If any of my solutions look wrong, please refer to the mark scheme. You can exit full-screen mode for the question paper and mark scheme by clicking the icon in the bottom-right corner or by pressing Esc on your keyboard.

GCSE Mathematics 2024 – Edexcel Foundation Paper 1 (Non-Calculator)

๐Ÿ’ก How to use this interactive exam:

  • Try it first: Attempt the question on paper before checking the solution.
  • Show Solution: Click the green button to reveal the step-by-step worked solution.
  • Understand “Why”: Read the cream-colored boxes to understand the mathematical reasoning.
  • Check “How”: Follow the calculation steps in the white boxes.
  • Diagrams: All visual elements are mathematically accurate SVGs.

Question 1 (1 mark)

Write the number \( 18475 \) correct to the nearest thousand.

Worked Solution

Step 1: Identifying the Place Value

Why we do this: We need to find which digit represents the “thousands” and look at the digit to its right to decide whether to round up or down.

In \( 18475 \):

  • \( 1 \) is the ten-thousands digit.
  • \( 8 \) is the thousands digit.
  • \( 4 \) is the hundreds digit (the decider).
Step 2: The Decision Rule

Rule: If the digit to the right (hundreds) is 5 or more, we round up. If it is 4 or less, we round down (keep the thousands digit the same).

The hundreds digit is \( 4 \). Since \( 4 < 5 \), we round down.

Keep the \( 8 \) as it is. All digits to the right become zero.

\[ 18475 \rightarrow 18000 \]

Final Answer:

\( 18000 \)

โœ“ (B1)

โ†‘ Back to Top

Question 2 (1 mark)

Write \( 0.3 \) as a percentage.

Worked Solution

Step 1: Converting Decimal to Percentage

How: To convert any decimal to a percentage, we multiply by \( 100 \).

\[ 0.3 \times 100 = 30 \]

Alternatively, think of \( 0.3 \) as \( 0.30 \), which is \( \frac{30}{100} \).

Final Answer:

\( 30\% \)

โœ“ (B1)

โ†‘ Back to Top

Question 3 (1 mark)

Write down the mathematical name for the type of angle marked \( y \).

y

Worked Solution

Step 1: Analyzing the Angle Size

Understanding Angle Types:

  • Acute: Between \( 0^\circ \) and \( 90^\circ \)
  • Right: Exactly \( 90^\circ \)
  • Obtuse: Between \( 90^\circ \) and \( 180^\circ \)
  • Reflex: Between \( 180^\circ \) and \( 360^\circ \)

Observation: The angle marked \( y \) goes “the long way round” the vertex, clearly indicating it is greater than a straight line (\( 180^\circ \)).

Final Answer:

Reflex angle

โœ“ (B1)

โ†‘ Back to Top

Question 4 (1 mark)

Write these numbers in order of size. Start with the smallest number.

\( 0.21 \quad 0.2 \quad 0.03 \quad 0.1 \quad 0.16 \)

Worked Solution

Step 1: Standardizing the Decimals

Strategy: To compare decimals easily, it helps to give them all the same number of decimal places. The longest decimal here has 2 places (e.g., \( 0.21 \)). We will add placeholder zeros to the others.

  • \( 0.21 \rightarrow 0.21 \)
  • \( 0.2 \rightarrow 0.20 \)
  • \( 0.03 \rightarrow 0.03 \)
  • \( 0.1 \rightarrow 0.10 \)
  • \( 0.16 \rightarrow 0.16 \)
Step 2: Comparing the Values

Now we can treat them like whole numbers (3, 10, 16, 20, 21):

Order from smallest to largest:

  1. \( 0.03 \) (Smallest)
  2. \( 0.10 \) (Original: \( 0.1 \))
  3. \( 0.16 \)
  4. \( 0.20 \) (Original: \( 0.2 \))
  5. \( 0.21 \) (Largest)

Final Answer:

\( 0.03, \ 0.1, \ 0.16, \ 0.2, \ 0.21 \)

โœ“ (B1)

โ†‘ Back to Top

Question 5 (1 mark)

Find the square root of \( 64 \).

Worked Solution

Step 1: Understanding Square Roots

Definition: The square root of a number is a value that, when multiplied by itself, gives that number. We are looking for a number \( x \) such that \( x \times x = 64 \).

