GCSE Mathematics (Higher) – Paper 1 Non-Calculator (June 2019)

What are we being asked to find?

We need to find the missing probabilities for red and yellow cubes. We know two important facts:

1. All the probabilities in a complete set of outcomes must add up to \(1\).
2. The number of red cubes equals the number of yellow cubes. Therefore, their probabilities must be exactly the same.

Why we do this:

First, we find out how much probability is “left over” after taking away the blue probability. Then, we split this remaining probability equally between red and yellow.

What this tells us:

The probability of picking red is 0.4 and the probability of picking yellow is 0.4. Let’s check: \(0.2 + 0.4 + 0.4 = 1.0\). Perfect!

Why we do this:

We are told there are \(12\) blue cubes. We also know the probability of a blue cube is \(0.2\) (which is \(\frac{1}{5}\) or \(20\%\)). This means \(12\) cubes represents \(\frac{1}{5}\) of the total box. To find the total, we can divide the amount (12) by the proportion (0.2), or multiply by 5.

Why we do this:

The recipe gives us ingredient amounts for 15 biscuits, but we need to make 60. We need to find out how many ‘batches’ of the recipe we are making by dividing the target amount by the original amount.

Why we do this:

First we find how much flour is needed for one batch (15 biscuits). We are told it’s three times the sugar. Once we have the flour for one batch, we multiply it by our batch multiplier (4) to get the total amount needed.

Why we do this:

We repeat the logic for butter. We need two times as much butter as sugar for one batch, and then we multiply by 4 batches. Finally, we divide the total butter by 250g (the size of a pack) to see how many whole packs we must buy.

Why we do this:

To find the Highest Common Factor (HCF), we can either use prime factorisation trees, or simply list the factor pairs of both numbers and look for the largest number that appears in both lists. Here, we’ll list the factors systematically to make sure we don’t miss any.

Why we do this:

Now we compare the two lists and circle the common factors: 1, 2, 3, 6, 9, 18. The largest number in both lists is our Highest Common Factor.

What this tells us:

18 is the largest integer that divides both 72 and 90 without leaving a remainder. Alternative check using prime factors: \(72 = 2^3 \times 3^2\) and \(90 = 2 \times 3^2 \times 5\). Shared prime factors are \(2\) and \(3^2\). \(2 \times 9 = 18\). Correct!

What are we looking at?

We are given three 2D views of a 3D object.

• Plan view (looking from above): It is a circle. This tells us the shape has a circular base.
• Front & Side elevations (looking from the front/side): They are rectangles of the exact same size.

A solid shape that looks like a circle from the top, and a rectangle from both the front and side, is a cylinder.

Why we do this:

The question asks us to include dimensions on our sketch. We need to read the lengths directly from the centimetre grid provided in the question.

Why we do this:

The problem states this is a two-step transformation. The first step is a reflection of shape A in the x-axis. We need to find the coordinates of this intermediate shape before we can figure out the translation to shape B.

Why we do this:

A translation vector \(\begin{pmatrix} c \\ d \end{pmatrix}\) moves a shape \(c\) units in the x-direction (left/right) and \(d\) units in the y-direction (up/down). By comparing a specific point on our intermediate shape to the corresponding point on the final shape B, we can see exactly how far it moved.

What this tells us:

The vector is \(\begin{pmatrix} -6 \\ -1 \end{pmatrix}\). Let’s check with another vertex. Take the tip \((3, -5)\). Translate by \(\begin{pmatrix} -6 \\ -1 \end{pmatrix}\): x = \(3 + (-6) = -3\), y = \(-5 + (-1) = -6\). Point is \((-3, -6)\). Looking at the graph, the tip of shape B is exactly at \((-3, -6)\). Our translation is correct!

What are we being asked to find?

We are given the ratio of packs sold, but the total given (\(212\)) is for individual pens. We need to convert our “packs ratio” into a “pens ratio” by multiplying the number of packs by the number of pens inside each pack.

Why we do this:

Now we have a ratio of actual pens sold (\(14 : 15 : 24\)). The total number of pens sold is \(212\). We add up the parts in our ratio to see how many parts make up the \(212\) pens, then divide to find out exactly how many pens one single “part” is worth.

Why we do this:

We need the total number of green pens. Looking at our pens ratio, green pens represent \(24\) parts. Since each part equals \(4\) pens, we multiply \(24\) by \(4\).

