GCSE November 2021 Edexcel Higher Paper 3 (Calculator)

Step 1: Understand what an outlier is

An outlier is a point that does not fit the general pattern or trend of the data. Look for a point that is far away from the main cluster of points.

Step 2: Locate the point

Most points follow an upward trend (positive correlation). There is one point at traffic volume 100 that is very high (level 18), far above the others.

Reading the scales: x-axis = 100, y-axis = 18.

Step 1: Draw a line of best fit

We ignore the outlier (100, 18) and draw a straight line through the middle of the remaining points.

Step 2: Read off the value for 370 cars

Find 370 on the x-axis (between 300 and 400). Go up to your line of best fit, then read across to the y-axis.

Typically, for 370 cars, the value is between 12.8 and 14.8.

Step 1: Analyze Alfie’s statement

Alfie claims the outlier means there is no correlation. However, if we ignore the single outlier, the rest of the points clearly show a pattern (as traffic increases, carbon monoxide increases).

Step 2: Formulate the reason

An outlier does not negate the trend of the majority of the data. The data still shows a positive correlation.

Why: The ingredients are shared in a ratio. We need to find the specific weight of the cheese portion.

How:

1. Find total parts in the ratio:

\[ 9 + 2 + 1 = 12 \text{ parts} \]

2. Calculate the weight of one part:

\[ 6000 \div 12 = 500 \text{ g} \]

3. Calculate the weight of cheese (2 parts):

\[ 2 \times 500 = 1000 \text{ g} \]

Why: Cheese is sold in 175g units. We need to know how many units cover 1000g.

How:

\[ 1000 \div 175 = 5.714… \]

Since she cannot buy partial amounts (implied by “costs £2.25 for 175g”), we interpret this as a proportional cost or needing to buy whole packets? The question asks for the “cost of the cheese needed”. This usually implies a proportional cost unless it says “bags of cheese”. However, usually in these money problems, if it’s “needed”, we calculate the exact cost of that amount.

Check Mark Scheme: “Process may lead to 5 or 6 instead of 5.71…”. It awards marks for \( 5.71 \times 2.25 \). This implies we calculate the exact proportional cost.

Why: Multiply the number of 175g units by the price per unit.

How:

\[ 5.714… \times 2.25 \]

Alternatively:

\[ \frac{1000}{175} \times 2.25 = 12.857… \]

Move the decimal point 5 places to the right.

\[ 4.5 \rightarrow 450,000 \]

Move the decimal point to create a number between 1 and 10.

\[ 0.007 = 7 \times 10^{-3} \]

Method 1: Convert to ordinary numbers

\[ 4.2 \times 10^3 = 4200 \] \[ 5.3 \times 10^2 = 530 \] \[ 4200 + 530 = 4730 \]

Method 2: Convert to Standard Form

\[ 4730 = 4.73 \times 10^3 \]

Convert time: \( 1 \) min \( 40 \) secs = \( 100 \) seconds = \( \frac{100}{60} \) mins = \( \frac{5}{3} \) mins.

Calculate Rate:

\[ \text{Rate A} = \frac{8 \text{ litres}}{100 \text{ seconds}} = 0.08 \text{ litres/sec} \] \[ \text{Rate A per min} = 0.08 \times 60 = 4.8 \text{ litres/min} \]

Convert gallons to litres.

\[ 2.2 \text{ gallons/min} \times 4.54 \text{ litres/gallon} \]

Time = Volume ÷ Rate

Why: In a Fibonacci sequence, the next term is the sum of the previous two terms.

How:

\[ \text{5th Term} = \text{3rd Term} + \text{4th Term} \] \[ \text{5th Term} = 3a + 5a = 8a \]

Sum of first five terms is 228.

\[ a + 2a + 3a + 5a + 8a = 228 \]

Step 1: Use the sum of probabilities

The sum of all probabilities must equal 1.

\[ 1 – (0.05 + 0.15) = 1 – 0.20 = 0.80 \]

So, \( P(\text{green}) + P(\text{pink}) = 0.80 \)

Step 2: Use the relationship given

Let \( x \) be the probability of pink.