Let’s check our times tables:

  • \( 5 \times 5 = 25 \)
  • \( 6 \times 6 = 36 \)
  • \( 7 \times 7 = 49 \)
  • \( 8 \times 8 = 64 \)

Final Answer:

\( 8 \)

โœ“ (B1)

(Note: While \( -8 \) is also a square root mathematically, in this GCSE context usually the positive root is required unless specified otherwise, but \( \pm 8 \) is also often accepted).

โ†‘ Back to Top

Question 6 (4 marks)

Ryan buys:

  • 4 cakes at \( ยฃ1.30 \) each
  • 2 identical tins of soup.

Ryan pays with a \( ยฃ10 \) note.
He gets \( ยฃ1.80 \) change.

How much does Ryan pay for each tin of soup?

Worked Solution

Step 1: Calculate the total amount Ryan actually spent

Why: We know he paid with \( ยฃ10 \) and got \( ยฃ1.80 \) change. The difference is the actual cost of all items.

  10.00
–  1.80
———
   8.20

Total spent = \( ยฃ8.20 \)

โœ“ (P1)

Step 2: Calculate the cost of the 4 cakes

Method: Multiply the cost of one cake (\( ยฃ1.30 \)) by 4.

\[ 4 \times 1.30 = 5.20 \]

Cost of cakes = \( ยฃ5.20 \)

โœ“ (P1)

Step 3: Find the cost of the 2 tins of soup

Method: Subtract the cost of the cakes from the total amount spent.

  8.20
–  5.20
———
   3.00

Cost of 2 tins = \( ยฃ3.00 \)

โœ“ (P1)

Step 4: Find the cost of one tin

Method: Divide the cost of 2 tins by 2.

\[ ยฃ3.00 \div 2 = ยฃ1.50 \]

Final Answer:

ยฃ1.50

โœ“ (A1)

โ†‘ Back to Top

Question 7 (4 marks)

The table shows the number of hours that Lena and Pavel worked on each of four days last week.

Wednesday Thursday Friday Saturday
Lena 6 9 8 6
Pavel 7 6 5 6

On the grid, draw a suitable diagram or chart for this information.

Day Hours

Worked Solution

Step 1: Choosing the Chart Type

Why: Since we are comparing two people (Lena and Pavel) over several days, a dual bar chart is the most suitable diagram. This allows side-by-side comparison for each day.

Step 2: Drawing the Axes

Vertical Axis (y-axis): Represents “Hours”. The maximum value is 9, so a scale from 0 to 10 is appropriate.

Horizontal Axis (x-axis): Represents the Days (Wed, Thu, Fri, Sat).

Key: We must distinguish between Lena and Pavel (e.g., different colors or shading).

Step 3: The Completed Diagram
0 4 8 12 Hours Worked Wed Thu Fri Sat Lena Pavel

Checklist for full marks:

  • Linear scale on y-axis (numbered correctly).
  • Days labeled on x-axis.
  • Key provided (or bars labeled) to distinguish Lena and Pavel.
  • Correct height for all bars.

โœ“ (C1)

โ†‘ Back to Top

Question 8 (3 marks)

\( OA \), \( OB \) and \( OC \) are three straight lines.

A B C O 220ยฐ 90ยฐ x

(i) Work out the size of the angle marked \( x \).

(ii) Give a reason for your answer.

Worked Solution

Step 1: Identifying the Geometric Rule

Concept: The angles meet at a single point \( O \). The sum of all angles around a point is a full circle.

Rule: Angles around a point add up to \( 360^\circ \).

Step 2: Calculation

We have two known angles: \( 220^\circ \) and \( 90^\circ \) (indicated by the square symbol).

We need to subtract these from \( 360^\circ \) to find \( x \).

\[ 220 + 90 = 310 \] \[ 360 – 310 = 50 \]

(i) Angle \( x \):

\( 50^\circ \)

โœ“ (M1 A1)

(ii) Reason:

Angles at a point add up to \( 360^\circ \).

โœ“ (C1)

โ†‘ Back to Top

Question 9 (5 marks)

Here is a number machine.

input ร— 2 โ€“ 10 output

(a) Work out the output when the input is \( 13 \).

(b) Work out the input when the output is \( 28 \).

(c) Show that there is a number for which the output is the same as the input.

Worked Solution

Part (a): Forward Calculation

Method: Follow the machine steps from left to right.

Input \( \rightarrow \times 2 \rightarrow – 10 \rightarrow \) Output

\[ 13 \times 2 = 26 \] \[ 26 – 10 = 16 \]

Answer: \( 16 \)

โœ“ (B1)

Part (b): Reverse Calculation

Method: To find the input, we work backwards from the output and do the inverse operations.

