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 Nov 2019 Edexcel Foundation Paper 2 (Calculator)
๐ก How to use this Interactive Exam
- Try it first: Attempt each question before looking at the solution.
- Show Solution: Click the green button to reveal the step-by-step worked solution.
- Calculator Paper: You may use a calculator. Key sequences are shown where helpful.
- Diagrams: All diagrams are mathematically accurate vector graphics (SVG).
Table of Contents
- Question 1 (Ordering Numbers)
- Question 2 (Rounding)
- Question 3 (Percentages)
- Question 4 (Calculator Skills)
- Question 5 (Powers)
- Question 6 (Problem Solving)
- Question 7 (Fractions)
- Question 8 (Number Machines)
- Question 9 (Averages & Probability)
- Question 10 (Time Calculations)
- Question 11 (Conversion Graphs)
- Question 12 (Fractions)
- Question 13 (Transformations)
- Question 14 (Best Buy)
- Question 15 (Scale Models)
- Question 16 (Simple Interest)
- Question 17 (Angles in Triangles)
- Question 18 (Algebraic Expressions)
- Question 19 (Ratio & Recipes)
- Question 20 (Formula Substitution)
- Question 21 (Frequency Polygons)
- Question 22 (Error Intervals)
- Question 23 (Ratio)
- Question 24 (Sampling)
- Question 25 (Graph Types)
- Question 26 (Sequences)
- Question 27 (Standard Form)
- Question 28 (Inverse Proportion)
- Question 29 (Surface Area & Volume)
- Question 30 (Vectors)
Question 1 (1 mark)
Write these numbers in order of size.
Start with the smallest number.
\[ 8 \quad -4 \quad 1 \quad -7 \quad -2 \]
Worked Solution
Step 1: Compare the numbers
Why we do this: Negative numbers are smaller than positive numbers. The larger the number after the negative sign, the smaller the value (further left on the number line).
The negative numbers are: \( -4, -7, -2 \)
The positive numbers are: \( 8, 1 \)
Ordering the negatives (smallest to largest): \( -7, -4, -2 \)
Ordering the positives (smallest to largest): \( 1, 8 \)
Final Answer:
\[ -7, -4, -2, 1, 8 \]
Question 2 (1 mark)
Write the number 8375 correct to the nearest thousand.
Worked Solution
Step 1: Identify the column
Why we do this: We need to look at the ‘thousands’ digit and the digit immediately to its right (the hundreds) to decide whether to round up or stay the same.
Number: \( 8375 \)
Thousands digit: \( 8 \)
Hundreds digit (decider): \( 3 \)
Step 2: Apply rounding rule
Rule: If the decider is 5 or more, round up. If it is less than 5, keep the rounding digit the same.
Since \( 3 \) is less than \( 5 \), we round down (keep the 8).
Final Answer:
8000
Question 3 (1 mark)
Write \( 0.23 \) as a percentage.
Worked Solution
Step 1: Convert decimal to percentage
Why we do this: To convert a decimal to a percentage, we multiply by 100.
Final Answer:
23%
Question 4 (1 mark)
Find the value of \( \sqrt{17.64} \)
Worked Solution
Step 1: Use calculator
How: Use the square root button on your calculator.
Calculator steps:
- Press
โ - Type
17.64 - Press
=
Check: Calculate \( 4.2 \times 4.2 = 17.64 \).
Final Answer:
4.2
Question 5 (1 mark)
Find the value of \( 6^5 \)
Worked Solution
Step 1: Calculate power
Why we do this: \( 6^5 \) means multiplying 6 by itself 5 times: \( 6 \times 6 \times 6 \times 6 \times 6 \).
Calculator steps:
- Type
6 - Press the power button (usually
x^yor^orx^โก) - Type
5 - Press
=
Final Answer:
7776
Question 6 (3 marks)
There are 14 rows of seats in a cinema.
There are 15 seats in each row.
A film was shown in the cinema on Saturday.
Each ticket for the film cost ยฃ6.50
The tickets that were sold cost a total of ยฃ1274
How many tickets were not sold?
Worked Solution
Step 1: Find the total number of seats
Why we do this: We need to know the full capacity of the cinema first.
Step 2: Calculate how many tickets were sold
Why we do this: We know the total money made and the cost per ticket. Dividing gives the number of people.
