Edexcel GCSE Mathematics – November 2017

What are we being asked to find?

We need to break down the number \(36\) into numbers that multiply together to make it, where every number must be a prime number (like \(2, 3, 5, 7, 11\), etc.).

Why we do this:

We can systematically divide \(36\) by the smallest prime numbers first. Let’s start by dividing by \(2\) because \(36\) is an even number.

What this tells us:

Our prime factors are \(2, 2, 3,\) and \(3\). We can check our answer by multiplying them back together: \(2 \times 2 = 4\), \(3 \times 3 = 9\), and \(4 \times 9 = 36\). Perfect!

What are we being asked to find?

We need to find the specific age of each person and then write them as a ratio. To do this, let’s turn the English sentences into algebra.

Why we do this:

Letting the youngest person’s age be a letter (like \(x\)) makes it easiest to construct the other ages without using fractions or subtraction.

Why we do this:

We are told the sum of all three ages is \(77\). We can add our three algebraic expressions together and set them equal to \(77\).

Why we do this:

Now that we know Jay’s age (\(x = 14\)), we can substitute this back into our expressions to find the other two ages, and finally write our answer in the requested ratio format (Jay : Kiaria : Martha).

What are we doing first?

We need to find the measure of angle \(\angle ABF\). To get there, we can use the properties of parallelograms to find the inner angles.

Why we do this:

In a parallelogram, opposite angles are equal. Since we are given \(\angle DCB = 75^\circ\), we know the opposite angle at vertex A must also be \(75^\circ\).

Why we do this:

Look at the lines crossing at point F (line AD and line BE). When two straight lines intersect, the angles opposite each other are equal. This will give us another angle inside triangle \(ABF\).

Why we do this:

Now we have two of the three angles inside triangle \(ABF\). We know \(\angle FAB = 75^\circ\) and \(\angle AFB = 35^\circ\). Since angles in a triangle always add up to \(180^\circ\), we can subtract the known angles from 180 to find \(\angle ABF\).

What this tells us:

We have successfully shown that \(\angle ABF = 70^\circ\) using geometric rules, exactly as the question asked us to show.

What are we being asked to find?

To check if Daisy is correct, we need to compare the shaded area to the total area of the logo. First, we need the exact radius of each circle.

Why we do this:

The area of any circle is \(\pi \times r^2\). We need the area of the whole logo, and the areas of the inner and middle circles to figure out just the shaded ring.

Why we do this:

The shaded part is a “ring” (an annulus). We find its area by taking the area of the middle circle and subtracting the empty inner circle that is cut out of it.

Why we do this:

We need to see if the shaded fraction is exactly \(\frac{1}{3}\). We take our shaded area and divide it by the total area.

What are we doing?

Because the earnings are grouped into ranges (classes), we don’t know exactly what each person earns. To estimate, we assume everyone in a group earns exactly the middle value of that group. This middle value is called the midpoint.

Why we do this:

If we assume \(11\) people earn exactly £\(300\), the total earned by that group is \(11 \times 300\). We must multiply each midpoint by its frequency, then add them all up to find the total estimated earnings for all 20 people.

Why we do this:

The mean is the total amount divided by the total number of people (\(20\)).

Why we do this:

We need to look at our data distribution. Most people (\(11 + 5 = 16\) people) earn between £250 and £450. But \(3\) people earn very high amounts (£550-£650). These extreme high values are called “outliers” and they drag the mean upwards, making it seem like the “average” worker earns more than they really do.

What are we being asked to find?

We need to prove that the top side (\(y\)) has a length of \(3\text{ cm}\). To get there, we first need to figure out what the side lengths actually are by finding \(x\).

Why we do this:

We know that opposite sides of a rectangle are exactly the same length. This means the left side (\(2x + 6\)) must be equal to the right side (\(5x – 9\)). We can turn this into an equation to find \(x\).

Why we do this:

Now that we know \(x = 5\), we can substitute it back into either expression to find the actual physical length of the side of the rectangle.

Why we do this:

We are told the area is \(48\text{ cm}^2\). The formula for the area of a rectangle is \(\text{base} \times \text{height}\) (or \(\text{width} \times \text{length}\)). We know the height is \(16\text{ cm}\), and the width is \(y\).

What this tells us:

We successfully worked backwards from the properties of the rectangle and its area to prove that the top side \(y\) must be \(3\text{ cm}\).

What are we looking for?

The equation given is \(y = x^2 + 1\). Any equation with an \(x^2\) term (and no higher powers) is called a quadratic equation. Quadratic equations always produce a smooth, “U”-shaped curve called a parabola.