What this tells us:

The shop sold \(96\) green pens. We can check by finding the others: Black = \(14 \times 4 = 56\). Red = \(15 \times 4 = 60\). Total = \(56 + 60 + 96 = 212\). The numbers match perfectly!

Why we do this:

We are given the area of rectangle \(PQRS\) and the length of one of its sides (\(QR\)). Since \(\text{Area} = \text{width} \times \text{height}\), we can work backwards to find the missing side, \(PQ\).

Why we do this:

The question tells us that \(BC = PQ\). This is the vital link between the two rectangles. Since we just found \(PQ\), we now know the height of rectangle \(ABCD\).

Why we do this:

We know the perimeter of \(ABCD\) is \(26\text{ cm}\). Perimeter is the total distance around the outside (\(AB + BC + CD + DA\)). In a rectangle, opposite sides are equal, so it’s \(2 \times AB + 2 \times BC\). We can set up an equation to find the missing width, \(AB\).

What this tells us:

The length of \(AB\) is \(8.5\text{ cm}\). Let’s verify: Perimeter = \(8.5 + 4.5 + 8.5 + 4.5 = 26\). Area of \(PQRS\) = \(4.5 \times 10 = 45\). Everything is mathematically consistent!

Why we do this:

When asked to “estimate”, we round the numbers to 1 significant figure, or to easily workable numbers, before performing any calculations. This makes the math simple enough to do without a calculator.

Why we do this:

We are given that \(2.3^6 = 148\). We need to find \(0.23^6\). Notice that \(0.23\) is exactly \(2.3 \div 10\). We can substitute this relationship into our power to see how the final answer changes.

Why we do this:

A negative index means “take the reciprocal”. In other words, \(x^{-n} = \frac{1}{x^n}\). Once we turn it into a fraction with a positive power, we just square the denominator.

Why we do this:

You cannot easily multiply mixed numbers as they are. You must convert them into “top-heavy” (improper) fractions first. You do this by multiplying the whole number by the denominator, and adding the numerator.

Why we do this:

To multiply fractions, multiply the top numbers (numerators) together, and multiply the bottom numbers (denominators) together. Then simplify the result by dividing top and bottom by their greatest common factor.

Why we do this:

The question specifically asks for the answer “as a mixed number in its simplest form”. We divide the numerator by the denominator to find the whole number, and the remainder becomes the new numerator.

What this tells us:

The product is exactly \(5.6\), which is \(5 \frac{3}{5}\). Pro-tip: You could also cross-cancel in Step 2 to make the math easier: \(\frac{7}{2} \times \frac{8}{5}\). Cancel the \(8\) and \(2\) to get \(\frac{7}{1} \times \frac{4}{5} = \frac{28}{5}\).

What are we looking for?

When solving simultaneous equations using a graph, the solution is the exact coordinate where the two lines intersect (cross each other). We need to read the x and y coordinates of this crossing point.

How to read it:

Locate the intersection. Then, track straight up to the x-axis to find the \(x\) value. Track straight to the left to hit the y-axis to find the \(y\) value. Each tiny square represents \(0.1\) units.

Why we do this:

We have an ordered list of 23 ages. We need to find the three key ‘dividing’ points of the data.

• Median is the exact middle value. For \(n\) pieces of data, it’s at position \(\frac{n+1}{2}\).
• Lower Quartile (LQ) is the middle of the lower half of the data.
• Upper Quartile (UQ) is the middle of the upper half of the data.

Why we do this:

When asked to compare if one group is “younger” or “older” on average, we must compare their medians (the average representing the middle person). We look at the median for Coach A from our table, and the median for Coach B from the given table.

Why we do this:

When asked if data “varies more”, we must look at a measure of spread. We can use either the Range (Greatest – Least) or the Interquartile Range (IQR) (Upper Quartile – Lower Quartile). The larger the number, the more the data varies.

What are we being asked to find?

We need to find the fraction \(\frac{P}{R}\). To do this, we need to convert the percentage increases into multipliers, and then link \(P\) to \(R\) directly.

“50% more” means \(100\% + 50\% = 150\%\). As a decimal multiplier, this is \(1.5\). As a fraction multiplier, this is \(\frac{3}{2}\).

Why we do this:

Substitute the expression for \(Q\) into the equation for \(R\). This removes \(Q\) completely and creates an equation with only \(P\) and \(R\).