Calculator: 1274 รท 6.5 =
Step 3: Calculate tickets NOT sold
Why we do this: Subtract the sold tickets from the total capacity.
Final Answer:
14 tickets
Question 7 (2 marks)
Harry has 20 sweets.
He gives 7 of the sweets to Nadia.
What fraction of the 20 sweets does Harry have now?
Worked Solution
Step 1: Calculate sweets remaining
Why we do this: Find out how many Harry keeps.
Step 2: Write as a fraction
Why we do this: The question asks for the fraction “of the 20 sweets”. This means 20 is the denominator.
Final Answer:
\[ \frac{13}{20} \]
Question 8 (2 marks)
Here is a number machine.
(a) Work out the output when the input is 6
Here is a different number machine.
When the input is 17, the output is 10
(b) Complete the number machine.
Worked Solution
Part (a): Calculate output
Why we do this: Follow the steps from left to right with the input 6.
Step 1: \( 6 \times 8 = 48 \)
Step 2: \( 48 – 5 = 43 \)
Part (b): Find missing operation
Why we do this: We know the input (17) and the output (10). We need to see what happens in between.
First box: Add 13. \( 17 + 13 = 30 \).
Second box: We need to get from 30 to 10.
To get from 30 to 10, we subtract 20 (since \( 30 – 20 = 10 \)).
Alternatively, we could divide by 3 (since \( 30 \div 3 = 10 \)).
Note: Either \( -20 \) or \( \div 3 \) is acceptable, but subtraction is often the most direct thought.
Final Answer:
(a) 43
(b) \( -20 \) (or \( \div 3 \))
Question 9 (6 marks)
Here is a list of numbers.
\[ 6 \quad 4 \quad 8 \quad 9 \quad 4 \quad 3 \]
(a) Work out the median.
Aisha picks at random one of the numbers.
(b) What is the probability that she picks an odd number?
Clara has five cards.
There is a number on each card.
Two of the numbers are hidden.
The mode of the five numbers is 3
The mean of the five numbers is 5
(c) Work out the two numbers that are hidden.
Worked Solution
Part (a): Median
Method: Put the numbers in order from smallest to largest and find the middle value.
Original list: \( 6, 4, 8, 9, 4, 3 \)
Ordered list: \( 3, 4, 4, 6, 8, 9 \)
There are 6 numbers. The middle is halfway between the 3rd and 4th numbers.
3rd number = 4
4th number = 6
Median = \( \frac{4 + 6}{2} = 5 \)
Part (b): Probability of Odd
Method: Count the odd numbers in the list: \( 3, 4, 4, 6, 8, 9 \)
Odd numbers: 3 and 9.
Total numbers: 6.
(This can be simplified to \( \frac{1}{3} \), but \( \frac{2}{6} \) is correct).
Part (c): Finding hidden numbers
Fact 1 (Mode is 3): 3 must be the most common number. We already have one 3. To make it the mode, we need at least one more 3.
Let’s assume one hidden card is 3.
Fact 2 (Mean is 5): The mean of 5 numbers is 5. This allows us to find the total sum.
Total Sum = Mean \( \times \) Count = \( 5 \times 5 = 25 \).
Current known cards: \( 3, 8, 5 \). Sum = \( 16 \).
Sum of two hidden cards = \( 25 – 16 = 9 \).
We know one hidden card is 3.
So, \( 3 + \text{Other Card} = 9 \).
Other Card = \( 9 – 3 = 6 \).
Check: List is \( 3, 3, 5, 6, 8 \). Mode is 3. Mean is \( 25 \div 5 = 5 \). Correct.
Final Answer:
(a) 5
(b) \( \frac{2}{6} \) (or \( \frac{1}{3} \))
(c) 3 and 6
Question 10 (3 marks)
Here is the charge at a car park in Spain.
Jon parked his car in this car park.
Jon drove into the car park at 10 45
When he drove out of the car park he had to pay 8.40 euros.
At what time did Jon drive out of the car park?
Worked Solution
Step 1: Calculate total minutes parked
Why we do this: Divide the total cost by the cost per minute to find the duration.
Calculator: 8.40 รท 0.024 =
Step 2: Convert minutes to hours and minutes
How: There are 60 minutes in an hour. Divide 350 by 60.
So, 5 hours and 50 minutes.
