Edexcel GCSE Mathematics – Higher Paper 2 (Calculator) – June 2019

Working:

Start with the original inequality:

\[ 14n > 11n + 6 \]

Subtract \(11n\) from both sides to collect the \(n\) terms on the left:

\[ 14n – 11n > 6 \]

✓ (M1) For a method to isolate terms in \(n\)

\[ 3n > 6 \]

Divide both sides by 3 to get \(n\) by itself:

\[ n > 2 \]

Working:

Start with:

\[ -2 < x + 3 \le 4 \]

Subtract 3 from all three parts to isolate \(x\):

\[ -2 – 3 < x \le 4 - 3 \] \[ -5 < x \le 1 \]

✓ (M1) For correct isolation steps giving \(-5 < x \le 1\)

Now, we draw this on the number line:

• At \(-5\), use an open circle (because the inequality is strictly \(<\), so \(x\) cannot equal -5).
• At \(1\), use a closed/solid circle (because the inequality is \(\le\), meaning \(x\) can equal 1).
• Draw a solid line connecting them.

✓ (M1) For open circle at -5 or closed circle at 1 or line drawn between -5 and 1

Working:

Let’s calculate the \(y\) values:

• When \(x = -2\), \(y = 2(-2) – 3 = -4 – 3 = -7\). Coordinate: \((-2, -7)\)
• When \(x = 0\), \(y = 2(0) – 3 = 0 – 3 = -3\). Coordinate: \((0, -3)\)
• When \(x = 2\), \(y = 2(2) – 3 = 4 – 3 = 1\). Coordinate: \((2, 1)\)
• When \(x = 4\), \(y = 2(4) – 3 = 8 – 3 = 5\). Coordinate: \((4, 5)\)

\(x\)	-2	-1	0	1	2	3	4
\(y\)	-7	-5	-3	-1	1	3	5

✓ (B1) For at least 2 correct points stated or plotted

Working:

✓ (B2) For a correct straight line segment through at least 3 of the points.

Working:

In the sample, 10 out of 30 students chose the Theme Park.

Fraction of sample = \(\frac{10}{30}\)

Multiply this fraction by the total number of students (195):

\[ \frac{10}{30} \times 195 \]

✓ (M1) For working with proportion.

Calculator Steps:

1. Press: 10 ÷ 30 × 195 =
2. Calculator shows: 65

\[ = 65 \]

Working:

State that the sample is representative of the whole population. If it isn’t representative (e.g., if there is bias), the actual number of students who want to go to the Theme Park could be higher or lower than 65.

✓ (C1) For an acceptable statement regarding the sample representing the population.

Working:

\[ \text{Volume} = 30 \times 6 \times 19 \]

✓ (P1) For working with the volume of the cuboid.

\[ \text{Volume} = 3420 \text{ cm}^3 \]

Working:

\[ \text{Volume of water} = 3420 \times \frac{2}{3} \]

✓ (P1) For a process to find the volume of the water.

Calculator Steps:

1. Press: 3420 × 2 ÷ 3 =
2. Calculator shows: 2280

\[ \text{Volume of water} = 2280 \text{ ml} \]

Working:

\[ 2280 \div 275 \]

✓ (P1) For a complete process to find the number of cups.

\[ = 8.2909… \]

Since we need the number of completely filled cups, we take the whole integer part of this answer.

\[ \text{Filled cups} = 8 \]

Working:

\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \]

Substitute our known values into the formula:

\[ \sin(38^{\circ}) = \frac{AB}{16} \]

✓ (M1) For setting up the correct trigonometric equation.

Working:

Multiply both sides by 16:

\[ AB = 16 \times \sin(38^{\circ}) \]

Calculator Steps:

1. Check calculator is in DEGREES mode (usually a small ‘D’ at the top of the screen).
2. Press: 16 × sin 38 =
3. Calculator shows: 9.850556…

\[ AB = 9.850556… \]

Round to 2 decimal places:

The third decimal digit is 0, so we do not round up.

\[ AB = 9.85 \text{ cm} \]

Working:

Lower Bound: What is the smallest number that begins with 8.3?
It is exactly 8.3 (or 8.3000…). So, \(y\) can be equal to 8.3.

