Edexcel GCSE Mathematics – June 2019 Higher Paper 3 (Calculator)

What are we being asked to do?

We need to place all the numbers from our universal set (\(1\) to \(9\)) into the correct regions of the Venn diagram.

Why we do this:

To ensure no numbers are placed twice, we should start from the innermost overlapping section (the intersection) and work our way outwards.

How to build it step-by-step:

• Intersection (middle): Find numbers that are in BOTH set \(A\) AND set \(B\). These are \(6\) and \(9\).
• Set \(A\) only (left circle): The numbers in \(A\) are \(1, 5, 6, 8, 9\). Since \(6\) and \(9\) are already in the middle, the remaining numbers for just the \(A\) circle are \(1, 5, 8\).
• Set \(B\) only (right circle): The numbers in \(B\) are \(2, 6, 9\). Since \(6\) and \(9\) are in the middle, only \(2\) goes in the \(B\)-only section.
• Outside the circles: The universal set contains all digits from \(1\) to \(9\). We have placed \(1, 2, 5, 6, 8, 9\). The remaining numbers are \(3, 4, 7\). These go outside both circles but inside the rectangle.

What are we being asked to find?

We need the probability of choosing a number from the set \(A \cap B\). This symbol ‘\(\cap\)’ means intersection (the overlap).

Why we do this:

We need to find the new total in the account before we can figure out how much is just the interest. We are dealing with compound interest, which means the interest multiplier is raised to the power of the number of years.

An increase of \(1.5\%\) means we calculate \(100\% + 1.5\% = 101.5\%\). As a decimal multiplier, this is \(1.015\).

Why we do this:

The question strictly asks for the total amount of interest, not the new total balance in the account. We must subtract the original £\(200\,000\) Katy invested to see how much extra money was made.

Why we do this:

The median is the middle value when the data is in order. We have \(80\) plants in total. The middle position is roughly the \(40^{\text{th}}\) to \(41^{\text{st}}\) value (\(\frac{n+1}{2} = 40.5\)). We need to keep a running total (cumulative frequency) of the plants to see which group contains the \(40^{\text{th}}\) plant.

How to draw a frequency polygon:

A frequency polygon is constructed by plotting points at the midpoint of each class interval, matching the frequency for that interval. Then, you join all points using a ruler (straight line segments).

First, find the midpoints:

• \(10\) to \(20 \rightarrow\) midpoint is \(15\)
• \(20\) to \(30 \rightarrow\) midpoint is \(25\)
• \(30\) to \(40 \rightarrow\) midpoint is \(35\)
• \(40\) to \(50 \rightarrow\) midpoint is \(45\)
• \(50\) to \(60 \rightarrow\) midpoint is \(55\)
• \(60\) to \(70 \rightarrow\) midpoint is \(65\)

The coordinates to plot are: \((15, 7), (25, 13), (35, 14), (45, 12), (55, 16), (65, 18)\).

What are we looking for?

We need to carefully inspect the axes (the numbering, labels, and starting points) and the way the data points are connected. Time series graphs have very specific mathematical rules.

Let’s break down the mistakes:

• Line connection: Time series graphs must use straight line segments to connect points. Sean has drawn a freehand curve.
• X-axis: The quarters listed for each year are “2, 3, 4”. The 1st quarter is completely missing for every year.
• Y-axis scale (part 1): The graph starts at \(6\) instead of \(0\). If a graph doesn’t start at \(0\), it must have a zig-zag ‘break’ line at the bottom to warn the reader, otherwise the visual difference is exaggerated.
• Y-axis scale (part 2): Look at the numbering between \(9\) and \(10\). It goes from \(9\) straight to \(10\), missing out the \(9.5\) label (making the scale non-linear or misleading at the top).

What do we know?

To find missing angles inside a polygon, we must first know what all the angles add up to.

How we do it:

The formula for the sum of interior angles is \((n – 2) \times 180^\circ\), where \(n\) is the number of sides. A hexagon has \(6\) sides.

What does symmetry tell us?

The problem states there is one line of symmetry. Because \(FA=BC\) and \(EF=CD\), the left side is a mirror image of the right side. This means the angles must match up:

• Angle \(A\) matches Angle \(B\). So, Angle \(A = 117^\circ\).
• Angle \(F\) (which is \(AFE\)) matches Angle \(C\) (which is \(BCD\)).
• Angle \(E\) matches Angle \(D\).

