Pearson Edexcel GCSE Mathematics (Foundation) – June 2020 (Paper 2)

✏ Working:

In \( 0.37 \), the 3 is in the tenths place and the 7 is in the hundredths place.

This means the decimal represents thirty-seven hundredths.

\[ 0.37 = \frac{37}{100} \]

✏ Working:

Look at the digit in the thousands place: \( 2\underline{9}381 \). The digit is 9 (representing 9000).

Now look at the “decider” digit in the hundreds place: \( 29\mathbf{3}81 \). The digit is 3.

Rule: If the decider digit is less than 5, we keep the target digit the same (round down). If it is 5 or more, we add 1 to the target digit (round up).

Since \( 3 < 5 \), we round down.

✏ Working:

Treat \( e \) as a single unit (like “apples”). Remember that a letter on its own like \( e \) has a coefficient of 1, so it means \( 1e \).

\[ 3e – 1e + 4e \] \[ (3 – 1 + 4)e \] \[ 3 – 1 = 2 \] \[ 2 + 4 = 6 \] \[ = 6e \]

✏ Working:

Method 1: Change denominator to 100.

\[ \frac{1 \times 25}{4 \times 25} = \frac{25}{100} = 25\% \]

Method 2: Multiply by 100%.

\[ \frac{1}{4} \times 100\% = 100 \div 4 = 25\% \]

✏ Working:

Let’s calculate the first few cube numbers:

• \( 1^3 = 1 \times 1 \times 1 = 1 \)
• \( 2^3 = 2 \times 2 \times 2 = 8 \)
• \( 3^3 = 3 \times 3 \times 3 = 27 \)
• \( 4^3 = 4 \times 4 \times 4 = 64 \)

Compare this list of results (\( 1, 8, 27, 64, \dots \)) to our given list: \( 3, 4, 9, 18, 27, 30, 36 \).

The only number that appears in both is 27.

✏ Working:

There are \( 60 \) minutes in \( 1 \) hour.

\[ 105 \text{ mins} = 60 \text{ mins} + 45 \text{ mins} \] \[ = 1 \text{ hour } 45 \text{ minutes} \]

✓ (M1) Correct time conversion

✏ Working:

Film end time: \( 14\,30 + 1 \text{ hr } 45 \text{ mins} \)

\[ 14\,30 + 1 \text{ hr} = 15\,30 \] \[ 15\,30 + 45 \text{ mins} = 16\,15 \]

Arrival at bus stop: \( 16\,15 + 20 \text{ mins} \)

\[ = 16\,35 \]

✓ (M1) Full method to find arrival time

✏ Working:

Liz arrives at \( 16\,35 \). The bus leaves at \( 16\,45 \).

\[ 16\,35 < 16\,45 \]

She arrives \( 10 \) minutes before the bus leaves.

✏ Working:

Check the vertical axis (y-axis):

• The scale goes from \( 0 \) straight to \( 71 \). This “jump” is not clearly marked or mathematically consistent with the rest of the scale (\( 1 \) unit per grid line).

Check the bars (x-axis):

• The bars for Farhad and George have the same width. However, Tom’s bar is twice as wide as the others. Bars in a bar chart should have equal widths.

✏ Working:

\[ x + 150^\circ = 180^\circ \] \[ x = 180^\circ – 150^\circ \] \[ x = 30^\circ \]

✓ (B1) Correct value

✏ Working:

The diagram shows a right angle (\( 90^\circ \)) and a reflex angle of \( 280^\circ \).

\[ 90^\circ + 280^\circ = 370^\circ \]

Since \( 370^\circ \neq 360^\circ \), the diagram is mathematically impossible.

✏ Working:

Find \( 40 \) on the bottom (kilometres) scale.

Move vertically up to the miles scale.

