Pearson Edexcel GCSE (9-1) Mathematics
November 2022 Higher Paper 2 (Calculator)

What we are looking for: An outlier is a point that does not fit the general pattern or trend of the data.

Looking at the graph, most points show that as rainfall decreases, sunshine increases (or vice versa). There is a general downward trend.

However, there is one point that sits very low down when the rainfall is low (2mm), whereas the other points at low rainfall have high sunshine.

What we are looking for: A description of the correlation.

As the values on the x-axis (rainfall) increase, the values on the y-axis (sunshine) generally decrease.

Strategy: We need to draw a line of best fit through the main cluster of points (ignoring the outlier) and use it to read off the value.

1. Draw a straight line that passes through the middle of the points.

2. Find \(7\) on the y-axis (Sunshine).

3. Go across to your line, then down to the x-axis (Rainfall).

The Task: Draw the Side Elevation from the direction of the arrow.

The arrow is pointing from the Right. This means we are looking at the object from the right-hand side.

Dimensions:

• The width of the side elevation will be the depth of the object (from the Plan). The Plan is 3 squares high, so the object is 3 squares deep.
  → Side Elevation Width = 3 squares.
• The height of the side elevation will be the maximum height of the object (from the Front Elevation). The highest point on the left is 5 squares high.
  → Side Elevation Total Height = 5 squares.

Looking from the right:

1. We see the nearest face first. In the Front Elevation, the right-hand wall is 3 squares high. This forms the bottom part of our view.
2. Behind that (in the middle), the object steps up. The middle vertical line goes up to 4 squares high. Since this is taller than the front face, we will see this edge.
3. At the very back (left in Front Elevation), the object goes up to 5 squares high. We will see this top section as well.

This means our drawing will be a rectangle (3 wide × 5 high) with horizontal lines across it at heights of 3 and 4.

Why we do this: An arithmetic sequence changes by the same amount each time. This amount gives us the coefficient of \(n\).

Strategy: Set the \(n\)th term expression equal to \(-58\) and solve for \(n\). If \(n\) is a whole number (integer), then it is a term in the sequence.

Strategy: Break the garden into 4 shapes: Rectangle ABCO, Rectangle DEFO, Triangle CDO, and Sector AFO.

Dimensions:

• Rectangle ABCO: Height \(BC = 7\), Width \(AB = 11\). So \(OA = 7\), \(OC = 11\).
• Rectangle DEFO: Height \(EF = 9\), Width \(ED = 7\). So \(OD = 9\), \(OF = 7\).
• Triangle CDO: Base \(OC = 11\), Height \(OD = 9\).
• Sector AFO: Radius \(r = 7\) (since \(OA=OF=7\)). Angle \(90^\circ\).

Sum the areas to find the total garden area, then divide by the coverage per bag (\(14 \text{ m}^2\)).

Strategy: This is a right-angled triangle, so we use SOH CAH TOA.

• The angle is \(53^\circ\).
• The side labelled \(x\) is next to the angle (and is not the longest side), so it is the Adjacent.
• The side labelled \(14.5\) is opposite the right angle, so it is the Hypotenuse.

We have Adjacent and Hypotenuse, so we use CAH (Cosine).

Strategy: To increase a number by a percentage, we multiply by \((1 + \text{rate})\).

• Year 1: \(3\% \text{ increase} \rightarrow 1 + 0.03 = 1.03\)
• Year 2: \(1.5\% \text{ increase} \rightarrow 1 + 0.015 = 1.015\)

What we are looking for: The point where the graph crosses the vertical y-axis (where \(x=0\)).

Looking at the graph at \(x=0\), or looking at the constant term in the equation \(y = x^2 – 6x + \mathbf{4}\).

What we are looking for: The lowest point (minimum) of the curve.

Looking at the graph, the symmetry line is at \(x=3\). The lowest point is at \(x=3\).

What we are looking for: The “roots” are the values of \(x\) where \(y = 0\). This corresponds to where the curve crosses the x-axis.

Why we do this: Percentage profit is based on the actual money gained compared to the original cost.

Formula: \(\frac{\text{Profit}}{\text{Original Cost}} \times 100\)

Concept: Two lines are parallel if they have the same gradient (slope).

We need to arrange both equations into the form \(y = mx + c\), where \(m\) is the gradient.

Concept: Depreciation decreases value. Multiplier = \(1 – \text{rate}\).

• Year 1 (23% loss): \(1 – 0.23 = 0.77\)
• Year 2 (19% loss): \(1 – 0.19 = 0.81\)

Let \(P\) be the original price.

Original Price \(\times\) Year 1 Multiplier \(\times\) Year 2 Multiplier = Final Value

What we need: A box plot requires 5 key values:

1. Minimum value (Least time)
2. Lower Quartile (LQ)
3. Median
4. Upper Quartile (UQ)
5. Maximum value (Greatest time)

Given values:

• Minimum = 24
• Maximum = 96

Method: Use the cumulative frequency total (60) to find the positions.

• Median Position: \(50\%\) of \(60 = 30\). Go across from 30 on the y-axis to the curve, then down to read the time.
• Lower Quartile Position: \(25\%\) of \(60 = 15\). Go across from 15 on the y-axis.
• Upper Quartile Position: \(75\%\) of \(60 = 45\). Go across from 45 on the y-axis.

Understanding the math: The population is multiplied by \(1.13\) each year (a \(13\%\) increase).

After \(n\) years, the population will be multiplied by \(1.13^n\).

