GCSE Nov 2019 Edexcel Higher Paper 2 (Calculator)

What we check: A frequency polygon should be plotted at the midpoints of the intervals. The y-value should match the frequency in the table.

First Interval: \(10 < w \le 20\). Midpoint is 15. Frequency is 6.

The Graph: The first point is plotted at \(x = 15\), which is correct for the midpoint. However, look at the y-axis (height). It is plotted at a height of 8 (4 grid squares up, where each square is 2 units). It should be at 6.

What we check: A frequency polygon connects the plotted points with straight lines. It should not be closed (joined back to the x-axis) unless specifically asked for (which is rare in GCSE).

The Graph: The polygon has been “closed” by adding lines connecting to the x-axis at 10 and 60. This is incorrect.

Why we do this: When a number is rounded to the “nearest unit”, the actual value could be half a unit below or half a unit above.

The degree of accuracy is 1 mm. We divide this by 2:

\[ \frac{1}{2} = 0.5 \text{ mm} \]

What this tells us: The length must be at least 127.5 but strictly less than 128.5 (because 128.5 would round up to 129).

Why we do this: We need to know how many stamps each person started with. The ratio is \(3:7\) and the total is 240.

Why we do this: Tom buys stamps from Adam. This means the Total (240) stays the same, but the distribution changes.

The new ratio is \(3:5\). We can find the new amounts using the same total.

What this tells us: The difference between Tom’s new amount and old amount is what he bought.

Why we do this: We use the sample results to estimate the preference for the whole population. In the sample of 50, 17 people wanted a sports bag.

Why we do this: Statistical estimates rely on the sample being unbiased and representative.

Assumption: The sample is representative of the whole population (no bias).

Effect: If the sample is biased (e.g., asked only people in the weights room), the estimate might be too high or too low.

What to look for: A cubic graph \(y = x^3\) passes through the origin (0,0). When \(x\) is positive, \(y\) is positive. When \(x\) is negative, \(y\) is negative. It has a characteristic ‘S’ shape.

Graph F matches this description.

What to look for: This is a reciprocal graph. It has asymptotes at \(x=0\) (y-axis) and \(y=0\) (x-axis). It exists in the top-right quadrant (positive/positive) and bottom-left quadrant (negative/negative).

Graph D matches this description.

(Note: Graph C is not a function of x as it has two y-values for positive x).

Why we do this: To find common numbers, we need to list the terms of each sequence to see if they overlap.

Method: Substitute \(n = 1, 2, 3, \dots\) into \(2n^2 – 1\).

Method: Substitute \(n = 1, 2, 3, \dots\) into \(40 – n^2\).

Note: This sequence decreases. We stop when terms become negative (unless specified otherwise, sequence terms are usually considered until they drop below zero or diverge away from the other sequence’s range).

What this tells us: We compare the two lists.

Why we do this: We can handle the number parts and the power of 10 parts separately, or simply type it into a calculator.

Why we do this: Standard form requires the first number to be between 1 and 10 (i.e., \(1 \le A < 10\)).

Why we do this: Substitute \(d = 40\) into the formula to find \(n\).

Why we do this: Substitute \(d = 30\) into the formula to find \(n\).

Why we do this: We are given the dimensions of the cuboid (Length=18, Width=8, Height=6). We need its total surface area to find information about the cube.

Formula: \(2(lw + lh + wh)\)

Why we do this: We know the surface area of the cube is also 600 cm². A cube has 6 identical square faces with side \(x\).

Formula: \(6x^2 = \text{Surface Area}\)

Why we do this: Janet claims the volumes are equal. We must calculate both volumes to verify this.

Why we do this: The term we want (\(k\)) is trapped inside the square root. We perform the inverse operation (squaring) to both sides.

Why we do this: We want \(k = \dots\). Currently \(k\) is negative.

Method 1: Add \(k\) to both sides, then subtract \(y^2\).

Method 2: Rearrange terms directly.

What we check: The median represents the “middle” value, where 50% of the data lies below it. Megan claims half her potatoes weigh less than 50g.

From the graph: Look at the vertical line inside Megan’s box. It is at 57 grams.

