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.

Pearson Edexcel GCSE Mathematics May 2019
Foundation Tier – Paper 1 (Non-Calculator)

Mark Scheme Legend

  • M: Method mark awarded for a correct method or partial method
  • P: Process mark awarded for a correct process as part of a problem solving question
  • A: Accuracy mark awarded after a correct method or process
  • B: Unconditional accuracy mark (no method needed)
  • cao: Correct Answer Only
  • oe: Or Equivalent

Question 1 (1 mark)

Write 180 minutes in hours.

Worked Solution

Step 1: Understanding the Conversion

What are we being asked to find?

We need to convert a time value from minutes to hours. To do this, we need to know how many minutes are in one hour.

Why we do this:

There are 60 minutes in 1 hour. To find the number of hours, we divide the total minutes by 60.

✏ Working:

    3
  ┌───
60│180
  -180
  ────
     0
                    

\[ 180 \div 60 = 3 \]

(B1) cao

Final Answer:

3 hours

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Question 2 (1 mark)

Write 0.73 as a percentage.

Worked Solution

Step 1: Decimal to Percentage

What are we being asked to find?

We need to express the decimal number 0.73 as a percentage.

Why we do this:

“Percent” means “out of 100”. To convert a decimal to a percentage, we multiply by 100.

✏ Working:

\[ 0.73 \times 100 = 73 \]

(B1) cao

Final Answer:

73%

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Question 3 (1 mark)

Work out \( 10 \times (3 + 5) \)

Worked Solution

Step 1: Order of Operations (BIDMAS)

What are we being asked to find?

We need to solve an arithmetic expression using the correct order of operations.

Why we do this:

We use BIDMAS (Brackets, Indices, Division/Multiplication, Addition/Subtraction). We must always calculate the content inside the Brackets first.

✏ Working:

First, calculate the brackets:

\[ 3 + 5 = 8 \]

Now multiply by 10:

\[ 10 \times 8 = 80 \]

(B1) cao

Final Answer:

80

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Question 4 (1 mark)

Write down a prime number that is between 20 and 30

Worked Solution

Step 1: Identify Prime Numbers

What are we being asked to find?

We need to find a number between 20 and 30 that has exactly two factors: 1 and itself.

Why we do this:

Let’s check the numbers between 20 and 30:

  • 21: \(3 \times 7\) (Not prime)
  • 22: Even (Not prime)
  • 23: Prime
  • 24: Even (Not prime)
  • 25: \(5 \times 5\) (Not prime)
  • 26: Even (Not prime)
  • 27: \(3 \times 9\) (Not prime)
  • 28: Even (Not prime)
  • 29: Prime

✏ Working:

The numbers 23 and 29 are prime.

(B1) for 23 or 29

Final Answer:

23 (or 29)

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Question 5 (1 mark)

Find the number that is exactly halfway between 7 and 15

Worked Solution

Step 1: Finding the Midpoint

What are we being asked to find?

We need the average (mean) of the two numbers.

Why we do this:

To find the halfway point, we add the two numbers together and divide by 2.

✏ Working:

\[ 7 + 15 = 22 \] \[ 22 \div 2 = 11 \]

(B1) cao

Final Answer:

11

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Question 6 (4 marks)

Harry is planning a holiday for 4 people for 7 days.
Here are the costs for the holiday for each person.

Travel: £150
Hotel: £50 for each day
Spending money: £250

Work out the total cost of the holiday for 4 people for 7 days.

Worked Solution

Step 1: Calculate the cost per person

What are we being asked to find?

We need to find the total bill for one person first. The travel and spending money are fixed, but the hotel cost depends on the number of days (7).

✏ Working:

Hotel cost for 7 days: \[ 7 \times £50 = £350 \]

Total for 1 person: \[ £150 + £350 + £250 = £750 \]

(P1) for finding cost of hotel or sum of two costs

Step 2: Calculate the total for 4 people

Why we do this:

Now that we know one person costs £750, we multiply this by 4 to get the grand total for the whole group.

