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GCSE Mathematics Paper 2 (Calculator) Higher Tier 2020
๐ก How to use this Interactive Exam
- Try it first: Solve the problem on paper before checking the solution.
- Calculator Allowed: This is a calculator paper. Show your button sequences.
- Three-Layer Solutions: We explain Why, then How, then Check.
- Diagrams: All visual elements are redrawn for clarity.
๐ Table of Contents
- Question 1 (Number – Prime Factors & LCM)
- Question 2 (Probability – Venn Diagrams)
- Question 3 (Ratio & Proportion)
- Question 4 (Algebra – Cubic Graphs)
- Question 5 (Trigonometry)
- Question 6 (Vectors)
- Question 7 (Geometry – Area)
- Question 8 (Percentages & Compound Interest)
- Question 9 (Statistics – Box Plots)
- Question 10 (Algebra – Indices)
- Question 11 (Combinatorics)
- Question 12 (Real Life Graphs)
- Question 13 (Geometry – Sine Rule)
- Question 14 (Algebra – Squares)
- Question 15 (Transformations)
- Question 16 (Sequences)
- Question 17 (Algebra – Turning Points)
- Question 18 (Geometry – Cones)
- Question 19 (Iteration)
- Question 20 (Probability)
- Question 21 (Functions & Graphs)
- Question 22 (Circle Geometry)
Question 1 (4 marks)
(a) Write \( 84 \) as a product of its prime factors.
(b) Find the lowest common multiple (LCM) of \( 60 \) and \( 84 \).
Worked Solution
Part (a): Prime Factors of 84
Why we do this: We need to break the number 84 down into its basic building blocks (prime numbers) that multiply together to make it.
Method: Factor Tree
We split 84 into factors until we reach prime numbers (circled).
- \( 84 = 2 \times 42 \)
- \( 42 = 2 \times 21 \)
- \( 21 = 3 \times 7 \)
The prime factors are 2, 2, 3, and 7.
Written as a product: \( 2 \times 2 \times 3 \times 7 \)
In index form: \( 2^2 \times 3 \times 7 \)
Part (b): LCM of 60 and 84
Why we do this: The LCM is the smallest number that both 60 and 84 divide into. We can use the prime factorization of both numbers to find it.
First, write 60 as a product of prime factors:
- \( 60 = 6 \times 10 = (2 \times 3) \times (2 \times 5) = 2^2 \times 3 \times 5 \)
We already have \( 84 = 2^2 \times 3 \times 7 \)
To find the LCM, we take the highest power of each prime factor present in either list:
- \( 2^2 \) (highest power of 2)
- \( 3^1 \) (highest power of 3)
- \( 5^1 \) (highest power of 5)
- \( 7^1 \) (highest power of 7)
LCM = \( 2^2 \times 3 \times 5 \times 7 \)
Calculation: \( 4 \times 3 \times 5 \times 7 = 12 \times 35 = 420 \)
Final Answer:
(a) \( 2^2 \times 3 \times 7 \)
(b) \( 420 \)
Question 2 (5 marks)
\( \mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)
\( A = \{\text{even numbers}\} \)
\( B = \{\text{factors of 10}\} \)
(a) Complete the Venn diagram for this information.
A number is chosen at random from the universal set, \( \mathcal{E} \).
(b) Find the probability that this number is in the set \( A \cap B \).
Worked Solution
Part (a): Completing the Venn Diagram
Why we do this: We need to list the members of each set and place them in the correct region of the diagram (Set A only, Set B only, Intersection, or Outside).
Lists:
- \( \mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)
- \( A (\text{even}) = \{2, 4, 6, 8, 10\} \)
- \( B (\text{factors of 10}) = \{1, 2, 5, 10\} \)
Intersection \( A \cap B \) (In both lists): \( \{2, 10\} \)
Set A only (In A but not B): \( \{4, 6, 8\} \)
Set B only (In B but not A): \( \{1, 5\} \)
Outside (In \( \mathcal{E} \) but not A or B): \( \{3, 7, 9\} \)
Correct Diagram:
Part (b): Probability of \( A \cap B \)
Why we do this: The probability is the number of items in the target region divided by the total number of items in the universal set.
