Edexcel GCSE Mathematics – May 2018 Paper 1 Higher (Non-Calculator)

Why we do this:

Before we can easily add these fractions, it helps to convert the mixed numbers into improper (top-heavy) fractions. Then, we must find a common denominator because we can only add fractions when their bottom numbers are exactly the same.

Why we do this:

When dividing fractions, we use the “Keep, Change, Flip” rule. First, we must change the mixed number into an improper fraction. Then we keep the first fraction, change the division sign to multiplication, and flip the second fraction upside down.

What are we being asked to find?

We need to find the actual total number of houses. We have two separate ratios linking Houses (H), Flats (F), and Bungalows (B), and we know the exact number of bungalows (50).

Why we do this:

To compare houses to bungalows directly, we need one combined ratio of H : F : B. The connecting variable is “Flats”, which appears in both ratios. We must make the “Flats” number the same in both ratios.

Why we do this:

We know the “5 parts” of bungalows actually represents 50 bungalows. We need to find what 1 ratio part is worth to find out what the 14 parts of houses represents.

What this tells us:

We successfully scaled our unified ratio up to reality using the known quantity. The village has 140 houses, 80 flats, and 50 bungalows.

What are we being asked to find?

We need to calculate the percentage profit. Percentage profit is found by dividing the actual profit by the original cost, then multiplying by 100.

To get there, we first need to know how many bags she makes, and how much total money she gets from selling them (revenue).

Why we do this:

To find total revenue, we need to know exactly how many 250 g bags can be made out of 5 kg of sweets. We must convert kg to g first so the units match.

Why we do this:

We need her total income to figure out how much actual profit (extra money) she made compared to her £10 initial spend.

Why we do this:

Percentage profit is calculated using the formula: \( \frac{\text{Profit}}{\text{Original Cost}} \times 100 \).

What this tells us:

Renee made a 30% return on her initial investment of £10.

Why we do this:

The word “estimate” is the critical instruction here. We must NOT calculate this exactly. On a non-calculator paper, when asked to estimate, we round numbers to 1 significant figure (or easily compatible numbers) to do the math in our heads.

Why we do this:

We need to state logically how an increased speed affects the real time taken, OR how it affects our estimated calculation. The mark scheme accepts either reasoning.

Why we do this:

The “front elevation” is what the 3D shape looks like when you look at it straight on from the front (flat 2D projection). From the front (looking at edge AB), we will see a flat triangle. The base of this triangle is the side AB (6 cm). The peak of this triangle is vertex V, which stands directly in the middle at a height of 4 cm.

Why we do this:

The total surface area is the sum of the areas of all the flat faces on the outside of the shape. A square-based pyramid has 5 faces: 1 square base, and 4 identical triangular sides.

What are we being asked to find?

We need to find the exact \( (x, y) \) coordinates of point C. C sits exactly in the middle of the pattern, at the corner connecting the second and third squares. To find it, we first need to figure out the side length of these identical squares.

Why we do this:

If we know how wide one square is, we can simply count squares from A to get to C. Looking at the horizontal distance (the x-coordinates), the pattern spans from A’s left edge to B’s right edge, covering exactly 4 full squares side-by-side.

Why we do this:

Now that we know every square is 8 units wide and 8 units tall, we can track the position of C starting from a known point. Let’s find the x-coordinate by moving right from A, and the y-coordinate by moving down from B.

What this tells us:

Point C is located at \( x = 22 \) and \( y = 20 \). Alternatively, because C is exactly halfway horizontally between A and B, we could have just found the midpoint of the x-coordinates: \( \frac{6 + 38}{2} = 22 \).

What are we being asked to find?

We are taking shape T and reflecting it twice across two perpendicular lines (\( x = -1 \) and \( y = -2 \)). We need to find the single transformation that would take us straight from the starting shape (T) to the final shape (S).

When you reflect a shape across a vertical line, and then a horizontal line, the combined effect is always a \( 180^\circ \) rotation. We need to find the center of this rotation.

Why we do this:

Let’s map out exactly where the shape goes to see the final result clearly. We’ll track the vertices of the triangle.

Why we do this:

To fully describe a rotation, we need three things: the transformation name (“rotation”), the angle (“\(180^\circ\)”), and the center of rotation. A neat trick for two perpendicular reflections: the center of rotation is always the point where the two mirror lines cross.

What are we being asked to find?

We need to find the area of the triangle. The formula for the area of a right-angled triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). To use this, we first need to figure out the actual lengths of the sides using the given perimeter and ratio.

Why we do this:

The sides are split into 3, 4, and 5 “parts”. The total perimeter is the sum of all sides, which is 72 cm. If we find out how many total “parts” make up the perimeter, we can determine how long one single part is.

Why we do this:

In a right-angled triangle, the two shorter sides (18 and 24) are the base and the perpendicular height. The longest side (30) is the hypotenuse, which we don’t need for the area formula.

Why we do this:

A power of \( \frac{1}{2} \) is the mathematical way of writing a square root. We just need to find the square root of 36.