  • Inverse of \( – 10 \) is \( + 10 \)
  • Inverse of \( \times 2 \) is \( \div 2 \)

Start with \( 28 \):

\[ 28 + 10 = 38 \] \[ 38 \div 2 = 19 \]

Answer: \( 19 \)

โœ“ (M1 A1)

Part (c): Input = Output

Method: Let the input be \( x \). The output is also \( x \).

We can write this as an equation based on the machine:

Input \( \times 2 – 10 = \) Output

\[ 2x – 10 = x \]

Subtract \( x \) from both sides:

\[ x – 10 = 0 \]

Add \( 10 \) to both sides:

\[ x = 10 \]

Verification: If input is \( 10 \):

\[ 10 \times 2 = 20 \] \[ 20 – 10 = 10 \]

Input \( (10) \) = Output \( (10) \). Correct.

โœ“ (M1 C1)

โ†‘ Back to Top

Question 10 (2 marks)

There are \( 24 \) cows and \( 36 \) sheep on a farm.

Write as a ratio the number of cows to the number of sheep.
Give your ratio in its simplest form.

Worked Solution

Step 1: Write the initial ratio

Order Matters: The question asks for “cows to sheep”.

Cows : Sheep

\[ 24 : 36 \]
Step 2: Simplify the ratio

How: Find the highest common factor (HCF) that divides both numbers.

Both divide by 12.

\[ 24 \div 12 = 2 \] \[ 36 \div 12 = 3 \]

Ratio: \( 2 : 3 \)

(Alternative: Divide by 2 first to get \( 12:18 \), then by 6 to get \( 2:3 \))

Final Answer:

\( 2 : 3 \)

โœ“ (M1 A1)

โ†‘ Back to Top

Question 11 (3 marks)

(a) Work out \( -12 \div -4 \)

(b) Find the value of \( 2^5 \)

(c) Write one pair of brackets in this calculation so that the answer is correct.

\( 30 \div 3 + 2 – 4 = 2 \)

Worked Solution

Part (a): Division with Negatives

Rule: A negative number divided by a negative number gives a positive result.

\[ 12 \div 4 = 3 \] \[ -12 \div -4 = 3 \]

Answer: \( 3 \)

โœ“ (B1)

Part (b): Powers

Meaning: \( 2^5 \) means multiplying 2 by itself 5 times.

\[ 2 \times 2 \times 2 \times 2 \times 2 \]
  • \( 2 \times 2 = 4 \)
  • \( 4 \times 2 = 8 \)
  • \( 8 \times 2 = 16 \)
  • \( 16 \times 2 = 32 \)

Answer: \( 32 \)

โœ“ (B1)

Part (c): Order of Operations

Trial: We need the result to be 2. Let’s try placing brackets around addition first, as usually division happens before addition (BIDMAS).

If we bracket \( (3 + 2) \):

\[ 30 \div (3 + 2) – 4 \] \[ = 30 \div 5 – 4 \] \[ = 6 – 4 \] \[ = 2 \]

This works.

Answer: \( 30 \div (3 + 2) – 4 = 2 \)

โœ“ (B1)

โ†‘ Back to Top

Question 12 (4 marks)

The diagram shows a triangle and a rectangle.

30 cm 36 cm 14 cm 4 cm length

The perimeter of the rectangle is a quarter of the perimeter of the triangle.

Work out the length of the rectangle.

Worked Solution

Step 1: Calculate the Perimeter of the Triangle

Why: The perimeter is the total distance around the outside of the shape. We add all three side lengths together.

\[ 36 + 14 + 30 = 80 \text{ cm} \]

โœ“ (P1)

Step 2: Calculate the Perimeter of the Rectangle

Why: The question states the rectangle’s perimeter is a “quarter” of the triangle’s perimeter.

\[ 80 \div 4 = 20 \text{ cm} \]

โœ“ (P1)

Step 3: Work out the length

Method: The perimeter of a rectangle is \( 2 \times (\text{length} + \text{width}) \) or just adding all 4 sides. We know the width is \( 4 \text{ cm} \).

Two widths = \( 4 + 4 = 8 \text{ cm} \).

The remaining perimeter must be the two lengths.

Remaining perimeter for lengths:

\[ 20 – 8 = 12 \text{ cm} \]

One length (divide by 2):

\[ 12 \div 2 = 6 \text{ cm} \]

โœ“ (P1)

Final Answer:

\( 6 \text{ cm} \)

โœ“ (A1)

โ†‘ Back to Top

Question 13 (3 marks)

There are only ยฃ10 notes and ยฃ20 notes in a wallet.

