GCSE Mathematics 2023 – Edexcel Higher Paper 1 (Non-Calculator)

💡 Why we do this: First, list all elements in each set from the numbers 1 to 10.

💡 Strategy:
1. Intersection: Numbers in BOTH A and B go in the middle. (\( \{1, 9\} \))
2. A Only: Remaining odd numbers. (\( \{3, 5, 7\} \))
3. B Only: Remaining square numbers. (\( \{4\} \))
4. Outside: Numbers in neither list. (\( \{2, 6, 8, 10\} \))

💡 What does B’ mean? \( B’ \) means “Not in B”. It is everything outside the B circle.

💡 What this tells us: As we move to the right (age increases), the crosses go up (weight increases).

💡 Strategy: To estimate a value, first draw a Line of Best Fit that goes through the middle of the data points. Then, find 5.8 on the y-axis, go across to the line, and down to the x-axis.

💡 Why we do this: The question tells us directly that \( 20\% \) of the original price is equal to \( £240 \). We need to find the original price, which represents \( 100\% \).

💡 Strategy: To get from \( 20\% \) to \( 100\% \), we can multiply by 5 (since \( 20 \times 5 = 100 \)).

💡 Why we do this: The pressure formula requires Area. We are given Volume and Height. For a cylinder (or any prism):

\( \text{Volume} = \text{Area of Base} \times \text{Height} \)

💡 Why we do this: Now we have the Force (90N) and the Area (30cm²). Use the given formula.

💡 Why we do this: The solution to a pair of simultaneous equations is the point where their graphs cross (intersect).

💡 Why we do this: A pentagon has 5 sides. The sum of the interior angles of any polygon is given by \( (n – 2) \times 180 \), where \( n \) is the number of sides.

💡 Strategy: Let Angle \( ABC = x \). Then Angle \( AED = 4x \).

We know the sum of all 5 angles is 540°.

💡 Why we do this: Solve the equation to find \( x \), which is angle \( ABC \).

💡 Why we do this: The question asks for Angle \( AED \), which is \( 4x \).

💡 Why we do this: Apply the power to everything inside the bracket. Remember \((x^m)^n = x^{mn}\).

💡 Why we do this: Multiply the terms. Add powers for the same base. Remember \(x = x^1\).

💡 Why we do this: Subtract powers when dividing. Divide the coefficients.

💡 Why we do this: “Lose at least one” includes:

• Win then Lose (W, L)
• Lose then Win (L, W)
• Lose then Lose (L, L)

The only outcome NOT included is Winning BOTH (W, W).

💡 Strategy: Since all probabilities add to 1, we can calculate \( 1 – P(\text{Win, Win}) \).

💡 Why we do this: “Directly proportional” means \( y = kx \). We need to find \( k \) using the given values.

💡 Strategy: Substitute \( x = 5 \) into our formula \( y = 16x \).

💡 Key Principle: A fraction \( \frac{1}{x^n} \) can be written as \( x^{-n} \).

💡 Key Principle: \( x^{\frac{a}{b}} = (\sqrt[b]{x})^a \). The bottom of the fraction is the root, the top is the power.

💡 Strategy: The line equation \( y = mx + c \) gives us the gradient \( m \).

💡 Key Principle: Perpendicular gradients multiply to make -1. (Negative reciprocal).

💡 Strategy: Rearrange \( 6y + kx – 12 = 0 \) into \( y = mx + c \) form to identify the gradient in terms of \( k \).

💡 Strategy: We know \( \frac{3}{8} \) of the area is \( 75\pi \). First, find \( \frac{1}{8} \), then find the full area (\( \frac{8}{8} \)).

💡 Strategy: Set the total area equal to the formula \( 4\pi r^2 \) and solve for \( r \).

💡 Why we do this: The question asks for diameter (which is \( 2r \)) in the form \( a\sqrt{b} \). We must simplify the surd.

💡 Strategy: Multiply both sides by the denominator \( (5x + 3) \) to remove the fraction.

💡 Strategy: We need all terms containing \( x \) on one side and everything else on the other.

💡 Strategy: Factor out \( x \) so it appears only once, then divide by the bracket.

💡 Strategy: Since the ratio of costs is \( 5:9 \), we can use algebra.

Let cost of carrots = \( 5x \)

Let cost of tomatoes = \( 9x \)

💡 Strategy: Total Cost = (Quantity × Unit Cost) + (Quantity × Unit Cost).