Conclusion: Since the median (50% mark) is 57g, then 50% of the potatoes weigh less than 57g. The number 50 is below the median, so less than half the potatoes are under 50g.

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

1. An average (Median).
2. A measure of spread (Interquartile Range – IQR, or Range).

Comparison 1 (Average): Megan’s median weight (57g) is higher than Amy’s median weight (42g). On average, Megan’s potatoes are heavier.

Comparison 2 (Spread): Megan’s IQR (26g) is larger than Amy’s IQR (16g). Megan’s potatoes have a greater variation/spread in weight.

Why we do this: Triangle \(BDC\) is a right-angled triangle. We know the hypotenuse \(BC = 8.6\) and angle \(C = 40^\). We need \(BD\) to locate point \(E\).

We use SOH CAH TOA. \(BD\) is Opposite to \(40^\).

Why we do this: \(E\) is the midpoint of \(DB\). We need length \(ED\) to work in the smaller triangle \(ADE\).

Why we do this: Triangle \(ADE\) is a right-angled triangle (since \(DEB\) is perpendicular to \(ADC\)). We know adjacent side \(AD = 4.4\) and opposite side \(ED \approx 2.764\).

We use \(\tan(x) = \frac{\text{Opp}}{\text{Adj}}\).

Why we do this: First, apply the 2.6% interest for 2 years using the multiplier \(1.026\).

Why we do this: Let \(x\) be the multiplier for the third year. We know the final amount is £3819.21.

What this tells us: The multiplier 1.022 means an increase of 2.2%.

Why we do this: The total number of combinations is the product of the number of choices for each category.

Let \(r\) be the number of roof colours.

Method: Multiply \((3x + 2)(2x + 1)\) first using FOIL (First, Outside, Inside, Last).

Method: Multiply the result \((6x^2 + 7x + 2)\) by \((x – 5)\).

Multiply every term in the first polynomial by \(x\), then every term by \(-5\).

Why we do this: The product of two numbers is even if at least one of the numbers is even. The product is odd only if both numbers are odd.

It is often easier to calculate \(1 – P(\text{Product Odd})\).

Method: We pick without replacement (he takes two cards).

Alternative Method: (Odd, Even) + (Even, Odd) + (Even, Even)

\((\frac{5}{9} \times \frac{4}{8}) + (\frac{4}{9} \times \frac{5}{8}) + (\frac{4}{9} \times \frac{3}{8}) = \frac{20}{72} + \frac{20}{72} + \frac{12}{72} = \frac{52}{72}\)

Reasoning: The tangent \(CAD\) is perpendicular to the radius \(OA\) at the point of contact \(A\).

Therefore, angle \(OAD = 90^\circ\).

In triangle \(OAD\), angles sum to \(180^\circ\).

Reasoning: \(BOD\) is a straight line. Angles on a straight line sum to \(180^\circ\).

Triangle \(AOB\) is isosceles because \(OA\) and \(OB\) are both radii.

Therefore, angles \(OAB\) and \(OBA\) are equal.

Reasoning: Angle \(OAC\) is \(90^\circ\) (Radius-Tangent). The line \(CAD\) is a straight line.

Alternatively, the angle \(CAB\) is the angle between the tangent and the chord.

From the diagram, we can see angle \(CAB = 90 – 29\).

(Wait, checking geometry: A is the vertex. Tangent is horizontal. OA is vertical. Angle OAC is 90. Angle OAB is inside this 90 degree sector?)

Yes, B is to the left of the vertical radius OA. C is to the left. Therefore Angle \(CAB = 90 – 29\).

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

What we know: 20% of the total trees are between 10 and 12.5 metres.

The missing bar covers the range 10 to 15.

Assuming the distribution is uniform within the 10-15 class (standard histogram assumption), the frequency for 10-12.5 (width 2.5) is exactly half of the frequency for 10-15 (width 5).

Let \(F\) be the frequency of the missing class (10-15).

Then trees in 10-12.5 \(= 0.5F\).

Note: A hemisphere is half a sphere. The formula is \(\frac{2}{3}\pi r^3\).

Why we do this: 10.9 is rounded to 3 s.f. (1 decimal place). The error is \(\pm 0.05\).