Why we do this:

This is a fundamental rule of indices: any non-zero number raised to the power of 0 is always exactly 1.

Why we do this:

This index has three separate instructions built into it:

1. The denominator 3 means “find the cube root” (\( \sqrt[3]{} \)).
2. The numerator 2 means “square it” (\( x^2 \)).
3. The negative sign means “find the reciprocal” (put it 1 over the number).

It is almost always easiest to apply the root first so the number gets smaller before we square it.

Why we do this:

A box plot represents the five-number summary from the table. We plot a vertical line segment for each of the five values, join the middle three to form a “box”, and draw horizontal “whiskers” connecting to the minimum and maximum values.

Why we do this:

A box plot represents data broken down into 4 equal quarters (quartiles) containing 25% of the data each.

• Minimum to Lower Quartile = 25%
• Lower Quartile to Median = 25%
• Median to Upper Quartile = 25%
• Upper Quartile to Maximum = 25%

We need the number of girls between 133 cm (the Minimum) and 157 cm (the Upper Quartile).

What are we being asked to find?

We need to find the angle at C (angle ACB) using algebra. We must also state the specific circle theorems and geometry rules we use at each step. Drawing a line from the centre O to the point B on the circumference will be extremely helpful here to create triangles we can work with.

Why we do this:

We are told angle ABO is \( x \). If we draw the radius OB, triangle AOB is formed. Since OA and OB are both radii of the same circle, they are equal in length. This makes triangle AOB an isosceles triangle.

Why we do this:

We are explicitly told that BC is a tangent. There is a fundamental circle theorem connecting a radius to a tangent at the point where they meet on the circumference.

Why we do this:

Now look at the large, complete triangle ABC. We know the angle at A (which is \( x \)) and we can find the entire angle at B. Then we can use the sum of angles in a triangle to find the final angle at C.

What this tells us:

By breaking the geometry down into smaller triangles and applying core theorems (isosceles radii, tangent-radius, triangle sum), we proved the relationship mathematically. The angle at C will always be \( 90^\circ \) minus twice the angle \( x \).

What are we being asked to find?

We need to use algebra to prove a rule for every odd number, not just test it with numbers like 3 or 5. First, we need an algebraic way to write “any odd number”, and then we need to show the final result can be factored to look like \( 4 \times (\text{something}) + 1 \).

Why we do this:

An even number is any integer multiplied by 2, which we write as \( 2n \). An odd number is always exactly one more (or one less) than an even number. Therefore, we can write any general odd number as \( 2n + 1 \).

Why we do this:

The question asks us to look at the “square of an odd number”. So we must multiply our expression \( (2n + 1) \) by itself.

Why we do this:

We need our final expression to clearly show it is a multiple of 4, plus 1. We do this by factorising out the number 4 from the relevant parts of our expression.

What are we being asked to find?

We need to expand the bracket containing surds (square roots), and simplify the resulting expression until it looks like a number multiplied by \( \sqrt{10} \). Then, we identify what that front number (\( a \)) is.

Why we do this:

When multiplying surds, we use the rule \( \sqrt{x} \times \sqrt{y} = \sqrt{xy} \). We multiply the \( \sqrt{5} \) outside by both terms inside the bracket.

Why we do this:

We want our final answer to have a \( \sqrt{10} \) in it. We need to look for square numbers that multiply by 10 to give 40 and 90, so we can pull whole numbers out of the square roots.

What this tells us:

We showed that by expanding and then simplifying the surds by factoring out square numbers, the complex expression cleanly reduces to \( 5\sqrt{10} \). (Note: You could also have simplified \( \sqrt{8} \) to \( 2\sqrt{2} \) and \( \sqrt{18} \) to \( 3\sqrt{2} \) inside the bracket first, giving \( \sqrt{5}(5\sqrt{2}) = 5\sqrt{10} \)).

What are we being asked to find?

We need to create a single equation that connects \( y \) directly to \( x \). We are given two separate relationships: one connecting \( y \) and \( d \), and another connecting \( d \) and \( x \). We must figure out the exact formulas for these two relationships and then substitute one into the other.

Why we do this:

“Inversely proportional to \( d^2 \)” mathematically translates to \( y = \frac{k}{d^2} \), where \( k \) is a constant number. We use the given values to find \( k \).

Why we do this:

“Directly proportional to \( x^2 \)” translates to \( d = c \times x^2 \), where \( c \) is another constant. We use the second set of given values to find \( c \).

Why we do this:

We need a formula for \( y \) in terms of \( x \). We currently have \( y \) in terms of \( d \). Since we know exactly what \( d \) equals in terms of \( x \) (\( d = 6x^2 \)), we can substitute this directly into our first equation.

Why we do this:

This is a standard algebraic identity called the “Difference of Two Squares”. Factoring it is a formula you are expected to memorize.

Why we do this:

The phrase “Hence” is a massive hint. Look at the structure of the expression: \( (\text{Something})^2 – (\text{Something else})^2 \). This is exactly the same structure as \( a^2 – b^2 \) from part (a).

