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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.
Table of Contents
- Question 1 (Factors)
- Question 2 (Fraction of Amount)
- Question 3 (Decimals to Fractions)
- Question 4 (Multiples)
- Question 5 (Unit Conversion)
- Question 6 (Ratio)
- Question 7 (Substitution)
- Question 8 (Sequences)
- Question 9 (Frequency Table & Bar Chart)
- Question 10 (Circle Parts)
- Question 11 (Money & Percentages)
- Question 12 (Proportion)
- Question 13 (Stem and Leaf)
- Question 14 (Geometry Errors)
- Question 15 (Percentages & Ratio)
- Question 16 (Probability)
- Question 17 (Linear Graphs)
- Question 18 (Reflection)
- Question 19 (Solving Equations)
- Question 20 (Venn Diagrams)
- Question 21 (Percentage Profit)
- Question 22 (Expand & Factorise)
- Question 23 (Calculator Use)
- Question 24 (Scatter Graphs)
- Question 25 (Mean from Frequency Table)
- Question 26 (Volume Conversion)
- Question 27 (Average Speed)
- Question 28 (Standard Form)
- Question 29 (Angles in Polygons)
- Question 30 (Area & Ratio)
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)
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)
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)
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)
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)
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)
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)
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)
Question 9 (6 marks)
Mrs Brown asked each child in her class which pet they liked best.
Here are her results.
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.
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)
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.
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.
Final Answer:
(a) Straight line drawn through the centre.
(b) Chord drawn and the region between chord and arc shaded.
✓✓ (B1, B1)
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)
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)
Question 13 (2 marks)
The stem and leaf diagram below gives information about the ages of people in a social club.
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)
Question 14 (2 marks)
Here is a rectangle.
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.
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)
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)
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)
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.
(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 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | -10 | -6 | -2 | 2 | 6 | 10 |
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)
Question 18 (2 marks)
Reflect shape P in the line \( y = 3 \)
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).
Final Answer:
Triangle with vertices at (1, 2), (2, 2) and (1, -1).
✓✓ (B2)
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)
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.
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: \( \{1, 5, 7, 11, 13\} \)
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)
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)
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)
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)
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.
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).
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)
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)
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)
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)
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)
Question 29 (4 marks)
The diagram shows a regular pentagon and a parallelogram.
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)
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.
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)