Number of items in \( \mathcal{E} \) = 10
Number of items in \( A \cap B \) (the intersection) = 2 (the numbers 2 and 10)
Probability = \( \frac{2}{10} \)
Simplified = \( \frac{1}{5} \)
Final Answer:
(a) See diagram above.
(b) \( \frac{2}{10} \) or \( \frac{1}{5} \) or \( 0.2 \)
Question 3 (5 marks)
Carlo puts tins into small boxes and into large boxes.
He puts 6 tins into each small box.
He puts 20 tins into each large box.
Carlo puts a total of 3000 tins into the boxes so that
number of tins in small boxes : number of tins in large boxes = \( 2 : 3 \)
Carlo says that less than 30% of the boxes filled with tins are large boxes.
Is Carlo correct?
You must show all your working.
Worked Solution
Step 1: Calculate tins in each type of box
Why we do this: The ratio \( 2:3 \) applies to the number of tins, not the boxes. We need to split the total 3000 tins according to this ratio.
Total parts in ratio = \( 2 + 3 = 5 \)
Value of one part = \( 3000 \div 5 = 600 \text{ tins} \)
- Tins in small boxes = \( 2 \times 600 = 1200 \text{ tins} \)
- Tins in large boxes = \( 3 \times 600 = 1800 \text{ tins} \)
Check: \( 1200 + 1800 = 3000 \). Correct.
Step 2: Calculate number of boxes
Why we do this: We know the capacity of each box type (6 for small, 20 for large), so we can find how many boxes were used.
Number of small boxes = \( \frac{\text{Total tins in small}}{\text{Capacity}} = \frac{1200}{6} = 200 \text{ boxes} \)
Number of large boxes = \( \frac{\text{Total tins in large}}{\text{Capacity}} = \frac{1800}{20} = 90 \text{ boxes} \)
Total boxes = \( 200 + 90 = 290 \text{ boxes} \)
Step 3: Calculate percentage of large boxes
Percentage of large boxes = \( \frac{\text{Number of large boxes}}{\text{Total boxes}} \times 100 \)
\( \text{Percentage} = \frac{90}{290} \times 100 \)
Calculator Steps: 90 รท 290 ร 100 = 31.034...
Percentage \( \approx 31\% \)
Step 4: Conclusion
What this tells us: Carlo said less than 30% are large boxes. We found it is 31%.
\( 31\% \) is greater than \( 30\% \).
Therefore, Carlo is incorrect.
Final Answer: No, Carlo is incorrect (Percentage is approx 31%).
Question 4 (4 marks)
(a) Complete the table of values for \( y = 5 – x^3 \)
| \( x \) | -2 | -1 | 0 | 1 | 2 |
| \( y \) | 6 |
(b) On the grid below, draw the graph of \( y = 5 – x^3 \) for values of \( x \) from -2 to 2.
Worked Solution
Part (a): Completing the Table
Why we do this: Substitute each \( x \) value into the equation \( y = 5 – x^3 \).
When \( x = -2 \): \( y = 5 – (-2)^3 = 5 – (-8) = 5 + 8 = 13 \)
When \( x = 0 \): \( y = 5 – 0^3 = 5 – 0 = 5 \)
When \( x = 1 \): \( y = 5 – 1^3 = 5 – 1 = 4 \)
When \( x = 2 \): \( y = 5 – 2^3 = 5 – 8 = -3 \)
Table Values: 13, 6, 5, 4, -3
Part (b): Drawing the Graph
How to do this: Plot the points from the table coordinates and draw a smooth curve through them.
Points: \( (-2, 13), (-1, 6), (0, 5), (1, 4), (2, -3) \)
Final Answer: (a) 13, 6, 5, 4, -3 (b) Correct smooth curve plotted.
Question 5 (2 marks)
Work out the value of \( x \).
Give your answer correct to 1 decimal place.
Worked Solution
Step 1: Identify Sides
Why we do this: To choose the correct trigonometric ratio (SOH CAH TOA), we must label the sides relative to the \( 34^\circ \) angle.
- The side opposite the right angle is the Hypotenuse = 178 mm.
- The side opposite the \( 34^\circ \) angle is the Opposite = \( x \).
Step 2: Choose Ratio
How to do this: We have Hypotenuse (H) and want Opposite (O). We use SOH.
\( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
\( \sin(34^\circ) = \frac{x}{178} \)
Step 3: Solve for x
\( x = 178 \times \sin(34^\circ) \)
Calculator Steps:
- Check mode is DEGREES (D).
- Type:
178รsin34= - Result:
99.5316...
Round to 1 decimal place: \( 99.5 \)
Final Answer: \( 99.5 \)
Question 6 (2 marks)
\( \mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \) \(\quad\) \( \mathbf{b} = \begin{pmatrix} 5 \\ -2 \end{pmatrix} \)
Find \( 2\mathbf{a} – 3\mathbf{b} \) as a column vector.
Worked Solution
Step 1: Substitute the vectors
Why we do this: We replace \( \mathbf{a} \) and \( \mathbf{b} \) with their column vector values.
Step 2: Multiply by scalars
How to do this: Multiply the top and bottom numbers of each vector by the number in front of it.
Step 3: Subtract the vectors
How to do this: Subtract the corresponding components (top minus top, bottom minus bottom). Be careful with the double negative.
Top: \( 6 – 15 = -9 \)
Bottom: \( 8 – (-6) = 8 + 6 = 14 \)
\[ \begin{pmatrix} -9 \\ 14 \end{pmatrix} \]Final Answer: \( \begin{pmatrix} -9 \\ 14 \end{pmatrix} \)
Question 7 (4 marks)
The diagram shows a right-angled triangle and a quarter circle.
The right-angled triangle \( ABC \) has angle \( ABC = 90^\circ \).
The quarter circle has centre \( C \) and radius \( CB \).
Work out the area of the quarter circle.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Find the radius (Length CB)
Why we do this: To find the area of the quarter circle, we first need its radius. The radius is the side \( CB \) of the right-angled triangle.
Method: Pythagoras’ Theorem \( a^2 + b^2 = c^2 \).
\( AB^2 + CB^2 = AC^2 \)
\( 6^2 + CB^2 = 9^2 \)
\( 36 + CB^2 = 81 \)
\( CB^2 = 81 – 36 \)
\( CB^2 = 45 \)
\( CB = \sqrt{45} \) (We can keep this exact value for now)
Step 2: Calculate Area of Quarter Circle
Formula: Area of a circle = \( \pi r^2 \). A quarter circle is \( \frac{1}{4} \pi r^2 \).
Radius \( r = \sqrt{45} \), so \( r^2 = 45 \).
Area = \( \frac{1}{4} \times \pi \times 45 \)
Area = \( \frac{45\pi}{4} \)
Calculator Steps: 45 ร ฯ รท 4 = 35.3429...
Step 3: Rounding
We need 3 significant figures.
35.3429… rounds to 35.3
Final Answer: \( 35.3 \, \text{m}^2 \)
Question 8 (5 marks)
Tariq buys a laptop.
He gets a discount of 5% off the normal price.
Tariq pays ยฃ551 for the laptop.
(a) Work out the normal price of the laptop.
Joan invests ยฃ6000 in a savings account.
The savings account pays compound interest at a rate of
- 2.4% for the first year
- 1.7% for each extra year.
(b) Work out the value of Joan’s investment at the end of 3 years.
Worked Solution
Part (a): Reverse Percentage
Why we do this: The price paid (ยฃ551) represents 95% of the original price (100% – 5%). We need to find 100%.
Multiplier for 5% decrease = \( 1 – 0.05 = 0.95 \)
\( \text{Normal Price} \times 0.95 = 551 \)
\( \text{Normal Price} = 551 \div 0.95 \)
Calculator: 551 รท 0.95 = 580
Part (b): Compound Interest
Why we do this: We apply different interest rate multipliers for each year.
- Year 1: 2.4% increase \(\rightarrow \times 1.024\)
- Year 2: 1.7% increase \(\rightarrow \times 1.017\)
- Year 3: 1.7% increase \(\rightarrow \times 1.017\)
Initial amount = ยฃ6000
Calculation: \( 6000 \times 1.024 \times 1.017 \times 1.017 \)
Or: \( 6000 \times 1.024 \times 1.017^2 \)
Calculator Steps:
6000ร1.024=6144(End of Year 1)6144ร1.017ร1.017=6354.671616...
