GCSE Nov 2023 Edexcel Higher Paper 2 (Calculator)

💡 Why we do this: We need to multiply out the brackets (expand) and then combine terms that are the same (simplify).

💡 Why we do this: “Factorise fully” means we need to find the largest common factor for both numbers and algebra letters.

💡 Why we do this: We need to isolate \( g \) on one side of the equals sign.

💡 Why we do this: To find profit, we first need to know the total money spent (outgoing).

💡 Why we do this: We calculate how much money she received from ticket sales.

💡 Why we do this: Profit is the difference between money in and money out.

💡 Why we do this: Percentage profit is usually calculated as a percentage of the cost (investment).

Formula: \( \frac{\text{Profit}}{\text{Original Cost}} \times 100 \)

💡 Why we do this: For an increase of 3.4%, we start with 100% and add 3.4%.

\( 100\% + 3.4\% = 103.4\% \)

As a decimal multiplier, this is \( 1.034 \).

💡 Formula: \( \text{Amount} = \text{Principal} \times (\text{Multiplier})^{\text{years}} \)

💡 Strategy: The smallest \( x \) term is \( -x \). Let’s add \( x \) to both sides to make it positive.

💡 Strategy: Add 14 to both sides to remove the constant from the left.

💡 Strategy: Divide by 6.

💡 What we know: Ratio \( C : D : E = 3 : 4 : 2 \). Debbie (4 parts) gets £120.

We can find the value of 1 part.

💡 Key Concept: The total amount of money (£270) hasn’t changed, it just moved around.

New Ratio \( C : D : E = 2 : 5 : 3 \)

❓ Question: “How much money did Chris give to Errol?”

We need to look at how Errol’s money changed.

💡 Key Concept: North lines are parallel. The angle inside the “C” shape (co-interior angle) adds up to \( 180^\circ \).

💡 Shortcut: To reverse a bearing, if the original is less than 180°, just add 180°.

💡 Formula: Volume of cylinder = \( \pi r^2 h \)

💡 Key Fact: \( 1000 \text{ cm}^3 = 1 \text{ litre} \)

💡 Reasoning: We need to find how many lots of 0.47 litres fit into the total volume.

\( \text{Time} = \frac{\text{Total Volume}}{\text{Rate}} \)

💡 How to find bounds:

The number is rounded to 2 significant figures (the tenths column here). The “gap” between 2 s.f. numbers is 0.1 (e.g., 1.8, 1.9, 2.0).

The boundary is halfway between the values.

Half of 0.1 is 0.05.

💡 Strategy: Ignore the third bracket for a moment. Just multiply \( (x+7) \) and \( (x-2) \).

💡 Strategy: Now multiply the result \( (x^2 + 5x – 14) \) by \( (x + 3) \).

💡 Rule: The probabilities on branches coming from a single point must add up to 1.

💡 Strategy: “At least one pass” includes:

• Pass, Pass
• Pass, Fail
• Fail, Pass

Shortcut Method: The only outcome we DON’T want is “Fail, Fail”. It’s easier to calculate 1 minus P(Fail, Fail).

• Graph A: A straight line through the origin. This represents direct proportion (\( y = kx \)).
• Graph B: A curve that gets closer to the axes but never touches (asymptotes). This represents inverse proportion (\( y = \frac{k}{x} \)).
• Graph C: A curve starting at origin, increasing quickly then slowing down. This is the square root graph (\( y = k\sqrt{x} \)).
• Graph D: A curve starting at origin, getting steeper. This is the square graph (\( y = kx^2 \)).

💡 Formula: Midpoint = \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

💡 Formula: \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)

💡 Why we do this: We want to create two equations where the recurring part matches, so we can subtract them and eliminate the decimal.

The dots are on the 7 and 2, so the pattern is “72”.

We need to move the decimal point past the non-recurring part (1 digit) and past the recurring part (2 digits).

Wait! The question asks for \( 0.0723723… \)?

Correction based on the fraction \( \frac{241}{3330} \):

Let’s check the fraction \( \frac{241}{3330} \approx 0.0723723 \). Ah, the dots in the question image for Q13 were on the 7 and the 3? Or 7, 2, 3? Let’s re-read the OCR.

OCR says: “0.0723 with dots on 7 and 3”. Pattern is 723.

Let’s re-calculate for \( 0.0\dot{7}2\dot{3} \) (pattern 723):

💡 Direct Proportion: \( y = kx^2 \)

💡 Inverse Proportion: \( x = \frac{c}{w} \)

We need to find \( y \) when \( w = 2 \). First, let’s find \( x \) when \( w = 2 \).

💡 Frequency Density = Frequency ÷ Class Width

From the first histogram, we compare the first bar (10-12) to the known bars.

The 12-16 bar has Freq=10, Width=4, so FD = 2.5.

Visually, the 10-12 bar is lower than the 2.5 bar. It corresponds to a height of 1.5.

💡 The Median: The value exactly in the middle of the data. For 43 students, the median is at position \( \frac{43+1}{2} = 22 \).

First, calculate frequencies from the second histogram areas.

💡 Triangle Sum: Angles in triangle \( ABD \) add to \( 180^\circ \).

💡 Sine Rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} \)

In triangle \( ABD \), we know side \( AB = 7.9 \) (opposite \( 42^\circ \)) and we want \( BD \) (opposite \( 112^\circ \)).

💡 Alternate Angles: Since \( AD \) is parallel to \( BC \), angle \( ADB \) matches angle \( DBC \).

💡 Area Formula: \( \text{Area} = \frac{1}{2}ab \sin C \)

We use sides \( BD \) and \( BC \), and the included angle \( \angle DBC \).

💡 Reasoning: If a continuous function goes from negative to positive (or vice versa), it must cross zero.

Since \( H \) is the midpoint of \( AD \) and \( \vec{AH} = \mathbf{a} \), then \( \vec{AD} = 2\mathbf{a} \).

Similarly, \( \vec{AB} = 2\mathbf{b} \) and \( \vec{DC} = 2\mathbf{c} \).

💡 Order matters: Work from the inside out. First calculate \( f(-3) \), then put the result into \( g \).

💡 Strategy: Swap \( x \) and \( y \), then solve for the new \( y \).

💡 Strategy: Factorise denominators and quadratics first to see what cancels.

💡 BODMAS: Division comes before Addition. To divide by a fraction, flip it and multiply.

💡 “Given that the medal is gold…”

This means we only care about the medals inside the Gold circle (Set B). The total we are looking at is the total of Set B.

\( n(B) = x + (5x – 3) = 6x – 3 \)

The probability that it is English (Set A) given it is Gold (Set B) is the intersection divided by the total of Gold.

\( P(A | B) = \frac{\text{Intersection}}{\text{Total B}} \)