GCSE Nov 2023 Edexcel Foundation Paper 2 (Calculator)

Let’s find the factor pairs of 18:

\( 1 \times 18 = 18 \)

\( 2 \times 9 = 18 \)

\( 3 \times 6 = 18 \)

The factors are: 1, 2, 3, 6, 9, 18.

The digit 9 is in the tenths column.

Therefore, \( 0.9 \) means “9 tenths”.

\( 1 \text{ metre} = 100 \text{ centimetres} \)

To go from metres to cm, we multiply by 100.

\( 7 \times 100 = 700 \)

Let’s list the first few square numbers:

• \( 1 \times 1 = 1 \)
• \( 2 \times 2 = 4 \)
• \( 3 \times 3 = 9 \)
• \( 4 \times 4 = 16 \) (Greater than 10)
• \( 5 \times 5 = 25 \)
• \( 6 \times 6 = 36 \)
• \( 7 \times 7 = 49 \) (Less than 50)
• \( 8 \times 8 = 64 \) (Too big)

\( 240 \div 2 = 120 \)

\[ \text{Pay per hour} = \frac{\text{Total Pay}}{\text{Total Hours}} \]

\[ \text{Pay per hour} = \frac{50}{4} \]

Half of 50 is 25.

Half of 25 is 12.5.

So, \( 50 \div 4 = 12.5 \)

✓ (M1) for \( 50 \div 4 \)

Since this is money, we write it as £12.50.

He needs 5 footballs at £9.50 each.

\( 5 \times 9.50 = 47.50 \)

Cost: £47.50

Number of pairs needed: \( 6 \div 2 = 3 \) pairs.

\( 3 \times 30 = 90 \)

Cost: £90.00

He needs 2 bats at £23 each.

\( 2 \times 23 = 46 \)

Cost: £46.00

He needs 4 balls. The price is “£5 for 4”.

Cost: £5.00

Total = \( 47.50 + 90.00 + 46.00 + 5.00 \)

✓ (M1) for summing individual costs

Total = £188.50

Budget = £200.00

\( 188.50 < 200 \)

White Shirt (W):

• WP (White, Plain)
• WS (White, Striped)
• WC (White, Checked)

Blue Shirt (B):

• BP (Blue, Plain)
• BS (Blue, Striped)
• BC (Blue, Checked)

Grey Shirt (G):

• GP (Grey, Plain)
• GS (Grey, Striped)
• GC (Grey, Checked)

Red : Blue

\( 24 : 40 \)

✓ (M1)

Divide both sides by 8:

\( 24 \div 8 = 3 \)

\( 40 \div 8 = 5 \)

Ratio: \( 3 : 5 \)

Substitute the values:

\( \text{Distance} = 37 \times 3 \)

\( 37 \times 3 = 111 \)

✓ (M1)

We know 3 kg of flour = £4.05

Cost of 1 kg = \( 4.05 \div 3 = 1.35 \)

So, flour costs £1.35 per kg.

✓ (P1)

We know: \( 5\text{ kg flour} + 6\text{ kg sugar} = 11.85 \)

Cost of 5 kg flour = \( 5 \times 1.35 = 6.75 \)

Substitute this into the equation:

\( 6.75 + 6\text{ kg sugar} = 11.85 \)

Cost of 6 kg sugar = \( 11.85 – 6.75 = 5.10 \)

Cost of 1 kg sugar = \( 5.10 \div 6 = 0.85 \)

So, sugar costs £0.85 per kg.

✓ (P1)

We need: 2 kg flour and 5 kg sugar.

Flour: \( 2 \times 1.35 = 2.70 \)

Sugar: \( 5 \times 0.85 = 4.25 \)

✓ (P1)

Total = \( 2.70 + 4.25 = 6.95 \)

Martha gets 72% of £120.

\( 0.72 \times 120 = 86.40 \)

So Martha receives £86.40.