✏ Working:

   750
 x   4
 ─────
  3000
  22
                    

\[ £750 \times 4 = £3000 \]

(P1) for full process to find total cost

What this tells us:

The total budget Harry needs to plan for the entire holiday group is £3000.

Final Answer:

£3000

(A1) cao

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Question 7 (3 marks)

In Adam’s garden, the flowers are only red or white or yellow or blue.

The chart shows the number of red flowers, the number of white flowers and the number of yellow flowers.

0 2 4 6 8 10 12 14 Number of flowers red white yellow Colour

The total number of flowers is 30

(a) Work out the number of blue flowers.

(b) Write down the mode.

Worked Solution

Part (a): Finding Blue Flowers

Why we do this:

We know the total is 30. If we subtract all the flowers we know about (Red, White, and Yellow), the leftover must be the Blue ones.

✏ Working:

From the chart:

  • Red = 8
  • White = 10
  • Yellow = 5

Sum of known flowers: \[ 8 + 10 + 5 = 23 \]

Number of blue: \[ 30 – 23 = 7 \]

(P1) for process to find sum (23) or subtract from 30

(A1) cao

Part (b): Finding the Mode

What this tells us:

The “mode” is the value that appears most often. In a bar chart, it is the tallest bar.

✏ Working:

Comparing the heights:

  • Red (8)
  • White (10) – Tallest
  • Yellow (5)
  • Blue (7)

(B1) for white (or ft from part a)

Final Answer:

(a) 7 blue flowers
(b) white

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Question 8 (2 marks)

Write the following fractions in order of size. Start with the smallest fraction.

\[ \frac{1}{3} \quad \frac{3}{4} \quad \frac{1}{4} \quad \frac{7}{12} \quad \frac{1}{2} \]

Worked Solution

Step 1: Use a Common Denominator

Why we do this:

It is difficult to compare fractions with different denominators. We find a number that 3, 4, 12, and 2 all go into. The Lowest Common Multiple is 12.

✏ Working:

  • \( \frac{1}{3} = \frac{4}{12} \)
  • \( \frac{3}{4} = \frac{9}{12} \)
  • \( \frac{1}{4} = \frac{3}{12} \)
  • \( \frac{7}{12} = \frac{7}{12} \)
  • \( \frac{1}{2} = \frac{6}{12} \)

(M1) for at least 2 correct conversions to common denominator

Step 2: Compare Numerators

What this tells us:

Now we order the fractions by looking at the top numbers (3, 4, 6, 7, 9) and then write back the original fractions.

Order: 3/12, 4/12, 6/12, 7/12, 9/12

Final Answer:

\( \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{7}{12}, \frac{3}{4} \)

(A1) cao

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Question 9 (4 marks)

Ruth left her home at 9 am and walked to the library.
She got to the library at 10 30 am.
Ruth walked at a speed of 4 mph.

(a) Work out the distance Ruth walked.

Ruth got to the library at 10 30 am.
She stayed at the library for 50 minutes.
Then she walked home.
Ruth took \( 1 \frac{1}{4} \) hours to walk home.

(b) At what time did Ruth get home?

Worked Solution

Part (a): Calculating Distance

How we find this:

Distance = Speed × Time. First, we find the duration of her walk from 9 am to 10:30 am.

✏ Working:

Time = 1 hour 30 minutes = 1.5 hours

\[ \text{Distance} = 4 \text{ mph} \times 1.5 \text{ hours} = 6 \text{ miles} \]

(M1) for 4 × time difference

(A1) cao

Part (b): Calculating Arrival Time

Why we do this:

We add the duration of her stay and the duration of her walk home to her arrival time at the library.