How do we find the missing angles?

We are given that “Angle \(BCD = 2 \times \text{Angle } CDE\)”. Let’s use some algebra to make this simpler.

Let Angle \(CDE = x\).

This means Angle \(BCD = 2x\).

Now apply our symmetry from Step 2 to label the rest of the shape:

• Angle \(D = x\)
• Angle \(E = x\) (due to symmetry)
• Angle \(C = 2x\)
• Angle \(F = 2x\) (due to symmetry)

Now we add up all the angles and set them equal to \(720^\circ\).

What is our final step?

The question asks for the size of angle \(AFE\) (which is Angle \(F\)). From our algebra in Step 3, we know that Angle \(F = 2x\).

We substitute our value of \(x = 81^\circ\) into this expression.

What are we calculating?

We are covering the OUTSIDE of the tanks with paint. This means we need to calculate the Surface Area, not the volume. A closed cylinder consists of three parts:

• The top circle
• The bottom circle
• The curved side (which rolls out into a rectangle)

How we do this:

First, we need the radius. The diameter is \(1.6\text{ m}\), so the radius is \(1.6 \div 2 = 0.8\text{ m}\).

Area of one circle = \(\pi \times r^2\)

Curved surface area = \(\pi \times d \times h\) (or \(2 \times \pi \times r \times h\))

What this tells us:

Now we know one tank needs roughly \(13\text{ m}^2\) of paint. Jeremy has \(3\) tanks. We multiply to find the total area needed, and then figure out how much area Jeremy’s paint can actually cover.

How we do this:

This question tests your ability to use a calculator correctly. It is safer to calculate the top (numerator) and the bottom (denominator) separately first to avoid typing errors, or you can use the fraction button on your calculator inside the square root.

What this tells us:

To round to \(3\) significant figures, we ignore any leading zeros. The first significant figure is the \(3\), the second is the \(1\), and the third is the \(9\).

Looking at the next digit (the fourth significant figure, which is a \(1\)): because it is less than \(5\), we round down (keep the \(9\) the same).

What is the rule for Pythagoras?

Pythagoras’ Theorem states that \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse (the longest side, which is always directly opposite the right angle).

Looking at the diagram, the right angle is at \(B\). The side opposite \(B\) is \(AC\). Therefore, \(AC\) is the hypotenuse, not \(BC\).

Why we do this:

An enlargement with scale factor \(1.5\) means that every side on the new shape (\(XYZ\)) must be exactly \(1.5\) times longer than the matching side on the old shape (\(PQR\)).

Let’s check the lengths by counting the squares on the grid.

How we measure the work:

This is an inverse proportion problem. A good way to solve these is to find the total “amount of work” required. We can measure work in “machine-days” (one machine working for one day).

If \(6\) machines working together can finish the job in \(9\) days, we multiply these numbers to find the total amount of work needed.

Why we do this:

The factory wasn’t running at full capacity for the first \(3\) days. We need to calculate how much work they actually managed to get done during that time, so we can see what’s left over.

What this tells us:

We subtract the work done from the total work needed. Then, because all \(6\) machines are working on “all other days”, we divide the remaining work by \(6\) to see how many extra days are required.

What are we trying to find?

We know Marie paid £\(28.80\) in tax, and we know this tax was exactly \(20\%\) of the total interest she earned. We need to work backwards to find what \(100\%\) of the interest was.

How we do this:

The interest rate is the amount of interest earned as a percentage of the original amount invested (£\(8000\)).

To turn any amount into a percentage, you put the part over the whole, and multiply by \(100\).

What are we trying to find?

We need to find the time it takes for a signal to travel a certain distance. The relationship between speed, distance, and time is:

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

Rearranging this to solve for time gives us:

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

What does this mean practically?

If something travels the exact same distance but moves at a slower speed, it will logically take more time to arrive.

Mathematically, in the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), if the denominator (Speed) gets smaller, the result (Time) gets bigger.

What is the rule for fractional powers?

A power (or index) is very different from multiplication. Patrick has confused a power of \(\frac{1}{4}\) with multiplying by \(\frac{1}{4}\) (or dividing by \(4\)).

The rule for a fractional index is \(x^{\frac{1}{n}} = \sqrt[n]{x}\).

