Edexcel GCSE Math June 2017 – Higher Paper 3 (Calculator)

What are we doing first?

Before we can place numbers into the Venn diagram, we need to list all the possible numbers. The universal set (\(\mathcal{E}\)) contains all “odd numbers less than \(30\)”. Let’s write them all out to make sure we don’t miss any.

What this tells us:

There are exactly \(15\) numbers in our entire universe. Every single one of these numbers must be placed somewhere inside the rectangular box.

How do we fill it in?

It’s always best to start from the center (the intersection) and work outwards.

1. Intersection (\(A \cap B\)): Look for numbers that are in BOTH set \(A\) and set \(B\). The only number in both lists is \(15\).
2. Only \(A\): Take set \(A\) and remove the \(15\). The remaining numbers go in the left circle only.
3. Only \(B\): Take set \(B\) and remove the \(15\). The remaining numbers go in the right circle only.
4. Outside the circles (\((A \cup B)’\)): Look at our master list from Step 1. Any number that hasn’t been placed in a circle goes outside them, but still inside the rectangle.

What does \(A \cup B\) mean?

The symbol \(\cup\) stands for “Union”. It means anything in \(A\), anything in \(B\), or anything in both. Simply put, it’s every number located inside the circles.

Probability is defined as: \(\frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}}\)

What this tells us:

If we blindly pick an odd number less than 30, there is a \(\frac{7}{15}\) chance it will belong to set \(A\), set \(B\), or both.

Why we do this:

We have two equations with two unknowns (\(x\) and \(y\)). We can’t solve for both at the same time. We need a strategy to get rid of one of them. Because the \(x\) terms are identical in both equations (\(3x\)), we can eliminate \(x\) entirely by subtracting the second equation from the first equation.

What’s the next step?

Now that we have a simple equation with only \(y\), we can solve it by dividing both sides by 5.

Why we do this:

We know \(y = -2\), but we still need to find \(x\). We can substitute this value back into either of our original equations. Equation (1) looks simpler because it has a single \(y\) term.

Checking our work:

Let’s plug both values into Equation (2) to see if it holds true: \(3(-\frac{2}{3}) – 4(-2) = -2 + 8 = 6\). Since it equals 6, we know our answers are perfectly correct!

What is a median and how do we find it from a table?

The median is the “middle” value when all the data points are arranged in order. Since we know there are \(n = 25\) women total, we can find the position of the middle person using the formula: \(\frac{n + 1}{2}\).

Why we do this:

We use a running total (cumulative frequency) to find where that \(13^{\text{th}}\) woman sits. We add up the women group by group until we pass the number \(13\).

Understanding Probability Rules:

Zoe simply added the two probabilities together. You are only allowed to add probabilities like this if the events are “Mutually Exclusive”. This means it is impossible for both events to happen at the same time.

Could a woman have a dress size of 14 AND a shoe size of 7? Yes, of course. It’s perfectly possible to be in both categories at the same time. If Zoe just adds the totals, she might be double-counting women who fit both descriptions!

Why we do this:

To find out how many lemon cakes there are, we first need to strip away the cakes we already have information for. We can convert fractions and percentages into actual quantities of cakes using the total (\(420\)). This is a calculator paper, so we can use it to speed up the process.

Why we do this:

Now that we know exactly how many vanilla and banana cakes Daniel baked, we subtract them from the total. What’s left over MUST be the combination of lemon and chocolate cakes.

How do we split up the remaining cakes?

The ratio of lemon to chocolate is \(4:5\). Think of this as putting the cakes into identical boxes. \(4\) boxes are for lemon, and \(5\) boxes are for chocolate. In total, we have \(9\) boxes to split the \(153\) cakes into evenly. Once we find the value of one “box” (or one part), we can multiply it by 4 to get the lemon cakes.

Checking our work:

If lemon is \(68\), then chocolate is \(5 \times 17 = 85\). Let’s check the sum: \(120 (\text{Vanilla}) + 147 (\text{Banana}) + 68 (\text{Lemon}) + 85 (\text{Chocolate}) = 420\). The total matches exactly!

