 \leq x \leq 6$ .

Fig. 3

DifficultyMedium-Easy

Worked solution

Approach

Plot the points from the completed table on the given Cartesian grid. Since the function is quadratic ( $x^2$ term is present), the graph is a parabola, so connect the points with a smooth U-shaped curve rather than straight lines.

Working

The points to plot are:

\begin{aligned} &(-2, -2), (-1, 1.5), (0, 4), (1, 5.5), (2, 6), \\ &(3, 5.5), (4, 4), (5, 1.5), (6, -2) \end{aligned}

Answer

A smooth parabola passing through all nine points, with its maximum (vertex) at $(2, 6)$ .

Point	Coordinates
1	$(-2, -2)$
2	$(-1, 1.5)$
3	$(0, 4)$
4	$(1, 5.5)$
5	$(2, 6)$
6	$(3, 5.5)$
7	$(4, 4)$
8	$(5, 1.5)$
9	$(6, -2)$

Final answer

A smooth downward-opening parabola with vertex at (2, 6) passing through (-2, -2) and (6, -2)

Detailed explanation

Walkthrough

The task is to draw the graph of $y = 4 + 2x - \frac{x^2}{2}$ for $-2 \leq x \leq 6$ . We already have all nine coordinate pairs from part (a) and the given table.

Set up the grid: The $x$ -axis runs from $-2$ to $6$ and the $y$ -axis from $-3$ to $7$ , with gridlines at every $0.2$ units.
Plot each point from the table: $(-2, -2)$ , $(-1, 1.5)$ , $(0, 4)$ , $(1, 5.5)$ , $(2, 6)$ , $(3, 5.5)$ , $(4, 4)$ , $(5, 1.5)$ , and $(6, -2)$ .
Draw a smooth curve through these points. Because the highest power of $x$ is $2$ , the graph is a parabola. It opens downwards (since the coefficient of $x^2$ is negative, $-\frac{1}{2}$ ), with the turning point (maximum) at $(2, 6)$ .

Key Takeaways

A table of values gives discrete points; for a quadratic function, these must be joined by a smooth curve, not straight line segments.
The vertex of a downward-opening parabola is its highest point.

Common Mistakes

Joining the points with straight line segments, which gives a polygonal shape instead of a parabola.
Misreading the grid: the gridlines are at $0.2$ units, so $y = 1.5$ is three small squares above the $x$ -axis, and $y = 5.5$ is three small squares below $y = 6$ .
Plotting $(-2)^2$ as $-4$ instead of $4$ , leading to an incorrect point at $(-2, 2)$ .

Things to Be Careful About

The mark scheme awards follow-through marks based on the number of correctly plotted points (B2FT for 8 or 9 points, B1FT for 6 or 7 points). Accuracy in plotting is critical.
Ensure the curve is smooth and does not have sharp corners at the plotted points.
The graph should only be drawn for the range $-2 \leq x \leq 6$ ; do not extend it beyond these $x$ -values.

Techniques used

plot the calculated table of values on the griddraw a smooth curve through the plotted points

(c)

Find the equation of the line of symmetry of the graph of $y = 4 + 2x - \frac{x^2}{2}$ .

______

DifficultyEasy

Worked solution

Approach

The graph is a parabola, so its line of symmetry passes through the vertex (turning point). This can be found by observing the symmetry in the table of values or by using the formula $x = -\frac{b}{2a}$ for $y = ax^2 + bx + c$ .

Working

Method 1: From the table
Notice that the $y$ -values are symmetric around $x = 2$ :

$x = 1$ and $x = 3$ both give $y = 5.5$
$x = 0$ and $x = 4$ both give $y = 4$
$x = -1$ and $x = 5$ both give $y = 1.5$
$x = -2$ and $x = 6$ both give $y = -2$

The axis of symmetry is exactly halfway between these pairs, which is $x = 2$ .

Method 2: Using the formula
Rewrite the equation in standard form $y = ax^2 + bx + c$ :

y = -\frac{1}{2}x^2 + 2x + 4

Here $a = -\frac{1}{2}$ and $b = 2$ . The axis of symmetry is:

x = -\frac{b}{2a} = -\frac{2}{2\left(-\frac{1}{2}\right)} = -\frac{2}{-1} = 2

Answer

x = 2

Final answer

x = 2

Detailed explanation

Walkthrough

The line of symmetry of a parabola is the vertical line that passes through its vertex, dividing the graph into two mirror-image halves.

From the table in part (a), we can see that for every $x$ -value less than $2$ , there is a corresponding $x$ -value greater than $2$ that gives the same $y$ -value. For example, $y = 5.5$ at both $x = 1$ and $x = 3$ . The midpoint between $1$ and $3$ is $2$ . Similarly, the midpoint between $-2$ and $6$ is $2$ . This confirms the axis of symmetry is $x = 2$ .

Alternatively, using the formula $x = -\frac{b}{2a}$ for the quadratic $y = -\frac{1}{2}x^2 + 2x + 4$ , we get $x = -\frac{2}{2(-1/2)} = 2$ .

Key Takeaways

The axis of symmetry of a parabola $y = ax^2 + bx + c$ is always $x = -\frac{b}{2a}$ .
You can also find it by looking for equal $y$ -values in a table of values and taking the average of the corresponding $x$ -values.

Common Mistakes

Writing $y = 2$ instead of $x = 2$ . The line of symmetry is a vertical line, so it must be in the form $x = \text{number}$ .
Calculating the vertex $y$ -coordinate instead of the axis of symmetry.

Things to Be Careful About

The answer must be an equation of a vertical line, so it must start with $x =$ .
No working is required for this 1-mark question, but ensuring accuracy is important.

Techniques used

identify the axis of symmetry from the table or formulause the vertex x-coordinate

(d)

On the grid, draw the line $3y = x + 6$ for $-2 \leq x \leq 6$ .

DifficultyMedium-Easy

Worked solution

Approach

Rearrange the equation $3y = x + 6$ into the form $y = mx + c$ to easily find coordinates. Then plot at least two points and draw a straight ruled line through them across the grid.

Working

Rearrange $3y = x + 6$ :

y = \frac{x + 6}{3} = \frac{1}{3}x + 2

Find coordinates by substituting convenient $x$ -values:

When $x = 0$ : $y = \frac{0 + 6}{3} = 2 \implies (0, 2)$
When $x = 3$ : $y = \frac{3 + 6}{3} = 3 \implies (3, 3)$
When $x = 6$ : $y = \frac{6 + 6}{3} = 4 \implies (6, 4)$
When $x = -2$ : $y = \frac{-2 + 6}{3} = \frac{4}{3} \approx 1.33 \implies (-2, 1.33)$

Plot these points on the grid and draw a straight ruled line through them from $x = -2$ to $x = 6$ .

Answer

A straight ruled line passing through $(0, 2)$ , $(3, 3)$ , and $(6, 4)$ with gradient $\frac{1}{3}$ .

$x$	$-2$	$0$	$3$	$6$
$y$	$1.33$	$2$	$3$	$4$

Final answer

A straight line passing through (0, 2), (3, 3), and (6, 4)

Detailed explanation

Walkthrough

The question asks to draw the line $3y = x + 6$ on the same grid used for the parabola.

First, rearrange the equation to make $y$ the subject:
$3y = x + 6 \implies y = \frac{1}{3}x + 2$

This tells us the line has a $y$ -intercept of $2$ and a gradient of $\frac{1}{3}$ . This means for every $3$ units moved to the right, the line goes up $1$ unit.

Choose easy $x$ -values to find exact coordinates:

At $x = 0$ , $y = 2$ . Point: $(0, 2)$ .
At $x = 3$ , $y = 3$ . Point: $(3, 3)$ .
At $x = 6$ , $y = 4$ . Point: $(6, 4)$ .

Plot these points on the grid. Use a ruler to draw a straight line through them, extending from $x = -2$ to $x = 6$ .

Key Takeaways

Rearranging equations into $y = mx + c$ form makes it easy to identify the $y$ -intercept and gradient.
Always use a ruler for straight lines in graphical questions.