✏ Working:

Arrived at library: 10:30 am

Stayed for 50 mins: 10:30 + 50 mins = 11:20 am

Walk home: \( 1 \frac{1}{4} \) hours = 1 hour 15 minutes

Total time added: 11:20 + 1 hour 15 minutes = 12:35 pm

(M1) for method to add time using consistent units

Final Answer:

(a) 6 miles
(b) 12 35 pm

(A1) cao

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Question 10 (4 marks)

(a) Solve \( t + t + t = 12 \)

(b) Solve \( x – 2 = 6 \)

(c) Solve \( 6w + 2 = 20 \)

Worked Solution

Part (a): Collecting Like Terms
\[ 3t = 12 \] \[ t = 12 \div 3 = 4 \]

(B1) cao

Part (b): Inverse Operations

To find \( x \), we perform the opposite of subtracting 2, which is adding 2 to both sides.

\[ x = 6 + 2 = 8 \]

(B1) cao

Part (c): Two-Step Equation

First, isolate the term with \( w \) by subtracting 2. Then, divide by 6.

Step 1: \( 6w = 20 – 2 \)

Step 1 result: \( 6w = 18 \)

Step 2: \( w = 18 \div 6 = 3 \)

(M1) for subtracting 2 or dividing by 6

(A1) cao

Final Answer:

(a) \( t = 4 \)
(b) \( x = 8 \)
(c) \( w = 3 \)

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Question 11 (2 marks)

Work out \( 74 \times 58 \)

Worked Solution

Step 1: Multiplication Process

What are we being asked to find?

We need to find the product of two 2-digit numbers. Since this is a non-calculator paper, we must show a clear arithmetic method like long multiplication or the grid method.

✏ Working (Long Multiplication):

      74
    x 58
    ────
     592  (8 x 74)
    3700  (50 x 74)
    ────
    4292
    1   
                    

(M1) for complete method with relative place value correct

What this tells us:

By breaking the multiplication into smaller parts (\( 8 \times 74 \) and \( 50 \times 74 \)) and adding them, we find that the total is 4292.

Final Answer:

4292

(A1) cao

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Question 12 (5 marks)

AB and BC are perpendicular lines.

A B C 25° 25°

(a) Find the value of \( x \).


RS and TU are parallel lines. PQ is a straight line.

R S T U P Q 125° a b c d e

(b) (i) Write down the letter of one other angle of size 125°. Give a reason for your answer.

(ii) Explain why \( a + b + c = 235^\circ \).

Worked Solution

Part (a): Solving for x

Why we do this:

The term “perpendicular” means the total angle between the lines is \( 90^\circ \). The three smaller angles must add up to this total.

✏ Working:

\[ 25 + x + 25 = 90 \] \[ 50 + x = 90 \] \[ x = 90 – 50 = 40 \]

(M1) for using 90, e.g. \( 90 – 25 – 25 \)

(A1) cao

Part (b)(i): Identifying equal angles

How we know:

On parallel lines, specific pairs of angles are equal due to geometric properties. We are looking for an angle that matches the \( 125^\circ \) provided.

✏ Possible Answers:

  • Angle a: Reason — Vertically opposite angles are equal.
  • Angle d: Reason — Corresponding angles are equal.

(B1) for b or d (or both)

(C1) for appropriate reason(s) with correct terminology underlined

Part (b)(ii): Angles around a point

What this tells us:

A full circle of angles around a single point adds up to \( 360^\circ \). The angles \( a, b, c \) and the given \( 125^\circ \) together form a complete turn.

✏ Working:

\[ a + b + c + 125 = 360 \] \[ a + b + c = 360 – 125 \] \[ a + b + c = 235 \]

(C1) for correct explanation using 360 around a point

Final Answer:

(a) \( x = 40 \)
(b)(i) \( a \) (vertically opposite) or \( d \) (corresponding)
(b)(ii) Angles around a point sum to \( 360^\circ \); \( 360 – 125 = 235 \)

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Question 13 (1 mark)

The length of a line is \( x \) centimetres.
Write down an expression, in terms of \( x \), for the length of the line in millimetres.

Worked Solution

Step 1: Unit Conversion Logic

Why we do this:

There are 10 millimetres (mm) in 1 centimetre (cm). To change from cm to mm, we must multiply by 10.