Step 3: Add time to arrival time
Start time: 10:45
Add 5 hours: 15:45
Add 50 minutes: 15:45 + 50 mins
15:45 + 15 mins = 16:00 (leaving 35 mins)
16:00 + 35 mins = 16:35
Final Answer:
16 35 (or 4:35 pm)
Question 11 (3 marks)
You can use this graph to change between stones and kilograms.
(a) Change 3 stones to kilograms.
(b) Change 80 kilograms to stones.
Worked Solution
Part (a): Reading the graph
Method: Find ‘3’ on the Stones (horizontal) axis. Go up to the line, then go across to the Kilograms (vertical) axis.
At \( x = 3 \), the line is at approximately \( y = 19 \).
(1 stone is roughly 6.35kg, so 3 stones is \( 19.05 \) kg. Reading from the graph, 19 is the expected answer).
Part (b): Extrapolating or Scaling
Problem: The graph only goes up to 70 kg, but we need 80 kg.
Method: We can find the value for 40 kg (or 8 kg) and multiply by 2 (or 10), or use the rate.
From graph: Find 40 kg on the vertical axis.
Go across to the line and down to the stones axis.
40 kg is approximately \( 6.3 \) stones.
Multiply by 2 for 80 kg:
\[ 6.3 \times 2 = 12.6 \text{ stones} \]Alternatively, 80 kg is \( \frac{80}{6.35} \approx 12.6 \).
Acceptable range: 12.4 to 12.8.
Final Answer:
(a) 19 kg
(b) 12.6 stones (approx)
Question 12 (2 marks)
Find the number that is exactly halfway between \( \frac{1}{10} \) and \( \frac{3}{5} \)
Worked Solution
Step 1: Make denominators the same
Why we do this: It is easier to compare and add fractions when they share a common denominator.
So we are finding the midpoint of \( \frac{1}{10} \) and \( \frac{6}{10} \).
Step 2: Find the midpoint
Method: Add the two numbers and divide by 2.
Alternatively, using decimals:
\[ \frac{1}{10} = 0.1, \quad \frac{3}{5} = 0.6 \] \[ \frac{0.1 + 0.6}{2} = \frac{0.7}{2} = 0.35 \]Final Answer:
\[ \frac{7}{20} \text{ or } 0.35 \]
Question 13 (2 marks)
On the grid, draw an enlargement of the shaded shape with a scale factor of 3.
Worked Solution
Step 1: Understand Scale Factor 3
What this means: Every side length of the new shape must be 3 times as long as the original shape.
- Vertical side: Original = 2 units. New = \( 2 \times 3 = 6 \) units.
- Bottom horizontal side: Original = 2 units. New = \( 2 \times 3 = 6 \) units.
- Top slanted side: Original goes “3 right, 1 up”. New goes “9 right, 3 up”.
Draw the new shape anywhere on the grid (unless a centre of enlargement is specified, which it isn’t).
Step 2: Verify the shape
Question 14 (3 marks)
A shop sells compost in 20 litre bags and in 40 litre bags.
One day the shop had two special offers for the compost.
Which offer is the better value for money?
You must show how you get your answer.
Worked Solution
Step 1: Calculate the cost per litre (or per 10 litres) for Offer 1
Offer 1: 2 bags of 20 litres = \( 40 \) litres total.
Cost = ยฃ3.50.
Step 2: Calculate the cost for the same amount for Offer 2
Offer 2: 3 bags of 40 litres = \( 120 \) litres total.
Cost = ยฃ9.00.
To compare fairly, we can find the cost of 40 litres in this deal, or the cost per litre.
Total Volume = 120 litres.
Cost per 40 litres = \( ยฃ9.00 \div 3 = ยฃ3.00 \).
Step 3: Compare
Offer 1 (20L bags): ยฃ3.50 for 40 litres.
Offer 2 (40L bags): ยฃ3.00 for 40 litres.
ยฃ3.00 is less than ยฃ3.50.
Final Answer:
The 40 litre bags (Offer 2) are better value.
Question 15 (3 marks)
The length of a plane is 19.2 metres.
Lukas buys a scale model of the plane.
The scale of the model is 1 : 24
Work out the length of the scale model of the plane.
Give your answer in centimetres.
Worked Solution
Step 1: Convert real length to centimetres
Why we do this: The question asks for the answer in cm, and it’s easier to work with whole numbers.