Common Mistakes

Forgetting to divide the entire right-hand side by $3$ when rearranging, e.g., writing $y = x + 2$ instead of $y = \frac{1}{3}x + 2$ .
Plotting the $y$ -intercept incorrectly as $(6, 0)$ instead of $(0, 2)$ .
Drawing a curved line or not extending the line across the full required range.

Things to Be Careful About

The mark scheme accepts a short or unruled line, or two correct coordinates, or a line with positive gradient through $(0, 2)$ , or gradient $\frac{1}{3}$ . However, a fully ruled line is best.
Ensure the line is drawn for $-2 \leq x \leq 6$ as specified.

Techniques used

rearrange the linear equation into y = mx + c formfind two or three points on the linedraw a ruled straight line through the points

(e)

Write down the $x$ -coordinates of the points of intersection of the graphs of $y = 4 + 2x - \frac{x^2}{2}$ and $3y = x + 6$ .

$x$ = ______ and $x$ = ______

DifficultyMedium-Easy

Worked solution

Approach

The points of intersection of the two graphs are where they cross each other. Read the $x$ -coordinates of these crossing points directly from the grid.

Working

Looking at the graph drawn in parts (b) and (d), the parabola and the straight line cross at two points.

By examining the grid carefully at the intersection points:

The left intersection is slightly to the right of $x = -1$ . Reading from the grid, $x \approx -0.9$ .
The right intersection is between $x = 4$ and $x = 5$ , closer to $4$ . Reading from the grid, $x \approx 4.3$ .

(Note: The exact algebraic solutions are $x = \frac{10 \pm \sqrt{244}}{6} \approx -0.94$ and $4.27$ . Graphical reading typically accepts values within $\pm 0.2$ of these.)

Answer

x \approx -0.9 \quad \text{and} \quad x \approx 4.3

(Acceptable range: $x = -0.8$ to $-1.0$ and $x = 4.2$ to $4.4$ )

Final answer

x = -0.9 and x = 4.3

Detailed explanation

Walkthrough

The question asks for the $x$ -coordinates of the points where the parabola $y = 4 + 2x - \frac{x^2}{2}$ and the line $3y = x + 6$ intersect. Since this is a graphical question, we read the values from the graph we drew.

Locate the two points where the curve and the straight line cross:

Left intersection: Between $x = -1$ and $x = 0$ . The line is at $y \approx 1.67$ and the curve is at $y \approx 1.75$ at $x = -1$ . They cross slightly to the right of $x = -1$ . Reading from the grid, this is approximately $x = -0.9$ .
Right intersection: Between $x = 4$ and $x = 5$ . At $x = 4$ , the curve is at $y = 4$ and the line is at $y = 3.33$ . At $x = 5$ , the curve is at $y = 1.5$ and the line is at $y = 3.67$ . They cross between these, closer to $x = 4$ . Reading from the grid, this is approximately $x = 4.3$ .

The mark scheme states STRICT FT, meaning the marks depend on the accuracy of the graphs drawn in parts (b) and (d). If a student's graphs are slightly off, their intersection readings will be adjusted accordingly within a reasonable tolerance.

Key Takeaways

Graphical solutions to simultaneous equations involve reading intersection points from the graph.
When exact solutions are irrational, graphical estimation is the intended method.
Always state the tolerance or range if the exact value cannot be read precisely.

Common Mistakes

Reading the $y$ -coordinates instead of the $x$ -coordinates. The question specifically asks for $x$ -coordinates.
Reading the intersection at $x = 0$ or $x = 3$ (common grid points) where the graphs do not actually cross.
Not allowing for the tolerance in graphical reading; giving answers to too many decimal places.

Things to Be Careful About

The mark scheme is STRICT FT on the graphs from parts (b) and (d). If the curves are drawn incorrectly, the intersection points will be wrong, and marks will be lost.
Give answers to 1 decimal place or as a simple fraction that can be read from the grid.
The exact algebraic solutions are $x = \frac{5 \pm \sqrt{61}}{3}$ , which are approximately $-0.94$ and $4.27$ . Values in the range $-0.8$ to $-1.0$ and $4.2$ to $4.4$ are typically accepted.

Techniques used

identify intersection points from the drawn graphsread x-coordinates from the grid

Q4Medium-EasyNumber

$\mathscr{E} = \{x: x \text{ is an integer } 1 \leq x \leq 15\}$
$A = \{x: x \text{ is a multiple of 3}\}$
$B = \{x: x \text{ is a factor of 30}\}$

(a)

Write down the elements of $A$ .

______

(b)

Complete the Venn diagram.

Fig. 4

DifficultyMedium-Easy

Worked solution

Approach

Determine which elements of $\mathscr{E}$ belong to $A$ , $B$ , both, or neither, then place them in the appropriate regions of the Venn diagram.

Working

First, list the elements of $B$ , the factors of 30 between 1 and 15:

B = \{1, 2, 3, 5, 6, 10, 15\}

The intersection $A \cap B$ contains elements in both sets:

A \cap B = \{3, 6, 15\}

The elements in $A$ but not in $B$ ( $A \cap B'$ ) are:

A \cap B' = \{9, 12\}

The elements in $B$ but not in $A$ ( $A' \cap B$ ) are:

A' \cap B = \{1, 2, 5, 10\}

The elements in $\mathscr{E}$ but not in $A \cup B$ ( $(A \cup B)'$ ) are the remaining integers:

(A \cup B)' = \{4, 7, 8, 11, 13, 14\}

Answer

Place 9, 12 in circle $A$ only; 3, 6, 15 in the intersection; 1, 2, 5, 10 in circle $B$ only; and 4, 7, 8, 11, 13, 14 outside both circles but inside $\mathscr{E}$ .

Final answer

Venn diagram: A only contains 9, 12; intersection contains 3, 6, 15; B only contains 1, 2, 5, 10; outside contains 4, 7, 8, 11, 13, 14.

Detailed explanation

Walkthrough

To complete the Venn diagram, we must place each integer from 1 to 15 into exactly one of the four regions. We start with the intersection $A \cap B$ , which contains numbers that are both multiples of 3 and factors of 30. These are 3, 6, and 15. Next, we find numbers in $A$ but not $B$ (multiples of 3 that are not factors of 30): 9 and 12. Then, numbers in $B$ but not $A$ (factors of 30 that are not multiples of 3): 1, 2, 5, and 10. Finally, any number from 1 to 15 not yet placed goes outside the circles: 4, 7, 8, 11, 13, and 14.

Key Takeaways

A Venn diagram partitions the universal set into regions based on set membership. The intersection contains elements satisfying both conditions, the exclusive regions contain elements satisfying only one, and the exterior contains elements satisfying neither.

Common Mistakes

Placing 30 in the diagram (it is outside the universal set). Forgetting that 1 is a factor of every number. Misclassifying 9 or 12 as factors of 30 (they are not, since 30 divided by 9 or 12 is not an integer).

Things to Be Careful About

Every element of the universal set must be placed exactly once. Do not leave any numbers out, and do not repeat any numbers. The diagram must be fully populated to earn full marks.

Techniques used

find the intersection of two setsidentify elements unique to each setidentify elements outside both sets

(c)

$x \in (A \cup B)'$

Find the smallest value of $x$ .

______

(d)

Find $n(A \cap B')$ .

______

Q5MediumAlgebra and Graphs

(a)

Solve.

3M

(i)

\frac{y}{4} = 8

$y$ = ______

(ii)

3 - 4x = 2x + 12

$x$ = ______

(b)

$w = 5x - 6y$

4M

(i)

Find the value of $w$ when $x = 6.2$ and $y = -1.8$ .

$w$ = ______

(ii)

Rearrange the formula to make $x$ the subject.

$x$ = ______

(c)

Factorise.

15y - x^2 - 3xy + 5x

______

(d)

Write as a single fraction in its simplest form.

