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GCSE Nov 2021 Edexcel Foundation Paper 3 (Calculator)
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Table of Contents
- Question 1 (Percentages)
- Question 2 (Factors)
- Question 3 (Time)
- Question 4 (Fractions)
- Question 5 (Midpoints)
- Question 6 (Algebraic Simplification)
- Question 7 (Combinations)
- Question 8 (Problem Solving)
- Question 9 (Two-way Tables)
- Question 10 (Best Buy)
- Question 11 (Speed/Time)
- Question 12 (Scale Drawings)
- Question 13 (Ratios and Percentages)
- Question 14 (Money and Fractions)
- Question 15 (Angles)
- Question 16 (Transformations)
- Question 17 (Algebra: Solving & Simplifying)
- Question 18 (Unit Conversion)
- Question 19 (Area of Squares)
- Question 20 (Stem and Leaf)
- Question 21 (Scatter Graphs)
- Question 22 (Proportion)
- Question 23 (Standard Form)
- Question 24 (Rate of Flow)
- Question 25 (Fibonacci Sequences)
- Question 26 (Probability)
- Question 27 (Area of Sector)
- Question 28 (Sketching Graphs)
Question 1 (1 mark)
Write 45% as a decimal.
Worked Solution
Step 1: Understanding the conversion
What does percent mean?
“Percent” means “out of 100”. To convert a percentage to a decimal, we divide the number by 100.
✏ Working:
\[ 45\% = \frac{45}{100} \] \[ 45 \div 100 = 0.45 \]Final Answer:
0.45
Question 2 (1 mark)
Write down two factors of 35.
Worked Solution
Step 1: Understanding factors
What is a factor?
A factor is a whole number that divides into another number exactly (without leaving a remainder).
✏ Working:
Let’s find pairs of numbers that multiply to make 35:
\[ 1 \times 35 = 35 \] \[ 5 \times 7 = 35 \]The factors of 35 are: 1, 5, 7, 35.
Final Answer:
Any two of: 1, 5, 7, 35
Question 3 (1 mark)
What is the time 2 hours 40 minutes after 8.05 am?
Worked Solution
Step 1: Adding the hours
Start by adding the full hours to the starting time.
✏ Working:
Start time: 8:05 am
\[ 8:05 + 2 \text{ hours} = 10:05 \text{ am} \]Step 2: Adding the minutes
Now add the 40 minutes to the result.
✏ Working:
\[ 10:05 + 40 \text{ minutes} = 10:45 \text{ am} \]Final Answer:
10:45 am
Question 4 (1 mark)
Work out \( \frac{1}{6} \) of 66.
Worked Solution
Step 1: Calculation
How to find a fraction of an amount:
Finding \( \frac{1}{6} \) of a number is the same as dividing that number by 6.
✏ Working:
\[ 66 \div 6 = 11 \]Final Answer:
11
Question 5 (1 mark)
\( AB \) is a straight line.
Mark with a cross (\( \times \)) the midpoint of \( AB \).
Worked Solution
Step 1: Finding the midpoint
Visual estimation:
The midpoint is the point exactly halfway between A and B. You can use a ruler to measure the total length and divide by 2 to find the exact position.
✏ Working:
Final Answer:
Cross marked in the center of the line (see diagram above).
Question 6 (3 marks)
(a) Simplify \( a \times b \times 4 \)
(b) Simplify \( 4x + 3 – x + 5 \)
Worked Solution
Part (a): Multiplying terms
Algebra convention:
When multiplying variables and numbers, we write the number first, followed by the letters in alphabetical order. We remove the multiplication signs.
✏ Working:
\[ a \times b \times 4 = 4ab \]Part (b): Collecting like terms
Strategy:
Group the terms with \( x \) together and the number terms together. Pay attention to the signs in front of each term.
✏ Working:
Terms with \( x \): \( 4x – x = 3x \)
Number terms: \( +3 + 5 = +8 \)
Combine them:
\[ 3x + 8 \]Final Answer:
(a) \( 4ab \)
(b) \( 3x + 8 \)
Question 7 (2 marks)
There are three cards in bag A and two cards in bag B. There is a letter on each card.
James takes a card from bag A and then a card from bag B.
List all the possible outcomes.
Worked Solution
Step 1: Systematic Listing
Strategy:
To make sure we don’t miss any, let’s take each card from Bag A and pair it with every possible card from Bag B.