1 metre = 100 centimetres.
Step 2: Apply the scale factor
Method: A scale of 1 : 24 means the real object is 24 times larger than the model. To find the model size, we divide the real size by 24.
Calculator: 1920 รท 24 = 80
Final Answer:
80 cm
Question 16 (2 marks)
Maria invests ยฃ4500 in a savings account for 3 years.
The account pays simple interest at a rate of 1.8% per year.
Work out the total amount of interest Maria gets by the end of the 3 years.
Worked Solution
Step 1: Calculate interest for one year
Why we do this: Simple interest is calculated on the original amount (principal) only.
Rate = 1.8% = 0.018
Calculator: 4500 ร 0.018 = 81
So, ยฃ81 per year.
Step 2: Calculate total interest for 3 years
Method: Multiply the annual interest by the number of years.
Calculator: 81 ร 3 = 243
Final Answer:
ยฃ243
Question 17 (4 marks)
\(ABC\) is a straight line.
\(AB = BC = BD\).
Angle \(DAB = 64^\circ\)
Work out the size of the angle marked \(x\).
Give a reason for each stage of your working.
Worked Solution
Step 1: Analyze Triangle ABD
Why: We are given \( AB = BD \). This means triangle \( ABD \) is an isosceles triangle.
Reason: Base angles of an isosceles triangle are equal.
So, \( \angle BDA = \angle DAB = 64^\circ \).
Step 2: Find Angle ABD
Reason: Angles in a triangle add up to \( 180^\circ \).
Step 3: Find Angle DBC
Reason: Angles on a straight line add up to \( 180^\circ \).
\( ABC \) is a straight line, so \( \angle ABD \) and \( \angle DBC \) are supplementary.
Step 4: Analyze Triangle BCD to find x
Why: We are given \( BC = BD \). This means triangle \( BCD \) is also an isosceles triangle.
Reason: Base angles of an isosceles triangle are equal.
Let the base angles be \( x \). So \( \angle BCD = x \) and \( \angle BDC = x \).
The sum of angles in triangle \( BCD \) is \( 180^\circ \).
Final Answer:
\( x = 26^\circ \)
Question 18 (4 marks)
Ben is \( n \) years old.
Chloe is twice as old as Ben.
Dan is five years younger than Ben.
The total of Benโs age, Chloeโs age and Danโs age is \( T \) years.
(a) Find a formula for \( T \) in terms of \( n \).
(b) In the table below, put a tick (โ) in the box next to the identity.
Worked Solution
Part (a): Write expressions for each age
Ben: \( n \)
Chloe: “Twice as old as Ben” means \( 2 \times n = 2n \).
Dan: “Five years younger than Ben” means \( n – 5 \).
Part (a): Sum the ages
Why: The question asks for the total age \( T \).
Part (a): Simplify the expression
Part (b): Identify the Identity
Definition: An identity is true for ALL values of the variable.
- \( 3h + 2 = 14 \): Equation (only true for \( h=4 \)).
- \( 3a + 4b – 2c \): Expression (no equals sign).
- \( A = \pi r^2 \): Formula (relates different variables).
- \( 5m – 3m = 2m \): Identity (always true, because 5 apples minus 3 apples is 2 apples).
- \( x + 7 \leq 12 \): Inequality.
Final Answer:
(a) \( T = 4n – 5 \)
(b) \( 5m – 3m = 2m \)
Question 19 (3 marks)
Here are the ingredients needed to make 16 biscuits.
Biscuits
Ingredients to make 16 biscuits
175 g of butter
75 g of sugar
250 g of flour
Anna has
- 500 g of butter
- 300 g of sugar
- 625 g of flour
Work out the greatest number of biscuits Anna can make.
Worked Solution
Step 1: Calculate how many ‘batches’ of each ingredient she has
Why we do this: One batch makes 16 biscuits. We need to see which ingredient runs out first (limiting factor).
Butter: \( 500 \div 175 = 2.857… \) batches
Sugar: \( 300 \div 75 = 4 \) batches
Flour: \( 625 \div 250 = 2.5 \) batches
Step 2: Identify limiting ingredient
The lowest number of batches is 2.5 (determined by the flour). She cannot make more than 2.5 batches.