Substitute the bounds of \(c\) into the formula \(d = \frac{c^3}{8}\).

Comparison:

\(d_{LB} = 159.661…\)

\(d_{UB} = 164.116…\)

We check to how many significant figures they agree.

• 1 s.f: 200 vs 200 (Yes)
• 2 s.f: 160 vs 160 (Yes)
• 3 s.f: 160 vs 164 (No)

Both round to 160 (2 significant figures).

Why we do this: The distance travelled is the area under the speed-time graph.

We can split the area into shapes: a triangle, a rectangle, and another triangle (or use the trapezium rule).

We need to find the time \(t\) when the accumulated area equals 450.

At \(t=30\), distance covered is 180m.

We need \(450 – 180 = 270\)m more distance.

This occurs during the constant speed section (speed = 12 m/s).

Acceleration is the gradient of the graph.

First part: Gradient is positive (speed increasing).

Last part: Gradient is negative (speed decreasing/deceleration).

We should compare the steepness (magnitude).

Start: End of March, \( P_0 = 200 \).

Apply the formula for the next month.

Since we are counting rabbits, we round to the nearest whole number (or truncate, but standard rounding is usually accepted unless context implies integers only).

163.6 rounds to 164.

Observation: The numbers are going down: 200 → 190 → 178 → 164.

Each month, the number of rabbits decreases.

Why we do this: The area of a parallelogram is given by \( \text{Area} = ab \sin C \), where \(a\) and \(b\) are adjacent sides and \(C\) is the included angle.

Method: Find the critical values (roots) by solving \( 2x^2 – 21x + 40 = 0 \).

Factorise: find two numbers that multiply to \(2 \times 40 = 80\) and add to -21. These are -5 and -16.

We need \( 2x^2 – 21x + 40 < 0 \). Since the coefficient of \(x^2\) is positive, the curve is U-shaped. It is less than 0 between the roots.

Reflection line: \(x = -1\).

Point A \((1, 4)\) is 2 units right of the line. Reflected point \(A’\) is 2 units left: \((-3, 4)\).

Point B \((3, 4)\) is 4 units right. Reflected \(B’\) is \((-5, 4)\).

Point C \((3, 2)\) -> \(C’ (-5, 2)\).

Point D \((1, 2)\) -> \(D’ (-3, 2)\).

We need to rotate the square at \((-3, 2), (-5, 2), (-5, 4), (-3, 4)\) so that two vertices land on the original vertices.

Original vertices: \((1,4), (3,4), (3,2), (1,2)\).

Look at \(D'(-3, 2)\) and \(A'(-3, 4)\). If we rotate 90° clockwise about \((-1, 2)\)? No.

Let’s check rotation about \((-1, 0)\).

Point \(D'(-3, 2)\): Vector from center \((-1,0)\) is \((-2, 2)\). Rotate 90° CW -> \((2, 2)\). Add to center -> \((1, 2)\). This is D.

Point \(A'(-3, 4)\): Vector from center \((-1,0)\) is \((-2, 4)\). Rotate 90° CW -> \((4, 2)\). Add to center -> \((3, 2)\). This is C.

So points D and C are invariant (D stays at D, A moves to C).

Rearrange \(3x + 2y = 17\) to \(y = mx + c\).

\(2y = -3x + 17 \Rightarrow y = -1.5x + 8.5\).

Gradient of L is \(-1.5\). The y-intercept is \(8.5\).

So, point \(B\) is \((0, 8.5)\).

Line M is perpendicular to L. Gradient of M is the negative reciprocal of \(-1.5\).

\( m_M = \frac{-1}{-1.5} = \frac{1}{1.5} = \frac{2}{3} \).

It passes through \(A(0, 2)\), so the y-intercept is 2.

Equation of M: \( y = \frac{2}{3}x + 2 \).

Points: \(A(0, 2)\), \(B(0, 8.5)\), \(C(3, 4)\).

The base \(AB\) is vertical along the y-axis.

Base Length \(= 8.5 – 2 = 6.5\).

Height is the perpendicular distance from C to the y-axis, which is the x-coordinate of C.

Height \(= 3\).