Green is \( 0.2 \) more than pink: \( P(\text{green}) = x + 0.2 \)

Step 3: Solve for x

\[ (x + 0.2) + x = 0.80 \] \[ 2x + 0.2 = 0.80 \] \[ 2x = 0.60 \] \[ x = 0.30 \]

So, Pink = \( 0.3 \), Green = \( 0.3 + 0.2 = 0.5 \)

Why: Probability = (Number of successful outcomes) ÷ (Total number of outcomes).

We know the number of blue counters is 18 and P(blue) is 0.15.

How:

\[ 0.15 = \frac{18}{\text{Total}} \] \[ \text{Total} = \frac{18}{0.15} \]

Why: The shape is a quarter circle (indicated by the right angle at \( O \)).

Formula: \( \text{Area} = \frac{\theta}{360} \times \pi r^2 \)

Here \( \theta = 90^\circ \) and \( r = 8 \).

Why: We need to remove the triangle area from the sector area to get the segment.

Formula: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

Subtract the triangle from the sector.

Theory: Direct proportion means \( y = kx \). This is the equation of a straight line passing through the origin.

Theory: Inverse proportion means \( y = \frac{k}{x} \). As \( x \) increases, \( y \) decreases. The graph is a curve that gets closer to the axes but never touches them (asymptotes).

Why: More people means less time. This is inverse proportion.

Step 1: Calculate Total Man-Hours

\[ \text{Total work} = 12 \text{ people} \times 5 \text{ hours} = 60 \text{ person-hours} \]

Step 2: Divide by new number of people

\[ \text{Time} = \frac{60 \text{ person-hours}}{15 \text{ people}} \]

If they don’t work at the same rate, the time could change.

Examples:

• If they work slower, it takes longer.
• If they work faster, it takes less time.

Why: Both triangles \( ACB \) and \( DEB \) share angle \( B \) and have a right angle (\( 90^\circ \)). Therefore, they are similar.

Find Scale Factor:

Length of hypotenuse \( AB = AD + DB = 9 + 6 = 15 \text{ cm} \).

Length of hypotenuse \( DB = 6 \text{ cm} \).

\[ \text{Scale Factor} = \frac{AB}{DB} = \frac{15}{6} = 2.5 \]

Alternatively, we can use trigonometry on the smaller triangle first.

Why: Use Pythagoras’ Theorem in \( \triangle DEB \).

\( DE = 2 \), \( DB = 6 \) (hypotenuse).

\[ EB^2 + DE^2 = DB^2 \]

Method: Multiply \( EB \) by the scale factor.

\[ CB = \text{Scale Factor} \times EB \]

Why: When rounding to 1 decimal place, the range of possible values spans half a unit (0.05) above and below the rounded value.

How:

• Lower Bound: \( 6.4 – 0.05 = 6.35 \)
• Upper Bound: \( 6.4 + 0.05 = 6.45 \)

Rule: \( (AB)^m = A^m B^m \) and \( (X^p)^q = X^{pq} \)

Apply:

\[ (ax^6)^{\frac{1}{n}} = a^{\frac{1}{n}} (x^6)^{\frac{1}{n}} \] \[ = a^{\frac{1}{n}} x^{\frac{6}{n}} \]

Compare with Right Side: \( 7x^3 \)

The power of \( x \) must match:

\[ \frac{6}{n} = 3 \]

The coefficient term must match:

\[ a^{\frac{1}{n}} = 7 \]

Substitute \( n = 2 \):

\[ a^{\frac{1}{2}} = 7 \]

\( a^{\frac{1}{2}} \) means square root of \( a \).

\[ \sqrt{a} = 7 \] \[ a = 7^2 \]

The pattern consists of 4 identical rectangles. Looking at the step pattern from A to B:

• There are 2 “flat” rectangles and 2 “tall” rectangles (or 4 steps involved).
• Let the short side be \( x \) and the long side be \( y \).
• Or let width be \( w \) and height be \( h \) for the flat one.