What is our strategy?

To identify polygon \(P\), we need to know its interior angle. If we look closely at point \(B\), we can see that three shapes meet there perfectly without any gaps: the square, the 12-sided polygon, and polygon \(P\). Because angles around a single point always sum to \(360^\circ\), if we find the angles for the square and the 12-gon, we can subtract them from \(360^\circ\) to find the angle for \(P\).

How do we calculate this?

There is a standard formula for finding the interior angle of any regular polygon. A fast method is to first find the exterior angle (\(360^\circ \div n\), where \(n\) is the number of sides) and then subtract it from \(180^\circ\) (since interior + exterior = \(180^\circ\) on a straight line).

Why we do this:

Now that we have two out of the three angles clustered around point \(B\), we can use subtraction to discover the missing angle, which belongs to Polygon \(P\).

What this tells us:

We now know that Polygon \(P\) has interior angles of \(120^\circ\). How many sides does a regular polygon with \(120^\circ\) interior angles have? We can reverse the trick we used in Step 2. If the interior is \(120^\circ\), the exterior is \(60^\circ\). Since exterior angles always sum to \(360^\circ\), dividing \(360^\circ\) by \(60^\circ\) will give us the number of sides.

What are we trying to find?

We need to find the overall density of the mixed drink. The key relationship between density, mass, and volume is:

\(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)

We already know the total volume is \(320 \text{ cm}^3\). However, we don’t have the total mass yet. To find the total mass, we need to calculate the mass of each individual ingredient first by rearranging the formula to: \(\text{Mass} = \text{Density} \times \text{Volume}\).

Why we do this:

By multiplying the given density and volume of each liquid, we find out exactly how heavy each component is. Then we can add them all together to get the total mass of the final drink.

Why we do this:

Now that we have the total mass (\(324.45 \text{ g}\)) and the total volume (\(320 \text{ cm}^3\)), we divide the mass by the volume to find the overall density of the mixture. Finally, we must follow the instruction to round to 2 decimal places.

What this tells us:

The final density (\(1.01\)) is between the lowest density (\(0.99\)) and the highest density (\(1.4\)). Since the mixture is mostly carbonated water (\(0.99\)), it makes perfect sense that the final density is very close to \(1.00\).

What are we being asked to find?

We need to find a missing side length in a right-angled triangle, and we are given an angle. This means we must use SOH CAH TOA.

First, label the sides relative to the given angle (\(23^\circ\)):

• Hypotenuse (H): The longest side opposite the right angle is \(AC\), which is \(15 \text{ cm}\).
• Opposite (O): The side across from the \(23^\circ\) angle is \(AB\), which is what we want to find.
• Adjacent (A): The side next to the angle is \(CB\).

We want the Opposite and we know the Hypotenuse. This tells us we need to use Sine (SOH).

How do we solve for AB?

To get \(AB\) by itself, we need to undo the division by \(15\). We do this by multiplying both sides of the equation by \(15\).

What is our strategy?

We are given the area of the circle. We can use the formula for the area of a circle (\(\text{Area} = \pi r^2\)) to find its radius, and then its diameter. Why do we want the diameter? Because if you draw a line straight across the circle connecting two opposite corners of the square, that line is exactly the diameter of the circle!

How do we find \(x\)?

Imagine the diagonal line drawn across the square. It cuts the square into two identical right-angled triangles. The hypotenuse of this triangle is the diameter of the circle we just found. The two shorter sides are the sides of the square, both length \(x\). We can use Pythagoras’ theorem (\(a^2 + b^2 = c^2\)).

Why we do this:

Now we simply solve the algebra to isolate \(x\). Remember to keep the unrounded numbers in your calculator as long as possible to avoid rounding errors.

What this tells us:

The length of the side of the square is \(5.59 \text{ cm}\). A quick mental check: if the area of the square is roughly \(5.5 \times 5.5 \approx 30.25\), this is comfortably less than the circle’s area of \(49\), which makes logical sense.