✏ Working:
If we pick E from A: (E, J), (E, K)
If we pick F from A: (F, J), (F, K)
If we pick G from A: (G, J), (G, K)
Final Answer:
EJ, EK, FJ, FK, GJ, GK
Question 8 (3 marks)
On Monday, Sandy pays for 2 plane tickets, 7 nights in a hotel and 2 theme park tickets.
| Item | Cost (dollars) |
|---|---|
| each plane ticket | 600 |
| each night in a hotel | 120 |
| each theme park ticket | 250 |
Show that Sandy pays more than 2500 dollars on Monday.
Worked Solution
Step 1: Calculate the cost for each category
Multiply the number of items by the cost per item.
✏ Working:
Plane tickets: \( 2 \times 600 = 1200 \)
Hotel nights: \( 7 \times 120 = 840 \)
Theme park tickets: \( 2 \times 250 = 500 \)
Step 2: Calculate total cost
✏ Working:
\[ \text{Total} = 1200 + 840 + 500 \] \[ \text{Total} = 2540 \text{ dollars} \]Step 3: Conclusion
Compare the total with 2500.
✏ Working:
\( 2540 > 2500 \)
Sandy pays 2540 dollars, which is more than 2500 dollars.
Final Answer:
Total = 2540 dollars. (Shown working above)
Question 9 (3 marks)
Vadim has 56 clocks.
The clocks are only red, only blue or only black.
- 32 of the clocks are plastic.
- 5 of the 14 blue clocks are plastic.
- 8 of the 12 red clocks are not plastic.
Use this information to complete the two-way table.
| Red | Blue | Black | Total | |
|---|---|---|---|---|
| Plastic | ||||
| Not plastic | ||||
| Total |
Worked Solution
Step 1: Enter the given information
Fill in the cells directly provided by the text.
✏ Working:
- Total clocks = 56 (Grand Total)
- Plastic Total = 32
- Blue Total = 14
- Blue Plastic = 5
- Red Total = 12
- Red Not Plastic = 8
Step 2: Calculate missing values
Use subtraction to fill the gaps. Rows and columns must sum to their totals.
✏ Working:
Red: Red Plastic = Total Red (12) – Red Not Plastic (8) = 4
Blue: Blue Not Plastic = Total Blue (14) – Blue Plastic (5) = 9
Plastic Row: Black Plastic = Total Plastic (32) – Red Plastic (4) – Blue Plastic (5) = 32 – 9 = 23
Not Plastic Row: First, find Total Not Plastic = Grand Total (56) – Total Plastic (32) = 24.
Now, Black Not Plastic = Total Not Plastic (24) – Red Not Plastic (8) – Blue Not Plastic (9) = 24 – 17 = 7
Black Column: Total Black = Black Plastic (23) + Black Not Plastic (7) = 30
Final Answer:
| Red | Blue | Black | Total | |
|---|---|---|---|---|
| Plastic | 4 | 5 | 23 | 32 |
| Not plastic | 8 | 9 | 7 | 24 |
| Total | 12 | 14 | 30 | 56 |
Question 10 (3 marks)
Corina has £300 to spend on books.
Each book costs £4.85.
Work out the greatest number of books Corina can buy.
Worked Solution
Step 1: Division
Divide the total money by the cost of one book to find how many fit into the budget.
✏ Working:
\[ 300 \div 4.85 \]Calculator: 300 ÷ 4.85 = 61.8556…
Step 2: Interpreting the result
You cannot buy a fraction of a book. Corina must buy a whole number of books. We must round down because she doesn’t have enough money for the 62nd book.
✏ Working:
61.855… means she can buy 61 books.
Check: \( 61 \times 4.85 = 295.85 \) (Affordable)
Check: \( 62 \times 4.85 = 300.70 \) (Too expensive)
Final Answer:
61
Question 11 (3 marks)
(a) Write 196 minutes in hours and minutes.
A train travels \( x \) miles in 2 hours.
(b) Write down an expression, in terms of \( x \), for the average speed of the train.
Worked Solution
Part (a): Converting minutes
Strategy:
There are 60 minutes in 1 hour. We need to find how many 60s fit into 196.
✏ Working:
\[ 196 \div 60 = 3 \text{ remainder } 16 \]Or:
\[ 60 + 60 + 60 = 180 \text{ (3 hours)} \] \[ 196 – 180 = 16 \text{ minutes} \]Part (b): Speed expression
Formula:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
✏ Working:
Distance = \( x \) miles
Time = 2 hours
\[ \text{Speed} = \frac{x}{2} \]Final Answer:
(a) 3 hours 16 minutes
(b) \( \frac{x}{2} \) miles per hour
Question 12 (4 marks)
The diagram shows two places on a map.