Horizontal Change (x):

From 3 to 11 is a change of 8.

The path goes: Long + Short + Long + Short? Or is it 2 Longs?

Looking at the diagram (Flat, Tall, Flat, Tall):

Total X width = \( L + S \)? No, from A to B.

Left of 1st to Right of 4th?

Let’s look at the corners.

A is bottom-left of Rect 1.

B is top-right of Rect 4.

Sequence: 1(Flat), 2(Tall), 3(Flat), 4(Tall).

X distance = \( L \) (rect 1) \( + S \) (rect 2) \( + L \) (rect 3) \( + S \) (rect 4)?

Actually, usually these patterns stack. Let’s assume:

Total X = 2L + 2S? No.

Wait, B is (11, 20). A is (3, 4).

Change in X = 8.

Change in Y = 16.

Looking at the pattern:

There are 2 “steps” in X and 2 “steps” in Y? Or 4?

If there are 4 rectangles, usually arranged in 2 pairs.

Equation 1: \( 2L = 8 \) (if S is negligible? No).

Let’s look at the equations from the mark scheme.

Mark scheme says: \( 2b – a = 8 \) or \( 2x + y = 8 \).

And \( 2b + a = 16 \) or \( 2x + 3y = 16 \).

Let’s decode this.

Let \( x \) be the width of the rectangle and \( y \) be the height (of the flat one).

Rect 1 (Flat): Width \( L \), Height \( S \).

Rect 2 (Tall): Width \( S \), Height \( L \).

Rect 3 (Flat): Width \( L \), Height \( S \).

Rect 4 (Tall): Width \( S \), Height \( L \).

Total X distance from A to B:

Left of 1 to Right of 4?

Usually B is aligned with the right edge. Let’s assume standard stacking.

Mark scheme \( 2b – a = 8 \) suggests X-dist involves subtraction?

Ah, look at the diagram in Q13 (page 14 of PDF).

Diagram shows:

Rect 1: Flat.

Rect 2: Standing on top of Rect 1’s right side.

Rect 3: Flat, sticking out from Rect 2’s right side.

Rect 4: Standing on top of Rect 3.

So, X-coordinates:

Start A (3). End B (11).

Gap = 8.

The gap consists of: (Length of 1 – Width of 2?) No.

Let’s trace the right-most edge.

A is bottom-left.

Right side of 1 is at \( 3 + L \).

Rect 2 is placed… where? Usually centered or aligned.

Let’s look at Mark Scheme Q13 again.

\( 2b – a = 8 \) and \( 2b + a = 16 \).

This implies:

\( 4b = 24 \implies b = 6 \). (Length)

\( 12 – a = 8 \implies a = 4 \). (Width? No, \( a \) usually width).

Wait, if \( b=6, a=4 \).

Let’s check Y.

\( 2b + a = 2(6) + 4 = 16 \). Correct.

\( 2b – a = 2(6) – 4 = 8 \). Correct.

So dimensions are 6 and 4.

Step 2: Find coordinates of C

C is marked on the diagram.

Usually C is the top-right corner of the first rectangle, or the join point.

Let’s assume C is the top-right of Rect 1.

A is (3, 4).

Rect 1 is flat (Length 6, Height 4? Or Length 6, Height 2?)

Wait, \( b=6, a=4 \). Is 4 the short side?

If Rect 1 is flat, width is 6, height is 4.

C = \( (3+6, 4+4) = (9, 8) \).

Let’s check Mark Scheme answer: (9, 8). Correct.

So Rect 1 is \( 6 \times 4 \).

Let Total(Fran + Gary) = \( T_{FG} \)

Let Total(Olivia + Jessica) = \( T_{OJ} \)

We are told: \( T_{OJ} = \frac{1}{2} T_{FG} \)

Or \( T_{FG} = 2 T_{OJ} \).