What is the Interquartile Range (IQR)?

The IQR measures the spread of the middle 50% of the data. It is calculated by subtracting the Lower Quartile (LQ) from the Upper Quartile (UQ).

\(\text{IQR} = \text{UQ} – \text{LQ}\)

Looking at the male box plot, the box represents the middle 50%. The left edge of the box is the LQ, and the right edge is the UQ. Each large grid square represents £100, and is divided into 10 small squares, meaning each small square represents £10.

How to plot:

We use the 5 key values from the table to draw the plot on the same axis scale (\(1 \text{ small square} = \text{£}10\)):

• Smallest value (Min): Plot a short vertical line at \(60\).
• Lower Quartile (LQ): Plot the left edge of the box at \(180\).
• Median: Plot a vertical line inside the box at \(300\).
• Upper Quartile (UQ): Plot the right edge of the box at \(350\).
• Largest value (Max): Plot a short vertical line at \(650\).
• Connect the Min to LQ and UQ to Max with horizontal lines (whiskers).

Is Chris correct?

When comparing “who spent more” on average using a box plot, we should look at the Median (the middle line inside the box), because it represents the typical amount spent and isn’t affected by extreme outliers.

• Female median = \(300\)
• Male median = \(250\)

What is the formula for compound interest?

\(\text{Final Amount} = \text{Initial Amount} \times (\text{Multiplier})^{\text{years}}\)

We know the Initial Amount (\(£6000\)), the Final Amount (\(£8029.35\)), and the number of years (\(5\)). Let’s call our unknown multiplier \(y\). The multiplier represents \(1 + \frac{x}{100}\).

Why we do this:

To find the interest rate, we first need to isolate the multiplier (\(y\)). We do this by dividing both sides by \(6000\), and then taking the 5th root to get rid of the power of 5.

What this tells us:

A multiplier of \(1.06\) represents \(100\%\) of the original amount PLUS an additional \(6\%\) of interest added on top. Therefore, the interest rate \(x\) must be \(6\).

What is the rule for combinations?

When you are choosing one item from group A and one item from group B, the total number of possible combinations is found by multiplying the number of options together.

\(\text{Total ways} = (\text{Number of shrubs}) \times (\text{Number of rose trees})\)

We know the total ways (\(215\)) and the number of shrubs (\(17\)). We need to find out how many rose trees there must be for Jeff to be correct.

Why we do this:

By dividing \(215\) by \(17\), we can find out how many rose trees are required to make \(215\) combinations. Since you cannot have a fraction of a tree type at a garden centre, the result MUST be a whole integer for Jeff’s statement to be possible.

What this tells us:

Because \(12.647…\) is not a whole number, it’s impossible to have exactly \(215\) combinations. If there were \(12\) rose trees, there would be \(17 \times 12 = 204\) ways. If there were \(13\) rose trees, there would be \(17 \times 13 = 221\) ways. \(215\) ways is mathematically impossible.

What is our strategy?

We have two different ratios that describe the same straight line from \(A\) to \(D\). To compare them, they need to be sharing the same “total parts” for the whole line \(AD\).

• From the first ratio: \(AB + BD = AD\). So, the line is split into \(1 + 5 = 6\) parts.
• From the second ratio: \(AC + CD = AD\). So, the line is split into \(7 + 11 = 18\) parts.

We need a common denominator-style approach. We will make both ratios represent the line \(AD\) in \(18\) total parts by multiplying the first ratio.

Why we do this:

Now that both ratios are operating on the same scale (\(18\) total parts for the line \(AD\)), we can directly read off the sizes of the segments and use simple subtraction to find the middle segment, \(BC\).

Checking our work:

Let’s verify using the other side of the line. Does \(BD – CD\) also equal \(4\)? Yes, \(15 – 11 = 4\). Let’s also check if all segments add up to \(18\): \(AB + BC + CD = 3 + 4 + 11 = 18\). The logic is perfectly sound.

What is our strategy?