So, a power of \(\frac{1}{4}\) actually means taking the fourth root of the number.

What’s the tricky part here?

The densities are given in grams per cubic centimetre (\(\text{g/cm}^3\)), but the volumes are given in litres. We must make sure the units match before calculating.

We know that \(1\text{ litre} = 1000\text{ cm}^3\). So, we must multiply the volumes by \(1000\).

Why we do this:

You cannot simply average two densities. To find the density of a mixed liquid, you must find the Total Mass and divide it by the Total Volume.

We use the formula: \(\text{Mass} = \text{Density} \times \text{Volume}\).

What this tells us:

Now we have the Total Mass and the Total Volume. We use the formula \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\) to find the final density.

What formula do we use?

We are given an angle and its two surrounding sides in the triangle \(AFE\). The side \(AF = 24\text{ cm}\), and the side \(AE\) is an unknown length.

The formula for the area of a triangle using trigonometry is: \(\text{Area} = \frac{1}{2} a b \sin(C)\).

Why we do this:

The question states: \(\text{Rectangle Area} = \text{Triangle } AFE + \text{Triangle } BCD\).

Since the triangles are congruent (identical), their areas are the same.

What this tells us:

The question tells us the ratio \(AB : AE = 1 : 3\). This means that the side \(AE\) is \(3\) times longer than the side \(AB\).

Now that we know \(AB = 12\text{ cm}\), finding \(AE\) is straightforward.

What do these function changes mean?

There are two transformations happening to the original function \(f(x)\):

1. \(f(-x)\) is a transformation inside the bracket. This affects the x-coordinates. Specifically, it is a reflection in the y-axis, meaning every x-coordinate gets multiplied by \(-1\).
2. \(- 3\) is a transformation outside the bracket. This affects the y-coordinates. Specifically, it is a vertical translation downwards by \(3\), meaning we subtract \(3\) from every y-coordinate.

What are we trying to find?

For a linear sequence, the differences between terms are constant. For a quadratic sequence (which has an \(n^2\) term), the first differences change, but the second differences (the differences between the differences) are constant.

How we use the second difference:

A rule for quadratic sequences is that the constant second difference is equal to \(2a\) (where the sequence starts with \(an^2\)).

We divide the second difference by \(2\) to find how many \(n^2\) we need.

What this tells us:

Now we compare our original sequence to the sequence generated purely by \(2n^2\). The difference between them will give us the rest of the formula.

What does each mathematical function look like?

• Inversely proportional: The equation looks like \(y = \frac{k}{x}\). You cannot divide by zero, so the graph has asymptotes at the \(x\) and \(y\) axes. It forms curves in the top-right and bottom-left quadrants. This perfectly matches Graph B.
• Trigonometrical function: Sin and Cosine graphs produce a repeating wave shape that oscillates up and down. This matches the wave seen in Graph A.
• Exponential function: The equation looks like \(y = k^x\). It starts close to the x-axis, crosses the y-axis, and then grows very rapidly upwards. This matches Graph D.
• Directly proportional to \(\sqrt{x}\): The equation looks like \(y = k\sqrt{x}\). You cannot square root a negative number, so it only exists on the positive side. It starts at \((0,0)\) and curves gently to the right. This matches Graph C.

How to tackle triple brackets:

You cannot expand all three at once. Choose two brackets, expand them completely using FOIL (First, Outer, Inner, Last) or a grid, and simplify. Then multiply that new expression by the third bracket.

Let’s start with the first two: \((2x + 1)(x + 3)\).

What is the strategy?

We need to solve \((1 – x)^2 < \frac{9}{25}\). Since the left side is a perfect square, we can square root both sides. BUT remember, when you square root an inequality, you must consider both the positive and negative roots (e.g. \(x^2 < 4\) means \(-2 < x < 2\)).

What are the upper and lower limits?

\(u = 26.2\) is rounded to \(1\) decimal place (which is \(3\) sig figs here). The limit is \(\pm 0.05\). So, the bounds are \(26.15 \le u < 26.25\).

\(a = 4.3\) is rounded to \(1\) decimal place (which is \(2\) sig figs here). The limit is \(\pm 0.05\). So, the bounds are \(4.25 \le a < 4.35\).

How to maximize a fraction:

To make the result of a division as LARGE as possible, you must divide the BIGGEST possible number by the SMALLEST possible number.