Fran : Gary = \( 2 : 3 \). Total parts = 5.

So \( T_{FG} \) must be a multiple of 5.

Olivia : Jessica = \( 9 : 1 \). Total parts = 10.

So \( T_{OJ} \) must be a multiple of 10.

We need \( T_{FG} = 2 T_{OJ} \).

Let \( T_{OJ} = 10 \) (since ratio is 9:1).

Then \( T_{FG} = 20 \).

Check if 20 is divisible by 5 (Fran:Gary ratio). Yes, \( 20 \div 5 = 4 \).

The rate of flow at a specific time is the gradient of the tangent to the curve at that point.

1. Locate \( t = 12 \) on the x-axis.

2. Go up to the curve.

3. Draw a straight line that just touches the curve at this point, following the slope.

Pick two points on your tangent line to calculate \( \frac{\text{change in y}}{\text{change in x}} \).

Example points from mark scheme guidance:

Change in volume \( \approx 28 \)

Change in time \( \approx 30 \)

Gradient \( = \frac{-28}{30} \approx -0.93 \)

The question asks for rate of flow out, so we give the positive value.

Why: To solve simultaneous equations, it’s easier to substitute one into the other. Let’s make \( y \) the subject of the line equation.

Equation L: \( y – 5x + 4 = 0 \)

\[ y = 5x – 4 \]

Why: At the point of intersection, the \( y \) values are the same.

Substitute: \( x^2 + 3x – 3 = 5x – 4 \)

Rearrange to form \( ax^2 + bx + c = 0 \).

\[ x^2 + 3x – 5x – 3 + 4 = 0 \] \[ x^2 – 2x + 1 = 0 \]

Since \( (x-1)^2 = 0 \), there is only one solution for \( x \) (which is \( x=1 \)).

One value for \( x \) means there is only one point of intersection.

Statement 1: \( x \propto y^2 \rightarrow x = k y^2 \)

Statement 2: \( y \propto z^3 \rightarrow y = c z^3 \)

Substitute the expression for \( y \) into the equation for \( x \).

\[ x = k (c z^3)^2 \] \[ x = k c^2 z^6 \]

We can combine the constants \( k \) and \( c^2 \) into a single constant, let’s call it \( A \).

\[ x = A z^6 \]

Use the given values: \( x = 32 \) when \( z = 2 \).

\[ 32 = A (2)^6 \] \[ 32 = A (64) \] \[ A = \frac{32}{64} = 0.5 \]

We need to find the path from \( D \) to \( E \).

Path: \( \vec{DE} = \vec{DB} + \vec{BE} \)

We need to express \( \vec{DB} \) and \( \vec{BE} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \).

Vector OB: Go from \( O \) to \( B \) via \( A \).

\[ \vec{OB} = \vec{OA} + \vec{AB} = \mathbf{a} + \mathbf{b} \]

Vector DB: \( D \) divides \( OB \) in ratio \( 2:3 \). This means \( D \) is \( \frac{2}{5} \) of the way along.

\( DB \) is the remaining \( \frac{3}{5} \) of the vector.

\[ \vec{DB} = \frac{3}{5} \vec{OB} = \frac{3}{5}(\mathbf{a} + \mathbf{b}) \]

Vector BC: Go from \( B \) to \( C \) via \( A \) and \( O \) (or \( B \to O \to C \)? No, easier path).

\( \vec{BC} = \vec{BA} + \vec{AO} + \vec{OC} \)

\( \vec{BA} = -\mathbf{b} \)

\( \vec{AO} = -\mathbf{a} \)

\( \vec{OC} = 3\mathbf{b} \)

\[ \vec{BC} = -\mathbf{b} – \mathbf{a} + 3\mathbf{b} = 2\mathbf{b} – \mathbf{a} \]

Vector BE: \( E \) divides \( BC \) in ratio \( 1:4 \). So \( BE \) is \( \frac{1}{5} \) of \( BC \).