Before we can write inequalities, we need to know the straight-line equations (\(y = mx + c\)) for the solid black lines that form the border of the shaded triangle.

How do we figure out the direction (\(\le\) or \(\ge\))?

We need to look at whether the shaded region is ABOVE or BELOW each line. If it’s above, we use \(\ge\) (because it’s greater). If it’s below, we use \(\le\). Because all the lines are solid (not dashed), we include the “or equal to” line underneath the inequality symbols.

A great way to verify is to pick a test point clearly inside the shaded region, like \((-2, -1)\), and plug it into the inequalities to see if it makes mathematical sense.

What this tells us:

These three inequalities trap all coordinates inside that specific triangle. Any \(x,y\) coordinate you pick inside that gray area will satisfy all three mathematical statements simultaneously.

What is our strategy?

To simplify an algebraic fraction, we must factorize both the top (numerator) and the bottom (denominator) fully. Then, if we spot identical brackets on the top and bottom, we can cancel them out.

How do we make \(v\) the subject?

Right now, \(v\) is stuck inside a bracket AND trapped in a denominator. Our goal is to manipulate the algebra until \(v\) is completely alone on one side of the equals sign. We do this step-by-step to “rescue” it.

What this tells us:

We’ve successfully created a formula where if you input the values for \(t\) and \(w\), the calculator will spit out the answer for \(v\) immediately.

What formula do we need?

We don’t have the perpendicular height of this triangle, so we can’t use \(\frac{1}{2} \times \text{base} \times \text{height}\). Instead, we must use the trigonometric area formula: \(\text{Area} = \frac{1}{2}ab\sin(C)\).

In this formula, \(a\) and \(b\) are the two sides that trap the included angle \(C\). Our sides are \((2x – 1)\) and \((x + 3)\), and our trapped angle is \(45^\circ\).

Why we do this:

We need to turn this messy expression into a standard quadratic equation (\(ax^2 + bx + c = 0\)) so we can solve for \(x\). We know from exact trig values that \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\).

How do we solve for x?

Because the question asks for the answer to 3 significant figures, this is a massive clue that it will not factorise easily into brackets. We must use the quadratic formula.

\(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Here, \(a = 2\), \(b = 5\), and \(c = -27\).

What this tells us:

If \(x = 2.63\), the sides of the triangle are \(5.63 \text{ m}\) and \(4.26 \text{ m}\). This creates a perfectly valid physical triangle that matches the required area.

How do iteration formulas work?

An iteration formula is like a loop. You take the starting number (\(x_0\)), plug it into the formula to get the next number (\(x_1\)). Then, you take that new answer (\(x_1\)) and plug it back into the exact same formula to get the next number (\(x_2\)), and so on. A good calculator trick is to type -2.5 then = to store it as “Ans”, and then type -2 - 4 / Ans².

Why did we do this iteration?

Iteration is a method used to find numerical solutions (roots) to complex equations that are hard to solve normally. We can prove that our iterative formula comes directly from rearranging the equation \(x^3 + 2x^2 + 4 = 0\).

What this tells us:

Because the iteration formula is just a rearranged version of the original equation, the values \(x_1, x_2, x_3\) are stepping closer and closer to the actual root of \(x^3 + 2x^2 + 4 = 0\). They are progressively better estimates of the solution.

✓ (C1) for stating iteration is an estimation of a solution[cite: 113].

What are the maximum and minimum possible values?

When a measurement is rounded, we find the bounds by halving the degree of accuracy and adding/subtracting it from the stated value.

• Time is rounded to nearest \(5\) minutes. Half of \(5\) is \(2.5\).
• Distance is assumed nearest \(10\) km. Half of \(10\) is \(5\).

How do we maximize a division?

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

To get the biggest possible answer from a fraction, you need the biggest possible numerator divided by the smallest possible denominator. So, we use Upper Bound Distance \(\div\) Lower Bound Time.

Also, the time is in minutes, but the speed needs to be in km/h. We must convert the minimum time into hours by dividing by 60.