Round to money (2 decimal places): ยฃ6354.67
Final Answer:
(a) ยฃ580
(b) ยฃ6354.67
Question 9 (2 marks)
Aisha recorded the heights, in centimetres, of some girls.
She used her results to work out the information in this table.
| Least height | 142 cm |
| Lower quartile | 154 cm |
| Interquartile range | 17 cm |
| Median | 162 cm |
| Range | 40 cm |
Aisha drew this box plot for the information in the table.
The box plot is not fully correct.
Write down the two things Aisha should do to make the box plot fully correct.
Worked Solution
Step 1: Check values against the plot
Why we do this: We calculate the correct positions and compare them to where the lines are drawn on the graph.
- Median: Table says 162. Plot shows 161. (Incorrect)
- Upper Quartile: LQ + IQR = \( 154 + 17 = 171 \). Plot shows 172. (Incorrect)
- Max Value: Min + Range = \( 142 + 40 = 182 \). Plot shows 182. (Correct)
Final Answer:
- Plot the median at 162 (currently at 161).
- Plot the upper quartile at 171 (currently at 172).
Question 10 (4 marks)
(a) Simplify \( \left(\frac{1}{m^2}\right)^0 \)
(b) Simplify \( \frac{8(x-4)}{(x-4)^2} \)
(c) Simplify \( (3n^4w^2)^3 \)
Worked Solution
Part (a)
Rule: Anything to the power of 0 is 1.
Part (b)
Strategy: Cancel the common factors in the numerator and denominator.
Cancel one \( (x-4) \) from top and bottom:
\[ \require{cancel} \frac{8\cancel{(x-4)}}{\cancel{(x-4)}(x-4)} = \frac{8}{x-4} \]Part (c)
Rule: Apply the power of 3 to every part inside the bracket. \( (ab)^n = a^n b^n \) and \( (a^m)^n = a^{mn} \).
- \( 3^3 = 27 \)
- \( (n^4)^3 = n^{4 \times 3} = n^{12} \)
- \( (w^2)^3 = w^{2 \times 3} = w^6 \)
Combine them: \( 27n^{12}w^6 \)
Final Answer:
(a) \( 1 \)
(b) \( \frac{8}{x-4} \)
(c) \( 27n^{12}w^6 \)
Question 11 (2 marks)
Jack is in a restaurant.
There are 5 starters, 8 main courses and some desserts on the menu.
Jack is going to choose one starter, one main course and one dessert.
He says there are 240 ways that he can choose his starter, his main course and his dessert.
Could Jack be correct?
You must show how you get your answer.
Worked Solution
Step 1: Understand the combinations
Why we do this: To find the total number of combinations, we multiply the number of choices for each course together.
Total Combinations = Starters ร Mains ร Desserts
Let \( d \) be the number of desserts.
\( 5 \times 8 \times d = 240 \)
\( 40 \times d = 240 \)
Step 2: Solve for desserts
\( d = \frac{240}{40} \)
\( d = 6 \)
Step 3: Conclusion
What this tells us: Since \( d = 6 \) is a whole number, it is possible for there to be 6 desserts on the menu.
Yes, Jack could be correct if there are exactly 6 desserts.
Final Answer: Yes, because \( 240 \div (5 \times 8) = 6 \), which is a possible number of desserts.
Question 12 (3 marks)
The graph gives information about the volume, \( v \) litres, of petrol in the tank of Jimโs car after it has travelled a distance of \( d \) kilometres.
(a) Find the gradient of the graph.
(b) Interpret what the gradient of the graph represents.
Worked Solution
Part (a): Find the Gradient
Why we do this: Gradient = \( \frac{\text{change in } y}{\text{change in } x} \). We pick two points on the line to calculate this.
Point 1: \( (0, 27) \)
Point 2: \( (300, 0) \)
Gradient = \( \frac{y_2 – y_1}{x_2 – x_1} \)
\( = \frac{0 – 27}{300 – 0} \)
\( = \frac{-27}{300} \)
Simplify: \( -0.09 \)
Part (b): Interpret the Gradient
What this tells us: The y-axis is fuel (litres) and x-axis is distance (km). The gradient is \( \frac{\text{litres}}{\text{km}} \).