\frac{2}{x - 3} - \frac{4}{x + 3} + 1

______

DifficultyMedium

Worked solution

Approach

Write each term over the common denominator $(x-3)(x+3)$ , combine the numerators, then expand and simplify.

Working

\frac{2}{x - 3} - \frac{4}{x + 3} + 1

Use the common denominator $(x - 3)(x + 3)$ :

\frac{2(x+3)}{(x-3)(x+3)} - \frac{4(x-3)}{(x-3)(x+3)} + \frac{(x-3)(x+3)}{(x-3)(x+3)}

Combine numerators:

\frac{2(x+3) - 4(x-3) + (x-3)(x+3)}{(x-3)(x+3)}

Expand the numerator:

2x + 6 - 4x + 12 + x^2 - 9

Simplify:

x^2 - 2x + 9

Therefore:

\frac{x^2 - 2x + 9}{(x-3)(x+3)} = \frac{x^2 - 2x + 9}{x^2 - 9}

Answer

\frac{x^2 - 2x + 9}{x^2 - 9}

Final answer

(x^2 - 2x + 9)/(x^2 - 9)

Detailed explanation

Walkthrough

The expression has two fractions and a whole number $1$ . To combine them, use the common denominator $(x-3)(x+3)$ . Rewrite each fraction over this denominator: the first becomes $\frac{2(x+3)}{(x-3)(x+3)}$ , the second becomes $-\frac{4(x-3)}{(x-3)(x+3)}$ , and $1$ becomes $\frac{(x-3)(x+3)}{(x-3)(x+3)}$ . Combine the numerators: $2(x+3) - 4(x-3) + (x-3)(x+3)$ . Expand to $2x+6 -4x+12 + x^2 -9$ , which simplifies to $x^2 - 2x + 9$ . Thus the result is $\frac{x^2 -2x +9}{(x-3)(x+3)}$ or equivalently $\frac{x^2 -2x +9}{x^2 -9}$ .

Key Takeaways

To add or subtract algebraic fractions, first find a common denominator, then combine numerators. A whole number is treated as a fraction over the common denominator. Expanding and simplifying gives the final fraction.

Common Mistakes

Forgetting to include the $+1$ as a fraction; only combining the first two fractions.
Incorrect expansion of $(x-3)(x+3)$ as $x^2 - 9$ is fine, but sign errors in $2(x+3)$ or $-4(x-3)$ are common.
Writing $-4(x-3) = -4x - 12$ instead of $-4x + 12$ .
Leaving the denominator as $x^2 - 9$ but not simplifying the numerator, or vice versa.

Things to Be Careful About

This is a 4-mark question. The mark scheme awards B2 for a correct expanded numerator, B1 for the common denominator $(x-3)(x+3)$ , and B1 for the final simplified fraction; if the final answer is correct you score full marks. Intermediate lines must show the expanded numerator to earn partial credit. The denominator may be left as $(x-3)(x+3)$ or expanded to $x^2 - 9$ ; both are accepted. Make sure the fraction is in its simplest form, so the numerator is collected as $x^2 - 2x + 9$ .

Techniques used

find a common denominatorrewrite each term as an equivalent fractionexpand brackets in the numeratorcollect like termssimplify the fraction

Q6MediumMensurationNumberGeometry

(a)

[Volume of sphere = $\frac{4}{3}\pi r^3$ ]

Fig. 5

The diagram shows a sphere inside a cube.
The sphere touches all 6 faces of the cube.
The volume of the cube is $343\,\text{cm}^3$ .

Calculate the volume of the sphere.

______ $\text{cm}^3$

DifficultyMedium-Easy

Worked solution

Approach

Find the side length of the cube from its volume. Since the sphere is inscribed and touches all six faces, its diameter equals the side length of the cube, giving the radius. Substitute this radius into the given sphere volume formula.

Working

The volume of the cube is $343\,\text{cm}^3$ , so its side length $s$ is:

\begin{aligned} s^3 &= 343 \\ s &= \sqrt[3]{343} \\ s &= 7\,\text{cm} \end{aligned}

The sphere touches all 6 faces of the cube, so its diameter is equal to the side length of the cube:

\text{Diameter} = 7\,\text{cm} \implies r = 3.5\,\text{cm}

Using the formula for the volume of a sphere:

\begin{aligned} V &= \frac{4}{3}\pi r^3 \\ &= \frac{4}{3}\pi (3.5)^3 \\ &= \frac{4}{3}\pi (42.875) \\ &= \frac{171.5}{3}\pi \\ &\approx 179.594\,\text{cm}^3 \end{aligned}

Answer

180\,\text{cm}^3

Final answer

180

Detailed explanation

Walkthrough

First, we find the side length of the cube by taking the cube root of its volume: $\sqrt[3]{343} = 7\,\text{cm}$ . Because the sphere is perfectly inscribed and touches all six faces, its diameter must be exactly the side length of the cube. Therefore, the radius is half of 7, which is $3.5\,\text{cm}$ . We then substitute this radius into the volume formula $\frac{4}{3}\pi r^3$ . Calculating $3.5^3$ gives $42.875$ , and multiplying by $\frac{4}{3}\pi$ yields approximately $179.594$ . Rounding to 3 significant figures gives $180$ .

Key Takeaways

The side length of a cube is the cube root of its volume.
An inscribed sphere has a diameter equal to the side length of the enclosing cube.
Always use the exact radius in the volume formula before rounding the final answer.

Common Mistakes

Forgetting to halve the side length to get the radius (using 7 instead of 3.5).
Rounding intermediate values too early, which can lead to an answer like $179.5$ or $179.6$ instead of the correct $180$ to 3 s.f.
Using the area formula $4\pi r^2$ instead of the volume formula $\frac{4}{3}\pi r^3$ .

Things to Be Careful About

The question is on the calculator component, so 3 significant figures is the standard accuracy unless stated otherwise. Both $180$ and $179.6$ are accepted by the mark scheme, but $180$ is the proper 3 s.f. answer.
Ensure your calculator is in radian mode if using trigonometric functions later, though not needed here. Just use the $\pi$ button for accuracy.

Techniques used

find the side length of a cube from its volumerelate the inscribed sphere's diameter to the cube's side lengthapply the sphere volume formula

(b)

Solid $A$ is mathematically similar to solid $B$ .
The volume of solid $A$ is $540\,\text{cm}^3$ and its height is $15\,\text{cm}$ .
The volume of solid $B$ is $1280\,\text{cm}^3$ .

Calculate the height of solid $B$ .

______ $\text{cm}$

DifficultyMedium-Easy

Worked solution

Approach

For mathematically similar solids, the ratio of their volumes is the cube of the linear scale factor. Find the volume scale factor, take its cube root to get the linear scale factor, and multiply the height of solid A by this factor to find the height of solid B.

Working

The volume scale factor (V.S.F.) from A to B is:

\text{V.S.F.} = \frac{\text{Volume of B}}{\text{Volume of A}} = \frac{1280}{540} = \frac{64}{27}

The linear scale factor (L.S.F.) is the cube root of the volume scale factor:

\text{L.S.F.} = \sqrt[3]{\frac{64}{27}} = \frac{4}{3}

The height of solid B is:

\begin{aligned} \text{Height of B} &= \text{Height of A} \times \text{L.S.F.} \\ &= 15 \times \frac{4}{3} \\ &= 20\,\text{cm} \end{aligned}

Answer

20\,\text{cm}

Final answer

20

Detailed explanation

Walkthrough

Similar solids have proportional linear dimensions, areas, and volumes. The volume ratio is the cube of the linear ratio. Here, the volume ratio is $\frac{1280}{540}$ , which simplifies to $\frac{64}{27}$ . Taking the cube root gives the linear scale factor $\frac{4}{3}$ . Multiplying the height of solid A ( $15\,\text{cm}$ ) by $\frac{4}{3}$ gives the height of solid B as $20\,\text{cm}$ .

Key Takeaways

Volume scale factor = (linear scale factor) $^3$ .
Linear scale factor = $\sqrt[3]{\text{volume scale factor}}$ .
Always apply the linear scale factor to linear dimensions like height, not the volume scale factor.