✓ (B1) For 8.3 in the correct position.

Upper Bound: What is the highest value it could reach before the display no longer begins with 8.3?
It could be 8.39, or 8.399, or 8.39999… As we add 9s, we get infinitely close to 8.4. Therefore, the upper boundary is 8.4. Because it must begin with 8.3, it must be strictly less than 8.4.

✓ (B1) For 8.4 in the correct position.

Working:

Let’s find the number of parts for each person:

• Abby (A) = 2 parts
• Ben (B) = 7 parts
• Chloe (C) gets 1.5 times Abby’s amount: \(1.5 \times 2 = 3\) parts
• Denesh (D) gets 1.5 times Abby’s amount: \(1.5 \times 2 = 3\) parts

✓ (P1) For working with ratio to find the amount for C or D (e.g., \(1.5 \times 2 = 3\)).

The unified ratio \(A : B : C : D\) is \(2 : 7 : 3 : 3\).

Working:

Find total number of parts:

\[ \text{Total parts} = 2 + 7 + 3 + 3 = 15 \]

✓ (P1) For finding the sum of the parts (\(2 + 7 + 3 + 3 = 15\)).

Divide the total money by the total number of parts:

\[ \text{Value of 1 part} = \frac{360}{15} \]

Calculator Steps:

1. Press: 360 ÷ 15 =
2. Calculator shows: 24

\[ \text{Value of 1 part} = £24 \]

Working:

\[ \text{Ben’s share} = 7 \times 24 \]

✓ (P1) For a complete process to find Ben’s money (e.g., \(360 \div 15 \times 7\)).

\[ = 168 \]

Working:

Original number: \(0.00562\)

To make this a number between 1 and 10, the decimal point needs to move after the 5, giving us \(5.62\).

We moved the decimal point 3 places to the right. Because the original number is very small (less than 1), our exponent \(n\) must be negative.

\[ = 5.62 \times 10^{-3} \]

✓ (B1) Correct standard form.

Working:

Original number: \(1.452 \times 10^3\)

Move the decimal point 3 places to the right:

\(1.452 \rightarrow 14.52 \rightarrow 145.2 \rightarrow 1452\)

\[ = 1452 \]

✓ (B1) Correct ordinary number.

Working:

Circumference of B is 90% of A. This means the 1D ratio is:
\(A : B = 100 : 90\), which simplifies to \(10 : 9\).

✓ (M1) For identifying the 1D scale factor (e.g., \(10:9\) or \(0.9\)).

To find the ratio of their areas, we square the 1D ratio:

\[ \text{Area Ratio } A : B = 10^2 : 9^2 \] \[ \text{Area Ratio } A : B = 100 : 81 \]

Working:

Area of E is 44% greater than F. This means if F is 100%, E is 144%.
Area Scale Factor (ASF) from F to E is \(1.44\).

So, the 2D Area ratio is:
\(E : F = 144 : 100\)

✓ (P1) For using 1.44 as the area scale factor (or writing 144 : 100).

To find the 1D Length ratio (\(e : f\)), we square root both sides:

\[ \text{Length Ratio } e : f = \sqrt{144} : \sqrt{100} \] \[ e : f = 12 : 10 \]

Simplify the ratio by dividing both sides by 2:

\[ e : f = 6 : 5 \]

Working:

\[ P(\text{not late}) = 1 – 0.15 = 0.85 \]

✓ (M1) For identifying the probability of ‘not late’ is 0.85.

We write 0.85 on all the “not late” branches, and 0.15 on the missing “late” branches.

✓ (A1) For a fully correct diagram.

Method 1: The “1 minus” rule (Fastest)

First, find the probability that she is NOT late on both days (following the bottom branch all the way across):

\[ P(\text{Not Late, Not Late}) = 0.85 \times 0.85 \] \[ = 0.7225 \]

✓ (M1) For calculating one correct product from the tree.