The gradient represents the rate of fuel consumption.
Specifically: The car uses 0.09 litres of petrol for every kilometre travelled.
Final Answer:
(a) \( -0.09 \)
(b) It is the rate of fuel consumption (litres per km).
Question 13 (3 marks)
Here is triangle \( ABC \).
Work out the length of \( AB \).
Give your answer correct to 1 decimal place.
Worked Solution
Step 1: Calculate the missing angle A
Why we do this: To use the Sine Rule, we need a known side and its opposite angle. We know side \( CB = 23.8 \) cm. The angle opposite \( CB \) is angle \( A \).
Angles in a triangle sum to \( 180^\circ \).
\( A = 180 – (34 + 26) \)
\( A = 180 – 60 = 120^\circ \)
Step 2: Apply the Sine Rule
Formula: \( \frac{a}{\sin A} = \frac{c}{\sin C} \)
Here, we want side \( AB \) (opposite angle \( C = 34^\circ \)). We know side \( CB \) (opposite angle \( A = 120^\circ \)).
\( \frac{AB}{\sin 34^\circ} = \frac{23.8}{\sin 120^\circ} \)
Step 3: Solve for AB
\( AB = \frac{23.8 \times \sin 34^\circ}{\sin 120^\circ} \)
Calculator Steps:
- Type:
23.8รsin34รทsin120= - Result:
15.368...
Round to 1 decimal place: \( 15.4 \) cm
Final Answer: \( 15.4 \) cm
Question 14 (4 marks)
Here are two squares, A and B.
The length of each side of square B is 4 cm greater than the length of each side of square A.
The area of square B is \( 70 \text{ cm}^2 \) greater than the area of square A.
Find the area of square B.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Set up the equations
Why we do this: We don’t know the side lengths, so let the side of square A be \( x \).
Side of A = \( x \)
Side of B = \( x + 4 \)
Area of A = \( x^2 \)
Area of B = \( (x + 4)^2 \)
We are told: Area of B – Area of A = 70
\( (x + 4)^2 – x^2 = 70 \)
Step 2: Solve for x
Expand the brackets: \( (x^2 + 8x + 16) – x^2 = 70 \)
Simplify: \( 8x + 16 = 70 \)
\( 8x = 70 – 16 \)
\( 8x = 54 \)
\( x = \frac{54}{8} = 6.75 \text{ cm} \)
Step 3: Calculate Area of Square B
Side of B = \( x + 4 = 6.75 + 4 = 10.75 \text{ cm} \)
Area of B = \( 10.75^2 \)
Calculation: \( 115.5625 \)
Step 4: Rounding
Round to 3 significant figures: \( 116 \)
Final Answer: \( 116 \text{ cm}^2 \)
Question 15 (2 marks)
Describe fully the single transformation that maps triangle A onto triangle B.
Worked Solution
Step 1: Identify the type of transformation
Observation: Triangle B is larger than Triangle A and inverted. This indicates an Enlargement with a negative scale factor.
Step 2: Calculate Scale Factor
Compare side lengths:
- Triangle A height: From \( y=2 \) to \( y=5 \), height = 3 units.
- Triangle B height: From \( y=-0.5 \) to \( y=-5 \), height = 4.5 units.
Scale Factor = \( \frac{\text{New Length}}{\text{Old Length}} = \frac{4.5}{3} = 1.5 \)
Since the shape is inverted (rotated \( 180^\circ \)), the scale factor is negative.
Scale Factor = \( -1.5 \)
Step 3: Find the Centre of Enlargement
Method: Draw ray lines through corresponding vertices (e.g., top vertex of A to bottom vertex of B). The intersection point is the centre.
Connecting \( (5,5) \) and \( (-5,-5) \) passes through \( (1,1) \).
Connecting \( (5,2) \) and \( (-5,-0.5) \) passes through \( (1,1) \).
Centre of Enlargement = \( (1, 1) \)
Final Answer: Enlargement, scale factor -1.5, centre (1, 1).
Question 16 (3 marks)
Here are the first five terms of a quadratic sequence.
10 21 38 61 90
Find an expression, in terms of \( n \), for the \( n \)th term of this sequence.
Worked Solution
Step 1: Find the Differences
Why we do this: A quadratic sequence \( an^2 + bn + c \) has a constant second difference. We find the differences to determine the coefficients.