Common Mistakes

Using the volume scale factor directly on the height instead of taking the cube root first.
Reversing the ratio (calculating $\frac{540}{1280}$ and multiplying by 15), which would give the wrong direction of scaling.

Things to Be Careful About

Ensure the ratio is set up correctly: height of B = height of A $\times$ $\frac{\text{Volume B}}{\text{Volume A}}$ raised to the $\frac{1}{3}$ power.

Techniques used

calculate the volume scale factorfind the linear scale factor by taking the cube rootapply the linear scale factor to the given height

(c)

[Curved surface area of a cone = $\pi rl$ ]

Fig. 6

The diagram shows a solid formed by joining a cone to a cylinder.
The cone and the cylinder each have radius $6.3\,\text{cm}$ .
The slant height of the cone is $8.7\,\text{cm}$ .
The ratio height of cone : height of cylinder = $2 : 3$ .

Calculate the total surface area of the solid.

______ $\text{cm}^2$

5M Similar

DifficultyMedium

Worked solution

Approach

The solid is a cone on top of a cylinder. The total surface area consists of the curved surface of the cone, the curved surface of the cylinder, and the circular base of the cylinder (the top of the cylinder is hidden by the cone). Find the height of the cone using Pythagoras' theorem, then use the ratio to find the height of the cylinder. Sum the relevant surface areas.

Working

1. Height of the cone:
Using Pythagoras' theorem with radius $r = 6.3$ and slant height $l = 8.7$ :

\begin{aligned} h_{\text{cone}}^2 &= 8.7^2 - 6.3^2 \\ &= 75.69 - 39.69 \\ &= 36 \\ h_{\text{cone}} &= 6\,\text{cm} \end{aligned}

2. Height of the cylinder:
The ratio of height of cone to height of cylinder is $2 : 3$ :

\begin{aligned} h_{\text{cylinder}} &= \frac{3}{2} \times h_{\text{cone}} \\ &= \frac{3}{2} \times 6 \\ &= 9\,\text{cm} \end{aligned}

3. Curved surface area of the cone:

\begin{aligned} \text{CSA}_{\text{cone}} &= \pi r l \\ &= \pi \times 6.3 \times 8.7 \\ &= 54.81\pi \end{aligned}

4. Curved surface area of the cylinder:

\begin{aligned} \text{CSA}_{\text{cylinder}} &= 2\pi r h_{\text{cylinder}} \\ &= 2\pi \times 6.3 \times 9 \\ &= 113.4\pi \end{aligned}

5. Area of the circular base:

\begin{aligned} \text{Base area} &= \pi r^2 \\ &= \pi \times 6.3^2 \\ &= 39.69\pi \end{aligned}

6. Total surface area:

\begin{aligned} \text{Total SA} &= 54.81\pi + 113.4\pi + 39.69\pi \\ &= 207.9\pi \\ &\approx 653.148\,\text{cm}^2 \end{aligned}

Answer

653\,\text{cm}^2

Final answer

653

Detailed explanation

Walkthrough

The total surface area of the composite solid includes three parts: the curved surface of the cone, the curved surface of the cylinder, and the flat circular base at the bottom. The internal circle where the cone meets the cylinder is not part of the surface area.

First, find the height of the cone. We have a right-angled triangle formed by the radius ( $6.3$ ), the height ( $h$ ), and the slant height ( $8.7$ ). Using Pythagoras: $h = \sqrt{8.7^2 - 6.3^2} = \sqrt{36} = 6\,\text{cm}$ .

Next, use the ratio $2:3$ for cone height to cylinder height. Since the cone height is 6, the cylinder height is $\frac{3}{2} \times 6 = 9\,\text{cm}$ .

Calculate the curved surface area of the cone: $\pi \times 6.3 \times 8.7 = 54.81\pi$ .
Calculate the curved surface area of the cylinder: $2\pi \times 6.3 \times 9 = 113.4\pi$ .
Calculate the base area: $\pi \times 6.3^2 = 39.69\pi$ .
Add them together: $54.81\pi + 113.4\pi + 39.69\pi = 207.9\pi \approx 653.148$ . Rounding to 3 significant figures gives $653$ .

Key Takeaways

For a compound shape, identify all exposed surfaces and sum their areas.
The internal face where two shapes join is not included in the total surface area.
Pythagoras' theorem is often needed to find missing dimensions like the height of a cone when the slant height and radius are given.

Common Mistakes

Including the area of the top circle of the cylinder (where it joins the cone) in the total surface area.
Forgetting to calculate the height of the cone using Pythagoras before applying the ratio.
Using the wrong formula for the curved surface area of the cylinder (e.g., forgetting the 2 in $2\pi rh$ ).

Things to Be Careful About

The question provides the formula for the curved surface area of a cone ( $\pi rl$ ) but not the cylinder; remember it is $2\pi rh$ .
Round only the final answer to 3 significant figures. Keep the answer in terms of $\pi$ during intermediate steps to avoid rounding errors.
Ensure units are consistent; all dimensions are in cm, so the area will be in $\text{cm}^2$ .

Techniques used

use Pythagoras' theorem to find the height of the coneuse the given ratio to find the height of the cylindercalculate the curved surface area of the cone and cylinderadd the base area of the cylinder to find the total surface area

Q7MediumTransformations and VectorsCoordinate Geometry

$ABCD$ is a parallelogram with sides $AB$ , $BC$ , $CD$ and $DA$ .
$A$ is the point $(-3, 7)$ and $B$ is the point $(2, 5)$ .

\overrightarrow{AD} = \begin{pmatrix}-1 \\ -6\end{pmatrix}

(a)

Find the coordinates of point $D$ .

( ______ , ______ )

(b)

Find $|\overrightarrow{AD}|$ .

$|\overrightarrow{AD}|$ = ______

(c)

Find $\overrightarrow{AC}$ .

$\overrightarrow{AC}$ = ______

DifficultyMedium-Easy

Worked solution

Approach

In parallelogram $ABCD$ , the diagonal $AC$ can be written as the sum of the two adjacent side vectors:

\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}

First find $\overrightarrow{AB}$ , then add $\overrightarrow{AD}$ .

Working

\overrightarrow{AB} = B - A = \begin{pmatrix}2 - (-3) \\ 5 - 7\end{pmatrix} = \begin{pmatrix}5 \\ -2\end{pmatrix}

Therefore

\overrightarrow{AC} = \begin{pmatrix}5 \\ -2\end{pmatrix} + \begin{pmatrix}-1 \\ -6\end{pmatrix} = \begin{pmatrix}4 \\ -8\end{pmatrix}

Answer

\overrightarrow{AC} = \begin{pmatrix}4 \\ -8\end{pmatrix}

Final answer

(4, -8)

Detailed explanation

Walkthrough

To get from $A$ to $C$ in a parallelogram, you can go along two adjacent sides: from $A$ to $B$ , then from $B$ to $C$ . Since $BC$ is parallel and equal to $AD$ in a parallelogram, $\overrightarrow{BC} = \overrightarrow{AD}$ . Hence

\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}

We first find $\overrightarrow{AB} = B - A = (5,-2)$ . Then we add $\overrightarrow{AD} = (-1,-6)$ component-wise: get $(4,-8)$ .

Key Takeaways

In a parallelogram, the diagonal vector from one vertex to the opposite vertex equals the sum of the two adjacent side vectors. This vector property is often the cleanest way to find diagonal lengths or coordinates.

Common Mistakes

Forgetting that $\overrightarrow{BC} = \overrightarrow{AD}$ , so the wrong vector is added.
Using $\overrightarrow{AC} = C - A$ with an incorrect $C$ .
Sign errors in component-wise addition, especially with negative components.

Things to be Careful About

The final answer must be given as a column vector here. If only one component is correct, the scheme may still give partial credit, but the complete vector is needed for full marks. Notice that $(-4,8)$ or $(8,-4)$ carry sign errors.