✓ (P1)

Using Method B:

Percentage left = \( 100\% – 30\% = 70\% \)

Calculate 70% of £86.40:

\( 0.70 \times 86.40 = 60.48 \)

✓ (P1)

	Romance	Adventure	Horror	Thriller	Total
Children	19	[A]	[B]	[C]	85
Adults	[D]	34	[E]	7	[F]
Total	33	76	?	?	160

1. Adults Total [F]: \( 160 – 85 = 75 \).

2. Adults Romance [D]: \( 33 – 19 = 14 \).

3. Children Adventure [A]: \( 76 – 34 = 42 \).

✓ (P1) for finding totals or missing cells

Now consider the Adults row. We know Total (75), Romance (14), Adventure (34), Thriller (7). We can find Horror [E].

\( \text{Adult Horror} = 75 – (14 + 34 + 7) \)

\( \text{Adult Horror} = 75 – 55 = 20 \).

✓ (P1)

“Number of children who chose thriller [C] was the same as adults who chose horror [E]”.

Since Adults Horror = 20, then Children Thriller = 20.

We need the Children Horror [B] to get the total.

Children Total = 85.

Children Romance (19) + Adventure (42) + Thriller (20) + Horror (B) = 85.

\( 19 + 42 + 20 = 81 \)

\( \text{Children Horror} = 85 – 81 = 4 \)

Total Horror = Adult Horror (20) + Children Horror (4).

\( 20 + 4 = 24 \)

✓ (P1)

First, write the numbers in order from smallest to largest:

160, 162, 168

172, 172, 173, 177, 178

180, 180, 183, 184, 186, 188

191, 197

Top: \( \sqrt{9.8 + 6.8} = \sqrt{16.6} = 4.0743… \)

Bottom: \( 4.2 \times 2.1 = 8.82 \)

Calculation: \( 4.0743… \div 8.82 = 0.4619398… \)

Wait, let me double check the numbers.

Correction on reading: The calculation is \( \frac{\sqrt{16.6}}{8.82} \)

\( 4.074309757… \div 8.82 = 0.461939881… \)

Wait, looking at the mark scheme for Q18(a) Answer is 1.882…

Let’s re-read the screenshot. Screenshot 12.

Ah, the image text is messy. It says \( \frac{9.8 + 6.8}{4.2 \times 2.1} \)? Or is there a square root?

OCR says: “18 (a) Work out 9 8 6 8 / 4 2 2 1”.

Mark Scheme (Screenshot 10) for Q18 says: “1.882(0861678…)”

Let’s reverse engineer 1.882.

\( \frac{16.6}{8.82} = 1.882086… \)

Conclusion: There is NO square root. It is just \( \frac{9.8 + 6.8}{4.2 \times 2.1} \).

Numerator: \( 9.8 + 6.8 = 16.6 \)

Denominator: \( 4.2 \times 2.1 = 8.82 \)

Division: \( 16.6 \div 8.82 = 1.8820861678… \)

\( 1.88\underline{2}0… \)

The 3rd digit is 2, so we round down (keep the 8).

Angle \( BCD = 60^\circ \)

Angle \( CBD = 60^\circ \)

Angle \( CDB = 60^\circ \)

✓ (M1)

Using \( BD \) as the transversal:

Angle \( CBD \) and Angle \( BDE \) are Alternate Angles (Z-angles).

Therefore, Angle \( BDE = 60^\circ \).

(Alternative: Using Co-interior angles with transversal CD extended, leading to the same result for the polygon interior).

✓ (M1)

\( x + 90 + 60 + 132 = 360 \)

\( x + 282 = 360 \)

\( x = 360 – 282 \)

\( x = 78^\circ \)

✓ (C2) for answer with reasons

Substitute the \( x \) values into \( y = x^2 – x – 2 \):

When \( x = -1 \):

\( y = (-1)^2 – (-1) – 2 \)

\( y = 1 + 1 – 2 = 0 \)

When \( x = 0 \):

\( y = 0^2 – 0 – 2 = -2 \)

When \( x = 2 \):

\( y = 2^2 – 2 – 2 \)

\( y = 4 – 2 – 2 = 0 \)

When \( x = 3 \):

\( y = 3^2 – 3 – 2 \)

\( y = 9 – 3 – 2 = 4 \)

Missing values: 0, -2, 0, 4

✓ (B2)

Compare corresponding points:

Point on A: \( (1, 2) \) maps to Point on B: \( (-2, -1) \)

Point on A: \( (1, 5) \) maps to Point on B: \( (-5, -1) \)

The coordinates have swapped positions and both signs have changed (\( x \rightarrow -y \) and \( y \rightarrow -x \)).