✏ Working:

\[ \text{Length in mm} = x \times 10 \]

(B1) for \( 10x \) oe

Final Answer:

\( 10x \)

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Question 14 (2 marks)

(a) Work out \( \frac{1}{5} \) of 70


Fiona has to work out the exact value of \( 48 \div \frac{1}{2} \).
She writes \( 48 \div \frac{1}{2} = 24 \).
Fiona’s reason is, “There are 2 halves in 1, so there will be 24 halves in 48”.

(b) Explain what is wrong with Fiona’s reason.

Worked Solution

Part (a): Finding a Fraction of a Number

To find one fifth, we divide the total by 5.

✏ Working:

\[ 70 \div 5 = 14 \]

(B1) for 14

Part (b): Evaluating Division by a Fraction

Why we do this:

When you divide a number by a value smaller than 1 (like \( \frac{1}{2} \)), the answer gets larger. Fiona correctly identifies there are 2 halves in every 1, but she then divided 48 by 2 instead of multiplying 48 by 2.

✏ Explanation:

If there are 2 halves in every 1, then in 48 there must be \( 48 \times 2 \) halves.

\[ 48 \div 0.5 = 96 \]

(C1) for explanation, e.g. “she should have multiplied by 2” or “there are 96 halves in 48”

Final Answer:

(a) 14
(b) She divided by 2 instead of multiplying by 2 (the answer should be 96).

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Question 15 (2 marks)

(a) Write down the value of \( \sqrt{64} \)

(b) Work out the value of \( 5^3 \)

Worked Solution

Part (a): Square Roots

We need a number that, when multiplied by itself, equals 64.

\[ 8 \times 8 = 64 \]

(B1) cao

Part (b): Cubing a Number

A “cube” means multiplying the number by itself, then by itself again (three times total).

\[ 5 \times 5 = 25 \] \[ 25 \times 5 = 125 \]

(B1) cao

Final Answer:

(a) 8
(b) 125

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Question 16 (2 marks)

(a) Expand \( 5(2m – 3) \)

(b) Factorise \( 3n + 12 \)

Worked Solution

Part (a): Expanding Brackets

Why we do this:

To “expand” means to multiply the term outside the bracket by every term inside the bracket.

✏ Working:

\[ 5 \times 2m = 10m \] \[ 5 \times (-3) = -15 \]

(B1) for \( 10m – 15 \)

Part (b): Factorising

How we find this:

Factorising is the opposite of expanding. we look for the Highest Common Factor of \( 3n \) and \( 12 \), which is 3, and place it outside the bracket.

✏ Working:

\[ 3n = 3 \times n \] \[ 12 = 3 \times 4 \]

Take 3 outside: \( 3(n + 4) \)

(B1) for \( 3(n + 4) \)

Final Answer:

(a) \( 10m – 15 \)
(b) \( 3(n + 4) \)

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Question 17 (2 marks)

Stuart throws a biased coin 10 times. He gets 7 Tails.

Maxine throws the same coin 50 times. She gets 30 Tails.

Prasha is going to throw the coin once.

(i) Whose results will give the better estimate for the probability that she will get Tails, Stuart’s or Maxine’s? You must give a reason for your answer.

(ii) Use Stuart’s and Maxine’s results to work out an estimate for the probability that Prasha will get Tails.

Worked Solution

Part (i): Sample Size and Reliability

Why we do this:

In probability, the more times an experiment is carried out (the larger the sample size), the more reliable the estimate becomes. This reduces the effect of random luck.

✏ Reasoning:

Maxine. She carried out many more trials (50 compared to Stuart’s 10).

(C1) for Maxine with reason (more trials/throws)

Part (ii): Combined Estimate

How we find this:

To get the best possible estimate, we combine all the data from both people to find the total number of Tails relative to the total number of throws.

✏ Working:

Total Tails: \( 7 + 30 = 37 \)

Total Throws: \( 10 + 50 = 60 \)

Estimate: \( \frac{37}{60} \)

(B1) for \( \frac{37}{60} \) oe

Final Answer:

(i) Maxine (more trials)
(ii) \( \frac{37}{60} \)

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Question 18 (4 marks)

The diagram shows a rectangular garden path.