Scale: 1 centimetre represents 20 kilometres.
(a) What is the actual distance, in kilometres, from Shelton to Trilby?
On a scale drawing, the scale is given as \( 1 : 1200 \).
(b) How many metres does 5 centimetres represent on this drawing?
Worked Solution
Part (a): Using the map scale
Strategy:
First, we measure the distance on the map (shown as 2.5 cm in the diagram). Then multiply by the scale factor.
✏ Working:
Measurement = 2.5 cm
Scale: 1 cm = 20 km
\[ 2.5 \times 20 = 50 \]Part (b): Using the ratio scale
Understanding 1:1200:
This means 1 cm on the drawing is 1200 cm in real life.
✏ Working:
Multiply the drawing length by the scale ratio:
\[ 5 \text{ cm} \times 1200 = 6000 \text{ cm} \]Convert to metres (divide by 100):
\[ 6000 \div 100 = 60 \text{ metres} \]Final Answer:
(a) 50 km (Accept 46-54 km depending on measurement)
(b) 60 metres
Question 13 (4 marks)
In the Northern hemisphere the ratio of the area of land to the area of water is \( 2 : 3 \).
(a) Work out what percentage of the area of the Northern hemisphere is land.
20% of the area of the Southern hemisphere is land.
(b) Work out the ratio of the area of land to the area of water in the Southern hemisphere.
Worked Solution
Part (a): Ratio to Percentage
Strategy:
Total parts in the ratio \( 2:3 \) is \( 2+3=5 \). Land is 2 parts out of 5.
✏ Working:
Fraction of land = \( \frac{2}{5} \)
Convert to percentage:
\[ \frac{2}{5} \times 100 = 40\% \]Part (b): Percentage to Ratio
Strategy:
If land is 20%, then water must be the rest (100% – 20%).
✏ Working:
Water = \( 100\% – 20\% = 80\% \)
Ratio of Land : Water = \( 20 : 80 \)
Simplify by dividing both sides by 20:
\[ 1 : 4 \]Final Answer:
(a) 40%
(b) 1 : 4 (or 20 : 80)
Question 14 (3 marks)
A stadium cost £600 million.
\( \frac{13}{15} \) of this cost was for the building.
The rest of the cost was for the land.
Work out the cost of the land.
Worked Solution
Step 1: Find the fraction for the land
The whole cost is 1. If \( \frac{13}{15} \) is for the building, the remaining fraction is for the land.
✏ Working:
\[ 1 – \frac{13}{15} = \frac{2}{15} \]Step 2: Calculate the cost
Find \( \frac{2}{15} \) of 600 million.
✏ Working:
\[ 600 \div 15 = 40 \] \[ 40 \times 2 = 80 \]Final Answer:
£80 million
Question 15 (1 mark)
Jenna measures all the angles around a point.
Her results are \( 23^\circ, 145^\circ, 23^\circ \) and \( 69^\circ \).
Explain why these results cannot be true.
Worked Solution
Step 1: Check the total
Key Fact: Angles around a point must add up to \( 360^\circ \).
✏ Working:
\[ 23 + 145 + 23 + 69 = 260 \]Step 2: Conclusion
✏ Working:
The angles add up to 260, but they should add up to 360.
Final Answer:
They sum to 260° but should sum to 360°.
Question 16 (2 marks)
Here is a diagram showing triangle \( ABC \) and triangle \( ADE \).
Describe fully the single transformation that maps triangle \( ABC \) onto triangle \( ADE \).
Worked Solution
Step 1: Identify the type of transformation
The shape has changed size (it has got bigger), but not orientation. This is an Enlargement.
Step 2: Find the Scale Factor
Compare the lengths of corresponding sides.
- Side \( AC \) has length 1 unit (from \( x=1 \) to \( x=2 \)).
- Side \( AE \) has length 4 units (from \( x=1 \) to \( x=5 \)).
Scale Factor = \( \frac{\text{New Length}}{\text{Old Length}} = \frac{4}{1} = 4 \).
Step 3: Find the Centre of Enlargement
Draw lines through corresponding points (e.g., \( B \) to \( D \) and \( C \) to \( E \)). They all meet at the centre.
In this case, point \( A \) is shared by both triangles and stays in the same place. Therefore, \( A(1, 1) \) is the centre.
✏ Working:
Lines connect through \( (1,1) \).
Final Answer:
Enlargement, scale factor 4, centre (1, 1).