What did she do wrong?

If you look closely at how the points (the crosses) are connected, you can see sharp corners. Brogan has used a ruler to draw straight lines between each point, rather than drawing a continuous, smooth curve freehand.

Because quadratic graphs are always smooth curves, joining the points with straight line segments makes the graph mathematically incorrect.

What are we being asked to do?

We need to order four decimal numbers. Three of them have dots above some digits. A dot means that digit (and any digits between dots) repeats forever. To compare them, the easiest method is to expand them out to a few decimal places so we can look at them side-by-side.

Why we do this:

It’s hard to compare recurring decimals just by looking at the dots. If we write out the first 5 or 6 decimal places, it becomes very easy to see which is largest.

Why we do this:

Just like putting words in alphabetical order, we look at the numbers column by column from left to right. The first time a column is different, the smallest number in that column belongs to the smallest overall number.

What this tells us:

Now we just need to write them down in this order using their original printed formats.

What are we being asked to find?

We eventually need Peter’s speed. To find speed, we need Distance and Time. We know Peter cycled \(15\text{ km}\) to catch up, but we don’t know his time yet. Let’s start by figuring out everything we can about James.

Why we do this:

We know James’s total distance (\(50\text{ km}\)) and total time (\(2.5\) hours). We can use the formula \(\text{Speed} = \text{Distance} \div \text{Time}\) to find his speed.

Why we do this:

Peter catches James at the \(15\text{ km}\) mark. If we know how long James had been cycling to get to that point, we can figure out Peter’s time.

Why we do this:

Peter caught up to James exactly when James had been cycling for \(45\) minutes. However, Peter started \(5\) minutes after James. This means Peter wasn’t cycling for the full \(45\) minutes.

Why we do this:

Now we have Peter’s distance (\(15\text{ km}\)) and his time (\(40\) minutes). The final answer needs to be in km/h, so we must first convert his \(40\) minutes into hours before dividing.

What are we doing?

A fractional power of \(\frac{1}{2}\) is exactly the same as finding the square root of the number.

Why we do this:

For a fractional power like \(\frac{2}{3}\), the fraction tells you two separate operations to perform on the number. The denominator (bottom) is the root, and the numerator (top) is the power.

• The \(3\) on the bottom means we take the cube root.
• The \(2\) on top means we then square the result.

It is always easier to do the root first so the numbers stay small.

What are we being asked to find?

We need to find two unknown prices (cost of a tea, cost of a coffee). Because we have two unknowns and two different scenarios, we can write this as a pair of simultaneous equations.

Why we do this:

Let’s use letters to represent the unknown costs: let \(t\) be the cost of one tea, and \(c\) be the cost of one coffee. We can translate the English sentences directly into algebra.

Why we do this:

To solve for one variable, we need to eliminate the other. If we look at the coffees, we have \(2c\) in the first equation and \(4c\) in the second. If we multiply the entire first equation by \(2\), both equations will have \(4c\). Then we can subtract one equation from the other to make the \(c\) terms disappear.

Why we do this:

Now that we know the cost of a tea (\(t = 1.40\)), we can substitute this value back into either of the original equations to find the cost of a coffee (\(c\)).

What are we doing?

To draw a box plot, we need 5 key values: the minimum, the lower quartile (LQ), the median, the upper quartile (UQ), and the maximum. We have three of these, but we need to calculate the UQ and the maximum using the range and interquartile range (IQR).

Why we do this:

The IQR is the difference between the Upper Quartile and Lower Quartile (\(UQ – LQ = IQR\)). The Range is the difference between the maximum and minimum (\(\text{Max} – \text{Min} = \text{Range}\)).

How we draw it:

We plot vertical lines for our 5 values: \(154\) (min), \(161\) (LQ), \(165\) (median), \(168\) (UQ), and \(174\) (max). We draw a box from the LQ to the UQ, and whiskers out to the min and max.

What are we comparing?

When asked to compare distributions, we MUST make two specific comparisons in context:

1. Compare an average (the Median). This tells us who is generally taller.
2. Compare the spread (the IQR or Range). This tells us which group’s heights are more consistent/varied.

What are we doing?

We know that the ratio of chicken pies in the small sample (\(15\) pies) is exactly the same as the ratio of chicken pies in the whole daily batch (\(450\) pies).

Why we do this:

We can set up equivalent fractions: \(\frac{\text{chicken in sample}}{\text{total in sample}} = \frac{\text{total chicken made}}{\text{total made}}\).