600 cm 120 cm

Wasim is going to cover the path with paving stones.
Each paving stone is a square of side 30 cm.
Each paving stone costs £2.50
Wasim has £220 to spend on paving stones.

Show that he has enough money to buy all the paving stones he needs.

Worked Solution

Step 1: Calculate how many stones are needed

Why we do this:

We need to figure out how many 30cm stones fit along the length and the width of the path.

✏ Working:

Stones along length: \( 600 \div 30 = 20 \text{ stones} \)

Stones along width: \( 120 \div 30 = 4 \text{ stones} \)

Total stones: \( 20 \times 4 = 80 \text{ stones} \)

(M1) for process to find number of tiles needed (e.g., finding area of path and tile)

(M1) for finding 80 tiles

Step 2: Calculate the total cost

How we find this:

We multiply the 80 stones by the cost of each stone (£2.50).

✏ Working:

\[ 80 \times £2.50 \]

Easy way: \[ 80 \times 2 = 160 \] and half of 80 is 40.

\[ 160 + 40 = £200 \]

(M1) for full method to find cost (80 x 2.5)

Step 3: Verification

What this tells us:

We compare the total cost to Wasim’s budget of £220.

✏ Comparison:

Wasim needs £200. He has £220.

Since £200 < £220, he has enough money.

(A1) for 200 and conclusion

Final Answer:

Yes, he has enough. The total cost is £200, which is less than his budget of £220.

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Question 19 (4 marks)

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

(b) Work out \( \frac{2}{3} \times \frac{3}{4} \)
Give your answer as a fraction in its simplest form.

Worked Solution

Part (a): Subtracting Fractions

Why we do this:

To subtract fractions, we must have a common denominator. The Lowest Common Multiple of 3 and 5 is 15.

✏ Working:

Convert fractions:

\[ \frac{2}{3} = \frac{10}{15} \] \[ \frac{1}{5} = \frac{3}{15} \]

Subtract:

\[ \frac{10}{15} – \frac{3}{15} = \frac{7}{15} \]

(M1) for common denominator, e.g. \( \frac{10}{15} – \frac{3}{15} \)

(A1) oe

Part (b): Multiplying Fractions

How we find this:

Multiplying is simpler: multiply the top numbers (numerators) together and multiply the bottom numbers (denominators) together. Then simplify.

✏ Working:

\[ \frac{2 \times 3}{3 \times 4} = \frac{6}{12} \]

Simplify by dividing both by 6:

\[ \frac{6 \div 6}{12 \div 6} = \frac{1}{2} \]

(M1) for method to multiply, e.g. \( \frac{6}{12} \) or simplifying during calc

(A1) cao

Final Answer:

(a) \( \frac{7}{15} \)
(b) \( \frac{1}{2} \)

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Question 20 (3 marks)

Here are two squares, A and B.

A B

The length of the side of square A is 50% of the length of the side of square B.
Express the area of the shaded region of square A as a percentage of the area of square B.

Worked Solution

Step 1: Compare lengths and areas

How we find this:

Let’s use simple numbers to represent the sides. If Square B has a side length of 2, then Square A (at 50%) has a side length of 1.

✏ Working:

Side B = 2 units \( \rightarrow \) Area B = \( 2 \times 2 = 4 \text{ units}^2 \)

Side A = 1 unit \( \rightarrow \) Area A = \( 1 \times 1 = 1 \text{ unit}^2 \)

(P1) for assigns lengths in ratio 1:2 and calculates at least one area

Step 2: Find the shaded region percentage

Why we do this:

The shaded part of Square A is exactly half of its area (\( 0.5 \text{ units}^2 \)). We need to find what percentage this is of Square B’s area (4).