The point aligns exactly halfway between \( 20 \) and \( 30 \).

\[ = 25 \text{ miles} \]

✓ (B1) 25 (accept 24-26)

✏ Working:

1. Divide by \( 10 \): \( 40 \div 10 = 4 \)
2. Multiply by \( 6 \): \( 4 \times 6 = 24 \)

\[ \text{Result} = 24 \text{ miles} \]

✓ (M1) for \( 40 \div 10 \times 6 \)

✓ (A1) cao

✏ Working:

Answer (a) = \( 25 \)

Answer (b) = \( 24 \)

The mathematical rule gives an answer that is \( 1 \text{ mile} \) lower than the scale reading.

✓ (C1) Valid comparison statement

✏ Working:

\[ 3m = 36 \] \[ m = 36 \div 3 \] \[ m = 12 \]

✓ (B1) cao

✏ Working:

\[ 7 – x = 3 \]

Rearrange by adding \( x \) to both sides:

\[ 7 = 3 + x \]

Now subtract \( 3 \) from both sides:

\[ 7 – 3 = x \] \[ x = 4 \]

✓ (B1) cao

✏ Working:

Identify the dimensions: Length = \( 10 \text{ cm} \), Width = \( 15 \text{ cm} \), Height = \( 4 \text{ cm} \).

\[ \text{Volume} = \text{length} \times \text{width} \times \text{height} \] \[ \text{Volume} = 10 \times 15 \times 4 \]

✓ (M1) for complete multiplication method

✏ Working:

\[ 10 \times 15 = 150 \] \[ 150 \times 4 = 600 \]

✓ (A1) for 600

✓ (B1) for correct units: \( \text{cm}^3 \)

✏ Working:

List numbers between 20 and 30: 21, 22, 23, 24, 25, 26, 27, 28, 29.

• 21 is \( 3 \times 7 \)
• 22, 24, 26, 28 are even
• 25 is \( 5 \times 5 \)
• 27 is \( 3 \times 9 \)

The prime numbers are 23 and 29.

✓ (B1) for identifying 23 and 29

✏ Working:

• Letter L: L23, L29
• Letter U: U23, U29

✏ Working:

\[ 4275 \div 28 = 152.678… \]

This tells us the machine can fill 152 full bags.

✓ (P1) for division method

✏ Working:

Sweets in full bags: \( 152 \times 28 = 4256 \)

\[ 4275 – 4256 = 19 \]

✓ (P1) for finding sweets used or remaining fraction

✏ Working:

Total goals = \( 50 + 45 + 25 = 120 \)

Degrees per goal = \( 360^\circ \div 120 = 3^\circ \)

• City: \( 50 \times 3^\circ = 150^\circ \)
• Rovers: \( 45 \times 3^\circ = 135^\circ \)
• United: \( 25 \times 3^\circ = 75^\circ \)

Check total: \( 150 + 135 + 75 = 360^\circ \) ✓

✓ (M1) for finding method for at least one angle

✓ (A1) for calculating all 3 angles correctly

✏ Working:

\[ T = 3(5) + 4(-7) \] \[ T = 15 + (-28) \] \[ T = 15 – 28 \] \[ T = -13 \]

✓ (M1) for substitution

✓ (A1) cao

✏ Working:

\[ 38 = 3(6) + 4y \] \[ 38 = 18 + 4y \]

Subtract 18 from both sides:

\[ 20 = 4y \]

Divide by 4:

\[ y = 5 \]

✓ (M1) for \( 38 = 18 + 4y \)

✓ (A1) cao

✏ Working:

Total marks available = \( 60 + 90 = 150 \)

Pass mark = \( \frac{2}{3} \times 150 \)

\[ 150 \div 3 = 50 \] \[ 50 \times 2 = 100 \text{ marks} \]

✓ (P1) for finding the pass mark

✏ Working:

Marks gained = \( 70\% \text{ of } 60 \)

\[ \frac{70}{100} \times 60 = 42 \text{ marks} \]

✓ (P1) for 70% of 60

✏ Working:

\[ \text{Required marks} = 100 – 42 = 58 \]

✓ (P1) for complete set of processes

✏ Working:

Scale factor for juice = \( 8 \text{ litres} \div 2 \text{ litres} = 4 \)

Total oranges required = \( 30 \times 4 = 120 \text{ oranges} \)