Now, subtract this from 1 (the total probability of all possible outcomes):

\[ P(\text{At least one late}) = 1 – 0.7225 \]

✓ (M1) For a complete method (e.g., \(1 – 0.7225\)).

\[ = 0.2775 \]

---

Method 2: Adding the target branches (Alternative)

We can add up the three paths where she is late at least once:

• Late, Late: \(0.15 \times 0.15 = 0.0225\)
• Late, Not Late: \(0.15 \times 0.85 = 0.1275\)
• Not Late, Late: \(0.85 \times 0.15 = 0.1275\)

\[ P(\text{Total}) = 0.0225 + 0.1275 + 0.1275 = 0.2775 \]

Working:

• Up to 20 mins: \(5\)
• Up to 40 mins: \(5 + 30 = 35\)
• Up to 60 mins: \(35 + 20 = 55\)
• Up to 80 mins: \(55 + 15 = 70\)
• Up to 100 mins: \(70 + 8 = 78\)
• Up to 120 mins: \(78 + 2 = 80\) (This matches our total of 80 workers!)

✓ (B1) For the correct values: 5, 35, 55, 70, 78, 80.

Working:

We plot the coordinates: \((20, 5)\), \((40, 35)\), \((60, 55)\), \((80, 70)\), \((100, 78)\), \((120, 80)\).

Connect them with a smooth S-shaped curve (an ogive).

✓ (M1) For 5 or 6 points plotted correctly.

✓ (A1) For a fully correct smooth curve or joined line segments.

Working:

From the graph (see red dotted line), at \(t = 90\) minutes, the cumulative frequency is roughly \(74\).

✓ (M1) For a clear method to read off the graph at 90 (sight of 74 implies this).

This means 74 workers took 90 minutes or less.

Workers taking more than 90 minutes: \(80 – 74 = 6\)

Now, convert 6 out of 80 into a percentage:

\[ \frac{6}{80} \times 100 \]

✓ (M1) For a full method to find the percentage.

\[ = 0.075 \times 100 = 7.5\% \]

Working:

The formula for the Area of a Sector is:

\[ \text{Area} = \frac{\theta}{360} \times \pi \times r^2 \]

✓ (P1) For identifying the area of the whole circle: \(\pi \times 7^2 = 153.94\)

Substitute our known values (\text{Area} = 40, \(r = 7\)):

\[ 40 = \frac{\theta}{360} \times \pi \times 7^2 \] \[ 40 = \frac{\theta}{360} \times 49\pi \]

Rearrange to find \(\theta\):

\[ \theta = \frac{40 \times 360}{49\pi} \]

✓ (P2) For a complete method to find the angle \(\theta\).

Calculator Steps:

1. Press: 40 × 360 ÷ ( 49 × π ) =
2. Calculator shows: 93.5444…

\[ \theta = 93.5444…^{\circ} \]

Working:

The formula for Arc Length is:

\[ \text{Arc Length} = \frac{\theta}{360} \times \pi \times d \]

Since \(d = 2 \times r = 14\):

\[ \text{Arc Length} = \frac{93.5444…}{360} \times \pi \times 14 \]

✓ (P1) For a process to find the arc length.

\[ \text{Arc Length} = 11.4285… \text{ cm} \]

Now, add the two straight sides (\(7\text{ cm}\) each) to find the total perimeter:

\[ \text{Perimeter} = 11.4285… + 7 + 7 \] \[ \text{Perimeter} = 25.4285… \]

Round to 3 significant figures:

\[ 25.4 \text{ cm} \]

Working:

Factorise the quadratic \(x^2 + 3x – 10\):

We need two numbers that multiply to -10 and add to 3. Those numbers are 5 and -2.

\[ x^2 + 3x – 10 = (x+5)(x-2) \]

✓ (B1) For factorising the quadratic denominator.

Rewrite the bracket part and change division to multiplication:

\[ (x+5) \div \frac{(x+5)(x-2)}{x-1} \] \[ = \frac{(x+5)}{1} \times \frac{x-1}{(x+5)(x-2)} \]

✓ (M1) For a method to divide by the algebraic fraction.

