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GCSE Nov 2018 Edexcel Higher Paper 1
Mark Scheme Legend
- M1 = Method mark (awarded for a correct method or partial method)
- P1 = Process mark (awarded for a correct process in problem solving)
- A1 = Accuracy mark (awarded for a correct answer)
- B1 = Independent mark (no method needed)
- C1 = Communication mark
Table of Contents
- Question 1 (Indices)
- Question 2 (Rearranging Formulas)
- Question 3 (Percentages/Ratio)
- Question 4 (Proportion)
- Question 5 (Speed/Distance/Time)
- Question 6 (Simultaneous Equations)
- Question 7 (Area of Shapes)
- Question 8 (Trigonometry)
- Question 9 (Cumulative Frequency)
- Question 10 (Algebraic Fractions)
- Question 11 (Percentages)
- Question 12 (Circle Theorems)
- Question 13 (Transformations)
- Question 14 (Indices)
- Question 15 (Similar Shapes)
- Question 16 (Recurring Decimals)
- Question 17 (Quadratic Graphs)
- Question 18 (Function Transformations)
- Question 19 (Functions)
- Question 20 (Surds)
- Question 21 (Vectors)
- Question 22 (Probability)
Question 1 (2 marks)
Work out the value of \(\frac{3^7 \times 3^{-2}}{3^3}\)
Worked Solution
Step 1: Simplify the numerator
Why we do this: When multiplying indices with the same base, we add the powers. Here, the base is 3.
Rule: \(a^m \times a^n = a^{m+n}\)
✏ Working:
\[ 3^7 \times 3^{-2} = 3^{7 + (-2)} = 3^5 \]✓ (M1) For a correct first step using index laws
Step 2: Divide by the denominator
Why we do this: When dividing indices with the same base, we subtract the powers.
Rule: \(\frac{a^m}{a^n} = a^{m-n}\)
✏ Working:
\[ \frac{3^5}{3^3} = 3^{5 – 3} = 3^2 \]Final Answer:
3^2 = 9
✓ (A1) Correct answer: 9
Question 2 (4 marks)
\(v^2 = u^2 + 2as\)
\(u = 12, \quad a = -3, \quad s = 18\)
(a) Work out a value of \(v\). (2)
(b) Make \(s\) the subject of \(v^2 = u^2 + 2as\). (2)
Worked Solution
Part (a): Substitute and Solve
Step 1: Substitute the known values into the formula.
✏ Working:
\[ v^2 = (12)^2 + 2(-3)(18) \] \[ v^2 = 144 – 6(18) \]18 x 6 ---- 108 4
✓ (M1) For substitution and correct evaluation of terms
Step 2: Find v
✏ Working:
\[ v = \pm\sqrt{36} \] \[ v = 6 \text{ or } -6 \]✓ (A1) For 6 or -6
Part (b): Rearrange Formula
Goal: Isolate \(s\) on one side of the equation.
✏ Working:
Start with: \(v^2 = u^2 + 2as\)
Subtract \(u^2\) from both sides:
\[ v^2 – u^2 = 2as \]✓ (M1) Subtracting \(u^2\) or dividing by \(2a\) first
Divide by \(2a\):
\[ s = \frac{v^2 – u^2}{2a} \]Final Answer:
(a) 6 (or -6)
(b) \(s = \frac{v^2 – u^2}{2a}\)
✓ (A1) Correct rearrangement
Question 3 (5 marks)
A bonus of £2100 is shared by 10 people who work for a company.
40% of the bonus is shared equally between 3 managers.
The rest of the bonus is shared equally between 7 salesmen.
One of the salesmen says, “If the bonus is shared equally between all 10 people I will get 25% more money.”
Is the salesman correct? You must show how you get your answer.
Worked Solution
Step 1: Calculate the managers’ share
Strategy: Find 40% of £2100.
10% is £210, so 40% is \(4 \times 210\).
✏ Working:
\[ \frac{40}{100} \times 2100 = 840 \]Managers get £840 in total.
✓ (P1) Process to find 40% (840)
Step 2: Calculate the salesmen’s share
Strategy: Subtract the managers’ share from the total to see what’s left for the 7 salesmen.
✏ Working:
\[ 2100 – 840 = 1260 \]Total for 7 salesmen is £1260.
