Pearson Edexcel GCSE (9-1) Mathematics Higher Paper 3H (Summer 2022)

💡 Why we do this: Since we have a right-angled triangle and we know two sides, we can use Pythagoras’ theorem to find the third side.

Pythagoras’ theorem is: \[ a^2 + b^2 = c^2 \]

Here, the longest side (the hypotenuse) is opposite the right angle. This is \(c = 8.5\).

The other two sides are \(a = 4\) and \(b = x\).

💡 How we do it: We rearrange the formula to find one of the shorter sides.

💡 Why we do this: We need to replace \(m\) with \(-3\) in the formula. Remember to put brackets around negative numbers when squaring!

💡 Why we do this: “Make \(p\) the subject” means we want to get \(p\) on its own, like \(p = \dots\).

We use inverse operations (doing the opposite) to move the other terms away from \(p\).

💡 Why we do this: We can use algebra to represent the number of counters each person has.

Let \(x\) be the number of counters Rick has.

💡 How we do it: We know the total is 54, so we sum the expressions and solve for \(x\).

💡 Why we do this: The question asks for the ratio Rick : Tony in the form \(1 : p\).

💡 Strategy: Jo needs 15 rolls. At Chic Decor, they are sold in sets of 3.

Number of sets needed = \(15 \div 3 = 5\).

💡 Strategy: At Style Papers, they are sold in packs of 5.

Number of packs needed = \(15 \div 5 = 3\).

First, find the normal price, then calculate the discounted price.

💡 Comparison:

• Chic Decor: £180.00
• Style Papers: £184.80

💡 Rule: For a frequency polygon, points must be plotted at the midpoint of the interval.

Check the last interval: \(40 < t \leqslant 50\).

• Midpoint = \((40 + 50) \div 2 = 45\).
• Amos plotted the point at \(t = 50\) (the end of the interval).

💡 Rule: Scales must be linear (going up by the same amount each time).

Look at the vertical (Frequency) axis.

It goes: 0, 10, 20, 30, … 50, 60.

The number 40 is missing from the scale.

💡 Why we do this: To compare Amy and Jessica, we first need to know the total distance Jessica ran.

Formula: \(\text{Distance} = \text{Speed} \times \text{Time}\).

Important: The speed is in miles per hour, but time is in minutes. We must convert minutes to hours by dividing by 60.

💡 Why we do this: Now we have the total distance (7.5 miles) and Amy’s time (45 minutes).

Formula: \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\).

💡 Strategy: To find the area of the shaded trapezium \(QUVR\), we need the lengths of its parallel sides (\(QU\) and \(RV\)) and its height (\(UV\)).

• The height of the rectangle is given as \(4x\), so \(UV = 4x\).
• One parallel side is \(RV = 5\).
• We need to find the length of the top side \(QU\).

💡 Formula: Area of trapezium = \(\frac{1}{2}(a + b)h\)

Here: \(a = x + 5\) (top), \(b = 5\) (bottom), \(h = 4x\) (height).

💡 Strategy: To find the gradient, pick two clear points on the line and use the formula: \[ \text{Gradient} = \frac{\text{Change in } y}{\text{Change in } x} \]

The line starts at \((0, 0)\). Let’s find another point. At \(x = 100\), the line is at \(y = 14\).

💡 Why we do this: The gradient represents the change in Y (Cost in £) divided by the change in X (Number of units).

Unit: £ per unit.

💡 Rules of Indices:

1. Multiplying: \(a^m \times a^n = a^{m+n}\)
2. Dividing: \(\frac{a^m}{a^n} = a^{m-n}\)
3. Square Root: \(\sqrt{a^m} = a^{m/2}\)

💡 Rule: Power of a power means you multiply the indices, you don’t square them.

💡 Why we do this: Depreciation means the value goes down. We calculate multipliers for each year.

• Year 1: Decrease by 12.5%. Multiplier = \(1 – 0.125 = 0.875\)
• Year 2 & 3: Decrease by 10%. Multiplier = \(1 – 0.10 = 0.9\)

💡 How we do it: Let the original value be \(x\). The value after 3 years is found by multiplying \(x\) by the multipliers in order.