Notice that \((x+5)\) appears on both the top and the bottom, so they cancel each other out:

\[ = \frac{x-1}{x-2} \]

Working:

The expression is now:

\[ 6 + \frac{x-1}{x-2} \]

Write 6 as a fraction with denominator \(x-2\):

\[ \frac{6(x-2)}{x-2} + \frac{x-1}{x-2} \]

✓ (M1) For finding 2 fractions with a common denominator.

Combine them over the single denominator:

\[ = \frac{6(x-2) + (x-1)}{x-2} \]

Expand the top bracket and collect like terms:

\[ = \frac{6x – 12 + x – 1}{x-2} \] \[ = \frac{7x – 13}{x-2} \]

Working:

1. Draw a tangent to the curve at \(t = 15\).

✓ (B1) For a tangent drawn correctly at \(t = 15\).

2. Pick two clear points on your tangent line. (Since drawings vary slightly, your points might be different). For example, a good tangent might pass through \((6, 10.5)\) and \((24, 25.5)\).

\[ \text{Gradient} = \frac{25.5 – 10.5}{24 – 6} = \frac{15}{18} \]

✓ (M1) For a full method to find the gradient using the tangent.

\[ \text{Gradient} = 0.833… \]

Working:

The gradient represents the acceleration of the car.

✓ (C1) For stating “acceleration” or “rate of change of speed”.

Working:

Read the speed (\(v\)) at each 5-second interval:

• At \(t = 0\), \(v = 0\)
• At \(t = 5\), \(v = 4\)
• At \(t = 10\), \(v = 12\)
• At \(t = 15\), \(v = 18\)
• At \(t = 20\), \(v = 20\)

Calculate area of each strip:

Strip 1 (Triangle): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4 = 10\)

✓ (P1) For splitting the area into strips and finding the area of at least one correctly.

Strip 2 (Trapezium): \(\frac{1}{2} \times (a+b) \times h = \frac{1}{2} \times (4 + 12) \times 5 = \frac{1}{2} \times 16 \times 5 = 40\)

Strip 3 (Trapezium): \(\frac{1}{2} \times (12 + 18) \times 5 = \frac{1}{2} \times 30 \times 5 = 75\)

Strip 4 (Trapezium): \(\frac{1}{2} \times (18 + 20) \times 5 = \frac{1}{2} \times 38 \times 5 = 95\)

Add them all together:

\[ \text{Total Distance} = 10 + 40 + 75 + 95 \]

✓ (P1) For a complete process to sum the 4 strips.

\[ = 220 \]

Working:

\[ f \times (m – 1) = 3m + 4 \]

Expand the bracket on the left side:

\[ fm – f = 3m + 4 \]

✓ (M1) For multiplying both sides by \((m-1)\).

Working:

Move \(3m\) to the left (by subtracting \(3m\)) and move \(-f\) to the right (by adding \(f\)):

\[ fm – 3m = f + 4 \]

✓ (M1) For a method to isolate terms in \(m\).

Now, factorise out the common \(m\) on the left side:

\[ m(f – 3) = f + 4 \]

Finally, divide both sides by the bracket \((f – 3)\) to get \(m\) entirely by itself:

\[ m = \frac{f+4}{f-3} \]

Working:

\[ 3y = 4x + 7 \]

Divide everything by 3 to get \(y\) on its own:

\[ y = \frac{4}{3}x + \frac{7}{3} \]

The gradient of line \(\mathbf{L}\) is \(\frac{4}{3}\).

✓ (M1) For identifying the gradient of line L as \(\frac{4}{3}\).

Working:

Gradient of L = \(\frac{4}{3}\)

Perpendicular gradient (\(m\)) = \(-\frac{3}{4}\)

✓ (M1) For beginning a method to find the perpendicular gradient.

Working:

Using \(y = mx + c\):

\[ -5 = \left(-\frac{3}{4}\right)(3) + c \] \[ -5 = -\frac{9}{4} + c \]

Add \(\frac{9}{4}\) to both sides (Note: \(-5 = -\frac{20}{4}\)):

\[ -\frac{20}{4} + \frac{9}{4} = c \] \[ c = -\frac{11}{4} \]

So the full equation is:

\[ y = -\frac{3}{4}x – \frac{11}{4} \]

(Note: Multiplying by 4 to get \(4y = -3x – 11\) or \(4y + 3x = -11\) is also perfectly acceptable).

