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GCSE November 2019 Edexcel Foundation Paper 3

Using this Interactive Exam

  • Attempt First: Try solving the question on paper before checking the solution.
  • Show Solution: Click the green button to reveal the step-by-step worked answer.
  • Calculator: This is a Calculator paper. You may use a calculator for all questions.

Question 1 (1 mark)

Write down two factors of 12.

Worked Solution

Step 1: Understanding Factors

What is a factor?

A factor is a whole number that divides into another number exactly, leaving no remainder.

We need to find numbers that divide into 12.

✏ Working:

Let’s list pairs of numbers that multiply to make 12:

\[ 1 \times 12 = 12 \] \[ 2 \times 6 = 12 \] \[ 3 \times 4 = 12 \]

The factors of 12 are: 1, 2, 3, 4, 6, 12

Final Answer:

Any two from: 1, 2, 3, 4, 6, 12

✓ (B1)

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Question 2 (1 mark)

Find \( \frac{1}{3} \) of 30

Worked Solution

Step 1: Method

How do we find a fraction of an amount?

To find \( \frac{1}{3} \) of a number, we divide the number by 3.

✏ Working:

\[ 30 \div 3 = 10 \]

Final Answer:

10

✓ (B1)

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Question 3 (1 mark)

Write 0.7 as a fraction.

Worked Solution

Step 1: Place Value

Understanding decimals

The first digit after the decimal point is the “tenths” column.

0.7 means “seven tenths”.

✏ Working:

\[ 0.7 = \frac{7}{10} \]

Final Answer:

\( \frac{7}{10} \)

✓ (B1)

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Question 4 (1 mark)

Here is a list of numbers.

7    8    15    16    18    22

Write down the number from the list that is a multiple of 6.

Worked Solution

Step 1: Understanding Multiples

What is a multiple of 6?

A multiple of 6 is a number that is in the 6 times table (6, 12, 18, 24…).

✏ Working:

Let’s check the multiples of 6:

6, 12, 18, 24…

Now look at the list: 7, 8, 15, 16, 18, 22.

The number 18 appears in our multiples of 6.

Final Answer:

18

✓ (B1)

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Question 5 (1 mark)

Change 4 kilometres into metres.

Worked Solution

Step 1: Conversion Factor

How many metres in a kilometre?

“Kilo” means 1000.

So, 1 kilometre = 1000 metres.

✏ Working:

\[ 4 \times 1000 = 4000 \]

Final Answer:

4000 metres

✓ (B1)

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Question 6 (1 mark)

Here is a grid of squares.

Write down the ratio of the number of shaded squares to the number of unshaded squares.

Worked Solution

Step 1: Count the squares

What are we counting?

We need to count the shaded (grey) squares separately from the unshaded (white) squares.

✏ Working:

Number of shaded squares = 3

Number of unshaded squares = 5

Step 2: Write the ratio

Order matters

The question asks for the ratio of shaded to unshaded.

✏ Working:

Ratio = Shaded : Unshaded

Ratio = 3 : 5

Final Answer:

3 : 5

✓ (B1)

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Question 7 (2 marks)

\( w = 4u + 3 \)

Find the value of \( w \) when \( u = 8 \)

Worked Solution

Step 1: Substitute the value

What does substitution mean?

We replace the letter \( u \) with the number 8.

Note: \( 4u \) means \( 4 \times u \).

✏ Working:

\[ w = 4(8) + 3 \] \[ w = 32 + 3 \]
Step 2: Calculate the final value

✏ Working:

\[ w = 35 \]

Final Answer:

35

✓✓ (M1, A1)

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Question 8 (2 marks)

Here are the first five terms of a sequence.

1    3    6    10    15

Write down the next two terms of the sequence.

Worked Solution

Step 1: Find the pattern

How is the sequence changing?

Let’s look at the difference between each term.

✏ Working:

\( 1 \to 3 \) (Add 2)

\( 3 \to 6 \) (Add 3)

\( 6 \to 10 \) (Add 4)

\( 10 \to 15 \) (Add 5)

The rule is: Add 2, Add 3, Add 4, Add 5…

Step 2: Apply the pattern

To find the next term, we add 6. To find the one after that, we add 7.