Question 17 (7 marks)
(a) Expand \( y(y + 5) \)
(b) Factorise \( 4a – 6 \)
(c) Solve \( 2(5x – 4) = 21 \)
(d) Simplify \( 4e^2f \times 5ef^3 \)
Worked Solution
Part (a): Expand
Multiply the term outside the bracket by each term inside.
✏ Working:
\[ y \times y = y^2 \] \[ y \times 5 = 5y \] \[ y^2 + 5y \]Part (b): Factorise
Find the highest common factor (HCF) of 4 and 6.
✏ Working:
HCF is 2.
\[ 4a \div 2 = 2a \] \[ -6 \div 2 = -3 \] \[ 2(2a – 3) \]Part (c): Solve
First expand the brackets, then rearrange to find \( x \).
✏ Working:
\[ 2(5x – 4) = 21 \] \[ 10x – 8 = 21 \]Add 8 to both sides:
\[ 10x = 29 \]Divide by 10:
\[ x = 2.9 \]Part (d): Simplify Indices
Multiply numbers together, and add powers for matching letters.
✏ Working:
Numbers: \( 4 \times 5 = 20 \)
e terms: \( e^2 \times e^1 = e^{2+1} = e^3 \)
f terms: \( f^1 \times f^3 = f^{1+3} = f^4 \)
\[ 20e^3f^4 \]Final Answer:
(a) \( y^2 + 5y \)
(b) \( 2(2a – 3) \)
(c) \( x = 2.9 \)
(d) \( 20e^3f^4 \)
Question 18 (1 mark)
Change \( 1\text{m}^2 \) into \( \text{cm}^2 \).
Worked Solution
Step 1: Understanding Area Conversion
Common Mistake: It is NOT 100.
Imagine a square that is 1m by 1m.
In cm, this square is 100cm by 100cm.
✏ Working:
\[ \text{Area} = 100\text{cm} \times 100\text{cm} \] \[ \text{Area} = 10,000\text{cm}^2 \]Final Answer:
10,000 \( \text{cm}^2 \)
Question 19 (4 marks)
This diagram shows two squares.
Work out the area of the square shown shaded in the diagram.
Worked Solution
Method 1: Subtraction
Calculate the area of the large outer square and subtract the area of the four white triangles.
Side of large square = \( 3 + 5 = 8 \text{ cm} \).
✏ Working:
Area of large square: \( 8 \times 8 = 64 \text{ cm}^2 \)
Area of one triangle: \( \frac{1}{2} \times 3 \times 5 = 7.5 \text{ cm}^2 \)
Area of 4 triangles: \( 4 \times 7.5 = 30 \text{ cm}^2 \)
Shaded Area: \( 64 – 30 = 34 \text{ cm}^2 \)
Method 2: Pythagoras
The side of the shaded square is the hypotenuse of the right-angled triangle.
✏ Working:
\[ a^2 + b^2 = c^2 \] \[ 3^2 + 5^2 = c^2 \] \[ 9 + 25 = c^2 \] \[ c^2 = 34 \]The area of a square is side squared (\( c^2 \)), so the area is 34.
Final Answer:
34 \( \text{cm}^2 \)
Question 20 (3 marks)
Here are the heights, in centimetres, of 15 plants.
15, 20, 25, 33, 17, 22, 25, 18, 22, 19, 32, 35, 24, 28, 19
Draw a stem and leaf diagram for these heights.
Worked Solution
Step 1: Order the data
It is easier to fill the diagram if we list the numbers in order first.
✏ Working:
15, 17, 18, 19, 19
20, 22, 22, 24, 25, 25, 28
32, 33, 35
Step 2: Draw the Stem and Leaf
The “Stem” is the tens digit. The “Leaf” is the units digit. Leaves must be ordered and spaced evenly.
✏ Working:
Final Answer:
See diagram above. Remember to include the Key.
Question 21 (4 marks)
The scatter graph shows information about the volume of traffic and the carbon monoxide level at a point on a road each day for 22 days.
One point is an outlier.
(a) Write down the coordinates of this point.
For another day, 370 cars pass the point on the road.
(b) Estimate the carbon monoxide level for this day.
Alfie says, “Because there is an outlier, there is no correlation.”
(c) Is Alfie correct? You must give a reason for your answer.
Worked Solution
Part (a): Identifying the Outlier
An outlier is a point that does not fit the pattern of the other points. Look for the point that is far away from the main group.
There is a point high up on the left at (100, 18) which does not fit the upward trend.
