GCSE 2023 Edexcel Higher Paper 3 (Calculator)

💡 Why we do this: When we raise a power to another power, we multiply the indices. This is the law \( (a^m)^n = a^{m \times n} \).

💡 Why we do this: When multiplying terms with the same base, we add the indices. This is the law \( a^m \times a^n = a^{m+n} \).

💡 Why we do this: To expand a single bracket, we multiply the term outside by every term inside.

💡 Why we do this: We need to chain the calculations: Total People → Coffee Drinkers → Total Cups → Total Grams.

💡 What this tells us: If the percentage increases (72% > 68%), the number of drinkers increases, so the total coffee needed increases.

💡 Why we do this: We need to find the angles inside triangle \(ACD\) to check if two are equal (which proves it is isosceles).

First, we use the fact that lines \(BCD\) and \(EFG\) are parallel.

💡 Why we do this: We use the straight line \(ADG\) to find the interior angle at \(D\).

💡 Why we do this: Find the third angle in the triangle. If two angles are equal, the triangle is isosceles.

💡 Why we do this: This is an inverse proportion problem. Fewer pumps means it will take more time.

We calculate the total “pump-hours” needed to do the job.

💡 Why we do this: We divide the total work by the new number of pumps.

💡 Why we do this: The HCF is the product of the lowest powers of common prime factors.

• Powers of 2: \(A\) has \(2^2\), \(B\) has none (\(2^0\)). Lowest is \(2^0\) (none).
• Powers of 3: \(A\) has \(3^4\), \(B\) has \(3^2\). Lowest is \(3^2\).
• Powers of 7: \(A\) has \(7^1\), \(B\) has \(7^2\). Lowest is \(7^1\).

💡 Why we do this: The LCM is the product of the highest powers of all prime factors present.

• Powers of 2: Highest is \(2^2\).
• Powers of 3: Highest is \(3^4\).
• Powers of 7: Highest is \(7^2\).

💡 Why we do this: We know the rate per second and the total volume. Dividing volume by rate gives the total time in seconds.

💡 Why we do this: The question asks for days. We convert seconds to minutes, then hours, then days.

Seconds in a minute = 60

Minutes in an hour = 60

Hours in a day = 24

Seconds in a day = \( 60 \times 60 \times 24 = 86 \, 400 \)

💡 What this tells us: The turning point is the lowest point (minimum) on this curve. We locate it visually on the graph.

💡 What this tells us: The roots are the x-values where the graph crosses the x-axis (where \( y = 0 \)).

💡 Why we do this: We use the density formula triangle.

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

Therefore, \( \text{Mass} = \text{Density} \times \text{Volume} \).

💡 Why we do this: We need to find 80% of the 70%. In probability, “of” usually means multiply.

💡 Why we do this: We have a right-angled triangle \(ABC\) where we know the hypotenuse (\(6.8\)) and an angle (\(41^\circ\)). We need the shared side \(AC\) to connect it to triangle \(ADC\).

Using SOH CAH TOA:

💡 Why we do this: Now we use the right-angled triangle \(ADC\). We know the opposite side (\(AC\)) and the angle (\(55^\circ\)). We want the adjacent side (\(DC\)).

Using SOH CAH TOA:

💡 How to do this: A box plot needs 5 key values:

1. Minimum (Least height): 169
2. Lower Quartile (LQ): 174
3. Median: 177
4. Upper Quartile (UQ): 180
5. Maximum (Greatest height): 186

Draw vertical lines at these values. The “box” goes from LQ to UQ. The “whiskers” go out to Min and Max.

💡 Why we do this: To compare distributions, we must compare two things:

1. Average: Compare the Medians.
2. Spread: Compare the Interquartile Ranges (IQR) or Ranges.

💡 Why we do this: It’s easier to multiply two brackets first, simplify, and then multiply by the third bracket.

Let’s expand \( (x – 1)(x + 3) \).

💡 Why we do this: Now multiply the quadratic result by \( (x – 5) \).

💡 Why we do this: Two consecutive terms are the \( n \)th term and the \( (n+1) \)th term (or \( (n-1) \)th and \( n \)th).