✏ Working:

Shaded region = \( 0.5 \)

Fraction of B = \( \frac{0.5}{4} \)

Multiply by 100 to get percentage:

\[ \frac{0.5}{4} \times 100 = 0.125 \times 100 = 12.5\% \]

(P1) for process to find ratio of shaded to B (e.g., 1:8 or 0.125)

Final Answer:

12.5%

(A1) cao

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Question 21 (4 marks)

There are 40 students in a class.
Each student walks to school or cycles to school or gets the bus to school.

There are 22 girls in the class.
9 of the girls walk to school.
7 of the boys cycle to school.
6 of the 10 students who get the bus to school are boys.

Find the number of these students who walk to school.

Worked Solution

Step 1: Organizing the Data (Finding Boys)

What are we being asked to find?

We need the total number of walkers (boys + girls). We already know 9 girls walk, so we need to find how many boys walk.

✏ Working:

Total students = 40. Girls = 22.

Total boys = \( 40 – 22 = 18 \)

(P1) for finding total number of boys (18)

Step 2: Finding Boys’ Travel Methods

Why we do this:

We have enough information to find how boys travel. If we know the total boys and how many bus and cycle, the remainder must walk.

✏ Working:

Total boys = 18

  • Boys on Bus = 6 (given)
  • Boys Cycle = 7 (given)

Boys who walk = \( 18 – 6 – 7 = 5 \)

(P1) for process to find boys who walk (5)

Step 3: Calculating Total Walkers

What this tells us:

Finally, we combine the walking girls and walking boys to answer the question.

✏ Working:

Total walkers = Girls walk + Boys walk

\[ 9 + 5 = 14 \]

(P1) for full process to find total walkers

Final Answer:

14

(A1) cao

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Question 22 (4 marks)

There are only blue cubes, red cubes and yellow cubes in a box.
The table shows the probability of taking at random a blue cube from the box.

Colour blue red yellow Probability 0.2

The number of red cubes in the box is the same as the number of yellow cubes in the box.

(a) Complete the table.

There are 12 blue cubes in the box.

(b) Work out the total number of cubes in the box.

Worked Solution

Part (a): Completing the Probability Table

Why we do this:

The sum of all probabilities in a mutually exclusive set must equal 1. Since Red and Yellow cubes are equal in number, their probabilities must also be equal.

✏ Working:

Remaining probability: \[ 1 – 0.2 = 0.8 \]

Since Red = Yellow, we divide by 2:

\[ 0.8 \div 2 = 0.4 \]

(P1) for process to find sum (0.8)

(A1) for 0.4 and 0.4

Part (b): Finding the Total Quantity

How we find this:

Probability is defined as \( \frac{\text{Quantity}}{\text{Total}} \). We know 12 cubes represent a probability of 0.2.

✏ Working:

\[ \frac{12}{\text{Total}} = 0.2 \] \[ \text{Total} = \frac{12}{0.2} = \frac{120}{2} = 60 \]

(P1) for complete process (12 ÷ 0.2 or 12 × 5)

Final Answer:

(a) red: 0.4, yellow: 0.4
(b) 60

(A1) cao

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Question 23 (5 marks)

Deon needs 50 g of sugar to make 15 biscuits.
She also needs:

  • three times as much flour as sugar
  • two times as much butter as sugar
Deon is going to make 60 biscuits.

(a) Work out the amount of flour she needs.

Deon has to buy all the butter she needs to make 60 biscuits.
She buys the butter in 250 g packs.

(b) How many packs of butter does Deon need to buy?

Worked Solution

Part (a): Scaling the Recipe (Flour)

How we find this:

First, we find the amount of flour for 15 biscuits. Then, we find the “scale factor” to get from 15 biscuits to 60 biscuits.

✏ Working:

Flour for 15 biscuits: \[ 3 \times 50\text{g} = 150\text{g} \]

Scale Factor: \[ 60 \div 15 = 4 \]

Flour for 60 biscuits: \[ 150\text{g} \times 4 = 600\text{g} \]

(P1) for starting process (e.g., 60 ÷ 15 = 4)

(P1) for complete process (150 × 4)

(A1) cao

Part (b): Butter Packs Calculation

Why we do this:

We need to find the total grams of butter needed and then see how many 250g packs are required. We must round up to the nearest whole pack.