Techniques used

write a vector between two coordinatesuse the parallelogram rule for diagonalsadd column vectors component-wise

(d)

Line $L$ is the line perpendicular to $AB$ that passes through point $D$ .

Find the equation of line $L$ .

______

DifficultyMedium

Worked solution

Approach

Find the gradient of $AB$ , take the negative reciprocal to obtain the gradient of the perpendicular line $L$ . Then substitute the known point $D=(-4,1)$ into the equation of a line to find its $y$ -intercept.

Working

\text{gradient}\,(AB) = \frac{5 - 7}{2 - (-3)} = \frac{-2}{5}

A line perpendicular to $AB$ has gradient

-\frac{1}{-2/5} = \frac{5}{2}

Now use $y = \frac{5}{2}x + c$ with $D = (-4,1)$ :

1 = \frac{5}{2}(-4) + c

1 = -10 + c

c = 11

Answer

\boxed{y = \frac{5}{2}x + 11}

Final answer

y = (5/2)x + 11

Detailed explanation

Walkthrough

The gradient of $AB$ , slope of line, is found by dividing the change in $y$ by the change in $x$ :

m_{AB} = \frac{5-7}{2-(-3)} = \frac{-2}{5}

The perpendicular line $L$ has gradient $m_L$ satisfying

$m_{AB} \cdot m_L = -1$

so

$m_L = \frac{-1}{-2/5} = \frac{5}{2}$

Then use $D=(-4,1)$ and the form $y=mx+c$ : substituting gives $c=11$ . Therefore $y=\frac{5}{2}x+11$ .

Key Takeaways

Gradient: $\frac{y_2-y_1}{x_2-x_1}$ .
Perpendicular slopes multiply to $-1$ .
The equation of a line is determined by a slope and a point, using $y=mx+c$ or $y-y_1=m(x-x_1)$ .

Common Mistakes

Forgetting to use the correct $x$ -difference $2-(-3)$ , which gives 5, not 1.
Taking the reciprocal but not changing the sign: the perpendicular slope should be positive $\frac{5}{2}$ , not $\frac{5}{2}$ ? Actually it must be $\frac{5}{2}$ .
Using the wrong point for substitution; here it must be $D=(-4,1)$ from part (a), not $A(-3,7)$ .
Solving for $y$ incorrectly when substituting: $1 = -10 + c$ , so $c=11$ , not $-9$ .

Things to be Careful About

The question asks for the equation of line $L$ ; the final answer must include $y = \dots$ , not just the gradient or a single point. The answer is accepted in equivalent forms, e.g. $y = 2.5x + 11$ , as long as it is clearly the equation of the line.

Techniques used

calculate the gradient from two pointstake the negative reciprocal for a perpendicularsubstitute a point and gradient into the line equation

Q8Medium-HardNumberTrigonometryGeometry

Fig. 7

$ABCD$ is a field.
$AB = 320\,\text{m}$ , $BC = 250\,\text{m}$ , $CD = 132\,\text{m}$ and $AD = 365\,\text{m}$ .
Angle $B\hat{C}D = 90^\circ$ .

(a)

Ray walks from $A$ to $B$ at an average speed of $1.6\,\text{m/s}$ .
He then runs from $B$ to $C$ at an average speed of $2.8\,\text{m/s}$ .

Calculate Ray's average speed from $A$ to $B$ to $C$ .

______ $\text{m/s}$

DifficultyMedium-Easy

Worked solution

Approach

Average speed over a journey is the total distance travelled divided by the total time taken. Calculate the time for each leg using $t = d/v$ , then sum the distances and times.

Working

Total distance from $A$ to $C$ via $B$ :

d = AB + BC = 320 + 250 = 570 \text{ m}

Time from $A$ to $B$ :

t_1 = \frac{320}{1.6} = 200 \text{ s}

Time from $B$ to $C$ :

t_2 = \frac{250}{2.8} = \frac{625}{7} \approx 89.2857 \text{ s}

Total time:

t = 200 + \frac{625}{7} = \frac{2025}{7} \approx 289.2857 \text{ s}

Average speed:

v = \frac{570}{\frac{2025}{7}} = \frac{570 \times 7}{2025} = \frac{3990}{2025} = \frac{266}{135} \approx 1.9704 \text{ m/s}

Answer

1.97 \text{ m/s}

Final answer

1.97

Detailed explanation

Walkthrough

The question asks for the average speed over the entire journey from $A$ to $B$ to $C$ . A common mistake is to average the two speeds directly; instead, we must use the definition: average speed = total distance / total time.

First, find the total distance: $320 + 250 = 570$ m.

Next, find the time for each leg. Time equals distance divided by speed. For $A$ to $B$ , $320 / 1.6 = 200$ seconds. For $B$ to $C$ , $250 / 2.8 = 625/7 \approx 89.2857$ seconds.

Add the times to get the total journey time: $200 + 625/7 = 2025/7 \approx 289.2857$ seconds.

Finally, divide the total distance by the total time: $570 / (2025/7) = 3990 / 2025 \approx 1.97$ m/s.

Key Takeaways

Average speed is never the arithmetic mean of the speeds unless the time spent at each speed is equal. Always compute total distance and total time separately.

Common Mistakes

Averaging the two speeds: $(1.6 + 2.8) / 2 = 2.2$ m/s. This is incorrect because Ray spends more time at the slower speed.
Forgetting to convert or mixing up units (though here both speeds are in m/s and distances in m, so no conversion is needed).
Rounding intermediate time values too early, which can lead to an answer like $1.96$ or $1.98$ instead of the correct $1.97$ .

Things to Be Careful About

The mark scheme accepts $1.97[0\ldots]$ , so give the answer to at least 3 significant figures.
Show the full fraction $\frac{320 + 250}{\frac{320}{1.6} + \frac{250}{2.8}}$ in your working to secure the method marks.
Keep exact fractions during calculation to avoid rounding errors.

Techniques used

calculate time from distance and speed for each legsum total distance and total timedivide total distance by total time for average speed

(b)

The bearing of $D$ from $A$ is $243^\circ$ .

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

______

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

The bearing of $D$ from $A$ is given as $243^\circ$ . To find the bearing of $B$ from $A$ , we need the angle $BAD$ between $AD$ and $AB$ . We can find this by first computing the length of diagonal $BD$ using Pythagoras' theorem in the right-angled triangle $BCD$ , and then applying the cosine rule in triangle $ABD$ .

Working

Step 1: Find $BD$ using Pythagoras' theorem in $\triangle BCD$ .

Since $\angle BCD = 90^\circ$ :

BD^2 = BC^2 + CD^2 = 250^2 + 132^2 = 62500 + 17424 = 79924

BD = \sqrt{79924} \approx 282.708 \text{ m}

Step 2: Find $\angle BAD$ using the cosine rule in $\triangle ABD$ .

\cos(BAD) = \frac{AD^2 + AB^2 - BD^2}{2 \times AD \times AB}

\cos(BAD) = \frac{365^2 + 320^2 - 79924}{2 \times 365 \times 320}

\cos(BAD) = \frac{133225 + 102400 - 79924}{233600} = \frac{155701}{233600} \approx 0.666535

\angle BAD = \cos^{-1}(0.666535) \approx 48.19^\circ

Step 3: Calculate the bearing of $B$ from $A$ .

From the diagram, $B$ lies clockwise from $D$ when viewed from $A$ (i.e., $B$ is to the right of $AD$ , meaning its bearing is smaller than $243^\circ$ ). Therefore:

\text{Bearing of } B \text{ from } A = 243^\circ - \angle BAD = 243^\circ - 48.19^\circ = 194.81^\circ

Answer

195^\circ

Final answer

195

Detailed explanation

Walkthrough

The problem requires finding the bearing of $B$ from $A$ , given the bearing of $D$ from $A$ and the side lengths of the quadrilateral.

First, we isolate triangle $BCD$ , which is right-angled at $C$ . We use Pythagoras' theorem to find the length of the diagonal $BD$ : $BD^2 = 250^2 + 132^2 = 79924$ .