Sequence: 10, 21, 38, 61, 90
1st Diff: 11, 17, 23, 29
2nd Diff: 6, 6, 6
Step 2: Find the \( n^2 \) coefficient (a)
Rule: The second difference is equal to \( 2a \).
\( 2a = 6 \)
\( a = 3 \)
So the first part of the formula is \( 3n^2 \).
Step 3: Find the linear part
Method: Subtract \( 3n^2 \) from the original sequence to find the remaining linear sequence.
| \( n \) | 1 | 2 | 3 | 4 | 5 |
| Original | 10 | 21 | 38 | 61 | 90 |
| \( 3n^2 \) | 3 | 12 | 27 | 48 | 75 |
| Difference | 7 | 9 | 11 | 13 | 15 |
The difference sequence is 7, 9, 11, 13…
This goes up by 2 each time, so it is \( 2n + \text{something} \).
When \( n=1 \), \( 2(1) = 2 \). To get to 7, we add 5.
So the linear part is \( 2n + 5 \).
Step 4: Combine
Total formula: \( 3n^2 + 2n + 5 \)
Final Answer: \( 3n^2 + 2n + 5 \)
Question 17 (1 mark)
Write down the coordinates of the turning point on the graph of \( y = (x + 12)^2 – 7 \)
Worked Solution
Step 1: Identify vertex from completed square form
Why we do this: For a quadratic in the form \( y = a(x – h)^2 + k \), the turning point (vertex) is at coordinates \( (h, k) \).
Equation: \( y = (x + 12)^2 – 7 \)
Rewrite as: \( y = (x – (-12))^2 + (-7) \)
\( h = -12 \)
\( k = -7 \)
Final Answer: \( (-12, -7) \)
Question 18 (4 marks)
The diagram represents a solid cone.
Formula Sheet: Curved surface area of cone = \( \pi r l \)
The cone has a base diameter of 20 cm and a slant height of 25 cm.
A circle is drawn around the surface of the cone at a slant height of 10 cm above the base.
The curved surface of the cone above the circle is painted grey.
Work out the area of the curved surface of the cone that is not painted grey.
Give your answer as a multiple of \( \pi \).
Worked Solution
Step 1: Analyze the Geometry
Why we do this: The unpainted part is a “frustum” (the bottom slice of the cone). The easiest way to find its area is: Area of Full Cone – Area of Small Top Cone.
We know:
- Full Cone: Slant height \( L = 25 \), Diameter = 20 (so Radius \( R = 10 \)).
- Small Cone (Painted): Length along slant from base is 10 cm. So slant height of small cone is \( l = 25 – 10 = 15 \).
Step 2: Find the radius of the small cone
Method: The small cone is similar to the large cone. We use the ratio of their slant heights.
Scale Factor = \( \frac{\text{Small Slant}}{\text{Large Slant}} = \frac{15}{25} = \frac{3}{5} = 0.6 \)
Radius of small cone \( r = \text{Scale Factor} \times R \)
\( r = 0.6 \times 10 = 6 \text{ cm} \)
Step 3: Calculate Curved Surface Areas
Formula: \( \text{Area} = \pi r l \)
Large Cone Area:
\( A_{large} = \pi \times 10 \times 25 = 250\pi \)
Small Cone Area (Painted):
\( A_{small} = \pi \times 6 \times 15 = 90\pi \)
Step 4: Calculate Unpainted Area
\( \text{Unpainted Area} = A_{large} – A_{small} \)
\( 250\pi – 90\pi = 160\pi \)
Final Answer: \( 160\pi \text{ cm}^2 \)
Question 19 (4 marks)
A hot air balloon is descending.
The height of the balloon \( n \) minutes after it starts to descend is \( h_n \) metres.
The height of the balloon \( (n + 1) \) minutes after it starts to descend, \( h_{n+1} \) metres, is given by
\[ h_{n+1} = K \times h_n + 20 \]where \( K \) is a constant.
The balloon starts to descend from a height of 1200 metres at 09:15.
At 09:16 the height of the balloon is 1040 metres.
Work out the height of the balloon at 09:18.