✏ Working:

Butter for 15 biscuits: \[ 2 \times 50\text{g} = 100\text{g} \]

Butter for 60 biscuits: \[ 100\text{g} \times 4 = 400\text{g} \]

Packs needed: \[ 400\text{g} \div 250\text{g} = 1.6 \text{ packs} \]

Since we cannot buy 0.6 of a pack, we need 2 packs.

(P1) for process to find butter (400g)

Final Answer:

(a) 600 g
(b) 2

(A1) cao

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Question 24 (2 marks)

Find the highest common factor (HCF) of 72 and 90

Worked Solution

Step 1: Listing Factors

Why we do this:

The HCF is the largest number that divides exactly into both 72 and 90. We can list the factors of each and pick the largest one they share.

✏ Working:

Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

The highest common factor is 18.

(M1) for listing factors of 72 and 90 (at least 4 correct for each)

Final Answer:

18

(A1) for 18 or \( 2 \times 3^2 \)

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Question 25 (2 marks)

The diagram shows the plan, front elevation and side elevation of a solid shape, drawn on a centimetre grid.

Plan Front elevation Side elevation

In the space below, draw a sketch of the solid shape.
Give the dimensions of the solid on your sketch.

Worked Solution

Step 1: Identify the Solid

How we find this:

We look at the three views. A “Plan” (view from the top) that is a circle, combined with “Elevations” (views from the side/front) that are rectangles, identifies the shape as a cylinder.

✏ Working (Determining Dimensions):

  • The Plan circle is 4 units wide \(\rightarrow\) Diameter = 4 cm (Radius = 2 cm).
  • The Elevations are 5 units tall \(\rightarrow\) Height = 5 cm.

(M1) for sketch of a cylinder

Step 2: Drawing the Sketch
4 cm (diameter) 5 cm

Final Answer:

A cylinder with a diameter of 4 cm and a height of 5 cm.

(A1) sketch with dimensions shown

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Question 26 (3 marks)

x y O 5 -5 5 -5 A B

Shape A can be transformed to shape B by a reflection in the \( x \)-axis followed by a translation \( \binom{c}{d} \).

Find the value of \( c \) and the value of \( d \).

Worked Solution

Step 1: Reflect Shape A in the \( x \)-axis

Why we do this:

We need to perform the first part of the transformation to see where the intermediate shape lands. Reflection in the \( x \)-axis flips the shape vertically, changing the sign of the \( y \)-coordinates.

✏ Working:

Original vertices of A: \( (3, 2), (5, 2), (3, 5) \)

Reflected vertices: \( (3, -2), (5, -2), (3, -5) \)

(M1) for reflection in x-axis shown or coordinates identified

Step 2: Find the Translation Vector

How we find this:

We compare a corresponding vertex of our reflected shape to shape B. Let’s look at the “right-angle” vertex.

✏ Working:

Reflected right-angle vertex is at \( (3, -2) \).

Shape B‘s right-angle vertex is at \( (-3, -3) \).

Horizontal shift (\( c \)): \( -3 – 3 = -6 \)

Vertical shift (\( d \)): \( -3 – (-2) = -1 \)

(A1) for \( c = -6 \) or \( d = -1 \)

(A1) for both correct

What this tells us:

The vector \( \binom{-6}{-1} \) tells us to move the shape 6 units left and 1 unit down.

Final Answer:

\( c = -6, \quad d = -1 \)

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Question 27 (4 marks)

A shop sells packs of black pens, packs of red pens and packs of green pens.
There are:

  • 2 pens in each pack of black pens
  • 5 pens in each pack of red pens
  • 6 pens in each pack of green pens
On Monday,

\[ \text{number of packs of black pens sold} : \text{number of packs of red pens sold} : \text{number of packs of green pens sold} = 7 : 3 : 4 \]

A total of 212 pens were sold.
Work out the number of green pens sold.