Next, we look at triangle $ABD$ , where we now know all three sides: $AB = 320$ , $AD = 365$ , and $BD = \sqrt{79924}$ . We use the cosine rule to find the angle at $A$ , which is $\angle BAD$ . The cosine rule rearranged for the angle is $\cos A = (b^2 + c^2 - a^2) / (2bc)$ . Substituting the values gives $\cos(BAD) = (365^2 + 320^2 - 79924) / (2 \times 365 \times 320) = 155701 / 233600$ . Taking the inverse cosine yields $\angle BAD \approx 48.19^\circ$ .

Finally, we use the bearing information. The bearing of $D$ from $A$ is $243^\circ$ . Looking at the diagram, $B$ is to the right of $AD$ (closer to South), so its bearing is less than $243^\circ$ . We subtract $\angle BAD$ from $243^\circ$ : $243 - 48.19 = 194.81^\circ$ , which rounds to $195^\circ$ .

Key Takeaways

Diagonals can split quadrilaterals into triangles where standard rules (Pythagoras, cosine rule) apply.
Bearings are measured clockwise from North. Subtracting an internal angle from a given bearing works when the target point lies clockwise from the reference line.

Common Mistakes

Using the sine rule instead of the cosine rule when all three sides are known (or two sides and the included angle is unknown).
Forgetting to square $BD$ correctly in the cosine rule (using $BD$ instead of $BD^2 = 79924$ ).
Adding $\angle BAD$ to $243^\circ$ instead of subtracting it, giving $291^\circ$ . Check the diagram: $B$ is to the right of $AD$ , so its bearing must be smaller than $243^\circ$ .
Rounding $\angle BAD$ to $48^\circ$ too early and getting $195^\circ$ (which happens to be correct here, but could fail in other cases). Keep at least 4 significant figures during intermediate steps.

Things to Be Careful About

The mark scheme accepts $195$ or $194.79$ to $194.81$ . Give the answer to 3 significant figures or as a whole number as appropriate for bearings.
Bearings must be given as three figures (e.g., $195^\circ$ , not $195$ ). The mark scheme accepts $195$ .
Show the substitution into the cosine rule clearly to earn the method marks: $\cos[ ] = \frac{365^2 + 320^2 - *their* BD^2}{2 \times 365 \times 320}$ .

Techniques used

apply Pythagoras' theorem to find diagonal BDapply the cosine rule to find angle BADsubtract angle BAD from the bearing of D to find the bearing of B

Q9MediumGeometryTrigonometry

(a)

Fig. 8

$ABD$ is a triangle.
$C$ is a point on $BD$ and $E$ is a point on $AD$ .
$AB$ is parallel to $EC$ and $CD = DE$ .

Find the value of $x$ and the value of $y$ .

$x$ = ______
$y$ = ______

DifficultyMedium-Easy

Worked solution

Approach

Use the parallel lines $AB \parallel EC$ to find alternate and corresponding angles. Then apply the angle sum property in $\triangle ABE$ to find $x$ , and use the isosceles triangle property in $\triangle CDE$ to find $y$ .

Working

Since $AB \parallel EC$ and $BE$ is a transversal, the alternate angles are equal:

\angle ABE = \angle BEC = 28^\circ

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

\angle BAE + \angle ABE + \angle AEB = 180^\circ

54^\circ + 28^\circ + x^\circ = 180^\circ

x = 180 - 54 - 28 = 98

Since $AB \parallel EC$ and $AD$ is a transversal, the corresponding angles are equal:

\angle CED = \angle BAD = 54^\circ

In $\triangle CDE$ , we are given $CD = DE$ , so the triangle is isosceles. The angles opposite the equal sides are equal:

\angle ECD = \angle CED = 54^\circ

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

\angle CDE + \angle CED + \angle ECD = 180^\circ

y^\circ + 54^\circ + 54^\circ = 180^\circ

y = 180 - 108 = 72

Answer

x = 98, \quad y = 72

Final answer

x = 98, y = 72

Detailed explanation

Walkthrough

First, we look at the parallel lines $AB$ and $EC$ . The line $BE$ cuts across them, creating alternate interior angles. Therefore, $\angle ABE$ must equal $\angle BEC$ , which is given as $28^\circ$ . Now we focus on $\triangle ABE$ . We know two of its angles: $\angle BAE = 54^\circ$ and $\angle ABE = 28^\circ$ . Since a triangle's angles add up to $180^\circ$ , we can find the third angle $\angle AEB$ (which is $x^\circ$ ) by subtracting the known angles from $180^\circ$ . This gives $x = 98$ .

Next, we use the parallel lines again. The line $AD$ acts as a transversal cutting $AB$ and $EC$ . The angles $\angle BAD$ and $\angle CED$ are in the same relative position at each intersection, making them corresponding angles. Thus, $\angle CED = 54^\circ$ . We are told that $CD = DE$ , which means $\triangle CDE$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal, so $\angle ECD = \angle CED = 54^\circ$ . Finally, using the triangle angle sum property on $\triangle CDE$ , we find $y^\circ = 180^\circ - 54^\circ - 54^\circ = 72^\circ$ .

Key Takeaways

Alternate angles are equal when a transversal crosses parallel lines.
Corresponding angles are equal when a transversal crosses parallel lines.
The interior angles of any triangle always sum to $180^\circ$ .
In an isosceles triangle, the base angles (opposite the equal sides) are equal.

Common Mistakes

Forgetting to identify the correct transversal line when using parallel line rules (e.g., using $AD$ to find alternate angles instead of corresponding angles).
Assuming $\triangle CDE$ is equilateral rather than just isosceles, or incorrectly assigning the equal angles.
Rounding intermediate angle values if they were not exact (though here they are exact integers).

Things to Be Careful About

Always explicitly state which geometric rule is being applied (alternate angles, corresponding angles, angle sum of a triangle, isosceles triangle properties) to secure method marks.
Ensure that the angles you identify as equal actually correspond to the correct sides in the isosceles triangle.
The question asks for the values of $x$ and $y$ ; make sure to provide both clearly.

Techniques used

identify alternate and corresponding angles with parallel linesuse the angle sum of a triangleapply isosceles triangle base angle properties

(b)

Fig. 9

$PQR$ is a triangle.
$P$ and $R$ are points on a circle, centre $O$ .
$O$ is a point on $PQ$ .
$QR$ is a tangent to the circle at $R$ .
$QR = 12\,\text{cm}$ and angle $R\hat{P}Q = 35^\circ$ .

Calculate the area of triangle $PQR$ .

______ $\text{cm}^2$

6M Similar

DifficultyMedium

Worked solution

Approach

Use the tangent-radius theorem to establish a right angle at $R$ . Use the fact that $OP = OR$ (radii) to find angles in $\triangle OPR$ , then determine angles in right-angled $\triangle OQR$ . Calculate the radius and use trigonometry to find the area of $\triangle PQR$ .

Working

Since $QR$ is tangent to the circle at $R$ and $OR$ is a radius, the tangent is perpendicular to the radius:

\angle ORQ = 90^\circ

In $\triangle OPR$ , $OP = OR$ (both are radii of the circle), so the triangle is isosceles:

\angle ORP = \angle OPR = 35^\circ

The exterior angle of $\triangle OPR$ at $O$ is:

\angle ROQ = \angle OPR + \angle ORP = 35^\circ + 35^\circ = 70^\circ

In right-angled $\triangle OQR$ :

\angle OQR = 90^\circ - 70^\circ = 20^\circ

Using the tangent ratio in $\triangle OQR$ :

\tan 20^\circ = \frac{OR}{QR} = \frac{OR}{12}

OR = 12 \tan 20^\circ \approx 4.369 \text{ cm}

The length $OQ$ is:

OQ = \frac{12}{\cos 20^\circ} \approx 12.754 \text{ cm}

Since $OP = OR$ :

OP = 12 \tan 20^\circ \approx 4.369 \text{ cm}

The total length $PQ$ is:

PQ = OP + OQ = 12 \tan 20^\circ + \frac{12}{\cos 20^\circ} \approx 17.123 \text{ cm}

The area of $\triangle PQR$ can be found using the formula $\frac{1}{2} ab \sin C$ :

\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR)

\text{Area} = \frac{1}{2} \times 17.123 \times 12 \times \sin 20^\circ

\text{Area} \approx 35.16 \text{ cm}^2

Rounding to 3 significant figures:

\text{Area} = 35.2 \text{ cm}^2

Answer

35.2 \text{ cm}^2

Final answer

35.2

Detailed explanation

Walkthrough

First, we apply the tangent-radius theorem: a tangent to a circle is always perpendicular to the radius at the point of contact. This means $\angle ORQ = 90^\circ$ , making $\triangle OQR$ a right-angled triangle.