Number of boxes = \( 120 \text{ oranges} \div 24 \text{ per box} \)

\[ 120 \div 24 = 5 \]

✓ (P1) for finding oranges required (120)

✓ (P1) for finding boxes (120 / 24)

✓ (A1) cao

✏ Working:

\[ \text{Apples} : \text{Bananas} = 1260 : 280 \]

Divide by 10:

\[ 126 : 28 \]

Divide by 2:

\[ 63 : 14 \]

Divide by 7:

\[ 9 : 2 \]

✓ (M1) for partially simplified ratio

✓ (A1) cao

✏ Working:

Look at corresponding points:

• The right angle in A is at \( (-5, 3) \)
• The corresponding right angle in B is at \( (3, -3) \)

The shape has been turned half-way around. This indicates a **rotation of \( 180^\circ \)**.

✏ Working:

Corresponding pair: \( (-5, 3) \) and \( (3, -3) \)

\[ \text{Midpoint } X = \frac{-5 + 3}{2} = \frac{-2}{2} = -1 \] \[ \text{Midpoint } Y = \frac{3 + (-3)}{2} = \frac{0}{2} = 0 \]

The centre is at \( (-1, 0) \).

✏ Working:

Let Rytis’s share be \( 1 \) unit.

• Linda gets \( 3 \times \) Rytis \( \rightarrow 3 \) units.
• Linda gets \( \frac{1}{2} \) Adam \( \rightarrow \) Adam gets \( 2 \times \) Linda \( \rightarrow 6 \) units.

✓ (P1) for establishing shares in correct proportions

✏ Working:

Total units = \( 1 \text{ (Rytis)} + 3 \text{ (Linda)} + 6 \text{ (Adam)} = 10 \text{ units} \)

\[ \text{Linda’s fraction} = \frac{3}{10} \]

✏ Working:

\[ \text{Cost per pencil} = £1.80 \div 12 \] \[ = £0.15 \text{ (or 15p)} \]

✓ (P1) for finding unit cost of a pencil

✏ Working:

1 part = \( £0.15 \div 3 = £0.05 \)

Cost per pen (7 parts) = \( 7 \times £0.05 = £0.35 \)

✓ (P1) for method to find cost of one pen

✓ (P1) for cost of one pen (£0.35)

✏ Working:

\[ \text{Total cost} = 5 \times £0.35 = £1.75 \]

✏ Working:

Starting with 84:

• \( 84 = 2 \times 42 \)
• \( 42 = 2 \times 21 \)
• \( 21 = 3 \times 7 \)

Prime factors are: \( 2, 2, 3, 7 \)

\[ 84 = 2^2 \times 3 \times 7 \]

✓ (M1) for correct factor tree or list

✓ (A1) cao

✏ Working:

Prime factors of \( 84 = 2 \times 2 \times 3 \times 7 \)

Prime factors of \( 60 = 2 \times 2 \times 3 \times 5 \)

To find the LCM, we take the highest power of each prime factor present in either list:

• Highest power of 2: \( 2^2 \) (appears in both)
• Highest power of 3: \( 3^1 \) (appears in both)
• Highest power of 5: \( 5^1 \) (appears in 60)
• Highest power of 7: \( 7^1 \) (appears in 84)

\[ \text{LCM} = 2^2 \times 3 \times 5 \times 7 \] \[ \text{LCM} = 4 \times 3 \times 5 \times 7 \] \[ \text{LCM} = 12 \times 35 = 420 \]

✓ (M1) for factorizing 60 or listing multiples

✓ (A1) cao

✏ Working:

• \( \mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)
• \( A \text{ (Even)} = \{2, 4, 6, 8, 10\} \)
• \( B \text{ (Factors of 10)} = \{1, 2, 5, 10\} \)
• \( A \cap B \text{ (Intersection)} = \{2, 10\} \)

✓ (M1) for correct numbers in at least one region

✓ (M1) for correct numbers in at least two regions

✓ (A1) for all regions correct

✏ Working:

Number in \( A \cap B = 2 \) (these are the numbers 2 and 10).