Therefore, we need the Upper Bound for the numerator (\(u^2\)) and the Lower Bound for the denominator (\(2a\)).

What does “suitable degree of accuracy” mean?

We have a Lower Bound of \(78.6003\) and an Upper Bound of \(81.0662\). The true value of \(D\) could be anywhere between these numbers.

To find a safe, accurate answer, we round both bounds until they match. The point where they are identical is our “suitable degree of accuracy”.

What is the strategy?

When you have one linear equation (\(3x + 4y = 7\)) and one quadratic equation (\(x^2 – 4y^2 = 9\)), you cannot use the standard elimination method. Instead, you must isolate one letter in the linear equation and substitute it into the quadratic equation.

Let’s make \(x\) the subject of the linear equation.

How we handle the squared fraction:

To square a fraction, you square the numerator (the top) and square the denominator (the bottom).

For the top, \((7 – 4y)^2 = (7 – 4y)(7 – 4y)\).

How we find the values of \(y\):

We can factorise \(5y^2 + 14y + 8 = 0\), or use the quadratic formula.

Let’s factorise. We need two numbers that multiply to \(40\) (\(5 \times 8\)) and add to \(14\). These numbers are \(4\) and \(10\).

What this tells us:

Because these are simultaneous equations, every \(y\) value belongs to a specific \(x\) value. We must pair them up correctly by putting our \(y\) values back into the linear equation.

What is the rule for histograms?

In a histogram, the area of each bar represents the frequency (the actual number of items). We calculate this by multiplying the class width by the frequency density (the height).

\(\text{Frequency} = \text{Class Width} \times \text{Frequency Density}\)

Why we do this:

The pie chart groups the data into three categories. We need to find the total number of onions that go to the food processing factory (which is the middle group: between \(60\text{g}\) and \(120\text{g}\)).

What this tells us:

A full pie chart is \(360^\circ\). We need to find what fraction of the total onions went to the factory, and then find that same fraction of \(360^\circ\).

What geometry rules apply?

A radius meets a tangent at exactly \(90^\circ\). Therefore, triangle \(OAB\) is a right-angled triangle, with the right angle at \(A\).

We know the length of \(OB\) is \(16\) (because it goes from \((0,0)\) to \((16,0)\)). This is the hypotenuse.

We can use SOH CAH TOA to find the radius \(OA\), which is Opposite the \(30^\circ\) angle.

How we link the radius to point P:

The general equation for a circle centred at the origin \((0,0)\) is \(x^2 + y^2 = r^2\).

We know the radius \(r = 8\), so the equation of our circle is \(x^2 + y^2 = 8^2 = 64\).

Because point \(P\,(3p, p)\) lies on the circle, its x and y coordinates must work in this equation.

What this tells us:

We now have a simple quadratic equation to solve. We combine the \(p^2\) terms and square root the result.

What are we trying to find?

To use trigonometry rules on triangle \(ABC\), we need to know the angle inside the triangle at \(B\) (Angle \(ABC\)).

All North lines are parallel. By drawing a line straight South from \(B\), we split the situation into two parts:

• The angle between South and the line \(BA\) is \(37^\circ\) (alternate interior / Z-angles with the \(37^\circ\) bearing at \(A\)).
• The angle between South and the line \(BC\) is \(180^\circ – 150^\circ = 30^\circ\) (angles on a straight line add to \(180^\circ\)).

Adding these together gives us the whole angle inside the triangle.

Why we do this:

We know two sides (\(8\) and \(9\)) and the angle trapped between them (\(67^\circ\)). This is the perfect setup for the Cosine Rule to find the third side (\(AC\)).

Cosine Rule: \(a^2 = b^2 + c^2 – 2bc \cos(A)\)

How we find the angle:

To find the final bearing from \(A\), we first need to know the angle inside the triangle at \(A\) (Angle \(BAC\)). We now have a known side-angle pair: Side \(AC = 9.42\) and Angle \(B = 67^\circ\). We can use the Sine Rule.

Sine Rule: \(\frac{\sin(A)}{a} = \frac{\sin(B)}{b}\)

What this tells us:

A bearing is always measured clockwise from North.

The bearing of \(C\) from \(A\) is the original \(37^\circ\) bearing PLUS the angle we just found inside the triangle (\(61.58^\circ\)).