Why we do this:

We want the fraction \(\frac{P}{R}\). Using our equation, we can rearrange it to get \(\frac{P}{R}\) on one side. Then, we must convert any decimals into a proper fraction in its simplest form.

What this tells us:

Volume P is exactly \(\frac{4}{9}\) of Volume R. Alternative fraction method: \(Q = \frac{3}{2}P\). \(R = \frac{3}{2}Q\). Therefore, \(R = \frac{3}{2} \times \frac{3}{2} P = \frac{9}{4} P\). If \(R = \frac{9}{4} P\), then \(P = \frac{4}{9} R\). This avoids decimals entirely!

Why we do this:

When asked to prove properties about divisibility, odds, or evens with an expression like \(n^2 – n\), factorising it usually reveals a clear mathematical relationship. It transforms subtraction into multiplication.

Why we do this:

Now look at what \(n(n – 1)\) actually means. If \(n\) is a whole number, then \((n – 1)\) is the whole number right before it. Two numbers next to each other are called consecutive integers. What happens when you multiply consecutive integers?

What this tells us:

This is a formal proof. Alternative method: Split into two cases. Case 1: \(n\) is even. \((\text{Even})^2 – \text{Even} = \text{Even} – \text{Even} = \text{Even}\). Case 2: \(n\) is odd. \((\text{Odd})^2 – \text{Odd} = \text{Odd} – \text{Odd} = \text{Even}\). Since it’s even in all cases, it’s never odd!

What are we being asked to do?

This is a Paper 1 (non-calculator) exam, so you must know the exact values for sine, cosine, and tangent of standard angles (\(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}\)).

Why we do this:

Multiply the two fractions together. Surds behave just like algebra. If you see the same term multiplied on the top and the bottom, they will cancel each other out.

What are we being asked to find?

To use the volume formulas, we need the radius (\(r\)) and height (\(h\)). The question gives us the diameter. The radius is half the diameter.

Why we do this:

We’ll calculate the volumes of the two distinct parts separately, leaving the answer in terms of \(\pi\) to make the math easier without a calculator.

Why we do this:

A hemisphere is exactly half of a sphere. We take the sphere volume formula and multiply it by \(\frac{1}{2}\).

Why we do this:

Add the two parts together. The question says the total volume is \(k\pi\). By comparing our answer (\(48\pi\)) to \(k\pi\), we can identify the value of \(k\).

What are we being asked to find?

We need to find the total possible combinations. To find the total number of combinations for independent choices, we multiply the number of options for each choice together.

Why we do this:

Now we have a restriction: all three digits must be different. We approach this dial by dial. For the first dial, we have all options available. But for the second dial, we can’t use the number we just picked. For the third dial, we can’t use the two numbers we’ve already picked.

What this tells us:

Out of the \(125\) total possible codes, \(60\) of them use three completely different numbers.

Why we do this:

A ratio \(a : b\) can be written as a fraction \(\frac{a}{b}\). So, we can rewrite the ratio equation as a fraction equation. This allows us to use standard algebraic methods to solve for \(x\).

Why we do this:

To solve for \(x\), we need to get rid of the fractions by cross-multiplying. Then, because we have an \(x^2\) term and an \(x\) term, we know this is a quadratic equation. To solve a quadratic, we must move all terms to one side so the equation equals zero: \(ax^2 + bx + c = 0\).

Why we do this:

We need to find two brackets that multiply to give \(2x^2 – 3x – 5\). Since the first term is \(2x^2\), the brackets must start with \((2x \dots)(x \dots)\). We are looking for factors of \(2 \times -5 = -10\) that add up to \(-3\). Those numbers are \(-5\) and \(2\).

Why we do this:

If two brackets multiply to give zero, then at least one of the brackets must equal zero. We set each bracket to zero to find the two possible values for \(x\).

What this tells us:

The possible values are \(-1\) and \(2.5\). We can verify by plugging \(x=-1\) back into the ratio: \((-1)^2 : (3(-1)+5) = 1 : 2\). It works!

What are we doing?

You can only add surds if the number inside the square root is the same. \(\sqrt{3}\) is already as simple as it gets, but we can simplify \(\sqrt{12}\) by finding a square number that divides into \(12\).

Why we do this:

We need to apply the power of \(7\) to both the numerator and denominator. \(1^7\) is just \(1\). Evaluating \((\sqrt{3})^7\) requires us to group the square roots in pairs, because \(\sqrt{3} \times \sqrt{3} = 3\).