Ali takes at random a note from the wallet.

(a) Write down the probability that Ali takes a note with a value of more than ยฃ5.


There are only 1p coins and 2p coins in a bag.

The total value of the coins in the bag is 40p.

The total value of the 1p coins is the same as the total value of the 2p coins.

Simon takes at random a coin from the bag.

(b) Find the probability that Simon takes a 1p coin.

Worked Solution

Part (a): Certainty

Reasoning: The wallet contains ยฃ10 and ยฃ20 notes. Both ยฃ10 and ยฃ20 are greater than ยฃ5. Therefore, it is certain he picks a note worth more than ยฃ5.

Answer: \( 1 \) (or \( 100\% \))

โœ“ (B1)

Part (b): Finding Coin Counts

Step 1: Split the value. The total value is 40p. The value is split equally between 1p coins and 2p coins.

Value of 1p coins = \( 40 \div 2 = 20\text{p} \)

Value of 2p coins = \( 40 \div 2 = 20\text{p} \)

Step 2: Calculate number of coins.

  • Number of 1p coins: \( 20\text{p} \div 1\text{p} = 20 \) coins.
  • Number of 2p coins: \( 20\text{p} \div 2\text{p} = 10 \) coins.

Step 3: Total coins.

\[ 20 + 10 = 30 \text{ coins} \]
Part (b): Probability

Formula: \( \text{Probability} = \frac{\text{Number of desired outcomes}}{\text{Total number of outcomes}} \)

\[ P(1\text{p}) = \frac{20}{30} \]

Simplifying the fraction:

\[ \frac{2}{3} \]

Final Answer:

\( \frac{2}{3} \)

โœ“ (M1 A1)

โ†‘ Back to Top

Question 14 (3 marks)

Work out \( 273 \times 54 \)

Worked Solution

Step 1: Set up Long Multiplication

Method: We will multiply \( 273 \) by \( 4 \), then by \( 50 \), and add the results.

    273
x    54
——-
   1092 (273 ร— 4)
  13650 (273 ร— 50)
——-
  14742
Step 2: Check the Working

Row 1 (\( \times 4 \)):

  • \( 4 \times 3 = 12 \) (write 2, carry 1)
  • \( 4 \times 7 = 28 \), plus 1 = 29 (write 9, carry 2)
  • \( 4 \times 2 = 8 \), plus 2 = 10
  • Result: \( 1092 \)

Row 2 (\( \times 50 \)): (Put a zero first)

  • \( 5 \times 3 = 15 \) (write 5, carry 1)
  • \( 5 \times 7 = 35 \), plus 1 = 36 (write 6, carry 3)
  • \( 5 \times 2 = 10 \), plus 3 = 13
  • Result: \( 13650 \)

Total: \( 1092 + 13650 = 14742 \)

Final Answer:

\( 14742 \)

โœ“ (M1 M1 A1)

โ†‘ Back to Top

Question 15 (4 marks)

Tessa recorded the times that 15 adults took to complete a run.

She showed her results in a stem and leaf diagram.

4 5 6 7 8 59 378 124577 267 1 Key: 4 | 5 = 45 min

(a) Find the median.

(b) Find the range.

Tessa also recorded the times that 15 children took to complete the run.

For the children, the median was 75 minutes.

(c) Compare the times that the adults took with the times that the children took.

Worked Solution

Part (a): Finding the Median

Definition: The median is the middle value. Since there are \( 15 \) adults, we use the formula \( \frac{n+1}{2} \) to find the position.

Position = \( \frac{15+1}{2} = 8 \text{th} \) value.

Counting the leaves from the top:

  • Row 4: 45, 49 (2 values)
  • Row 5: 53, 57, 58 (3 values) &rightarrow; Total 5 values seen.
  • Row 6: 61 (6th), 62 (7th), 64 (8th)

Median: \( 64 \) minutes

โœ“ (B1)

Part (b): Finding the Range

Formula: Range = Highest Value – Lowest Value.

  • Highest: \( 81 \)
  • Lowest: \( 45 \)
\[ 81 – 45 = 36 \]

Range: \( 36 \) minutes

โœ“ (M1 A1)

Part (c): Comparison

Requirement: Compare the medians. We need to say who was faster (took less time).