Worked Solution
Step 1: Find the value of K
Why we do this: We are given the height at 09:15 (\( h_0 = 1200 \)) and 09:16 (\( h_1 = 1040 \)). We can plug these into the formula to solve for \( K \).
\( h_1 = K \times h_0 + 20 \)
\( 1040 = K \times 1200 + 20 \)
\( 1040 – 20 = 1200K \)
\( 1020 = 1200K \)
\( K = \frac{1020}{1200} \)
Calculator: 1020 รท 1200 = 0.85
Step 2: Find the height at 09:17 (\( h_2 \))
Method: Use the formula again with \( K = 0.85 \) and the height from 09:16 (\( h_1 \)).
\( h_2 = 0.85 \times 1040 + 20 \)
Calculator: 0.85 ร 1040 + 20 = 904 metres
Step 3: Find the height at 09:18 (\( h_3 \))
\( h_3 = 0.85 \times 904 + 20 \)
Calculator: 0.85 ร 904 + 20 = 788.4 metres
Final Answer: \( 788.4 \) m
Question 20 (8 marks)
There are only red sweets and yellow sweets in a bag.
There are \( n \) red sweets in the bag.
There are 8 yellow sweets in the bag.
Sajid is going to take at random a sweet from the bag and eat it.
He says that the probability that the sweet will be red is \( \frac{7}{10} \).
(a) Show why the probability cannot be \( \frac{7}{10} \).
After Sajid has taken the first sweet from the bag and eaten it, he is going to take at random a second sweet from the bag.
Given that the probability that both the sweets he takes will be red is \( \frac{3}{5} \),
(b) work out the number of red sweets in the bag.
You must show all your working.
Worked Solution
Part (a): Proving Impossible Probability
Why we do this: Probability = (Number of desired outcomes) รท (Total outcomes). We set up the equation and solve for \( n \). \( n \) represents red sweets, so it must be a whole number.
Number of Red = \( n \)
Total Sweets = \( n + 8 \)
Equation: \( \frac{n}{n+8} = \frac{7}{10} \)
Cross multiply: \( 10n = 7(n + 8) \)
\( 10n = 7n + 56 \)
\( 3n = 56 \)
\( n = \frac{56}{3} = 18.66… \)
Since \( n \) is not an integer (you can’t have 0.66 of a sweet), the probability cannot be \( \frac{7}{10} \).
Part (b): Probability of Two Red Sweets
Why we do this: Since the sweet is eaten (not replaced), the total number of sweets decreases by 1 for the second pick.
Prob(First Red) = \( \frac{n}{n+8} \)
Prob(Second Red) = \( \frac{n-1}{n+7} \)
\( \frac{n}{n+8} \times \frac{n-1}{n+7} = \frac{3}{5} \)
Multiply numerators and denominators:
\[ \frac{n(n-1)}{(n+8)(n+7)} = \frac{3}{5} \]Expand brackets:
\[ \frac{n^2 – n}{n^2 + 7n + 8n + 56} = \frac{3}{5} \] \[ \frac{n^2 – n}{n^2 + 15n + 56} = \frac{3}{5} \]Step 2: Solve the Quadratic Equation
Cross multiply:
\( 5(n^2 – n) = 3(n^2 + 15n + 56) \)
\( 5n^2 – 5n = 3n^2 + 45n + 168 \)
Rearrange to form \( ax^2 + bx + c = 0 \):
\( 5n^2 – 3n^2 – 5n – 45n – 168 = 0 \)
\( 2n^2 – 50n – 168 = 0 \)
Divide by 2 to simplify:
\( n^2 – 25n – 84 = 0 \)
Step 3: Factorise and Solve
Method: Find two numbers that multiply to -84 and add to -25.
Factors of 84: 1, 84; 2, 42; 3, 28; 4, 21…
Using -28 and +3: \( -28 \times 3 = -84 \) and \( -28 + 3 = -25 \).
\( (n – 28)(n + 3) = 0 \)
Solutions: \( n = 28 \) or \( n = -3 \)
Since we cannot have a negative number of sweets, \( n = 28 \).
Final Answer: \( n = 28 \)
Question 21 (4 marks)
The graph of the curve with equation \( y = f(x) \) is shown on the grid below.