Why we do this:

On Tuesday, the number of steak pies in the sample was rounded to \(6\). We need the lower bound of this number to find the minimum possible probability. A number that rounds to \(6\) to the nearest whole number could be anything from \(5.5\) up to just below \(6.5\).

Why we do this:

Because the proportion in the sample is identical to the whole factory floor, the probability of picking a steak pie from the whole batch is identical to the proportion in the sample.

However, probability is usually not written with decimals inside fractions. We need to scale it up to the total factory numbers (\(450\)).

What are we being asked to do?

We need to rearrange a ratio statement into a specific algebraic formula where \(y\) is the subject (on its own on one side).

Why we do this:

A ratio \(A : B = C : D\) can be written as fractions: \(\frac{A}{B} = \frac{C}{D}\). This makes it much easier to do algebra.

Why we do this:

We have \(y\) stuck on both the top and bottom of a fraction. To fix this, we multiply both sides of the equation by the denominator \((y – x)\) to clear the fraction, and then expand the brackets.

Why we do this:

Because we want the final equation to say “\(y = …\)”, we need all terms with a \(y\) in them on one side, and everything else on the other side.

Why we do this:

We have \(ky – y\). To get a single \(y\), we factorise it out. We do the same for the \(x\) on the other side so it matches the format we’ve been asked to show.

What are we doing?

We want to convert a repeating decimal into a fraction. The trick is to create two equations where the infinitely repeating tail (the \(3636…\) part) perfectly lines up after the decimal point, so we can subtract them and make the infinite part vanish.

Why we do this:

By subtracting Equation 1 from Equation 2, the infinitely repeating “.3636…” tails cancel each other out completely, leaving us with a simple whole number equation.

Why we do this:

We divide by \(990\) to find \(x\) as a fraction. Then we simplify it down until we reach the target fraction of \(\frac{24}{55}\).

What are we being asked to find?

We need to find the specific value of \(y\) for a new \(x\). To do this, we first need to build the exact equation that links \(y\) and \(x\).

Why we do this:

“Directly proportional” means that \(y\) is equal to some constant multiplier (we call it \(k\)) times the other expression. We write this as an equation with \(k\) to help us find the multiplier.

Why we do this:

We are given a pair of values (\(x=8, y=1\frac{1}{6}\)) that we know work together. By substituting these into our equation, we can calculate the secret multiplier \(k\). It’s easier to work with improper fractions, so let’s convert \(1\frac{1}{6}\) into \(\frac{7}{6}\).

Why we do this:

Now that we have our complete formula, we can just plug in the new \(x\) value to find the required \(y\).

What are we being asked to do?

We need to add (“sum”) these two algebraic expressions together and simplify them. The final result should look like something squared (like \(x^2\) or \((something)^2\)).

Why we do this:

We can’t prove it’s a square number while it’s in two separate messy pieces. We need to combine them into one neat expression.

Why we do this:

Notice that both terms share a common factor: they both have a \(\frac{1}{2}\) and an \((n + 1)\). We can pull this common factor out to the front. (Alternatively, you could expand all the brackets, but factorising is quicker and less prone to errors!)

Why we do this:

We multiply the pieces together to show the final form, then explain why that form proves it’s a square number.

What this tells us:

Since the question states \(n\) is an integer (a whole number), \(n + 1\) must also be an integer. When you square an integer, the result is a square number. We have proven it!

What are we doing?

We need to enlarge shape P, but the scale factor is \(-\frac{1}{2}\). A negative scale factor does two things:

• The “\(\frac{1}{2}\)” part means the new shape will be exactly half the size of the original.
• The “negative” part means the shape is projected through the centre of enlargement (the origin) to the opposite side of the grid, turning it upside down.

Why we do this:

To safely draw the new shape, we take the coordinates of the corners of shape P, and multiply both the \(x\) and \(y\) values by our scale factor \(-\frac{1}{2}\).

Why we do this:

We plot these three new points on the grid and join them up. Because the scale factor was negative, notice how the shape is now inverted and on the opposite side of the origin \((0,0)\).

What are we looking for?

We need the equation of line \(M\). To get it, we need a point it passes through (like \(A\)) and its gradient. Let’s start by using line \(L\)’s equation to find the exact coordinates of \(E\) and \(B\).

Why we do this:

Looking at the diagram, \(E\) crosses the y-axis, meaning its x-coordinate is \(0\). Point \(B\) crosses the x-axis, meaning its y-coordinate is \(0\). We can substitute these into \(x + 2y = 12\).