Next, we look at $\triangle OPR$ . Since $O$ is the centre and $P, R$ are on the circle, $OP$ and $OR$ are both radii and therefore equal. This makes $\triangle OPR$ an isosceles triangle, so the base angles are equal: $\angle ORP = \angle OPR = 35^\circ$ . The exterior angle at $O$ , $\angle ROQ$ , is the sum of the two opposite interior angles, giving $35^\circ + 35^\circ = 70^\circ$ .

Now in right-angled $\triangle OQR$ , we can find the remaining angle: $\angle OQR = 90^\circ - 70^\circ = 20^\circ$ . We know $QR = 12$ cm. Using the tangent ratio, $\tan 20^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{OR}{12}$ , so the radius $OR = 12 \tan 20^\circ \approx 4.369$ cm. We can also find $OQ = \frac{12}{\cos 20^\circ} \approx 12.754$ cm.

Since $P, O, Q$ are collinear, $PQ = OP + OQ$ . Because $OP$ is also a radius, $OP = 4.369$ cm, so $PQ \approx 17.123$ cm.

Finally, we calculate the area of $\triangle PQR$ using the sine area formula: $\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR)$ . Substituting the values gives $\frac{1}{2} \times 17.123 \times 12 \times \sin 20^\circ \approx 35.16$ cm $^2$ . Rounded to 3 significant figures, this is $35.2$ cm $^2$ .

Key Takeaways

The tangent to a circle is perpendicular to the radius at the point of tangency.
Radii of the same circle are equal, often creating isosceles triangles.
The exterior angle of a triangle equals the sum of the two opposite interior angles.
The area of a triangle can be calculated using $\frac{1}{2}ab\sin C$ when two sides and the included angle are known.

Common Mistakes

Forgetting that $\angle ORQ = 90^\circ$ and trying to use the sine rule directly on $\triangle PQR$ without establishing the right angle first.
Incorrectly assuming $\triangle PQR$ is a right-angled triangle.
Adding $OP$ and $OQ$ incorrectly or forgetting that $PQ = OP + OQ$ since $O$ lies on $PQ$ .
Rounding intermediate values (like the radius or $OQ$ ) too early, which can lead to a final answer outside the acceptable range.

Things to Be Careful About

The question asks for the area to 3 significant figures (standard for calculator papers unless stated otherwise). Ensure your final answer is $35.2$ .
Keep full precision in your calculator for all intermediate steps. Use $12 \tan 20^\circ + \frac{12}{\cos 20^\circ}$ directly in the area formula rather than rounded decimals.
Explicitly state $\angle ORQ = 90^\circ$ using the tangent-radius theorem to secure method marks.

Techniques used

apply the tangent-radius theoremuse isosceles triangle properties and exterior anglescalculate the radius using tangent ratiocompute the area by splitting the triangle or using the sine area formula

Q10MediumAlgebra and Graphs

Bag $A$ contains red balls and green balls.
The total number of balls in the bag is $x$ .
The number of green balls in the bag is 6 more than the number of red balls.

(a)

Show that the fraction of the balls in bag $A$ that are red is $\frac{x - 6}{2x}$ .

DifficultyMedium-Easy

Worked solution

Approach

Let $r$ be the number of red balls in bag $A$ . Since there are 6 more green balls than red balls, the total is $r+(r+6)=x$ . Solve this for $r$ , then divide by the total $x$ .

Working

Let $r$ be the number of red balls. Then the number of green balls is $r+6$ , so

r+(r+6)=x

Hence

2r+6=x

Subtract 6 from both sides and divide by 2:

r=\frac{x-6}{2}

The fraction of balls that are red is the number of red balls divided by the total number of balls:

\frac{r}{x}=\frac{\frac{x-6}{2}}{x}=\frac{x-6}{2x}

Answer

The fraction is $\dfrac{x-6}{2x}$ , which matches the printed result.

Final answer

(x - 6)/(2x)

Detailed explanation

Walkthrough

Let $r$ represent the number of red balls. There are then $r+6$ green balls, so the total number of balls is $r+(r+6)=2r+6$ . Since this total is $x$ , we solve $2r+6=x$ for $r$ . This gives $r=\frac{x-6}{2}$ . A fraction is found by dividing the part by the whole, so the fraction of red balls is $\frac{r}{x}$ , which simplifies to $\frac{x-6}{2x}$ .

Key Takeaways

This part asks you to translate a worded relationship into an algebraic equation and then express one quantity as a fraction of a total. The phrase "6 more than" means add 6, and the fraction is always the chosen part divided by the whole.

Common Mistakes

Rearranging $2r+6=x$ incorrectly: $r$ is $\frac{x-6}{2}$ , not $\frac{x+6}{2}$ .
Stopping at $r=\frac{x-6}{2}$ without dividing by the total $x$ .
Mixing the numbers: green rods are 6 more than red rods, so green is $r+6$ , not $r-6$ .

Things to Be Careful About

For a "show that" part, every step of the algebra must be present. The final answer must come after the equation and its rearrangement; writing only the given fraction scores no marks because no reasoning is shown.

Techniques used

form an equation for the total number of ballssolve for the number of red ballswrite the number of red balls as a fraction of the total

(b)

Bag $B$ also contains red balls and green balls.
The number of red balls in bag $B$ is $x$ .
The number of green balls in bag $B$ is 4 times the number of green balls in bag $A$ .

Show that the fraction of the balls in bag $B$ that are red is $\frac{x}{3x + 12}$ .

DifficultyMedium

Worked solution

Approach

Use part (a) to find the number of green balls in bag $A$ . The number of green balls in bag $B$ is then 4 times that amount, while the number of red balls in bag $B$ is $x$ . Add these to obtain the total in $B$ and hence the fraction red.

Working

From part (a), the number of red balls is $\frac{x-6}{2}$ , so the number of green balls in bag $A$ is

x-\frac{x-6}{2}=\frac{x+6}{2}=\frac{x}{2}+3

In bag $B$ , the number of green balls is 4 times this:

4\left(\frac{x}{2}+3\right)=2x+12

Bag $B$ also contains $x$ red balls, so the total number of balls in bag $B$ is

x+(2x+12)=3x+12

Therefore the fraction that are red is

\frac{x}{3x+12}

Answer

The fraction is $\dfrac{x}{3x+12}$ , which matches the printed result.

Final answer

x/(3x + 12)

Detailed explanation

Walkthrough

We first find the number of green balls in bag A. The red balls are $\frac{x-6}{2}$ , so the green balls are the remainder: $x-\frac{x-6}{2}=\frac{x}{2}+3$ . Bag B has 4 times as many green balls, so its green balls are $2x+12$ . Since bag B has $x$ red balls, the total is $x+2x+12=3x+12$ . The red fraction is then red/total, or $\frac{x}{3x+12}$ .

Key Takeaways

This continues the “forming expressions” idea: once you have the number of greens in one bag, scaling them and adding the reds gives the new total. The fraction is then part over total.

Common Mistakes

Forgetting that the green balls in bag $B$ are $4\times$ the greens in bag $A$ , not $4\times$ the greens in $A$ plus the reds.
Using the total of bag $A$ as the denominator for bag $B$ instead of the newly calculated bag $B$ total.
Missing the division by 2 sign in $\frac{x}{2}+3$ .