We want the population to double, so we need the multiplier to be \(2\) (or more).

We need to solve \(1.13^n > 2\).

Reasoning: If \(k\) increases, the population grows faster each year.

If it grows faster, it will reach the doubled size sooner (in less time).

Strategy: The length of a line segment between two coordinates \((x_1, y_1)\) and \((x_2, y_2)\) is found using the distance formula (derived from Pythagoras):

Strategy: Let \(x\) equal the recurring decimal. We need to multiply \(x\) by powers of 10 to shift the decimal point so that the recurring parts align and can be subtracted.

The decimal is \(1.062222…\) (only the 2 recurs).

Strategy: The proportion of tagged rabbits in the second sample should be roughly the same as the proportion of tagged rabbits in the whole population.

Let \(N\) be the total population.

Reasoning: Albie caught \(55\) rabbits in one go. The total population MUST be at least as big as the number of rabbits he actually held in his hands.

Method: Look for lines parallel to the axes.

• The top boundary is a horizontal line passing through \(y=6\). Since the shaded region is below this line, the inequality is \(y \le 6\).
• The left boundary is a vertical line passing through the x-axis halfway between -2 and -4, which is \(x=-3\). Since the region is to the right of this line, the inequality is \(x \ge -3\).

Method: Find the equation \(y=mx+c\) for each line by finding the gradient \(m\) and y-intercept \(c\).

Concept: Since the prisms are similar, the ratio of their volumes is the cube of the linear scale factor (\(k^3\)).

We are given the area and base of Triangle B. We can find its height.

Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)

Since Prism B is 3 times larger than Prism A, we divide the height of B by the scale factor \(k\).

Strategy: We have two sides and the included angle (SAS). The formula for the area is:

Method 1: Factorising

Multiply \(a \times c = 6 \times (-6) = -36\).

Find two numbers that multiply to \(-36\) and add to \(5\).

The numbers are \(+9\) and \(-4\).

Strategy: Consider the front face (ADEF). We have a right-angled triangle \(AFD\) (or using diagonal \(FD\)).

\(AD\) is the width (\(9\)). \(AF\) is the height. \(FD\) is the diagonal (\(13\)).

Strategy: Look at the base (floor) triangle \(FGH\). This is a right-angled triangle at \(G\) (back-left corner).

We need the hypotenuse \(FH\).

Side \(GH\) is parallel to the width \(AD\), so \(GH = 9\).

Angle \(GHF = 49^\circ\). \(GH\) is the Adjacent side to angle \(H\). \(FH\) is the Hypotenuse.

Strategy: Consider the triangle \(FAH\) which cuts through the cuboid. \(AF\) is vertical, \(FH\) is horizontal (on the floor). Therefore, triangle \(FAH\) is right-angled at \(F\).

We want angle \(A\) (FAH).

• Opposite = \(FH \approx 13.718\)
• Adjacent = \(AF \approx 9.381\)

Strategy: To find the gradient of a curve at a specific point, we must draw a tangent (a straight line touching the curve) at that point and calculate its slope.

1. Locate \(t = 17.5\) on the horizontal axis.
2. Go up to the curve.
3. Place a ruler to touch the curve at that exact point and draw a long straight line.
4. Pick two points on your tangent line to calculate \(\frac{\text{Change in } y}{\text{Change in } x}\).

Concept: The gradient is \(\frac{\text{Change in } y}{\text{Change in } x}\).

Here, \(y\) is Volume (litres) and \(x\) is Time (seconds).

So Gradient = \(\frac{\text{Litres}}{\text{Seconds}}\).

Method: \(fg(x)\) means we put the function \(g(x)\) inside the function \(f(x)\).

Replace \(x\) in \(f(x)\) with \((2x+3)\).

Method:

1. Let \(y = h(x)\).
2. Swap \(x\) and \(y\).
3. Rearrange the equation to solve for \(y\).

Distance (D): \(10.6\) (nearest \(0.1\)).

Interval is \(0.1\). Half is \(0.05\). Bounds are \(10.6 \pm 0.05\).

Time (T): \(31 \text{ min } 48 \text{ s}\) (nearest \(1 \text{ s}\)).

First convert to seconds: \(31 \times 60 + 48 = 1908 \text{ seconds}\).

Interval is \(1\). Half is \(0.5\). Bounds are \(1908 \pm 0.5\).

Formula: \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)

The speed must be in km/h. We need to convert Time from seconds to hours by dividing by \(3600\).

\(T_{hours} = \frac{T_{seconds}}{3600}\).

To find the Lower Bound of Speed (\(V_{LB}\)), divide the Smallest Distance by the Largest Time.

To find the Upper Bound of Speed (\(V_{UB}\)), divide the Largest Distance by the Smallest Time.

Compare the two bounds:

• \(V_{LB} = 19.900…\)
• \(V_{UB} = 20.099…\)

We need to round them so they give the same value.

To 1 decimal place: \(19.9\) vs \(20.1\) (No match)

To 2 significant figures: \(20\) vs \(20\) (Match!)

Concept: The centre of the circle \(x^2+y^2=r^2\) is \((0,0)\).

The gradient of the radius joining \((0,0)\) to \(P(2.1, 2.8)\) is given by \(\frac{y}{x}\).

Concept: The tangent is perpendicular to the radius.

Therefore, \(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}}\) (negative reciprocal).

Use \(y – y_1 = m(x – x_1)\) with point \(P(2.1, 2.8)\).