  • Adults Median: \( 64 \) mins
  • Children Median: \( 75 \) mins

Since \( 64 < 75 \), the adults took less time on average.

Statement: “The adults were faster because they had a lower median time.”

โœ“ (C1)

โ†‘ Back to Top

Question 16 (3 marks)

Batteries are sold in packs of 4, in packs of 8 and in packs of 12.

ยฃ1.80 ยฃ3.20 12 Batteries ยฃ6.00

A pack of 4 batteries costs ยฃ1.80
A pack of 8 batteries costs ยฃ3.20
A pack of 12 batteries costs ยฃ6.00

Which pack gives the best value for money?
You must show how you get your answer.

Worked Solution

Step 1: Choose a Method

Why: To compare values, we need to compare like with like. We can either:

  • Find the cost of 1 battery (Unit Cost) for each pack.
  • Or find the cost for a common number of batteries (e.g., 24).

Let’s find the cost per battery (Unit Cost) as it’s a standard method.

Step 2: Calculate Unit Costs

Pack of 4 (ยฃ1.80):

\[ 1.80 \div 4 \]

Half of \( 1.80 \) is \( 0.90 \). Half of \( 0.90 \) is \( 0.45 \).

Cost per battery = 45p


Pack of 8 (ยฃ3.20):

\[ 3.20 \div 8 \]

\( 32 \div 8 = 4 \), so \( 3.20 \div 8 = 0.40 \).

Cost per battery = 40p


Pack of 12 (ยฃ6.00):

\[ 6.00 \div 12 \]

\( 60 \div 12 = 5 \), so \( 6.00 \div 12 = 0.50 \).

Cost per battery = 50p

Step 3: Comparison and Conclusion

Compare: 45p vs 40p vs 50p.

The lowest cost per battery is 40p.

Final Answer:

Pack of 8

โœ“ (P1 P1 A1)

โ†‘ Back to Top

Question 17 (3 marks)

Solve \( 2(4x – 5) = 18 \)

Worked Solution

Step 1: Expand the Brackets

Method: Multiply the term outside the bracket (\( 2 \)) by each term inside.

\[ 2 \times 4x = 8x \] \[ 2 \times -5 = -10 \]

Equation becomes:

\[ 8x – 10 = 18 \]
Step 2: Isolate the x term

Method: Add 10 to both sides to remove the -10.

\[ 8x = 18 + 10 \] \[ 8x = 28 \]
Step 3: Solve for x

Method: Divide by 8. Note that 28 doesn’t divide by 8 perfectly into a whole number, so we expect a decimal or fraction.

\[ x = \frac{28}{8} \]

Simplify the fraction (divide top and bottom by 4):

\[ x = \frac{7}{2} \]

Convert to decimal:

\[ x = 3.5 \]

Final Answer:

\( x = 3.5 \) (or \( 3.5 \))

โœ“ (M1 M1 A1)

โ†‘ Back to Top

Question 18 (1 mark)

Write down the value of \( 10^0 \)

Worked Solution

Step 1: Zero Index Law

Rule: Any non-zero number raised to the power of 0 is always 1.

\( a^0 = 1 \)

Final Answer:

\( 1 \)

โœ“ (B1)

โ†‘ Back to Top

Question 19 (2 marks)

Describe fully the single transformation that maps triangle A onto triangle B.

x y A B

Worked Solution

Step 1: Identify the Transformation Type

Observation:

  • Is it rotated? No, the orientation is identical.
  • Is it a reflection? No, it hasn’t flipped.
  • Is it resized? No, same size.

Conclusion: It has simply moved position. This is a Translation.

Step 2: Calculate the Vector

Method: Pick one corner on A and find the path to the matching corner on B.

Let’s use the top-left corner (the right angle).

  • Triangle A corner: \( (-3, 3) \)
  • Triangle B corner: \( (2, -1) \)

Horizontal change (x): From -3 to 2 is a move of +5 (Right).

Vertical change (y): From 3 to -1 is a move of -4 (Down).

Vector format: \( \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix} \)

Final Answer:

Translation by vector \( \begin{pmatrix} 5 \\ -4 \end{pmatrix} \)

โœ“ (B1 B1)

โ†‘ Back to Top

Question 20 (2 marks)

Here are the first four terms of an arithmetic sequence.

\( 1 \quad 5 \quad 9 \quad 13 \)

Find an expression, in terms of \( n \), for the \( n \)th term of this sequence.