✏ Working:

Next term: \( 15 + 6 = 21 \)

Term after that: \( 21 + 7 = 28 \)

Final Answer:

21, 28

✓✓ (B1, B1)

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Question 9 (6 marks)

Mrs Brown asked each child in her class which pet they liked best.

Here are her results.

dog      rabbit    cat       dog       dog      hamster
cat      dog       rabbit    hamster   cat      cat
dog      dog       cat       dog       rabbit   dog

(a) Complete the frequency table for this information.

Pet Tally Frequency
dog
rabbit
cat
hamster

(b) On the grid below, draw a bar chart for this information.

(c) Write down the most popular pet.

Worked Solution

Step 1: Tally the results (Part a)

Strategy:

Go through the list one by one and count them. Crossing them off as you go helps avoid mistakes.

✏ Working:

Dog: 8

Rabbit: 3

Cat: 5

Hamster: 2

Completed Table:

Dog |||| ||| (8)
Rabbit ||| (3)
Cat |||| | (5)
Hamster || (2)
Step 2: Draw the Bar Chart (Part b)

Rules for Bar Charts:

  • Label the axes (Frequency on vertical, Pet on horizontal).
  • Choose a suitable scale (e.g., 1 square = 1 pet).
  • Bars must be equal width.
  • Gaps between bars must be equal.
Frequency Pet 0 2 4 6 8 Dog Rabbit Cat Hamster
Step 3: Identify the most popular (Part c)

Look for the highest frequency or the tallest bar.

Dog has the highest frequency (8).

Final Answer:

(a) Table completed (8, 3, 5, 2)

(b) Bar chart drawn correctly

(c) Dog

✓✓✓✓✓✓ (Total: 6 marks)

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Question 10 (2 marks)

(a) On the diagram below, draw a diameter of the circle.

(b) On the diagram below, draw a segment of the circle. Shade the segment.

Worked Solution

Step 1: Diameter (Part a)

Definition: A diameter is a straight line passing from side to side through the centre of a circle.

Diameter
Step 2: Segment (Part b)

Definition: A segment is the area between a chord and an arc.

To draw this, draw any straight line across the circle (not through the center, though that creates a semi-circle which is technically a segment, usually we draw a smaller one) and shade the area enclosed.

Segment

Final Answer:

(a) Straight line drawn through the centre.

(b) Chord drawn and the region between chord and arc shaded.

✓✓ (B1, B1)

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Question 11 (5 marks)

Dylan buys 13 bicycle lights for £7.50 each.

He pays with five £20 notes.

(a) How much change should Dylan get?

(3 marks)


The normal price of a bicycle is £120.

In a sale, there is \( \frac{1}{5} \) off the normal price of the bicycle.

(b) Work out the price of the bicycle in the sale.

(2 marks)

Worked Solution

Part (a): Calculating Change

Step 1: Calculate the total cost of the lights.

We need to multiply the number of lights by the cost per light.

✏ Working:

\[ 13 \times 7.50 = 97.50 \]

Total cost is £97.50

Step 2: Calculate how much money he hands over.

He uses five £20 notes.

✏ Working:

\[ 5 \times 20 = 100 \]

He pays £100.

Step 3: Calculate the change.

Change = Money Paid – Total Cost

✏ Working:

\[ 100.00 – 97.50 = 2.50 \]
Part (b): Sale Price

Step 1: Calculate the discount.

We need to find \( \frac{1}{5} \) of £120.

✏ Working:

\[ 120 \div 5 = 24 \]

The discount is £24.

Step 2: Calculate the sale price.

Subtract the discount from the normal price.

✏ Working:

\[ 120 – 24 = 96 \]

Final Answer:

(a) £2.50

(b) £96

✓✓✓✓✓ (Total: 5 marks)

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Question 12 (3 marks)

Cornflakes are sold in two sizes of box.

Size of box Weight of cornflakes
small 450 g
large 750 g

Rae buys 3 small boxes of cornflakes and some large boxes of cornflakes.

In total she buys 5850 g of cornflakes.