The mark scheme requires the green balls in $B$ and the total in $B$ to be clearly found before the final fraction.

Techniques used

find the number of green balls in bag Amultiply the green balls by 4 for bag Bform the total number of balls in bag Bwrite the required fraction of red balls

(c)

\frac{x - 6}{2x} = \frac{x}{3x + 12}

Show that $x^2 - 6x - 72 = 0$ .

DifficultyMedium

Worked solution

Approach

Start from the equality of the two fractions in parts (a) and (b). Cross-multiply to remove the denominators, then expand the product and rearrange all terms onto one side.

Working

The two fractions are equal, so

\frac{x-6}{2x}=\frac{x}{3x+12}

Cross-multiply:

(x-6)(3x+12)=2x \cdot x

so

(x-6)(3x+12)=2x^2

Expand the left-hand side:

x(3x+12)-6(3x+12)=3x^2+12x-18x-72

Simplify:

3x^2-6x-72

Thus the equation is

3x^2-6x-72=2x^2

Subtract $2x^2$ from both sides:

x^2-6x-72=0

Answer

$x^2-6x-72=0$ , which is the required result.

Final answer

x^2 - 6x - 72 = 0

Detailed explanation

Walkthrough

You start with the equality of two fractions. Multiply both sides by both denominators to get rid of the fractions: this is called cross-multiplication. Then expand the product $(x-6)(3x+12)$ carefully, collecting like terms. The result is $3x^2-6x-72=2x^2$ . Moving $2x^2$ to the left gives $x^2-6x-72=0$ .

Key Takeaways

This part shows how an equation with fractions can be cleared by cross-multiplication. After expanding and simplifying, it becomes a quadratic equation that can be solved in the next part.

Common Mistakes

Expanding $(x-6)(3x+12)$ incorrectly: the constant term must be $-72$ , not zero.
Forgetting the middle term when multiplying: $12x-18x=-6x$ .
Subtract only one of the $2x^2$ terms, failing to get $x^2$ on the left.

The mark scheme has an A1 for the final correct equation; errors in earlier expansion prevent the mark.

Things to Be Careful

When subtracting $2x^2$ from both sides, make sure you take it from $3x^2$ to leave exactly $x^2$ . All terms must be placed on one side before writing the final zero after moving.

Techniques used

cross-multiply to eliminate the fraction denominatorsexpand a product of binomialsrearrange a quadratic equation

(d)

Solve by factorisation $x^2 - 6x - 72 = 0$ .

$x$ = ______ or $x$ = ______

(e)

$x$ is the total number of balls in bag $A$ .

Use your answer to part (d) to find the number of green balls in bag $A$ .

______

Q11MediumProbability

Mia has 25 shapes.
She uses their properties to sort them into groups.
The table shows the number of shapes in each group.

	Triangle	Quadrilateral
Line symmetry	4	9
No line symmetry	5	7

(a)

Mia takes one of the triangles at random, notes its properties and replaces it.

Find the probability that it has line symmetry.

______

DifficultyEasy

Worked solution

Approach

The question specifies that Mia takes one of the triangles at random. This restricts our sample space to only the shapes classified as triangles. We need to find the probability that this chosen triangle has line symmetry.

Working

First, determine the total number of triangles.
From the table, the columns under "Triangle" are:

Triangles with line symmetry: $4$
Triangles with no line symmetry: $5$

Total number of triangles = $4 + 5 = 9$ .

The number of favourable outcomes (triangles with line symmetry) is $4$ .

The probability $P$ is given by:

P(\text{Line symmetry} | \text{Triangle}) = \frac{\text{Number of triangles with line symmetry}}{\text{Total number of triangles}}

P = \frac{4}{9}

This fraction cannot be simplified further.

Answer

\frac{4}{9}

Final answer

4/9

Detailed explanation

Walkthrough

The key to this part is identifying the correct group. The problem states Mia picks from the triangles, not from all 25 shapes. Therefore, we ignore the quadrilateral column entirely.

Looking at the "Triangle" column:

There are $4$ triangles with line symmetry.
There are $5$ triangles without line symmetry.

The total number of possible outcomes is the sum of these two groups: $4 + 5 = 9$ .
The number of successful outcomes is the count of triangles with line symmetry, which is $4$ .

So, the probability is $\frac{4}{9}$ .

Key Takeaways

Always check the condition specified in the question (e.g., "takes one of the triangles") to define your denominator correctly. Do not use the grand total ( $25$ ) unless the question asks for a pick from the entire set.

Common Mistakes

Using the total number of shapes ( $25$ ) as the denominator instead of the total number of triangles ( $9$ ).
Only counting the triangles with symmetry ( $4$ ) but forgetting to add the ones without symmetry ( $5$ ) to get the total sample size.

Things to Be Careful About

Ensure you are looking at the correct column (Triangle vs Quadrilateral). The mark scheme awards a B1 for seeing an unsimplified fraction like $\frac{4}{9}$ or intermediate forms if they were more complex, but here it is straightforward.

Techniques used

identify the total number of outcomes in the reduced sample spaceidentify the number of favourable outcomesform the probability fraction and simplify

(b)

Mia takes one of the 25 shapes at random, notes its properties and replaces it.
She then takes a second shape at random, notes its properties and replaces it.

Find the probability that both shapes are quadrilaterals.

______

DifficultyMedium-Easy

Worked solution

Approach

Mia picks two shapes with replacement. This means the two events are independent; the outcome of the first pick does not affect the probabilities for the second pick. We need the probability that both shapes are quadrilaterals.

Working

First, determine the total number of quadrilaterals.
From the table, the columns under "Quadrilateral" are:

Quadrilaterals with line symmetry: $9$
Quadrilaterals with no line symmetry: $7$

Total number of quadrilaterals = $9 + 7 = 16$ .

The total number of shapes available to pick from is $25$ .

The probability of picking a quadrilateral on the first draw is:

P(Q_1) = \frac{16}{25}

Since the shape is replaced, the total remains $25$ and the number of quadrilaterals remains $16$ . The probability of picking a quadrilateral on the second draw is:

P(Q_2) = \frac{16}{25}

Because the events are independent, we multiply the probabilities:

P(\text{Both are quadrilaterals}) = P(Q_1) \times P(Q_2)

P = \frac{16}{25} \times \frac{16}{25}

P = \frac{256}{625}

The fraction $\frac{256}{625}$ cannot be simplified further (as $256 = 2^8$ and $625 = 5^4$ , they share no common factors).

Answer

\frac{256}{625}

Final answer

256/625

Detailed explanation

Walkthrough

The phrase "with replacement" is crucial. It tells us that after the first shape is picked and noted, it is put back into the pile. This resets the conditions for the second pick.

Step 1: Find the total number of quadrilaterals. Add the values in the Quadrilateral column: $9 + 7 = 16$ .
Step 2: Identify the total population. The problem states there are $25$ shapes in total.
Step 3: Calculate the probability of one event. The chance of picking a quadrilateral is $\frac{16}{25}$ .
Step 4: Combine the events. Since the picks are independent (due to replacement), we multiply the probability of the first event by the probability of the second event: $\frac{16}{25} \times \frac{16}{25}$ .

Key Takeaways

With replacement = Independent events. Multiply the same probability twice.
Without replacement = Dependent events. The denominators decrease, and potentially the numerators too.

Common Mistakes

Treating the events as dependent (without replacement) when the question says "with replacement".
Adding the probabilities instead of multiplying them (a common error for students confusing "and" with "or").
Miscalculating the total number of quadrilaterals.

Things to Be Careful About

Check if the resulting fraction can be simplified. In this case, $\frac{256}{625}$ is in its simplest form. The mark scheme accepts equivalent decimals if calculated, but fractions are preferred in exact form unless specified otherwise.

Techniques used

calculate the total number of quadrilateralsmultiply probabilities for independent eventssimplify the resulting fraction

(c)

Mia takes three of the 25 shapes at random without replacement.

Find the probability that only one of the shapes is a triangle with line symmetry.

______