Worked Solution

Step 1: Find the common difference

Why: In an arithmetic sequence, the difference between consecutive terms is constant.

\[ 5 – 1 = 4 \] \[ 9 – 5 = 4 \]

The difference is \( +4 \). This means the rule relates to the 4 times table (\( 4n \)).

Step 2: Adjust to find the rule

Method: Compare our sequence to the 4 times table.

\( n = 1 \): \( 4 \times 1 = 4 \). We want \( 1 \). So subtract \( 3 \).

\( n = 2 \): \( 4 \times 2 = 8 \). We want \( 5 \). So subtract \( 3 \).

Rule: \( 4n – 3 \)

Step 3: Verification

Check 3rd term (\( n=3 \)):

\[ 4(3) – 3 = 12 – 3 = 9 \]

Correct.

Final Answer:

\( 4n – 3 \)

โœ“ (B2)

โ†‘ Back to Top

Question 21 (3 marks)

(a) Work out \( 3 \frac{4}{5} – 1 \frac{2}{3} \)


Kevin was asked to work out \( 2 \frac{1}{3} \times \frac{5}{8} \)

Here is his working and his answer.

\[ 2 \frac{1}{3} \times \frac{5}{8} = \frac{7}{3} \times \frac{5}{8} = \frac{35}{24} = 1 \frac{9}{24} \]

Kevin’s answer is wrong.

(b) What mistake has Kevin made?

Worked Solution

Part (a): Subtracting Mixed Numbers

Method: Convert mixed numbers to improper fractions first, then find a common denominator.

Step 1: Convert to improper fractions

\[ 3 \frac{4}{5} = \frac{(3 \times 5) + 4}{5} = \frac{19}{5} \] \[ 1 \frac{2}{3} = \frac{(1 \times 3) + 2}{3} = \frac{5}{3} \]

Step 2: Find common denominator (15)

\[ \frac{19}{5} = \frac{19 \times 3}{15} = \frac{57}{15} \] \[ \frac{5}{3} = \frac{5 \times 5}{15} = \frac{25}{15} \]

Step 3: Subtract

\[ \frac{57}{15} – \frac{25}{15} = \frac{32}{15} \]

Step 4: Convert back to mixed number

15 goes into 32 twice (remainder 2).

\[ 2 \frac{2}{15} \]

Answer (a):

\( 2 \frac{2}{15} \)

โœ“ (M1 A1)

Part (b): Identifying the Error

Check his conversion: \( \frac{35}{24} \rightarrow 1 \frac{9}{24} \). Is this correct?

\( 35 – 24 = 11 \). The remainder is 11, not 9.

Answer (b):

He subtracted incorrectly when converting the improper fraction to a mixed number. The remainder should be 11, so the answer is \( 1 \frac{11}{24} \).

โœ“ (C1)

โ†‘ Back to Top

Question 22 (5 marks)

The diagram shows a plan of a floor.

10 m 8 m 5 m 6 m

Petra is going to cover the floor with paint.

  • Petra has 3 tins of paint.
  • There are 2.5 litres of paint in each tin.
  • Petra thinks 1 litre of paint will cover \( 10\text{m}^2 \) of floor.

(a) Assuming Petra is correct, does she have enough paint to cover the floor? You must show all your working.


Actually, 1 litre of paint will cover \( 11\text{m}^2 \) of floor.

(b) Does this affect your answer to part (a)? You must give a reason for your answer.

Worked Solution

Step 1: Calculate Total Area

Method: Split the compound shape into two rectangles.

Option A: Split vertically.

  • Rectangle 1 (Left): Height \( 8\text{m} \), Width \( 6\text{m} \) (Wait, if bottom is 6m and top is 10m…). Let’s re-read diagram logic.
  • If Left is 8m and Right is 5m, the missing vertical is 3m.
  • If Top is 10m and the shape closes… usually the bottom width is derived. If the label “6m” is on the bottom edge, then the shape is likely an L-shape where the bottom leg is 6m wide?
  • Let’s assume standard splitting: Top rectangle \( 10 \times 5 \) and bottom rectangle?
  • Let’s look at the labels again. Top=10, Right=5. Left=8. Bottom=6.
  • Area = Area of large rectangle minus missing piece? Or sum of two parts.
  • Let’s split horizontally at the 5m mark.
  • Top Rectangle: Width \( 10\text{m} \), Height \( 5\text{m} \). Area = \( 50\text{m}^2 \).
  • Bottom Rectangle: Total height is \( 8\text{m} \), top part used \( 5\text{m} \), so remaining height is \( 3\text{m} \). The label says \( 6\text{m} \) on bottom. So area = \( 6 \times 3 = 18\text{m}^2 \).
  • Total Area: \( 50 + 18 = 68\text{m}^2 \).
\[ 10 \times 5 = 50 \] \[ (8 – 5) \times 6 = 3 \times 6 = 18 \] \[ \text{Total Area} = 50 + 18 = 68 \text{ m}^2 \]