Work out the number of large boxes of cornflakes Rae buys.

Worked Solution

Step 1: Calculate weight of small boxes

First, find out how much weight comes from the 3 small boxes she definitely bought.

✏ Working:

\[ 3 \times 450 = 1350 \text{ g} \]
Step 2: Calculate remaining weight

Subtract the weight of the small boxes from the total weight to find out how much weight is made up of large boxes.

✏ Working:

\[ 5850 – 1350 = 4500 \text{ g} \]
Step 3: Calculate number of large boxes

Divide the remaining weight by the weight of a single large box.

✏ Working:

\[ 4500 \div 750 \]

We can simplify this by dividing both by 10 first:

\[ 450 \div 75 \]

How many 75s in 450?

\[ 75 \times 2 = 150 \] \[ 150 \times 3 = 450 \] \[ \text{So, } 75 \times 6 = 450 \]

Answer: 6 boxes

Final Answer:

6

✓✓✓ (P1, P1, A1)

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Question 13 (2 marks)

The stem and leaf diagram below gives information about the ages of people in a social club.

3 4 5 6 7 145 02256 017789 3459 04 Key: 4|2 represents 42

Find the range of these ages.

Worked Solution

Step 1: Identify Highest and Lowest

What is range?

Range = Highest Value – Lowest Value

Using the Stem and Leaf diagram:

  • The lowest value is the first number: Stem 3, Leaf 1 -> 31
  • The highest value is the last number: Stem 7, Leaf 4 -> 74

✏ Working:

Highest = 74

Lowest = 31

Step 2: Calculate Range

✏ Working:

\[ 74 – 31 = 43 \]

Final Answer:

43

✓✓ (M1, A1)

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Question 14 (2 marks)

Here is a rectangle.

7 cm 3 cm

Coby has to find the perimeter of this rectangle.

He writes,    Perimeter = \( 7 \times 3 \)

(a) What mistake has Coby made?

(1 mark)


Here is a triangle.

(x + 8) cm (x + 7) cm x cm

Iram solves a problem about this triangle to find the value of \( x \).

Her answer is \( x = -2 \).

(b) Explain why Iram’s answer must be wrong.

(1 mark)

Worked Solution

Part (a): Perimeter Error

What is Perimeter?

Perimeter is the distance around the outside of the shape (Add all sides).

What did Coby do?

He did \( 7 \times 3 \). This is Length multiplied by Width.

Length \( \times \) Width is the formula for Area, not Perimeter.

Answer: He has calculated the area instead of the perimeter (or he should have added the sides).

✓ (C1)

Part (b): Triangle Error

Analyze the answer \( x = -2 \)

Look at the bottom side of the triangle. Its length is labeled as \( x \text{ cm} \).

If \( x = -2 \), then the length of the bottom side is \(-2 \text{ cm}\).

Can a length be negative?

No, a physical length must be a positive number.

Answer: A length cannot be negative (or you cannot have a side of length -2).

✓ (C1)

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Question 15 (4 marks)

There are 800 students at a school.

Each student has either a school dinner or a packed lunch.

  • 31% of the students have packed lunches.
  • 55% of the students are boys.
  • 60% of the boys have school dinners.

How many girls have packed lunches?

You must show all your working.

Worked Solution

Step 1: Create a plan

There is a lot of information here. A two-way table or “tree” logic is the best way to organize it.

Goal: Find “Girls with Packed Lunches”.

We know: Total = 800.

Step 2: Find Total Packed Lunches

31% of all students have packed lunches.

✏ Working:

\[ 0.31 \times 800 = 248 \]

Total Packed Lunches = 248

Step 3: Analyze the Boys

First, find total number of boys (55% of students).

✏ Working:

\[ 0.55 \times 800 = 440 \text{ Boys} \]

Now, find how many boys have packed lunches.

We know 60% of boys have school dinners.

This means 40% of boys must have packed lunches (100% – 60% = 40%).

✏ Working:

Boys with Packed Lunch = 40% of 440

\[ 0.40 \times 440 = 176 \]
Step 4: Find Girls with Packed Lunches

Total Packed Lunches = Boys with Packed Lunch + Girls with Packed Lunch

We know Total (248) and Boys (176).

