Edexcel GCSE Mathematics – Higher Paper 3 (June 2018)

Why we do this:

We need to observe the general pattern of the points on the graph. As the age (x-axis) increases, the time taken (y-axis) tends to decrease. A line of best fit would slope downwards from left to right.

Why we do this:

We are told Kristina’s point (11 years, 12 seconds) is an outlier. If we plot \((11, 12)\) on the grid, it sits far below the main cluster of points for 11-year-olds. An outlier is a data point that differs significantly from other observations.

What this tells us:

Debbie is 15. Look at the x-axis: the data only goes up to around 14 years of age. Predicting a value outside the range of given data is called extrapolation, and it is unreliable because we don’t know if the trend continues.

Why we do this:

To simplify the expression, we first need to multiply the terms outside the brackets by each term inside the brackets. Watch out for the negative sign when multiplying \(-2\) by \(-2p\).

Why we do this:

Now we combine the \(p\) terms together and the number terms together to fully simplify the expression.

Why we do this:

We first need to know the target area. We count the squares to find the lengths of the parallel sides (let’s call them \(a\) and \(b\)) and the perpendicular height (\(h\)). Then we use the formula: Area = \(\frac{1}{2}(a+b)h\).

Why we do this:

We need a triangle with an area of 18. The formula for the area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). We need to choose a base and height so that \(\frac{1}{2}bh = 18\), meaning \(b \times h = 36\).

Why we do this:

In a probability tree diagram, the probabilities on branches that originate from the same point must always add up to exactly 1. We inspect the first set of branches.

Why we do this:

The probability of the dice landing on 4 is fixed at 0.65 for every throw. We check if the second set of branches reflects this correct probability.

Why we do this:

We need to find the size of the angle at B. Looking from angle B, we know the Adjacent side (BC = 7 cm) and the Hypotenuse (AB = 11 cm). Because we have Adjacent and Hypotenuse, we use the Cosine ratio: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\).

Why we do this:

We are asked how the value of \(\cos ABC\) changes. The new hypotenuse AB is \(11 – 1 = 10 \text{ cm}\). The adjacent BC is still 7 cm.

Why we do this:

We know that the probability of picking a blue counter is \(0.45\), and there are exactly 18 blue counters. This means that \(0.45\) (or 45%) of the total counters equals 18. We can use this to find the total number of counters in the bag.

Why we do this:

All probabilities in an experiment must add up to \(1\). We can subtract the known probabilities for blue and yellow to find the combined probability of drawing red or white.

Why we do this:

Now that we know the probability of picking a red counter (\(0.2\)) and the total number of counters (\(40\)), we can find exactly how many red counters there are.

Why we do this:

If the probability of picking a silver marble is \(0.5\) (or exactly half), the number of silver marbles must be exactly half of the total number of marbles. Since we cannot have a fraction of a marble, the total must be divisible by 2.

Why we do this:

To solve an equation with a fraction, it is easiest to multiply both sides by the denominator to eliminate it. Here, we multiply both sides by \(2\).

Why we do this:

We want all the terms involving \(x\) on one side, and all the constant numbers on the other side. We can achieve this by adding \(x\) to both sides, and then adding \(14\) to both sides.

Why we do this:

Now we just divide by 5 to find the value of a single \(x\).

Why we do this:

To find the missing angles inside the shape, we first need to know what all the angles inside a pentagon add up to. The formula for the sum of interior angles in an \(n\)-sided polygon is \((n – 2) \times 180^\circ\).

Why we do this:

We know three of the angles: \(125^\circ\), \(115^\circ\), and \(90^\circ\) (the right angle at D). We can subtract these from the total \(540^\circ\) to find what the remaining two angles (ABC and BCD) must add up to.

Why we do this:

The problem tells us that angle BCD is twice as big as angle ABC. We can set up an equation. If we let angle ABC be \(x\), then angle BCD must be \(2x\).

What this tells us:

The question specifically asks for angle BCD, which we established is \(2x\).

Why we do this:

We need to substitute the given standard form values into the formula and calculate the result carefully using our calculator.

Why we do this:

To prove Lottie wrong, we can mathematically check the impact of the percentage increases using decimal multipliers. Increasing by 10% is a multiplier of \(1.1\). Increasing by 5% is a multiplier of \(1.05\).

Why we do this:

The lamps will flash together at a time (in seconds) that is a common multiple of their individual flashing intervals: 20, 45, and 120. We need to find the Lowest Common Multiple (LCM) to find out how many seconds it takes for all three to sync up again.

Why we do this:

We need to know how many 360-second intervals fit into one hour. First, convert one hour into seconds, then divide by 360.

Why we do this:

We need to work backwards from 2012 to 2003. First, we find out what Mia originally paid Jerry in 2007. Because Mia made a 10% loss, the £162,000 she sold it for represents 90% of what she paid (100% – 10% = 90%). We use reverse percentages (dividing by the decimal multiplier) to find the original 100%.

Why we do this:

Now we know Jerry sold the house for £180,000. He made a 20% profit on his original 2003 purchase price. This means the £180,000 represents 120% of what he paid (100% + 20% = 120%). Again, we divide by the decimal multiplier to find his starting price.

Why we do this:

The gradient of a straight line is a measure of its steepness, calculated as “change in \(y\)” divided by “change in \(x\)”. First, we identify two clear points on the line. The line starts at \((0, 4)\) and ends cleanly at \((8, 16)\).

What this tells us:

The gradient tells us how much the y-value changes for every 1 unit of the x-value. Here, it is the change in Volume (litres) per unit of Time (seconds). Therefore, the gradient represents the rate at which liquid is filling the container in litres per second.