Worked Solution

Step 1: Convert the Pack Ratio to a Pen Ratio

Why we do this:

The 212 total is for individual pens, but the ratio is for packs. We need to multiply each part of the ratio by the number of pens in that pack.

✏ Working:

Black: \( 7 \times 2 = 14 \text{ pens} \)

Red: \( 3 \times 5 = 15 \text{ pens} \)

Green: \( 4 \times 6 = 24 \text{ pens} \)

The ratio of pens sold is \( 14 : 15 : 24 \).

(P1) for finding ratios of pen numbers sold: 14, 15, 24

Step 2: Find the Value per Ratio Part

How we find this:

We add the parts of the pen ratio and divide the total number of pens (212) by this sum.

✏ Working:

Total parts = \( 14 + 15 + 24 = 53 \)

Value of 1 part = \( 212 \div 53 = 4 \)

(P1) for dividing 212 by sum of pen parts

Step 3: Calculate Green Pens

What this tells us:

Multiply the “Green” part of the pen ratio (24) by the value per part (4).

✏ Working:

\[ 24 \times 4 = 96 \text{ pens} \]

(P1) for complete process to find green pens

Final Answer:

96

(A1) cao

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Question 28 (4 marks)

Here are two rectangles.

A B C D P Q R S

\[ QR = 10 \text{ cm} \]

\[ BC = PQ \]

The perimeter of \( ABCD \) is \( 26 \text{ cm} \)
The area of \( PQRS \) is \( 45 \text{ cm}^2 \)

Find the length of \( AB \).

Worked Solution

Step 1: Solve for Rectangle PQRS

Why we do this:

We know the area and one side (\( QR = 10 \)) of \( PQRS \). We can find the other side length, \( PQ \).

✏ Working:

\[ \text{Area } PQRS = PQ \times QR \] \[ 45 = PQ \times 10 \] \[ PQ = 45 \div 10 = 4.5 \text{ cm} \]

(P1) for finding \( PQ \), e.g. \( 45 \div 10 = 4.5 \)

Step 2: Transfer Information to Rectangle ABCD

How we find this:

The question states \( BC = PQ \). So we now know the height of the first rectangle.

✏ Working:

\[ BC = 4.5 \text{ cm} \]
Step 3: Solve for Perimeter

What this tells us:

Perimeter is the sum of all four sides. For a rectangle, it is \( 2 \times (\text{Length} + \text{Width}) \).

✏ Working:

\[ 2 \times (AB + 4.5) = 26 \] \[ AB + 4.5 = 26 \div 2 = 13 \] \[ AB = 13 – 4.5 = 8.5 \text{ cm} \]

(P1) for process to use perimeter of \( ABCD \) with \( BC \)

(P1) for process to find \( AB \)

Final Answer:

8.5 cm

(A1) cao

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Question 29 (3 marks)

Here is the graph of \( y = x^2 – 2x – 3 \)

O 2 4 -2 4 -2 -4

(a) Write down the coordinates of the turning point on the graph of \( y = x^2 – 2x – 3 \)

(b) Use the graph to find the roots of the equation \( x^2 – 2x – 3 = 0 \)

Worked Solution

Part (a): Identifying the Turning Point

What are we being asked to find?

The turning point is the vertex of the parabola—the exact point where the curve stops going down and starts going up (the minimum point).

✏ Working:

Looking at the lowest point of the blue curve:

  • It is exactly halfway between the \( x \)-intercepts (\( x = 1 \)).
  • The corresponding \( y \)-value is \( -4 \).

(B1) cao

Part (b): Finding the Roots

How we find this:

The “roots” of the equation are the values of \( x \) where the graph crosses the \( x \)-axis (where \( y = 0 \)).

✏ Working:

The curve crosses the horizontal axis at:

  • \( x = -1 \)
  • \( x = 3 \)

(B2) for both correct (or B1 for one correct)

Final Answer:

(a) \( (1, -4) \)
(b) \( x = -1 \) and \( x = 3 \)

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