✏ Working:

\[ 248 – 176 = 72 \]

Final Answer:

72

✓✓✓✓ (P1, P1, P1, A1)

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Question 16 (4 marks)

In a bag there are only red counters, blue counters, green counters and yellow counters.

A counter is taken at random from the bag.

The table shows the probabilities of getting a red counter or a yellow counter.

Colour red blue green yellow
Probability 0.4 0.25

The number of blue counters : the number of green counters = 3 : 4

Complete the table.

Worked Solution

Step 1: Find remaining probability

Total Probability Rule: All probabilities must add up to 1.

We know Red (0.4) and Yellow (0.25).

✏ Working:

\[ 1 – (0.4 + 0.25) \] \[ 1 – 0.65 = 0.35 \]

So, the probability of (Blue + Green) is 0.35.

Step 2: Share in ratio

We need to split 0.35 in the ratio 3 : 4.

Total parts = \( 3 + 4 = 7 \).

✏ Working:

Value of one part:

\[ 0.35 \div 7 = 0.05 \]

Blue (3 parts):

\[ 3 \times 0.05 = 0.15 \]

Green (4 parts):

\[ 4 \times 0.05 = 0.20 \]

Final Answer:

Blue: 0.15

Green: 0.20

✓✓✓✓ (P1, P1, P1, A1)

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Question 17 (4 marks)

(a) Complete the table of values for \( y = 4x – 6 \)

\(x\) -1 0 1 2 3 4
\(y\) -2 10

(2 marks)

(b) On the grid, draw the graph of \( y = 4x – 6 \) for values of \( x \) from -1 to 4.

-1 O 1 2 3 4 x 2 4 6 8 10 12 14 -2 -4 -6 -8 -10 -12 y

(2 marks)

Worked Solution

Step 1: Complete the table (Part a)

Substitute the \( x \) values into the equation \( y = 4x – 6 \).

✏ Working:

When \( x = -1 \): \( 4(-1) – 6 = -4 – 6 = -10 \)

When \( x = 0 \): \( 4(0) – 6 = 0 – 6 = -6 \)

When \( x = 2 \): \( 4(2) – 6 = 8 – 6 = 2 \)

When \( x = 3 \): \( 4(3) – 6 = 12 – 6 = 6 \)

Completed Table:

x-101234
y-10-6-22610
Step 2: Plot the points (Part b)

Plot the coordinates from your table: (-1, -10), (0, -6), (1, -2), (2, 2), (3, 6), (4, 10).

Draw a straight line through them.

Final Answer:

(a) Table completed correctly

(b) Straight line drawn from x = -1 to x = 4

✓✓✓✓ (Total: 4 marks)

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Question 18 (2 marks)

Reflect shape P in the line \( y = 3 \)

O 1 2 3 x 1 2 3 4 5 6 7 8 y P

Worked Solution

Step 1: Draw the reflection line

Draw a horizontal line at \( y = 3 \).

This is the “mirror line”.

Step 2: Reflect the vertices

Count how far each corner of the triangle is from the line \( y = 3 \).

  • Point (1, 4) is 1 unit above. Reflected point is 1 unit below: (1, 2).
  • Point (2, 4) is 1 unit above. Reflected point is 1 unit below: (2, 2).
  • Point (1, 7) is 4 units above. Reflected point is 4 units below: (1, -1).
y = 3 P

Final Answer:

Triangle with vertices at (1, 2), (2, 2) and (1, -1).

✓✓ (B2)

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Question 19 (2 marks)

Solve \( 4(x – 6) = 44 \)

Worked Solution

Step 1: Divide or Expand

Method 1: Divide first

Since 4 is multiplying the whole bracket, we can divide both sides by 4.

✏ Working:

\[ x – 6 = \frac{44}{4} \] \[ x – 6 = 11 \]
Step 2: Isolate x

✏ Working:

\[ x = 11 + 6 \] \[ x = 17 \]

Alternative Method (Expand):

\( 4x – 24 = 44 \)

\( 4x = 68 \)

\( x = 17 \)

Final Answer:

\( x = 17 \)

✓✓ (M1, A1)

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Question 20 (4 marks)

\( \mathscr{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\} \)

\( A = \{\text{multiples of 3}\} \)

\( B = \{\text{even numbers}\} \)

Complete the Venn diagram for this information.