What this tells us:

The y-intercept is where the line crosses the y-axis, which corresponds to \(t = 0\) (time is zero). At this exact starting moment, the volume is already 4 litres. This represents the initial amount of liquid in the container before any more was added.

Why we do this:

We are given the ratio of the surface areas (Area Scale Factor). To convert between areas and volumes, we must first find the linear (Length) Scale Factor. We do this by square-rooting the Area Scale Factor.

Why we do this:

Once we have the linear ratio, we cube it to find the Volume Scale Factor. We then apply this to the volume of shape B to find the volume of shape A.

Why we do this:

There are 16 teams. Each team has to play the 15 other teams. If we multiply 16 by 15, we get the total number of fixtures if Team A playing Team B is considered a separate fixture from Team B playing Team A (which perfectly matches the condition “each plays two matches”).

Why we do this:

The distance travelled is equal to the area under a speed-time graph. We are asked to estimate this area using 4 strips over 20 seconds. The width of each strip is \( \frac{20}{4} = 5 \) seconds. We need to read the speeds (y-values) at \( t = 0, 5, 10, 15, \) and \( 20 \).

Why we do this:

We use the trapezium rule to find the area of each strip. The area of a trapezium is \( \frac{1}{2}(a+b)h \), where \(a\) and \(b\) are the parallel vertical heights (the speeds) and \(h\) is the strip width (5).

What this tells us:

When you draw straight lines between the points on this specific curve to make trapezia, the straight lines sit slightly below the actual curve (because the curve is bending downwards or is concave). This means the area of the trapezia misses a small sliver of area under the curve.

Why we do this:

We are given the general formula \(an^2 + bn\) and two specific terms. By substituting \(n=2\) and \(n=4\) into the formula, we can create two equations with two unknowns (\(a\) and \(b\))[cite: 109].

Why we do this:

To find \(a\) and \(b\), we can eliminate \(b\) by making the coefficients of \(b\) the same. Let’s multiply Equation 1 by 2.

What this tells us:

Now we know the full \(n\)th term formula is \(2n^2 – 5n\). We substitute \(n = 6\) into this formula to answer the question.

Why we do this:

To find the formula for a quadratic sequence, we first look at the differences between consecutive terms. The second difference will be constant, and half of this second difference gives us the coefficient of \(n^2\).

Why we do this:

We need to find the length of AD, which is in triangle ABD. To solve a triangle, we need at least three pieces of information. Right now, in triangle ABD, we only know one side (\(11.4\)) and one angle (\(86^\circ\)). However, triangle ABD shares the side BD with triangle BCD. We can use the information in triangle BCD to find the length of BD first.

Why we do this:

In triangle BCD, we know a side (\(12.5\)) and its opposite angle (\(109^\circ\)). We want to find the side BD, and we know its opposite angle is \(34^\circ\). Because we have matching “side-opposite angle” pairs, the Sine Rule is the perfect tool[cite: 110].

Why we do this:

Now look at triangle ABD. We know two sides (\(AB = 11.4\) and \(BD = 7.392…\)) and the angle between them (\(86^\circ\)). When you know two sides and the included angle, use the Cosine Rule to find the third side (\(AD\))[cite: 110].

Why we do this:

To prove a solution exists between two numbers, we substitute both numbers into the left side of the equation. If one result is below 7 and the other is above 7, the graph must cross 7 in between them[cite: 111].

Why we do this:

We need to manipulate the equation to isolate the \(x^3\) term first, and then undo the cube by taking the cube root.

Why we do this:

Iteration means repeating a process. We put our starting value (\(x_0 = 2\)) into the formula to get \(x_1\). Then we take \(x_1\) and put it back into the formula to get \(x_2\), and repeat this one more time to find \(x_3\)[cite: 111].

Why we do this:

We use SOH CAH TOA. The tangent of an angle is the Opposite side divided by the Adjacent side (\(\tan \theta = \frac{O}{A}\)). We can write an expression for \(\tan e\) and \(\tan f\) using the algebraic side lengths, and set them equal because we are told \(\tan e = \tan f\).

Why we do this:

To solve an equation with fractions on both sides, we “cross-multiply” to get rid of the denominators. This involves multiplying the top of one side by the bottom of the other[cite: 112].

Why we do this:

We now have \(x^2\) terms, \(x\) terms, and numbers. To solve a quadratic equation, we must move all terms to one side so the equation equals zero.

Why we do this:

We can solve this using the quadratic formula, or by factorising. Let’s use factorisation. We need two numbers that multiply to \(12 \times -5 = -60\) and add to \(-17\). Those numbers are \(-20\) and \(3\).

Why we do this:

To find how many people speak ONLY Spanish, we need to map out the overlapping groups. A Venn Diagram is the best way to sort this[cite: 114]. Let’s break down the clues:

Why we do this:

We need to figure out the empty regions to eventually find “Spanish ONLY”. We can deduce the F&G region using the German total, and the F ONLY region using the French total.

Why we do this:

We are picking two people at random. This means we calculate the probability of picking one “Spanish only” speaker, and multiply it by the probability of picking a second one. Because the first person isn’t replaced, the total number of people drops by 1.

Why we do this:

To prove two triangles are congruent (identical in shape and size), we need to find three matching features. This can be SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), or RHS (Right angle-Hypotenuse-Side). Let’s list the facts we know about triangles ADP and CBQ.

What this tells us:

We have two pairs of matching angles and one matching pair of included sides. We must state the final condition (ASA) to complete the proof formally.

Why we do this:

Now we look at the larger shape APCQ. We need to prove its opposite sides AQ and PC are parallel. We can do this by proving APCQ is a parallelogram. A quadrilateral is a parallelogram if one pair of opposite sides is both parallel and equal in length.