โœ“ (P1)

Step 2: Calculate Paint Needed

Rule: 1 litre covers \( 10\text{m}^2 \).

\[ 68 \div 10 = 6.8 \text{ litres needed} \]

โœ“ (P1)

Step 3: Calculate Paint Available

3 tins of 2.5 litres each.

\[ 3 \times 2.5 = 7.5 \text{ litres} \]

โœ“ (P1)

Step 4: Comparison

Compare what she has vs what she needs.

\[ 7.5 \text{ litres} > 6.8 \text{ litres} \]

Conclusion: Yes, she has enough paint.

โœ“ (A1)

Part (b): Effect of change

Scenario: Paint covers MORE area (\( 11\text{m}^2 \)) per litre.

This means the paint is more efficient, so she will need LESS paint than calculated in part (a).

Answer: No, it does not change the answer. She already had enough, and now she will have even more spare paint.

โœ“ (C1)

โ†‘ Back to Top

Question 23 (3 marks)

Here is a Venn diagram.

E P Q 12 15 18 10 14 11 13 16 17

(a) Write down the numbers that are in set \( P’ \)

(b) Find the probability that a number chosen at random is in the set \( P \cup Q \)

Worked Solution

Part (a): Understanding \( P’ \)

Definition: \( P’ \) means the “complement of P”. It includes everything NOT inside the circle P.

Numbers inside P: 12, 15, 18

Numbers NOT inside P (check Q only and Outside):

  • Q only: 10, 14
  • Outside: 11, 13, 16, 17

List: 10, 11, 13, 14, 16, 17

โœ“ (B1)

Part (b): Probability of \( P \cup Q \)

Definition: \( P \cup Q \) means “P union Q”, or everything inside EITHER circle P OR circle Q (or both).

Step 1: Count total numbers

Inside P: 12, 15

Intersection: 18

Inside Q: 10, 14

Outside: 11, 13, 16, 17

Total count = \( 2 + 1 + 2 + 4 = 9 \) numbers.

Step 2: Count numbers in \( P \cup Q \)

These are the numbers in the circles: 12, 15, 18, 10, 14.

Count = 5.

Step 3: Probability

\[ \frac{5}{9} \]

โœ“ (M1 A1)

โ†‘ Back to Top

Question 24 (4 marks)

Sophie drives a distance of \( 513 \) kilometres on a motorway in France.

She pays \( 0.81 \) euros for every \( 10 \) kilometres she drives.

(a) Work out an estimate for the total amount that Sophie pays.

(b) Is your answer to part (a) an underestimate or an overestimate? Give a reason for your answer.

Worked Solution

Part (a): Estimating

Rule: To estimate, we round numbers to 1 significant figure.

Round \( 513 \) km \(\rightarrow 500 \) km.

Round \( 0.81 \) euros \(\rightarrow 0.8 \) euros.


Calculation:

How many “10 km” chunks are in 500 km?

\[ 500 \div 10 = 50 \]

Cost = \( 50 \times 0.8 \)

\[ 5 \times 8 = 40 \]

Estimate: \( 40 \) euros.

โœ“ (P1 P1 A1)

Part (b): Evaluation

Analyze the rounding:

  • We rounded \( 513 \) down to \( 500 \).
  • We rounded \( 0.81 \) down to \( 0.8 \).

Since we rounded both numbers down, our result will be lower than the actual cost.

Answer: Underestimate, because we rounded both the distance and the cost down.

โœ“ (C1)

โ†‘ Back to Top

Question 25 (4 marks)

Here is a straight line \( L \) drawn on a grid.

L x y O 3 -2

(a) Find an equation for \( L \).

\( M \) is a different straight line with equation \( y = 5x \)

(b) Write down the equation of a straight line parallel to \( M \).

Worked Solution

Part (a): Finding the Equation \( y = mx + c \)

Step 1: Find the y-intercept (\( c \)).

Look where the line crosses the vertical y-axis. It crosses at \( 3 \).

So, \( c = 3 \).

Step 2: Find the gradient (\( m \)).