A B

Worked Solution

Step 1: List the sets

First, write down the members of set A and set B from the universal set.

✏ Working:

\( A = \{3, 6, 9, 12\} \) (Multiples of 3)

\( B = \{2, 4, 6, 8, 10, 12\} \) (Even numbers)

Step 2: Find the Intersection

Find numbers that are in BOTH A and B (Multiples of 3 AND Even).

✏ Working:

\( A \cap B = \{6, 12\} \)

These go in the middle overlap.

Step 3: Fill the rest

Set A only: Numbers in A but not B.

\( \{3, 9\} \)

Set B only: Numbers in B but not A.

\( \{2, 4, 8, 10\} \)

Outside: Numbers in \( \mathscr{E} \) but not in A or B.

Check off numbers used: 2, 3, 4, 6, 8, 9, 10, 12.

Remaining: \( \{1, 5, 7, 11, 13\} \)

A B 6 12 3 9 2 4 8 10 1 5 7 11 13

Final Answer:

Venn diagram completed with {6, 12} in intersection, {3, 9} in A only, {2, 4, 8, 10} in B only, and {1, 5, 7, 11, 13} outside.

✓✓✓✓ (Total: 4 marks)

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Question 21 (3 marks)

Franco buys a house for £146 500.

He sells the house for £158 220.

Calculate the percentage profit Franco makes.

Worked Solution

Step 1: Calculate the Profit

What is profit?

Profit is the difference between the selling price and the cost price.

✏ Working:

\[ 158220 – 146500 = 11720 \]

The profit is £11 720.

Step 2: Calculate Percentage Profit

Formula:

\[ \text{Percentage Profit} = \frac{\text{Profit}}{\text{Original Cost}} \times 100 \]

Always divide by the original amount he bought it for.

✏ Working:

\[ \frac{11720}{146500} \times 100 \]

Using a calculator:

\[ 0.08 \times 100 = 8 \]

Final Answer:

8%

✓✓✓ (M1, M1, A1)

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Question 22 (4 marks)

(a) Expand and simplify \( (x + 5)(x – 9) \)

(2 marks)


(b) Factorise fully \( 9x^2 + 6x \)

(2 marks)

Worked Solution

Part (a): Expand and Simplify

Method: FOIL (First, Outside, Inside, Last)

Multiply each term in the first bracket by each term in the second bracket.

✏ Working:

\( x \times x = x^2 \)

\( x \times -9 = -9x \)

\( 5 \times x = 5x \)

\( 5 \times -9 = -45 \)

Combine: \( x^2 – 9x + 5x – 45 \)

Simplify: \( x^2 – 4x – 45 \)

Part (b): Factorise Fully

Look for common factors in numbers and algebra.

Numbers: 9 and 6 both divide by 3.

Algebra: \( x^2 \) and \( x \) both contain x.

Common factor = \( 3x \).

✏ Working:

Pull out \( 3x \):

\[ 9x^2 \div 3x = 3x \] \[ 6x \div 3x = 2 \]

Result: \( 3x(3x + 2) \)

Final Answer:

(a) \( x^2 – 4x – 45 \)

(b) \( 3x(3x + 2) \)

✓✓✓✓ (Total: 4 marks)

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Question 23 (3 marks)

(a) Use your calculator to work out \( \frac{29^2 – 4.6}{\sqrt{35 – 1.9^3}} \)

Write down all the figures on your calculator display.

(2 marks)


(b) Write your answer to part (a) correct to 4 significant figures.

(1 mark)

Worked Solution

Part (a): Calculator Use

Tip: Calculate the numerator (top) and denominator (bottom) separately if you find entering it all at once difficult, or use the fraction button.