Total numbers in \( \mathcal{E} = 10 \).

\[ P(A \cap B) = \frac{2}{10} \]

✓ (M1) for \( a/10 \) or \( 2/b \)

✓ (A1) oe

✏ Working:

Total parts in ratio = \( 2 + 3 = 5 \)

Value of one part = \( 3000 \div 5 = 600 \text{ tins} \)

Tins in small boxes = \( 2 \times 600 = 1200 \)

Tins in large boxes = \( 3 \times 600 = 1800 \)

✓ (P1) for finding tins in each type of box

✏ Working:

Number of small boxes = \( 1200 \div 6 = 200 \text{ boxes} \)

Number of large boxes = \( 1800 \div 20 = 90 \text{ boxes} \)

✓ (P1) for division to find number of boxes

✓ (P1) for finding 200 and 90

✏ Working:

Total boxes = \( 200 + 90 = 290 \)

Percentage of large boxes = \( \frac{90}{290} \times 100 \)

\[ = 31.03…\% \]

✓ (P1) for full method to find percentage

✏ Working:

• \( x = -2: y = 5 – (-2)^3 = 5 – (-8) = 13 \)
• \( x = -1: y = 5 – (-1)^3 = 5 – (-1) = 6 \) (given)
• \( x = 0: y = 5 – (0)^3 = 5 \)
• \( x = 1: y = 5 – (1)^3 = 5 – 1 = 4 \)
• \( x = 2: y = 5 – (2)^3 = 5 – 8 = -3 \)

✓ (B2) for all 4 values correct

✏ Working:

Points: \( (-2, 13), (-1, 6), (0, 5), (1, 4), (2, -3) \)

✓ (M1) for plotting at least 4 points correctly

✓ (A1) for a smooth curve through all points

✏ Working:

Opposite and Hypotenuse \( \rightarrow \) use **Sine (SIN)**.

\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \] \[ \sin(34^\circ) = \frac{x}{178} \]

✓ (M1) for correct trigonometric equation

✏ Working:

\[ x = 178 \times \sin(34^\circ) \] \[ x = 178 \times 0.55919… \] \[ x = 99.536… \]

Rounding to 1 decimal place:

\[ x = 99.5 \]

✓ (A1) for answer in range 99.5 to 99.7

✏ Working:

\[ 2\mathbf{a} = 2 \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix} \] \[ 3\mathbf{b} = 3 \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 15 \\ -6 \end{pmatrix} \]

✓ (M1) for either correct scalar multiplication

✏ Working:

\[ \begin{pmatrix} 6 \\ 8 \end{pmatrix} – \begin{pmatrix} 15 \\ -6 \end{pmatrix} \]

Top: \( 6 – 15 = -9 \)

Bottom: \( 8 – (-6) = 8 + 6 = 14 \)

\[ = \begin{pmatrix} -9 \\ 14 \end{pmatrix} \]

✓ (A1) cao

✏ Working:

\[ AC^2 = AB^2 + CB^2 \] \[ 9^2 = 6^2 + CB^2 \] \[ 81 = 36 + CB^2 \] \[ CB^2 = 81 – 36 = 45 \] \[ CB = \sqrt{45} \]

✓ (P1) for starting Pythagoras process: \( 9^2 – 6^2 \)

✓ (P1) for finding \( r^2 = 45 \)

✏ Working:

\[ \text{Area} = \frac{\pi \times r^2}{4} \] \[ \text{Area} = \frac{\pi \times 45}{4} \] \[ \text{Area} = 35.3429… \]

✓ (P1) for stating or using \( \pi \times r^2 \div 4 \)

✏ Working:

\[ \text{Number of sides } (n) = \frac{360^\circ}{\text{exterior angle}} \] \[ n = 360 \div 15 \] \[ n = 24 \]

✓ (M1) for complete method \( 360 \div 15 \)

✓ (A1) cao

✏ Working:

Comparing \( y = 2x + 3 \) to \( y = mx + c \):

\[ m = 2 \] \[ c = 3 \]