This is the characteristic of a reflection in the line \( y = -x \).

Multiply out the brackets:

\( 3(2y – 5) = 6y – 15 \)

\( 7(y + 2) = 7y + 14 \)

Combine them:

\( 6y – 15 + 7y + 14 \)

Collect like terms:

\( (6y + 7y) + (-15 + 14) \)

\( 13y – 1 \)

✓ (A1)

Expression: \( 6x^2 + 15x \)

Look for common factors in numbers and letters.

Numbers: 6 and 15 are both divisible by 3.

Letters: \( x^2 \) and \( x \) are both divisible by \( x \).

Common factor: \( 3x \).

\( 3x(2x + 5) \)

✓ (B2)

Equation: \( f = 3g + 11 \)

Subtract 11 from both sides:

\( f – 11 = 3g \)

Divide by 3:

\( g = \frac{f – 11}{3} \)

✓ (A1)

\( \text{Costs} = 100 + 120 + 80 = 300 \)

Total Cost: £300

✓ (P1)

\( \text{Income} = 54 \text{ tickets} \times 7.50 \)

\( 54 \times 7.50 = 405 \)

Total Income: £405

✓ (P1)

\( \text{Profit} = \text{Income} – \text{Costs} \)

\( 405 – 300 = 105 \)

Profit: £105

✓ (P1)

\( \frac{105}{300} \times 100 \)

\( 0.35 \times 100 = 35\% \)

Use the formula: \( \text{Amount} = \text{Principal} \times \text{Multiplier}^{\text{years}} \)

\( \text{Amount} = 4500 \times 1.034^2 \)

✓ (M1)

Calculator Steps:

4500 × 1.034 ² = 4811.202

\( 5x + x – 14 = 52 \)

\( 6x – 14 = 52 \)

✓ (M1)

\( 6x = 52 + 14 \)

\( 6x = 66 \)

✓ (M1)

Divide by 6:

\( x = \frac{66}{6} \)

\( x = 11 \)

Value of 1 part = \( 120 \div 4 = 30 \)

Initial Amounts:

Chris: \( 3 \times 30 = 90 \)

Debbie: \( 4 \times 30 = 120 \)

Errol: \( 2 \times 30 = 60 \)

Total Money: \( 90 + 120 + 60 = 270 \)

✓ (P1)

Value of 1 new part = \( 270 \div 10 = 27 \)

✓ (P1)

New Amounts:

Chris: \( 2 \times 27 = 54 \)

Debbie: \( 5 \times 27 = 135 \)

Errol: \( 3 \times 27 = 81 \)

✓ (P1)

Errol had £60.

Errol now has £81.

Difference: \( 81 – 60 = 21 \)

Chris gave Errol £21.

\( 147^\circ + 180^\circ = 327^\circ \)

✓ (M1)

\( V = \pi \times 15^2 \times 43 \)

\( V = \pi \times 225 \times 43 \)

\( V = 30394.908… \text{ cm}^3 \)

✓ (P1)

\( 30394.908… \div 1000 = 30.3949… \text{ litres} \)

✓ (P1)

\( \text{Time} = \frac{30.3949…}{0.47} \)

\( \text{Time} = 64.669… \text{ minutes} \)

✓ (P1)

\( \tan(49) = \frac{\text{Opposite}}{\text{Adjacent}} \)

\( \tan(49) = \frac{AB}{28} \)

✓ (M1)

\( AB = 28 \times \tan(49) \)

\( AB = 32.2098… \)

(1) \( 3x + y = -4.5 \)

(2) \( 4x + 3y = -3.5 \)

Multiply (1) by 3:

(3) \( 9x + 3y = -13.5 \)

✓ (M1)

Subtract (2) from (3):

\( (9x + 3y) – (4x + 3y) = -13.5 – (-3.5) \)

\( 5x = -13.5 + 3.5 \)

\( 5x = -10 \)

\( x = -2 \)

Substitute \( x = -2 \) into equation (1):

\( 3(-2) + y = -4.5 \)

\( -6 + y = -4.5 \)

\( y = -4.5 + 6 \)

\( y = 1.5 \)

✓ (M1)