✏ Working:

Numerator: \( 29^2 – 4.6 = 841 – 4.6 = 836.4 \)

Denominator: \( \sqrt{35 – 1.9^3} = \sqrt{35 – 6.859} = \sqrt{28.141} \approx 5.3048… \)

Calculation: \( 836.4 \div 5.3048… \)

Calculator Display: 157.668255…

Part (b): Significant Figures

Count from the first non-zero digit.

1 (1st), 5 (2nd), 7 (3rd), 6 (4th).

Look at the next digit (6) to decide whether to round up.

✏ Working:

157.668…

The 5th digit is 6 (5 or more), so round up the 4th digit.

157.6 becomes 157.7

Final Answer:

(a) 157.668255…

(b) 157.7

✓✓✓ (Total: 3 marks)

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Question 24 (2 marks)

The scatter graph shows information about the marks a group of students got in a Science test and in a Maths test.

Science test mark Maths test mark 0 10 20 30 40 50 60 0 10 20 30 40 50 60

Jamie got a mark of 34 in the Science test.

Using the scatter graph, find an estimate for Jamie’s mark in the Maths test.

Worked Solution

Step 1: Line of Best Fit

Method: Draw a straight line that goes through the middle of the points.

It should follow the general trend of the data (upwards).

34 ~39
Step 2: Read the value

1. Find 34 on the Science (horizontal) axis.

2. Go up to your line of best fit.

3. Go across to the Maths (vertical) axis.

✏ Working:

Reading from the line gives a value between 35 and 42.

Final Answer:

Any answer in the range 35 to 42.

✓✓ (M1, A1)

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Question 25 (3 marks)

The table gives information about the times taken, in seconds, by 18 students to run a race.

Time (t seconds) Frequency
\( 5 < t \le 10 \) 1
\( 10 < t \le 15 \) 2
\( 15 < t \le 20 \) 7
\( 20 < t \le 25 \) 8

Work out an estimate for the mean time.

Give your answer correct to 3 significant figures.

Worked Solution

Step 1: Find Midpoints

We don’t know the exact times, so we use the midpoint of each group.

Midpoint of \( 5 < t \le 10 \) is \( \frac{5+10}{2} = 7.5 \).

Group Midpoint (x) Frequency (f) f × x
5 – 10 7.5 1 7.5
10 – 15 12.5 2 25.0
15 – 20 17.5 7 122.5
20 – 25 22.5 8 180.0
Step 2: Calculate Totals and Mean

Total Frequency: \( 1 + 2 + 7 + 8 = 18 \)

Total Time (\( \sum fx \)): \( 7.5 + 25.0 + 122.5 + 180.0 = 335 \)

✏ Working:

\[ \text{Mean} = \frac{\text{Total Time}}{\text{Total Students}} \] \[ \text{Mean} = \frac{335}{18} \] \[ \text{Mean} = 18.6111… \]
Step 3: Rounding

✏ Working:

Round 18.6111… to 3 significant figures.

First three digits are 1, 8, 6.

Answer: 18.6

Final Answer:

18.6 seconds

✓✓✓ (M1, M1, A1)

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Question 26 (1 mark)

Write \( 37 \text{ cm}^3 \) in \( \text{mm}^3 \).

Worked Solution

Step 1: Understand Volume Conversion

Linear vs Volume:

\( 1 \text{ cm} = 10 \text{ mm} \)

For volume (cubic units), we must cube the conversion factor.

\( 1 \text{ cm}^3 = 10 \times 10 \times 10 \text{ mm}^3 = 1000 \text{ mm}^3 \)

✏ Working:

\[ 37 \times 1000 = 37\,000 \]

Final Answer:

\( 37\,000 \text{ mm}^3 \)

✓ (B1)

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Question 27 (4 marks)

Nimer was driving to a hotel.

He looked at his Sat Nav at 13:30.

Time 13 30
Distance to destination 65 miles

Nimer arrived at the hotel at 14:48.

Work out the average speed of the car from 13:30 to 14:48.

You must show all your working.

Worked Solution

Step 1: Calculate Time Taken

Find the difference between the start time (13:30) and arrival time (14:48).

✏ Working:

From 13:30 to 14:30 is 1 hour.

From 14:30 to 14:48 is 18 minutes.

Total time = 1 hour 18 minutes.