(a) On the grid above, sketch the graph of the curve with equation \( y = f(-x) \)
The curve \( C \) with equation \( y = 5 + 2x – x^2 \) is transformed by a translation to give the curve \( S \) such that the point \( (1, 6) \) on \( C \) is mapped to the point \( (4, 6) \) on \( S \).
(b) Find an equation for \( S \).
Worked Solution
Part (a): Sketching \( y = f(-x) \)
Why we do this: The transformation \( f(-x) \) represents a reflection in the y-axis. Every x-coordinate changes its sign (e.g., \( -2 \) becomes \( +2 \)).
Key Points to Reflect:
- \( (-4, 0) \to (4, 0) \)
- \( (-2, -2) \to (2, -2) \)
- \( (0, 0) \to (0, 0) \) (Invariant point)
- \( (2, 2) \to (-2, 2) \)
The curve basically flips horizontally around the y-axis.
Part (b): Finding the Equation of S
Step 1: Identify the Transformation
Point maps from \( (1, 6) \) to \( (4, 6) \).
- x-coordinate: \( 1 \to 4 \) (Increase by 3)
- y-coordinate: \( 6 \to 6 \) (No change)
This is a translation by vector \( \begin{pmatrix} 3 \\ 0 \end{pmatrix} \).
For a function \( f(x) \), a translation of \( +3 \) in the x-direction becomes \( f(x – 3) \).
Original Equation: \( y = 5 + 2x – x^2 \)
Replace every \( x \) with \( (x – 3) \):
\( y = 5 + 2(x – 3) – (x – 3)^2 \)
Simplified form (Optional but good practice):
\( y = 5 + (2x – 6) – (x^2 – 6x + 9) \)
\( y = 5 + 2x – 6 – x^2 + 6x – 9 \)
\( y = -x^2 + 8x – 10 \)
Note: The vertex form is easier to check. Original vertex \( (1,6) \) means \( y = -(x-1)^2 + 6 \). New vertex \( (4,6) \) means \( y = -(x-4)^2 + 6 \).
Final Answer:
(a) Reflection of the curve in the y-axis.
(b) \( y = 5 + 2(x – 3) – (x – 3)^2 \) OR \( y = -(x-4)^2 + 6 \) OR \( y = -x^2 + 8x – 10 \)
Question 22 (5 marks)
\( C \) is a circle with centre the origin.
A tangent to \( C \) passes through the points \( (-20, 0) \) and \( (0, 10) \).
Work out an equation of \( C \).
You must show all your working.
Worked Solution
Step 1: Find the equation of the tangent
Why we do this: We know two points on the tangent line, so we can find its gradient and equation.
Points: \( (-20, 0) \) and \( (0, 10) \)
Gradient \( m = \frac{\text{change in } y}{\text{change in } x} = \frac{10 – 0}{0 – (-20)} = \frac{10}{20} = 0.5 \)
Y-intercept \( c = 10 \)
Equation of tangent: \( y = 0.5x + 10 \)
Step 2: Find the equation of the radius
Rule: The radius meets the tangent at \( 90^\circ \) (perpendicular). The product of perpendicular gradients is \( -1 \).
Gradient of tangent = \( 0.5 \)
Gradient of radius = \( \frac{-1}{0.5} = -2 \)
The radius passes through the origin \( (0,0) \), so its equation is \( y = -2x \).
Step 3: Find the point of intersection
Why we do this: The intersection of the radius and tangent is a point on the circle. We can calculate the distance from the origin to this point to find the radius length \( r \).
Equate the two lines:
\( 0.5x + 10 = -2x \)
Add \( 2x \) to both sides:
\( 2.5x + 10 = 0 \)
\( 2.5x = -10 \)
\( x = \frac{-10}{2.5} = -4 \)
Find \( y \):
\( y = -2(-4) = 8 \)
Point of contact: \( (-4, 8) \)
Step 4: Find the equation of the circle
Formula: \( x^2 + y^2 = r^2 \)
Calculate \( r^2 \) using the point \( (-4, 8) \):
\( r^2 = (-4)^2 + 8^2 \)
\( r^2 = 16 + 64 \)
\( r^2 = 80 \)
Equation: \( x^2 + y^2 = 80 \)
Final Answer: \( x^2 + y^2 = 80 \)