Step 3: Calculate total biscuits
Total Biscuits = Batches \( \times \) Biscuits per batch
\[ 2.5 \times 16 = 40 \]Final Answer:
40 biscuits
Question 20 (4 marks)
An estimate of the height, \( H \) metres, of a tall building can be found using the formula
\[ H = 4f + 12 \]
where the building is \( f \) floors high.
A tall building is 110 floors high.
The real height of the building is 442 m.
Seb uses the formula to find an estimate of the height of this building.
He then finds the difference between his estimate and the real height.
Show that this difference is less than 5% of the real height.
Worked Solution
Step 1: Calculate the estimated height
Method: Substitute \( f = 110 \) into the formula.
Step 2: Find the difference
Step 3: Calculate 5% of the real height
Why: The question asks to compare the difference to “5% of the real height”.
Real height = 442 m.
Calculator: 0.05 ร 442 = 22.1
Alternatively: \( 10\% = 44.2 \), so \( 5\% = 22.1 \).
Step 4: Compare
Difference = 10 m
5% of real height = 22.1 m
Since \( 10 < 22.1 \), the difference is indeed less than 5%.
Final Answer:
Show calculations: Estimate = 452m, Diff = 10m, 5% of Real = 22.1m. \( 10 < 22.1 \). Valid.
Question 21 (2 marks)
The table shows some information about the weights of 50 potatoes.
| Weight (\(w\) grams) | Frequency |
|---|---|
| \( 10 < w \le 20 \) | 6 |
| \( 20 < w \le 30 \) | 21 |
| \( 30 < w \le 40 \) | 13 |
| \( 40 < w \le 50 \) | 7 |
| \( 50 < w \le 60 \) | 3 |
Iveta drew this frequency polygon for the information in the table.
The frequency polygon is not fully correct.
Write down two things that are wrong with the frequency polygon.
Worked Solution
Step 1: Check the plotting of points
Method: Check each midpoint against the frequency table.
- 10-20 (Midpoint 15): Frequency is 6. Graph shows plotted at 8. (Error 1)
- 20-30 (Midpoint 25): Frequency is 21. Plotted correctly.
Step 2: Check the rules for Frequency Polygons
Rule: Frequency polygons are lines connecting the midpoints at the correct height. They should NOT be closed loops (connected back to the x-axis) unless specifically requested.
The graph connects the first and last points back to the axis, forming a closed shape. (Error 2)
Final Answer:
- The first point (for 10-20) is plotted at a frequency of 8, but it should be 6.
- The polygon should not be closed (it should not be connected to the x-axis at the ends).
Question 22 (2 marks)
The length of a pencil is 128 mm correct to the nearest millimetre.
Complete the error interval for the length of the pencil.
\[ \ldots\ldots\ldots\ldots \le \text{length} < \ldots\ldots\ldots\ldots \]
Worked Solution
Step 1: Identify the rounding unit
Why: “Correct to the nearest millimetre” means the rounding unit is 1 mm.
To find the bounds, we halve the rounding unit: \( 1 \div 2 = 0.5 \).
Step 2: Calculate Upper and Lower Bounds
Lower Bound = \( 128 – 0.5 = 127.5 \)
Upper Bound = \( 128 + 0.5 = 128.5 \)
Step 3: Write the error interval
The error interval includes the lower bound but goes up to (not including) the upper bound.
Final Answer:
\( 127.5 \le \text{length} < 128.5 \)
Question 23 (4 marks)
Tom and Adam have a total of 240 stamps.
The ratio of the number of Tom’s stamps to the number of Adam’s stamps is \( 3:7 \)
Tom buys some stamps from Adam.
The ratio of the number of Tom’s stamps to the number of Adam’s stamps is now \( 3:5 \)
How many stamps does Tom buy from Adam?
You must show all your working.
Worked Solution
Step 1: Calculate original number of stamps
Why: Find how many each person has at the start.
Ratio \( 3:7 \). Total parts = \( 3 + 7 = 10 \).
Value of one part = \( 240 \div 10 = 24 \).
Tom’s original stamps = \( 3 \times 24 = 72 \).
Adam’s original stamps = \( 7 \times 24 = 168 \).
Step 2: Calculate new number of stamps
Fact: When Tom buys stamps from Adam, the total number of stamps (240) stays the same.