💡 Why we do this: Solving for \( x \) gives us the multiplier for our ratio.

💡 Strategy: Substitute \( x = 6 \) back into our expressions \( 5x \) and \( 9x \).

💡 Key Principle: To find the total combinations, we multiply the number of choices for each option.

\( \text{Total Combinations} = \text{Starters} \times \text{Mains} \times \text{Desserts} \)

💡 Strategy: Divide 420 by 60.

💡 Strategy:
1. Replace \( g(x) \) with \( y \).
2. Swap \( x \) and \( y \).
3. Solve for \( y \).

💡 Strategy: Put \( f(x) \) INSIDE \( g(x) \). So substitute \( (3x + 4) \) wherever there is an \( x \) in \( g \).

💡 Strategy: Set equal to 3 and solve.

💡 Strategy: We know \( \frac{3}{8} \) of the area is \( 75\pi \). First, find \( \frac{1}{8} \), then find the full area (\( \frac{8}{8} \)).

💡 Strategy: Set the total area equal to the formula \( 4\pi r^2 \) and solve for \( r \).

💡 Why we do this: The question asks for diameter (which is \( 2r \)) in the form \( a\sqrt{b} \). We must simplify the surd.

💡 Strategy: Multiply both sides by the denominator \( (5x + 3) \) to remove the fraction.

💡 Strategy: We need all terms containing \( x \) on one side and everything else on the other.

💡 Strategy: Factor out \( x \) so it appears only once, then divide by the bracket.

💡 Strategy: Since the ratio of costs is \( 5:9 \), we can use algebra.

Let cost of carrots = \( 5x \)

Let cost of tomatoes = \( 9x \)

💡 Strategy: Total Cost = (Quantity × Unit Cost) + (Quantity × Unit Cost).

💡 Why we do this: Solving for \( x \) gives us the multiplier for our ratio.

💡 Strategy: Substitute \( x = 6 \) back into our expressions \( 5x \) and \( 9x \).

💡 Key Principle: To find the total combinations, we multiply the number of choices for each option.

\( \text{Total Combinations} = \text{Starters} \times \text{Mains} \times \text{Desserts} \)

💡 Strategy: Divide 420 by 60.

💡 Strategy:
1. Replace \( g(x) \) with \( y \).
2. Swap \( x \) and \( y \).
3. Solve for \( y \).

💡 Strategy: Put \( f(x) \) INSIDE \( g(x) \). So substitute \( (3x + 4) \) wherever there is an \( x \) in \( g \).

💡 Strategy: Set equal to 3 and solve.

💡 Circle Theorem: \( OA \) and \( OB \) are radii, so triangle \( OAB \) is isosceles.

The angle at \( B \) (Angle \( OBA \)) is \( 51^{\circ} \).

💡 Strategy: We need to find the angles in triangle \( OAD \). First, find the central angle \( AOD \).

💡 Circle Theorem: The tangent to a circle is perpendicular to the radius at the point of contact.

So, angle \( ODC = 90^{\circ} \).

💡 Strategy: The angle between the line \( FC \) and the plane \( ABCD \) is the angle \( FCA \).

We form a right-angled triangle \( FAC \), where \( AF \) is the vertical height (Opposite) and \( FC \) is the hypotenuse.

💡 Why we do this: We have the Opposite (\( 6.8 \)) and the Hypotenuse (\( 13.6 \)). We use SOH CAH TOA.

\( \sin(x) = \frac{\text{Opp}}{\text{Hyp}} \)

💡 Strategy: Multiply top and bottom by \( 4 + \sqrt{3} \).

💡 Strategy: Multiply top and bottom by \( \sqrt{3} \).

💡 Strategy: Find a common denominator (39).

💡 Strategy: \( 4x^2 – 25 < 0 \) is a difference of two squares. Find the critical values where it equals 0.

💡 Strategy: Rearrange or factorise \( 12 – 5x – 3x^2 > 0 \). It’s often easier to work with a positive \( x^2 \) term: \( 3x^2 + 5x – 12 < 0 \).

💡 Strategy: We need values that satisfy BOTH conditions.

• Condition A: \( -2.5 < x < 2.5 \)
• Condition B: \( -3 < x < 1.33... \)

Look at the number line. The lower limit is determined by the tighter constraint (-2.5 is tighter than -3). The upper limit is determined by the tighter constraint (1.33 is tighter than 2.5).