Pick two clear points on the line: \( (-2, 0) \) and \( (0, 3) \).

Gradient = \( \frac{\text{Change in } y}{\text{Change in } x} \)

\[ m = \frac{3 – 0}{0 – (-2)} = \frac{3}{2} \]

So, \( m = 1.5 \).

Step 3: Write equation

\[ y = 1.5x + 3 \quad \text{or} \quad y = \frac{3}{2}x + 3 \]

โœ“ (M1 M1 A1)

Part (b): Parallel Lines

Rule: Parallel lines have the same gradient.

The line \( y = 5x \) has a gradient of 5.

Any line with \( y = 5x + c \) is correct (as long as \( c \) is any number).

Example Answer:

\( y = 5x + 1 \)

โœ“ (B1)

โ†‘ Back to Top

Question 26 (5 marks)

Kasim has some small jars, some medium jars and some large jars.

He has a total of \( 400 \) jars.

\( \frac{3}{8} \) of the \( 400 \) jars are empty.

For the empty jars,

  • number of small jars : number of medium jars = \( 3 : 4 \)
  • number of medium jars : number of large jars = \( 1 : 2 \)

Work out the percentage of Kasim’s jars that are empty small jars.

Worked Solution

Step 1: Calculate Total Empty Jars

Method: Find \( \frac{3}{8} \) of 400.

\[ 400 \div 8 = 50 \] \[ 50 \times 3 = 150 \]

There are \( 150 \) empty jars.

โœ“ (P1)

Step 2: Combine the Ratios

We have two separate ratios for the empty jars:

  • Small : Medium = \( 3 : 4 \)
  • Medium : Large = \( 1 : 2 \)

The “Medium” part must be the same number to combine them. Currently, it is 4 in the first ratio and 1 in the second.

Multiply the second ratio by 4:

Medium : Large = \( 4 : 8 \)

Combined Ratio (Small : Medium : Large):

\[ 3 : 4 : 8 \]

โœ“ (P1)

Step 3: Calculate number of small empty jars

Total parts in ratio: \( 3 + 4 + 8 = 15 \) parts.

Value of one part: Total empty jars (\( 150 \)) divided by 15.

\[ 150 \div 15 = 10 \text{ jars per part} \]

Number of small jars (3 parts):

\[ 3 \times 10 = 30 \text{ small jars} \]

โœ“ (P1)

Step 4: Calculate the percentage of total jars

Question asks: Percentage of Kasim’s jars (all 400) that are empty small jars (30).

\[ \frac{30}{400} \times 100 \]

Simplify fraction:

\[ \frac{3}{40} \times 100 \] \[ = \frac{300}{40} \] \[ = \frac{30}{4} \] \[ = 7.5 \% \]

Final Answer:

\( 7.5 \% \)

โœ“ (P1 A1)

โ†‘ Back to Top

Question 27 (2 marks)

In a sale, normal prices are reduced by \( 30\% \)

The sale price of a TV is \( ยฃ280 \)

Work out the normal price of the TV.

Worked Solution

Step 1: Determine Percentage Represented

Logic: If the price is reduced by \( 30\% \), the sale price represents \( 100\% – 30\% = 70\% \) of the original price.

So, \( 70\% = ยฃ280 \).

Step 2: Reverse Percentage Calculation

Find \( 10\% \):

\[ 280 \div 7 = 40 \]

So \( 10\% = ยฃ40 \).


Find \( 100\% \) (Original Price):

\[ 40 \times 10 = 400 \]

Final Answer:

\( ยฃ400 \)

โœ“ (M1 A1)

โ†‘ Back to Top

Question 28 (3 marks)

Solve \( x + 11 \leq 5 – \frac{1}{2}x \)

Worked Solution

Step 1: Eliminate the Fraction

Method: Multiply every term by 2 to get rid of the \( \frac{1}{2} \). This makes it easier to solve.

\[ 2(x) + 2(11) \leq 2(5) – 2(\frac{1}{2}x) \] \[ 2x + 22 \leq 10 – x \]
Step 2: Collect x terms

Add \( x \) to both sides.

\[ 3x + 22 \leq 10 \]
Step 3: Isolate x

Subtract \( 22 \) from both sides:

\[ 3x \leq 10 – 22 \] \[ 3x \leq -12 \]

Divide by 3:

\[ x \leq -4 \]

Final Answer:

\( x \leq -4 \)

โœ“ (M1 M1 A1)

โ†‘ Back to Top