New ratio \( 3:5 \). Total parts = \( 3 + 5 = 8 \).
Value of one new part = \( 240 \div 8 = 30 \).
Tom’s new stamps = \( 3 \times 30 = 90 \).
Adam’s new stamps = \( 5 \times 30 = 150 \).
Step 3: Find the difference
Tom started with 72 and ended with 90.
\[ 90 – 72 = 18 \](Check: Adam went from 168 to 150. \( 168 – 150 = 18 \). Correct).
Final Answer:
18 stamps
Question 24 (3 marks)
Each person in a fitness club is going to get a free gift.
Stan is going to order the gifts.
Stan takes a sample of 50 people in the fitness club.
He asks each person to tell him the gift they would like.
The table shows information about his results.
| Gift | Number of people |
|---|---|
| sports bag | 17 |
| gym towel | 7 |
| headphones | 11 |
| voucher | 15 |
There are 700 people in the fitness club.
(i) Work out how many sports bags Stan should order.
(ii) Write down any assumption you made and explain how this could affect your answer.
Worked Solution
Part (i): Scale up the sample
Why: The sample size is 50. The population size is 700.
Scale factor = \( 700 \div 50 = 14 \).
Number wanting sports bag in sample = 17.
\[ 17 \times 14 = 238 \]Alternatively: \( \frac{17}{50} \times 700 = 238 \)
Part (ii): Assumption
Reasoning: Scaling up a sample only works if the sample accurately reflects the whole group.
Assumption: The sample is representative of the whole fitness club.
Effect: If the sample is biased (e.g., asked only young people), the answer might be too high or too low.
Final Answer:
(i) 238
(ii) Assumed sample is representative. If not, the estimate could be wrong.
Question 25 (2 marks)
Here are six graphs.
Write down the letter of the graph that could have the equation
(a) \( y = x^3 \)
(b) \( y = \frac{1}{x} \)
Worked Solution
Part (a): Identify \( y = x^3 \)
Features: A cubic graph of this form goes from bottom-left to top-right and has an inflection point at the origin (0,0). It looks like an ‘S’ shape.
Graph F matches this shape.
Part (b): Identify \( y = \frac{1}{x} \)
Features: This is a reciprocal graph (hyperbola). As \( x \) gets close to 0, \( y \) shoots to infinity. The graph has two separate parts (branches) in the 1st and 3rd quadrants (top-right and bottom-left).
Graph D matches this shape.
Final Answer:
(a) F
(b) D
Question 26 (3 marks)
The \( n \)th term of a sequence is \( 2n^2 – 1 \)
The \( n \)th term of a different sequence is \( 40 – n^2 \)
Show that there is only one number that is in both of these sequences.
Worked Solution
Step 1: Generate terms for the first sequence
Formula: \( 2n^2 – 1 \). Substitute \( n = 1, 2, 3… \)
\( n=1: \quad 2(1)^2 – 1 = 2(1) – 1 = 1 \)
\( n=2: \quad 2(2)^2 – 1 = 2(4) – 1 = 7 \)
\( n=3: \quad 2(3)^2 – 1 = 2(9) – 1 = 17 \)
\( n=4: \quad 2(4)^2 – 1 = 2(16) – 1 = 31 \)
\( n=5: \quad 2(5)^2 – 1 = 2(25) – 1 = 49 \)
\( n=6: \quad 2(6)^2 – 1 = 2(36) – 1 = 71 \)
Sequence 1: \( 1, 7, 17, 31, 49, 71, \dots \)
Step 2: Generate terms for the second sequence
Formula: \( 40 – n^2 \). Substitute \( n = 1, 2, 3… \)
Note: Since we are subtracting \( n^2 \), the terms will decrease and eventually become negative.
\( n=1: \quad 40 – 1^2 = 39 \)
\( n=2: \quad 40 – 2^2 = 36 \)
\( n=3: \quad 40 – 3^2 = 31 \)
\( n=4: \quad 40 – 4^2 = 24 \)
\( n=5: \quad 40 – 5^2 = 15 \)
\( n=6: \quad 40 – 6^2 = 4 \)
\( n=7: \quad 40 – 7^2 = -9 \) (Stop here as Sequence 1 terms are increasing)
Sequence 2: \( 39, 36, 31, 24, 15, 4, -9, \dots \)
Step 3: Find common numbers
Sequence 1: \( 1, 7, 17, \mathbf{31}, 49 \dots \)
Sequence 2: \( 39, 36, \mathbf{31}, 24, 15, 4, \dots \)
The number 31 appears in both.