Step 2: Convert Time to Decimals

To use the speed formula, time must be in hours (as a decimal).

Divide the minutes by 60.

✏ Working:

\[ 18 \div 60 = 0.3 \text{ hours} \]

Total time = 1.3 hours.

Step 3: Calculate Speed

Formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)

✏ Working:

\[ \text{Speed} = \frac{65}{1.3} \]

Calculation: \( 65 \div 1.3 = 50 \)

Final Answer:

50 mph

✓✓✓✓ (B1, P1, P1, A1)

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Question 28 (3 marks)

(a) Write 32 460 000 in standard form.

(1 mark)


(b) Write \( 4.96 \times 10^{-3} \) as an ordinary number.

(1 mark)


Asma was asked to compare the following two numbers.

\( A = 6.212 \times 10^8 \quad \text{and} \quad B = 4.73 \times 10^9 \)

She says,

“6.212 is bigger than 4.73 so A is bigger than B.”

(c) Is Asma correct? You must give a reason for your answer.

(1 mark)

Worked Solution

Part (a): Standard Form

Place the decimal after the first non-zero digit (3) and count the places moved.

✏ Working:

3.246…

Moved 7 places to the left.

Answer: \( 3.246 \times 10^7 \)

Part (b): Ordinary Number

The power is negative (-3), so move the decimal 3 places to the left.

✏ Working:

0.00496

Part (c): Comparison

In standard form, the power of 10 is the most important part for size.

Answer: No, Asma is incorrect.

Reason: B has a higher power of 10 (\(10^9\)) than A (\(10^8\)), so B is much larger.

✓ (C1)

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Question 29 (4 marks)

The diagram shows a regular pentagon and a parallelogram.

x 117°

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

You must show all your working.

Worked Solution

Step 1: Interior Angle of a Pentagon

Formula: Sum of interior angles = \( (n-2) \times 180 \)

For a pentagon (\( n=5 \)):

\[ (5-2) \times 180 = 3 \times 180 = 540^\circ \]

Since it is regular, all angles are equal.

\[ 540 \div 5 = 108^\circ \]

So, the total angle at the vertex is \( 108^\circ \).

Step 2: Angles in the Parallelogram

We are given an angle of \( 117^\circ \) in the parallelogram.

Co-interior angles (angles between parallel lines) add to \( 180^\circ \).

The angle of the parallelogram at the vertex shared with the pentagon is:

✏ Working:

\[ 180 – 117 = 63^\circ \]
Step 3: Calculate x

Angle \( x \) and the parallelogram angle (\( 63^\circ \)) make up the full interior angle of the pentagon (\( 108^\circ \)).

✏ Working:

\[ x = 108 – 63 \] \[ x = 45^\circ \]

Final Answer:

\( 45^\circ \)

✓✓✓✓ (P1, P1, P1, A1)

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Question 30 (3 marks)

A is in the shape of a quarter circle of radius 15 cm.

B is in the shape of a circle.

A 15 cm B

The area of A is 9 times the area of B.

Show that the radius of B is 2.5 cm.

Worked Solution

Step 1: Area of Shape A

Shape A is a quarter circle with radius \( r = 15 \).

Area of full circle = \( \pi r^2 \).

Area of quarter circle = \( \frac{\pi r^2}{4} \).

✏ Working:

\[ \text{Area A} = \frac{\pi \times 15^2}{4} \] \[ \text{Area A} = \frac{225\pi}{4} = 56.25\pi \]
Step 2: Find Area of Shape B

We are told Area A is 9 times Area B.

So, \( \text{Area B} = \text{Area A} \div 9 \).

✏ Working:

\[ \text{Area B} = \frac{56.25\pi}{9} \] \[ \text{Area B} = 6.25\pi \]
Step 3: Find Radius of B

Let radius of B be \( r \).

\( \pi r^2 = 6.25\pi \)

Divide both sides by \( \pi \).

✏ Working:

\[ r^2 = 6.25 \] \[ r = \sqrt{6.25} \] \[ r = 2.5 \]

Final Answer:

Radius is 2.5 cm (Shown).

✓✓✓ (M1, M1, C1)

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