What this tells us:

Even if the train travelled the absolute furthest possible distance in the absolute shortest possible time, it could only reach \(153.5 \text{ km/h}\). This is safely below \(160 \text{ km/h}\).

✓ (C1) for concluding “No” supported by the correct maximum figure[cite: 113].

What happens if the distance was more accurately measured?

If the distance is now correct to the nearest \(5\) km, the error margin shrinks. Half of \(5\) is \(2.5\).

What do we know about tangents and radii?

A crucial circle theorem states that a tangent (like line \(DA\)) meets a radius (like line \(OA\)) at exactly \(90^\circ\). Therefore, triangle \(OAD\) is a right-angled triangle.

We know the hypotenuse \(OD = 9 \text{ cm}\) (given) and the adjacent side \(OA = 5 \text{ cm}\) (radius). We can use trigonometry to find the angle at the centre, \(\angle AOD\).

Which angle do we need for the arc length?

The arc \(ABC\) is the long path around the back of the circle. To find its length, we need the “reflex angle” \(\angle AOC\) (the large angle on the left side of point \(O\)).

First, by symmetry, the bottom triangle \(OCD\) is identical to \(OAD\). So, the full interior angle \(\angle AOC = 2 \times 56.251…^\circ = 112.502…^\circ\).

Angles around a point sum to \(360^\circ\). To get the reflex angle for the outer arc, we subtract from \(360^\circ\).

How do we find arc length?

An arc is just a fraction of the full circumference. The formula is: \(\frac{\text{Angle}}{360} \times \pi \times \text{Diameter}\).

The radius is \(5\), so the full circumference is \(\pi \times 10\).

What this tells us:

A full circle with radius \(5\) has a circumference of \(10\pi \approx 31.4 \text{ cm}\). The reflex angle is about two-thirds of the circle. \(21.6\) is roughly two-thirds of \(31.4\), so our answer makes perfect geometric sense.

How do we start solving an inequality?

First, pretend it’s a normal equation with an equals sign (\(2x^2 + 3x – 2 = 0\)). We solve this to find the “critical values”—the exact points where the graph crosses the x-axis. To solve this quadratic, we can factorise it.

We need two numbers that multiply to \(2 \times -2 = -4\) and add to \(3\). Those numbers are \(4\) and \(-1\).

How do we handle the > sign?

Think about what the graph of \(y = 2x^2 + 3x – 2\) looks like. Because the \(x^2\) term is positive (\(2\)), it’s a happy, U-shaped parabola. It cuts through the x-axis at \(-2\) and \(0.5\).

The inequality asks: where is this graph greater than zero (\(> 0\))? This means we want the parts of the U-shape that are ABOVE the x-axis. These “arms” point outwards from the roots, in opposite directions.

Checking our work:

Let’s pick a test value in our “safe” zone. Let’s try \(x = 1\) (which is \(> 0.5\)). Plugging it in: \(2(1)^2 + 3(1) – 2 = 2 + 3 – 2 = 3\). Is \(3 > 0\)? Yes! The inequality holds true.

How do we find where a graph crosses the y-axis?

Every single point on the y-axis has an x-coordinate of \(0\). To find where ANY equation crosses the y-axis, simply substitute \(x = 0\) into the equation.

What is the original circle, and where does it go?

The equation \(x^2 + y^2 = 16\) represents a circle locked onto the origin \((0,0)\). Since the formula is \(x^2 + y^2 = r^2\), the radius \(r\) is \(\sqrt{16} = 4\).

The translation vector \(\binom{3}{0}\) tells us to slide the entire circle \(3\) units horizontally (to the right) and \(0\) units vertically. Therefore, the new centre moves from \((0,0)\) to \((3,0)\). The radius remains \(4\).

Why we do this:

We compile our findings into a clear visual sketch, ensuring we explicitly label the three requested coordinate pairs.

What this tells us:

Translations just shift an object without changing its size. A circle with radius 4 simply shifted its centre right by 3, making its span on the x-axis reach from \(-1\) to \(7\).