Working:

From the size ratio (\(4 : 7\)), the total number of parts is \(4 + 7 = 11\). Therefore, the total number of cubes must be a multiple of 11.

From the colour ratio (\(3 : 5\)), the total number of parts is \(3 + 5 = 8\). Therefore, the total number of cubes must be a multiple of 8.

The lowest common multiple (LCM) of 11 and 8 is \(11 \times 8 = 88\).

✓ (C1) For stating the LCM of (4+7) and (3+5) is 88.

Working:

Using the total of 88 cubes:

• Number of small cubes = \(\frac{4}{11} \times 88 = 32\)
• Number of large cubes = \(\frac{7}{11} \times 88 = 56\)
• Number of red cubes = \(\frac{3}{8} \times 88 = 33\)
• Number of yellow cubes = \(\frac{5}{8} \times 88 = 55\)

✓ (P1) For using a scale factor appropriately (e.g., \(4 \times 8 = 32\)).

✓ (P1) For finding the number of large cubes and red/yellow cubes.

We are told all the small cubes are yellow. This means there are 32 small yellow cubes.

We know the total number of yellow cubes is 55. If 32 of them are small, the rest must be large.

\[ \text{Large Yellow Cubes} = \text{Total Yellow} – \text{Small Yellow} \] \[ \text{Large Yellow Cubes} = 55 – 32 = 23 \]

Working:

\[ \angle BAD = \frac{180^{\circ} – 40^{\circ}}{2} = \frac{140^{\circ}}{2} = 70^{\circ} \]

✓ (M1) For finding angle \(BAD = 70^{\circ}\).

Reason: Base angles of an isosceles triangle are equal, and angles in a triangle add up to 180°.

Working:

\[ \angle BCD = 180^{\circ} – \angle BAD \] \[ \angle BCD = 180^{\circ} – 70^{\circ} = 110^{\circ} \]

✓ (M1) For finding angle \(BCD = 110^{\circ}\).

Reason: Opposite angles of a cyclic quadrilateral add up to 180°.

Working:

\[ \angle CDB = \frac{180^{\circ} – 110^{\circ}}{2} = \frac{70^{\circ}}{2} = 35^{\circ} \]

Reason: Base angles of an isosceles triangle are equal, and angles in a triangle add up to 180°.

Working:

First, from Step 1, we know \(\angle BDA = 70^{\circ}\) (it’s the other base angle in the first triangle).

The total angle inside the quadrilateral at point \(D\) is:

\[ \angle CDA = \angle CDB + \angle BDA = 35^{\circ} + 70^{\circ} = 105^{\circ} \]

Because \(CDE\) is a straight line:

\[ \angle ADE = 180^{\circ} – \angle CDA \] \[ \angle ADE = 180^{\circ} – 105^{\circ} = 75^{\circ} \]

✓ (A1) For finding angle \(ADE = 75^{\circ}\).

Reason: Angles on a straight line add up to 180°.

✓ (C2) For listing “Opposite angles of a cyclic quadrilateral” AND one other valid reason correctly linked to the working.

Working:

\[ \tan(35^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BE}{15} \] \[ BE = 15 \times \tan(35^{\circ}) \]

✓ (P1) For using trigonometry to find BE.

Calculator Steps:

1. Press: 15 × tan 35 =
2. Calculator shows: 10.5031…

\[ BE = 10.503… \text{ cm} \]

Working:

The ratio \(DM : MA\) is \(2 : 3\). The total length of \(DA\) is 15 cm.

There are \(2 + 3 = 5\) parts in total. One part is \(15 \div 5 = 3\) cm.

\(MA\) is 3 parts, so \(MA = 3 \times 3 = 9\) cm.

Now use Pythagoras’ theorem in \(\triangle MAB\):

\[ MB^2 = MA^2 + AB^2 \] \[ MB^2 = 9^2 + 15^2 \] \[ MB^2 = 81 + 225 = 306 \] \[ MB = \sqrt{306} = 17.492… \text{ cm} \]

✓ (P1) For finding \(MB = \sqrt{306}\) or \(17.49…\).