Why we do this:

The required format \(\frac{\sqrt{b}}{c}\) has an integer on the bottom, not a surd. To clear a square root from the denominator (rationalise it), we multiply the top and bottom of the fraction by that square root.

What this tells us:

Comparing to \(\frac{\sqrt{b}}{c}\), we have $b=3$ and $c=81$. The question just asks for the expression in that form, so our final answer is \(\frac{\sqrt{3}}{81}\).

Why we do this:

The form \((x – a)^2 – b\) is called “completed square” form. To put a quadratic into this form, we take half of the coefficient of \(x\) (which is \(-6\)), and put it inside the squared bracket. Then we must subtract the square of that number so the equation stays balanced.

What this tells us:

Comparing \((x – 3)^2 – 8\) to the formula \((x – a)^2 – b\), we can clearly see that \(a = 3\) and \(b = 8\).

✓ (A1) For \(a = 3, b = 8\).

Why we do this:

The main reason we complete the square is to find the minimum point (turning point) of a parabola. The lowest value \((x – a)^2\) can ever be is zero (because any number squared is positive). This happens when \(x = a\). When the bracket is zero, the y-value is just \(-b\).

What are we doing?

We are given two statements about proportion. “Inversely proportional” means \(y = \frac{k}{x}\), and “directly proportional” means \(y = kx\). Let’s write these as algebra using two different constants, \(k_1\) and \(k_2\).

Why we do this:

We are given a snapshot in time when \(h=10\), \(p=6\), and \(t=144\). By plugging these numbers into our equations, we can calculate the exact values of \(k_1\) and \(k_2\).

Why we do this:

The question wants a formula for \(h\) in terms of \(t\). This means \(p\) shouldn’t be in our final answer. We take our equation for \(h\) and replace the \(p\) with our expression containing \(t\).

What this tells us:

We’ve created a direct bridge between \(h\) and \(t\). Let’s double check it: if \(t=144\), \(\sqrt{t} = 12\). So \(h = \frac{120}{12} = 10\). This perfectly matches the numbers we were given at the start!

What are we doing?

\(f^{-1}(x)\) means the inverse function of \(f(x)\). A normal function takes \(x\) and turns it into \(y\). The inverse function works backwards, taking \(y\) and turning it back into \(x\). To find it algebraically, we set \(y = 3x – 1\) and rearrange the formula to make \(x\) the subject.

Why we do this:

A composite function means putting one function inside another. \(fg(x)\) means put the whole \(g(x)\) equation into the \(x\) spot of the \(f(x)\) equation. \(gf(x)\) means put \(f(x)\) inside \(g(x)\). We need to work both of these out before we can solve the equation.

Why we do this:

The question says \(fg(x) = 2gf(x)\). We will set our two expressions equal to each other (remembering to multiply \(gf(x)\) by 2), expand all brackets, and move everything to one side to show it simplifies to \(15x^2 – 12x – 1 = 0\).

What this tells us:

We have successfully shown that substituting the composite functions into the given equation leads directly to the required quadratic equation. The “show that” question is complete.

What are we doing?

Initially, the probability of picking a green counter is \(\frac{3}{7}\). This means for every \(7\) counters in total, \(3\) are green. Therefore, \(4\) must be red. This gives us a ratio of Green : Red = \(3 : 4\). To write this algebraically without knowing the exact numbers, we can say there are \(3x\) green counters and \(4x\) red counters, making a total of \(7x\) counters.

Why we do this:

Counters are added to the bag, which changes the probability. We must write a new fraction for the probability of picking green, using our algebraic expressions. The new probability is \(\frac{\text{New Green}}{\text{New Total}}\).

Why we do this:

We solve the algebraic equation by cross-multiplying to remove the fractions, and then grouping the \(x\) terms. Once we find \(x\), we can use it to calculate the exact number of red and green counters we started with.

What this tells us:

Our multiplier \(x\) is \(3\). Now we substitute this back into our original formulas: Green = \(3x\) and Red = \(4x\). Let’s calculate the final values: Green = \(3(3) = 9\). Red = \(4(3) = 12\). We can verify: Initial total is \(21\), \(P(\text{green}) = \frac{9}{21} = \frac{3}{7}\). Add counters: Green is \(12\), Total is \(26\). \(P(\text{green}) = \frac{12}{26} = \frac{6}{13}\). Perfect!