Answer: (100, 18)
Part (b): Estimation using Line of Best Fit
Method:
- Draw a straight line through the middle of the main group of points (ignoring the outlier).
- Go to 370 on the x-axis (Traffic).
- Go up to your line, then across to the y-axis (CO level).
✏ Working:
A Line of Best Fit would pass roughly through (200, 8) and (400, 14).
At \( x = 370 \), the line is typically between 12.8 and 14.8.
Answer: Approx 13.5 mg/m³ (Range 12.8 – 14.8 accepted)
Part (c): Understanding Correlation
Reasoning:
Correlation describes the general trend of the data. One “bad” point (outlier) does not mean the trend doesn’t exist.
The rest of the points clearly show that as traffic increases, CO level increases (positive correlation).
Answer: No, Alfie is incorrect. Ignoring the outlier, there is a clear positive correlation.
Question 22 (4 marks)
Natalie makes potato cakes in a restaurant.
She mixes potato, cheese and onion so that
weight of potato : weight of cheese : weight of onion = \( 9 : 2 : 1 \)
Natalie needs to make 6000g of potato cakes.
Cheese costs £2.25 for 175g.
Work out the cost of the cheese needed to make 6000g of potato cakes.
Worked Solution
Step 1: Calculate the amount of cheese needed
First, find the total number of parts in the ratio.
✏ Working:
\[ 9 + 2 + 1 = 12 \text{ parts} \]Cheese is 2 parts out of 12.
\[ \text{Weight of cheese} = \frac{2}{12} \times 6000 \]Calculator: 2 ÷ 12 × 6000 = 1000g
Step 2: Calculate how many packs of cheese to buy
Cheese is sold in packs of 175g. We need to find how many packs cover 1000g.
✏ Working:
\[ 1000 \div 175 = 5.714… \]You cannot buy 0.7 of a pack, so Natalie must buy 6 packs.
Step 3: Calculate total cost
✏ Working:
\[ 6 \text{ packs} \times £2.25 \]Calculator: 6 × 2.25 = 13.50
Final Answer:
£13.50
Question 23 (4 marks)
(a) Write \( 4.5 \times 10^5 \) as an ordinary number.
(b) Write 0.007 in standard form.
(c) Work out \( 4.2 \times 10^3 + 5.3 \times 10^2 \)
Give your answer in standard form.
Worked Solution
Part (a): Standard Form to Ordinary
Move the decimal point 5 places to the right.
✏ Working:
\[ 4.5 \times 100,000 = 450,000 \]Part (b): Ordinary to Standard Form
Move the decimal point so the number is between 1 and 10, then count the jumps. Moving right means a negative power.
✏ Working:
\[ 0.007 = 7 \times 10^{-3} \]Part (c): Addition in Standard Form
Convert both to ordinary numbers first, add them, then convert back. Or use your calculator’s standard form button (usually \( \times 10^x \) or EXP).
✏ Working:
\[ 4.2 \times 10^3 = 4200 \] \[ 5.3 \times 10^2 = 530 \] \[ 4200 + 530 = 4730 \]Convert 4730 to standard form:
\[ 4.73 \times 10^3 \]Final Answer:
(a) 450 000
(b) \( 7 \times 10^{-3} \)
(c) \( 4.73 \times 10^3 \)
Question 24 (4 marks)
A water tank is empty.
Anil needs to fill the tank with 2400 litres of water.
- Company A supplies water at a rate of 8 litres in 1 minute 40 seconds.
- Company B supplies water at a rate of 2.2 gallons per minute.
1 gallon = 4.54 litres
Company A would take more time to fill the tank than Company B would take to fill the tank.
How much more time?
Give your answer in minutes correct to the nearest minute.
Worked Solution
Step 1: Rate for Company A
Convert the time to seconds or decimal minutes.
1 min 40 sec = 100 seconds.
✏ Working:
Rate A = 8 litres per 100 seconds.
Total needed = 2400 litres.
Number of “8 litre” chunks = \( 2400 \div 8 = 300 \).
Time for A = \( 300 \times 100 \) seconds = 30,000 seconds.
Convert to minutes: \( 30,000 \div 60 = 500 \) minutes.
Step 2: Rate for Company B
Convert gallons to litres first.
✏ Working:
Rate B = 2.2 gallons/min.
In litres: \( 2.2 \times 4.54 = 9.988 \) litres/min.
Time for B = \( \frac{2400}{9.988} \)
Calculator: 2400 ÷ 9.988 = 240.288… minutes.