Working:

\[ \tan(\angle EMB) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{10.503…}{17.492…} \]

✓ (P1) For using an appropriate trig ratio to set up an equation in angle EMB.

Calculator Steps:

1. Press: Shift tan ( 10.503 ÷ 17.492 ) =
2. Calculator shows: 30.984…

\[ \angle EMB = 30.984…^{\circ} \]

Round to 1 decimal place. The 8 rounds the 9 up to 10, carrying over.

\[ \angle EMB = 31.0^{\circ} \]

Working:

\[ \vec{FE} = \vec{FC} + \vec{CD} + \vec{DE} \]

Substitute the known vectors:

\[ \vec{FE} = (\mathbf{a} – \mathbf{b}) + \mathbf{a} + \mathbf{b} \]

✓ (M1) For a method to find \(\vec{FE}\).

Collect like terms: The \(-\mathbf{b}\) and \(+\mathbf{b}\) cancel out.

\[ = 2\mathbf{a} \]

Method Path 1: Down the line CE

First, find the total vector \(\vec{CE}\):

\[ \vec{CE} = \vec{CD} + \vec{DE} = \mathbf{a} + \mathbf{b} \]

Since \(C, X, E\) are on a straight line, \(\vec{CX}\) is a multiple of \(\vec{CE}\):

\[ \vec{CX} = k(\mathbf{a} + \mathbf{b}) = k\mathbf{a} + k\mathbf{b} \quad \text{— (Equation 1)} \]

Method Path 2: Finding \(\vec{CX}\) via \(F\) and \(M\)

We can also travel: \(\vec{CX} = \vec{CF} + \vec{FX}\).

We know \(\vec{CF} = -\vec{FC} = -(\mathbf{a} – \mathbf{b}) = \mathbf{b} – \mathbf{a}\).

We need \(\vec{FX}\). The ratio \(FX : XM = n : 1\) means \(FX\) is \(\frac{n}{n+1}\) of the total line \(\vec{FM}\).

Let’s find \(\vec{FM}\) first. \(M\) is the midpoint of \(DE\), so \(\vec{DM} = 0.5\mathbf{b}\).

\[ \vec{FM} = \vec{FD} + \vec{DM} = (\vec{FC} + \vec{CD}) + 0.5\mathbf{b} \] \[ \vec{FM} = (\mathbf{a} – \mathbf{b} + \mathbf{a}) + 0.5\mathbf{b} = 2\mathbf{a} – 0.5\mathbf{b} \]

✓ (P1) For a process to find \(\vec{FM}\) (or \(\vec{CE}\) etc).

So, \(\vec{FX} = \frac{n}{n+1}(2\mathbf{a} – 0.5\mathbf{b})\).

Now put it back into Path 2:

\[ \vec{CX} = (\mathbf{b} – \mathbf{a}) + \frac{n}{n+1}(2\mathbf{a} – 0.5\mathbf{b}) \]

✓ (P1) For a valid vector expression for \(\vec{CX}\).

Working:

Let’s group the \(\mathbf{a}\)’s and \(\mathbf{b}\)’s in our Path 2 equation:

\[ \vec{CX} = \mathbf{a}\left( \frac{2n}{n+1} – 1 \right) + \mathbf{b}\left( 1 – \frac{0.5n}{n+1} \right) \]

Comparing this to Equation 1 (\(\vec{CX} = k\mathbf{a} + k\mathbf{b}\)), both brackets must equal \(k\). Therefore, they must equal each other!

\[ \frac{2n}{n+1} – 1 = 1 – \frac{0.5n}{n+1} \]

✓ (P1) For a complete process to equate the coefficients.

Solve for \(n\). Add \(\frac{0.5n}{n+1}\) to both sides, and add 1 to both sides:

\[ \frac{2.5n}{n+1} = 2 \]

Multiply both sides by \((n+1)\):

\[ 2.5n = 2(n + 1) \] \[ 2.5n = 2n + 2 \] \[ 0.5n = 2 \] \[ n = 4 \]