Since Sequence 2 is decreasing and Sequence 1 is increasing (and all other terms are different), there are no other common numbers.
Final Answer:
Show lists: \( 1, 7, 17, 31, 49… \) and \( 39, 36, 31, 24, 15… \). The only common number is 31.
Question 27 (2 marks)
Work out \( (3.42 \times 10^{-7}) \div (7.5 \times 10^{-6}) \)
Give your answer in standard form.
Worked Solution
Step 1: Perform the division
Method: Divide the numbers first, then divide the powers of 10.
Use your calculator for the whole calculation or split it.
Calculator Input: (3.42 EXP -7) รท (7.5 EXP -6)
Or: \( 3.42 \div 7.5 = 0.456 \)
\( 10^{-7} \div 10^{-6} = 10^{-7 – (-6)} = 10^{-1} \)
Result: \( 0.456 \times 10^{-1} \)
Step 2: Convert to Standard Form
Rule: Standard form must be \( A \times 10^n \) where \( 1 \le A < 10 \).
\( 0.456 \) is not between 1 and 10.
\( 0.456 = 4.56 \times 10^{-1} \)
So, \( 0.456 \times 10^{-1} = (4.56 \times 10^{-1}) \times 10^{-1} \)
\( = 4.56 \times 10^{-2} \)
Final Answer:
\[ 4.56 \times 10^{-2} \]
Question 28 (3 marks)
The number of days, \( d \), that it will take to build a house is given by
\[ d = \frac{720}{n} \]
where \( n \) is the number of workers used each day.
Ali’s company will take 40 days to build the house.
Hayley’s company will take 30 days to build the house.
Hayley’s company will have to use more workers each day than Ali’s company.
How many more?
Worked Solution
Step 1: Calculate workers for Ali
Formula: \( d = \frac{720}{n} \). Rearrange to find \( n \): \( n = \frac{720}{d} \).
Ali’s time \( d = 40 \).
Step 2: Calculate workers for Hayley
Hayley’s time \( d = 30 \).
Step 3: Find the difference
Final Answer:
6 workers
Question 29 (5 marks)
The diagram shows a cube and a cuboid.
The total surface area of the cube is equal to the total surface area of the cuboid.
Janet says,
“The volume of the cube is equal to the volume of the cuboid.”
Is Janet correct?
You must show how you get your answer.
Worked Solution
Step 1: Calculate Surface Area of the Cuboid
Dimensions: \( L=18, W=8, H=6 \).
Formula: \( 2(LW + LH + WH) \)
Area 1 (Top/Bottom): \( 18 \times 8 = 144 \)
Area 2 (Front/Back): \( 18 \times 6 = 108 \)
Area 3 (Sides): \( 8 \times 6 = 48 \)
Total SA = \( 2 \times (144 + 108 + 48) \)
\( = 2 \times 300 = 600 \text{ cm}^2 \)
Step 2: Find side length of the Cube
Fact: A cube has 6 identical square faces.
Let side length = \( x \). Surface Area = \( 6x^2 \).
We know SA = 600.
Step 3: Compare Volumes
Volume of Cube: \( x^3 = 10^3 = 1000 \text{ cm}^3 \)
Volume of Cuboid: \( L \times W \times H = 18 \times 8 \times 6 = 864 \text{ cm}^3 \)
\( 1000 \neq 864 \)
Final Answer:
No, Janet is not correct.
(Volume of Cube = 1000, Volume of Cuboid = 864)
Question 30 (3 marks)
Here are two column vectors.
\[ \mathbf{a} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} \quad \mathbf{b} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} \]
On the grid below, draw and label the vector \( \mathbf{a} – 2\mathbf{b} \)
Worked Solution
Step 1: Calculate the resultant vector
Formula: \( \mathbf{a} – 2\mathbf{b} \)
Step 2: Draw the vector
Instruction: Start anywhere on the grid. Move 1 unit LEFT (because x is -1) and 4 units UP (because y is 4). Draw an arrow.
Final Answer:
Vector drawn: \( \begin{pmatrix} -1 \\ 4 \end{pmatrix} \)