\[ \vec{BE} = \frac{1}{5} \vec{BC} = \frac{1}{5}(2\mathbf{b} – \mathbf{a}) \]

Add the components together.

\[ \vec{DE} = \vec{DB} + \vec{BE} \] \[ \vec{DE} = \left(\frac{3}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) + \left(\frac{2}{5}\mathbf{b} – \frac{1}{5}\mathbf{a}\right) \]

Start 2017: \( P_{2017} = 4000 \)

Start 2018: \( P_{2018} = k \times P_{2017} = 4000k \)

Start 2019: \( P_{2019} = k \times P_{2018} = k(4000k) = 4000k^2 \)

Strategy: It is easier to calculate the probability of the opposite events and subtract from 1.

The total probability space includes:

• All Heads
• All Tails
• Mix of Heads and Tails (At least 1 head and at least 1 tail)

So, \( P(\text{mix}) = 1 – P(\text{all heads}) – P(\text{all tails}) \)

For a fair coin, \( P(H) = \frac{1}{2} \) and \( P(T) = \frac{1}{2} \).

\( P(\text{all } n \text{ heads}) = \left(\frac{1}{2}\right)^n \)

\( P(\text{all } n \text{ tails}) = \left(\frac{1}{2}\right)^n \)

Step 1: Define the Trapeziums

Width of each trapezium = 1 unit (from 1 to 2, 2 to 3, 3 to 4).

Read the height (\( y \)-values) from the graph at \( t = 1, 2, 3, 4 \).

• At \( t=1 \), \( y \approx 4 \)
• At \( t=2 \), \( y \approx 6 \)
• At \( t=3 \), \( y \approx 7.2 \)
• At \( t=4 \), \( y \approx 7.8 \)

Step 2: Calculate Area of Each Trapezium

Formula: \( \text{Area} = \frac{h}{2}(a + b) \)

Trapezium 1 (t=1 to 2): \( \frac{1}{2}(4 + 6) = 5 \)

Trapezium 2 (t=2 to 3): \( \frac{1}{2}(6 + 7.2) = 6.6 \)

Trapezium 3 (t=3 to 4): \( \frac{1}{2}(7.2 + 7.8) = 7.5 \)

Step 3: Sum the Areas

\[ 5 + 6.6 + 7.5 = 19.1 \]

The area under a speed-time graph represents distance traveled.

Why: Dividing by a fraction is the same as multiplying by its reciprocal.

\[ \frac{6x^3}{9x^2 – 144} \times \frac{3(x-4)}{2x^4} \]

Factorise \( 9x^2 – 144 \):

Take out common factor 9: \( 9(x^2 – 16) \)

Difference of two squares: \( 9(x-4)(x+4) \)

Cancel \( (x-4) \): Top and bottom.

Cancel Numbers: \( 6 \times 3 = 18 \) on top. \( 9 \times 2 = 18 \) on bottom. They cancel to 1.

Cancel \( x \): \( x^3 \) on top, \( x^4 \) on bottom. Leaves \( x \) on bottom.

In \( \triangle ABD \):

\[ \frac{AB}{\sin(\angle ADB)} = \frac{BD}{\sin x} \]

Rearrange for \( AB \):

\[ AB = \frac{BD \sin(\angle ADB)}{\sin x} \]

In \( \triangle ADC \):

\[ \frac{AC}{\sin(\angle ADC)} = \frac{DC}{\sin x} \]

Rearrange for \( AC \):

\[ AC = \frac{DC \sin(\angle ADC)}{\sin x} \]

Angles on a straight line add to \( 180^\circ \).

\( \angle ADC = 180^\circ – \angle ADB \)

Property of Sine: \( \sin(180^\circ – \theta) = \sin(\theta) \)

Therefore, \( \sin(\angle ADC) = \sin(\angle ADB) \).

Since the numerators \( \sin(\angle ADB) \) and \( \sin(\angle ADC) \) are equal, and the denominators \( \sin x \) are equal:

\[ \frac{AB}{BD} = \frac{AC}{DC} \]