If we treat \( a = (x^2 + 4) \) and \( b = (x^2 – 2) \), we can use our factorised formula from part (a) to break this down much faster than multiplying out the huge brackets.

Alternative Method:

If you didn’t see the “Difference of Two Squares” trick, you could also expand both brackets manually: \( (x^4 + 8x^2 + 16) – (x^4 – 4x^2 + 4) \), which also simplifies to \( 12x^2 + 12 \).

What are we being asked to find?

We need to find the exact probability of picking a red counter. We know the bag only contains red, blue, and purple counters. A fundamental rule of probability is that the sum of all possible outcomes must add up to 1 (or 100%).

Why we do this:

We are given the probability of picking a purple counter (\( 0.2 \)). If we subtract this from 1, we will find the combined probability of picking either a red OR a blue counter.

Why we do this:

The remaining probability of \( 0.8 \) must be split between red and blue in the ratio \( 3 : 17 \). We can treat the probability of \( 0.8 \) just like a physical amount we need to share into parts.

What this tells us:

We found that there is a 12% chance (0.12) of drawing a red counter. As a check: Red = 0.12. Blue would be \( 17 \times 0.04 = 0.68 \). Purple is 0.20. \( 0.12 + 0.68 + 0.20 = 1.00 \). The math is perfectly verified.

What are we being asked to find?

We need to simplify this fraction. The most common mistake is trying to cross off the \( x^2 \)s right away. We cannot cancel terms that are added or subtracted. We must fully factorise both the top (numerator) and the bottom (denominator) into brackets first. Only then can we cancel identical brackets.

Why we do this:

The top is a quadratic expression: \( 3x^2 – 8x – 3 \). We need to put it into two brackets. Because the \( x^2 \) coefficient is not 1, we can use the “ac method” or trial and error.

Why we do this:

The bottom expression is \( 2x^2 – 6x \). This isn’t a full quadratic trinomial (it has no number at the end). Instead, we look for the highest common factor (HCF) that divides into both terms.

Why we do this:

Now we rewrite our fraction with the factorised versions. If there is an identical bracket on the top and bottom, it divides to equal 1, effectively cancelling out.

What this tells us:

The expression has been simplified to its most compact form. Notice that you cannot simplify \( \frac{3x}{2x} \) inside this answer because the \( 3x \) is bound to the \( + 1 \). The simplification stops here.

What are we being asked to find?

We need to draw a new graph based on the original one provided. The equation \( y = \sin x^\circ – 2 \) takes the original function \( \sin x^\circ \) and subtracts 2 from the final result. In graph transformations, adding or subtracting a number on the outside of the function translates (moves) the entire graph vertically up or down.

Why we do this:

Because it’s a ” – 2″ at the end, every single y-coordinate drops straight down by 2 units. The x-coordinates stay exactly the same. We identify the peak, the trough, and the intercepts of the original graph, move them all down by 2, and then draw a smooth curve through the new points.

What are we being asked to find?

We need an equation that says “\( b = \dots \)” using the letter \( a \). We are given a specific line equation that is perpendicular (at 90 degrees) to the line joining P and Q. If we find the gradient of the given line, we can deduce the gradient of line PQ. Then we can use the gradient formula with coordinates P and Q to build our equation.

Why we do this:

To find the gradient (slope) of the line \( 3x + 2y = 7 \), we must rearrange it into the standard format \( y = mx + c \), where ‘\( m \)’ is the gradient.

Why we do this:

Perpendicular lines have gradients that multiply together to make -1 (the “negative reciprocal”). We flip the fraction upside down and change the sign.

Why we do this:

The formula for a gradient between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \frac{y_2 – y_1}{x_2 – x_1} \). We know the coordinates P\( (3, 4) \) and Q\( (a, b) \), and we know the gradient is \( \frac{2}{3} \). We can plug these in to form an equation linking \( a \) and \( b \).

What are we being asked to find?

We are looking for specific integers (whole numbers: … -2, -1, 0, 1, 2 …) that make BOTH of these mathematical statements true at the same time. We will solve each inequality separately to find the allowed ranges, and then see where those ranges overlap.

Why we do this:

This is a simple linear inequality. We solve it exactly like a normal equation to find the maximum possible value for \( n \).

Why we do this:

The second inequality has algebra on the denominator. Usually, multiplying by an unknown is dangerous with inequalities (because multiplying by a negative flips the sign). However, \( n^2 \) is always positive, and \( n^2 + 5 \) is therefore strictly positive. We can safely multiply both sides by \( n^2 + 5 \) without flipping the inequality sign.

Why we do this:

We now have two conditions that must both be true:

1. \( n \le 4 \)
2. \( 1 < n < 5 \)

We list the whole numbers that satisfy the strict second condition, and ensure they also obey the first condition.

Check your answer:

Let’s test \( n=3 \):

Condition 1: \( 3(3) + 2 = 11 \). Is 11 \( \le \) 14? Yes.

Condition 2: \( \frac{6(3)}{3^2 + 5} = \frac{18}{9+5} = \frac{18}{14} = 1.28 \). Is 1.28 > 1? Yes.

The logic holds.