Why we do this:

We are told \(AE = EB\). This means point \(E\) is exactly halfway between \(A\) and \(B\) (it’s the midpoint). We can use the midpoint logic backwards to find \(A\).

Why we do this:

Because \(ABCD\) is a rectangle, the corners are \(90^\circ\) right angles. This means line \(M\) (passing through AD) is exactly perpendicular to line \(L\) (passing through AB). If we know the gradient of \(L\), we can find the perpendicular gradient for \(M\).

Why we do this:

We now have everything we need for line \(M\): its gradient (\(m = 2\)) and a point it passes through (\(A(-12, 12)\)). We use the formula \(y = mx + c\) to find the full equation.

What are we being asked to do?

We need to find \(y\) at \(x=45^\circ\). But right now, our equation \(y = a \cos x^\circ + b\) has two unknown constants, \(a\) and \(b\). We must find these first using the table.

Why we do this:

We look for an \(x\) value in the table where \(\cos x\) is \(0\). We know that \(\cos 90^\circ = 0\). If we plug \(x=90\) into our equation, the ‘\(a\)’ will multiply by \(0\) and vanish, letting us find \(b\) easily.

Why we do this:

Now we know the equation is \(y = a \cos x^\circ + 1\). To find \(a\), we just need to plug in another pair of values from the table. Using \(x = 0\) is best because \(\cos 0^\circ = 1\).

Why we do this:

Now we have our complete equation: \(y = 2 \cos x^\circ + 1\). We simply substitute \(x = 45\) into it. We must know our exact trigonometric values for this (Non-Calculator paper).

What this tells us:

We successfully worked out the full formula from the data points, and applied our exact trig knowledge to find the specific missing value.

What are we being asked to do?

We need to remove the square root from the bottom of the fraction. This process is called “rationalising the denominator”.

Why we do this:

To eliminate a surd like \(\sqrt{2} – 1\) from the denominator, we multiply the top and the bottom of the fraction by its “conjugate”. The conjugate is exactly the same, but with the sign in the middle flipped. So, we multiply by \(\sqrt{2} + 1\).

Why we do this:

We must carefully expand the double brackets on both the top and the bottom, just like we would with algebraic expressions like \((x-y)(x+y)\).

Why we do this:

We need to collect like terms and simplify any surds that can be broken down to match the required format \(a + b\sqrt{2}\).

What this tells us:

We have successfully written the expression in the required format. Here, \(a = 2\) and \(b = 4\).

What are we being asked to find?

We need to find the length \(CB\) (which is \(x\)). We are told the large triangle \(CDA\) is similar to the smaller internal triangle \(BEA\). However, we are not told how they align. This is why there are two possible answers—there are two ways the triangles can be similar.

Let’s map out the total lengths of the large triangle’s sides first:

• Side \(AC = x + 8\)
• Side \(AD = 12 + 3 = 15\)

Why we do this:

Assumption 1: We assume that line \(BE\) is parallel to line \(CD\). This means the triangles are perfectly aligned (\(\angle ABE = \angle ACD\) and \(\angle AEB = \angle ADC\)).

Why we do this:

Assumption 2: We assume that the small triangle is inverted relative to the large one. This means \(\angle ABE = \angle ADC\) and \(\angle AEB = \angle ACD\).

What are we doing?

We are told about the relationship between their areas. So, first, we must write down algebraic expressions for the area of both shapes.

Why we do this:

The question states: “Area of rectangle > Area of triangle”. We insert our expressions into this inequality, expand the brackets, and rearrange it so that it is greater than zero. This turns it into a standard quadratic inequality that we can solve.

Why we do this:

To solve \(2x^2 – 5x + 2 > 0\), we first find the “critical values” by pretending it’s an “equals” sign and factorising it. These critical values tell us where the quadratic curve crosses the x-axis.

Why we do this:

A positive \(x^2\) quadratic makes a U-shape. It is greater than zero (\(>0\)) on the outside of the critical values. So mathematically, \(x < \frac{1}{2}\) or \(x > 2\).

However, this is a real-world geometry problem. Lengths cannot be negative or zero. Look at the rectangle’s height: \(x – 1\). If \(x\) was less than \(\frac{1}{2}\) (for example, \(0\)), the height would be \(0 – 1 = -1\text{ cm}\). This is impossible! Therefore, \(x\) must be large enough to make all sides positive.

What this tells us:

For the rectangle’s area to be strictly greater than the triangle’s area, while ensuring the physical shapes can actually exist, \(x\) must be any value greater than \(2\).