Step 3: Difference in time
✏ Working:
\[ \text{Difference} = 500 – 240.288… \] \[ \text{Difference} = 259.71… \]Round to nearest minute: 260.
Final Answer:
260 minutes
Question 25 (3 marks)
The first four terms of a Fibonacci sequence are
\( a \quad 2a \quad 3a \quad 5a \)
The sum of the first five terms of this sequence is 228.
Work out the value of \( a \).
Worked Solution
Step 1: Find the 5th term
Fibonacci Rule: Each term is the sum of the two previous terms.
✏ Working:
1st: \( a \)
2nd: \( 2a \)
3rd: \( a + 2a = 3a \)
4th: \( 2a + 3a = 5a \)
5th: \( 3a + 5a = 8a \)
Step 2: Form an equation
The sum of the first five terms is 228.
✏ Working:
\[ a + 2a + 3a + 5a + 8a = 228 \] \[ 19a = 228 \]Step 3: Solve for a
✏ Working:
\[ a = 228 \div 19 \]Calculator: 228 ÷ 19 = 12
Final Answer:
12
Question 26 (4 marks)
In a bag there are only red counters, blue counters, green counters and pink counters.
A counter is going to be taken at random from the bag.
The table shows the probabilities of taking a red counter or a blue counter.
| Colour | red | blue | green | pink |
|---|---|---|---|---|
| Probability | 0.05 | 0.15 | ……. | ……. |
The probability of taking a green counter is 0.2 more than the probability of taking a pink counter.
(a) Complete the table.
There are 18 blue counters in the bag.
(b) Work out the total number of counters in the bag.
Worked Solution
Part (a): Using Algebra for Probability
Key Concept: The sum of all probabilities must be 1.
Let the probability of Pink be \( x \).
Then the probability of Green is \( x + 0.2 \).
✏ Working:
\[ 0.05 + 0.15 + (x + 0.2) + x = 1 \] \[ 0.4 + 2x = 1 \] \[ 2x = 0.6 \] \[ x = 0.3 \]So, Pink = 0.3
Green = 0.3 + 0.2 = 0.5
Check: \( 0.05 + 0.15 + 0.5 + 0.3 = 1.0 \) (Correct)
Part (b): Finding the Total
We know that 0.15 of the total is 18 counters.
✏ Working:
\[ 0.15 \times \text{Total} = 18 \] \[ \text{Total} = 18 \div 0.15 \]Calculator: 18 ÷ 0.15 = 120
Final Answer:
(a) Green: 0.5, Pink: 0.3
(b) 120
Question 27 (4 marks)
The diagram shows a sector \( OPQR \) of a circle, centre \( O \) and radius 8 cm.
\( OPR \) is a triangle.
Work out the area of the shaded segment \( PQR \).
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Understand the Area of a Segment
The area of the shaded segment is the Area of the Sector minus the Area of the Triangle.
Step 2: Area of Sector
The angle at \( O \) is \( 90^\circ \) (indicated by the square symbol).
\( \text{Fraction of circle} = \frac{90}{360} = \frac{1}{4} \)
✏ Working:
\[ \text{Area}_{\text{sector}} = \frac{1}{4} \times \pi \times r^2 \] \[ \text{Area}_{\text{sector}} = \frac{1}{4} \times \pi \times 8^2 \] \[ \text{Area}_{\text{sector}} = 16\pi \approx 50.265… \]Step 3: Area of Triangle
Triangle \( OPR \) is a right-angled triangle with base 8 and height 8.
✏ Working:
\[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ \text{Area}_{\text{triangle}} = 0.5 \times 8 \times 8 = 32 \]Step 4: Subtraction
✏ Working:
\[ \text{Segment} = 50.265… – 32 = 18.265… \]Round to 3 significant figures:
\[ 18.3 \]Final Answer:
18.3 \( \text{cm}^2 \)
Question 28 (2 marks)
Sketch the graph of \( y = \frac{1}{x} \).
Worked Solution
Step 1: Recognizing the Reciprocal Graph
The function \( y = \frac{1}{x} \) is a reciprocal graph.
Key Features:
- It has two separate curves (hyperbolas).
- One curve is in the top-right quadrant (positive \( x \), positive \( y \)).
- One curve is in the bottom-left quadrant (negative \( x \), negative \( y \)).
- The curves get closer and closer to the axes but never touch them (asymptotes).
✏ Working:
Final Answer:
Two curves in the 1st and 3rd quadrants approaching the axes.