Let’s use the \( n \)th term and the \( (n+1) \)th term.

💡 Why we do this: We need to add them together and show the result is a perfect square (like \( k^2 \)).

💡 Why we do this: To find the area of sector \(OBC\), we need the radius \(OB\) (or \(OC\)).

We can use the Cosine Rule on triangle \(OAC\). We know:

• \(OA = 9\) cm
• \(AC = 6\) cm
• Angle \(OAC = 35^\circ\)

The Cosine Rule is \( a^2 = b^2 + c^2 – 2bc \cos A \).

💡 Why we do this: Use the formula for the area of a sector.

💡 Why we do this: This is the “Difference of Two Squares” identity.

💡 Why we do this: We can treat \( 2^{40} – 1 \) as a difference of two squares where \( a = 2^{20} \) and \( b = 1 \).

💡 Why we do this: Enlargement by scale factor \( k \) from a centre \( (x_c, y_c) \) multiplies the vector from the centre to each point by \( k \).

Centre: \( (-2, -2) \)

Scale Factor: \( -2 \)

💡 Why we do this: The gradient of a curve at a specific point is found by drawing a tangent line at that point and calculating its gradient (Rise over Run).

💡 What this tells us: The y-axis is Distance and x-axis is Time. Gradient is Distance divided by Time.

💡 Why we do this: The small cone and large cone are similar shapes. We use ratios to find the missing radius.

Ratio of slant heights = \( \frac{6}{10} = 0.6 \).

💡 Why we do this: The curved area of the frustum is the curved area of the large cone minus the curved area of the small cone.

Formula: \( \pi r l \)

💡 Why we do this: Total Surface Area includes the top circular face, the bottom circular face, and the curved surface.

💡 Why we do this: Check the value of \( y \) when \( x = 0 \).

The graph should pass through \( (0, 1) \), not the origin \( (0, 0) \).

💡 Why we do this: Let \( x \) equal the recurring decimal. We want to align the recurring parts so we can subtract them away.

\( x = 0.1232323… \)

💡 Why we do this: Subtracting the two equations eliminates the infinite recurring tail.

💡 Why we do this: To solve for \(x\), we need to eliminate the denominators. First, we find a common denominator.

Common denominator = \( (x + 4)(2 – 2x) \)

💡 Why we do this: Multiply both sides by the denominator to get a linear equation equal to a quadratic, then rearrange to \(ax^2 + bx + c = 0\).

💡 Why we do this: Factorise or use the quadratic formula.

\( 2 \times 6 = 12 \). Find factors of 12 that add to 7: 3 and 4.

💡 Why we do this: Combine the components into a single vector expression.

💡 Why we do this: If two vectors are parallel, one is a scalar multiple of the other. The ratio of their x and y components must be the same.

Ratio of parallel vector \( \begin{pmatrix} 13 \\ 6 \end{pmatrix} \) is \( \frac{x}{y} = \frac{13}{6} \).

So, \( \frac{2a + 8b}{6a + 2b} = \frac{13}{6} \).

💡 Why we do this: The tangent is perpendicular to the radius. First, we find the gradient of the radius from the origin \( (0,0) \) to \( P(-3, 4) \).

💡 Why we do this: The gradient of the tangent is the negative reciprocal of the radius gradient.

Now use \( y – y_1 = m(x – x_1) \) with point \( P(-3, 4) \).

💡 Why we do this: Find where the line crosses the x-axis by setting \( y = 0 \).

💡 Why we do this: We need the number of each colour in terms of \( x \).

Given \( x : y = 1 : 3 \), so \( y = 3x \). Total = \( 3x \).

Pink = \( x \)

Blue = 5

Green = Total – Pink – Blue = \( 3x – x – 5 = 2x – 5 \)

💡 Why we do this: “Same colour” means (Pink AND Pink) OR (Blue AND Blue) OR (Green AND Green).

Sampling is without replacement (takes two counters).

Prob = \( \frac{n}{Total} \times \frac{n-1}{Total-1} \)