💡 Why we do this: When choosing one item from each of several independent groups, we multiply the number of choices in each group to find the total combinations.

💡 Why we do this: The side \(AC\) is shared by both triangles. We know \(DC=8\) and the angle \(D=45^\circ\). We need \(AC\) (Opposite) and we have \(DC\) (Adjacent).

Formula: \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)

💡 Why we do this: Now we know \(AC=8\). In triangle \(ABC\), relative to angle \(B=20^\circ\):

• \(AC\) is the Opposite side.
• \(AB\) is the Hypotenuse (side opposite the right angle).

Formula: \(\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\)

💡 How we do it: Multiply each component of vector \(\mathbf{a}\) by 3.

💡 Strategy: We have \(3\mathbf{a} – 2\mathbf{b} = \begin{pmatrix} 8 \\ -17 \end{pmatrix}\).

Rearrange to find \(2\mathbf{b}\): \(2\mathbf{b} = 3\mathbf{a} – \begin{pmatrix} 8 \\ -17 \end{pmatrix}\).

💡 Strategy: Look for a common factor first, then check for Difference of Two Squares.

1. Take out the factor 4: \(4(p^2 – 9)\)
2. \(p^2 – 9\) is a difference of two squares: \((p – 3)(p + 3)\)

💡 Strategy: Multiply the first two brackets together, simplify, then multiply the result by the third bracket.

💡 Reason: Angles formed by intersecting lines are equal.

💡 Reason: Angles subtended by the same arc at the circumference are equal.

💡 Strategy: To make a fraction (division) as large as possible (Upper Bound), we need the numerator to be big and the denominator to be small.

Find the bounds for \(e\) and \(f\):

• \(e = 6.8\) (1 d.p.) \(\rightarrow\) Bounds are \(6.75\) and \(6.85\).
• \(f = 0.05\) (1 s.f.) \(\rightarrow\) Bounds are \(0.045\) and \(0.055\).

💡 Formula: \(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}\)

💡 Key Idea: In a histogram, Area = Frequency. We compare the area of the bar for 1-2 miles with the area of the bar for 3-5 miles.

💡 Strategy: The cross section \(SRCD\) is a trapezium. We know its area is 147 cm\(^2\) and width \(CD = 14\) cm.

Formula: \(\text{Area} = \frac{SD + CR}{2} \times CD\)

💡 Strategy: We need the distance \(DT\) on the base to form a right-angled triangle \(\Delta SDT\). \(D\) is a corner of the square base.

First, find \(TC\). Ratio \(BT:TC = 4:3\), so \(TC = \frac{3}{7}\) of \(BC\).

💡 Strategy: In the vertical triangle \(\Delta SDT\), angle \(D\) is \(90^\circ\). We want angle \(\angle STD\).

\(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{SD}{DT}\)

💡 Strategy: The common denominator is \((x+2)(x-2)\), which is \(x^2 – 4\).

We write each term with this denominator.

⚠️ Watch Out: Be careful with the minus signs! We are subtracting the entire second and third expressions.

💡 Strategy: Use the values for 2018 (\(P_n\)) and 2019 (\(P_{n+1}\)) to solve for \(a\).

💡 Strategy: To get an even sum from 3 numbers, we need to consider Odds (O) and Evens (E).

Numbers 1 to 9:

• Odd: 1, 3, 5, 7, 9 (5 numbers)
• Even: 2, 4, 6, 8 (4 numbers)
• Total: 9 numbers

Combinations that sum to Even:

• Even + Even + Even (E, E, E)
• Even + Odd + Odd (E, O, O) (in any order)

💡 Method: Selection is without replacement.

💡 Strategy: There are 3 arrangements: (E,O,O), (O,E,O), (O,O,E).

All have the same probability value.

💡 Strategy: Substitute \(y = 2x – 5\) into \(y^2 = 6x^2 – 25x – 8\).

💡 Strategy: Factorise or use quadratic formula. \(2x^2 – 5x – 33 = 0\).

\(2 \times -33 = -66\). Factors of -66 adding to -5 are -11 and 6.