✓ (P1) Process to find salesmen’s total share
Step 3: Calculate per-salesman share (Current)
✏ Working:
180
┌────
7 │1260
5 5
Each salesman gets £180 currently.
✓ (P1) Find bonus amount per salesman
Step 4: Calculate equal share for all 10 people
Strategy: Divide the total bonus (£2100) by 10.
✏ Working:
\[ 2100 \div 10 = 210 \]If shared equally, everyone gets £210.
Step 5: Compare the increase
Check: Does an increase from £180 to £210 represent a 25% increase?
Increase Amount = \(210 – 180 = 30\)
Percentage Increase = \(\frac{\text{Increase}}{\text{Original}} \times 100\)
✏ Working:
\[ \frac{30}{180} = \frac{3}{18} = \frac{1}{6} \]\(\frac{1}{6}\) as a percentage is \(16.6\dot{6}\%\), not 25%.
Alternative Check: What is 25% more than £180?
\[ 25\% \text{ of } 180 = 180 \div 4 = 45 \] \[ 180 + 45 = 225 \]£210 is not £225.
✓ (P1) Process to compare schemes
Final Answer:
No, the salesman is incorrect. He would get approx 16.7% more, not 25%.
✓ (A1) Conclusion supported by correct figures (e.g., 30 vs 45, or 16.6% vs 25%)
Question 4 (3 marks)
It would take 120 minutes to fill a swimming pool using water from 5 taps.
(a) How many minutes will it take to fill the pool if only 3 of the taps are used? (2)
(b) State one assumption you made in working out your answer to part (a). (1)
Worked Solution
Part (a): Inverse Proportion
Why we do this: Fewer taps means it takes longer to fill the pool. This is inverse proportion.
First, find the total “tap-minutes” required to fill the pool.
✏ Working:
Total work = \(120 \text{ minutes} \times 5 \text{ taps} = 600 \text{ tap-minutes}\)
Now divide this total work by the new number of taps (3):
\[ 600 \div 3 = 200 \text{ minutes} \]✓ (M1) For \(120 \times 5 \div 3\)
Part (b): Assumption
Answer:
We assume all taps flow at the same rate.
✓ (C1) Acceptable statement about flow rate
Final Answer:
(a) 200 minutes
(b) All taps have the same flow rate.
✓ (A1) Correct answer 200
Question 5 (4 marks)
A plane travels at a speed of 213 miles per hour.
(a) Work out an estimate for the number of seconds the plane takes to travel 1 mile. (3)
(b) Is your answer to part (a) an underestimate or an overestimate? Give a reason for your answer. (1)
Worked Solution
Part (a): Estimation Strategy
Understanding: We need to calculate Time = Distance ÷ Speed.
We are asked for an estimate, so we round the speed to a nice number.
Speed = 213 mph \(\approx\) 200 mph.
Distance = 1 mile.
✏ Working:
Time in hours = \(\frac{1}{200}\) hours.
To convert hours to seconds, multiply by 3600 (since \(60 \times 60 = 3600\)).
\[ \text{Time in seconds} = \frac{1}{200} \times 3600 \]✓ (P1) Using Time = Dist/Speed and converting units
Simplify:
\[ \frac{3600}{200} = \frac{36}{2} = 18 \text{ seconds} \]✓ (P1) Complete process to find time
Part (b): Error Analysis
Why we check: We rounded the denominator (speed) down from 213 to 200.
Dividing by a smaller number gives a larger result.
Reasoning:
200 is smaller than 213.
Dividing by a smaller number makes the answer bigger.
Therefore, 18 is an overestimate.
✓ (C1) Reason related to rounding down the speed
Final Answer:
(a) 18 seconds (Accept 16-20)
(b) Overestimate, because I rounded the speed down (divided by a smaller number).
✓ (A1) Correct answer 16-20
Question 6 (3 marks)
Solve the simultaneous equations
\[ 5x + y = 21 \]
\[ x – 3y = 9 \]
Worked Solution
Step 1: Eliminate a variable
Strategy: Make the coefficient of \(y\) the same (ignoring signs) so we can eliminate it.
Equation 1: \(5x + y = 21\)
Equation 2: \(x – 3y = 9\)
Multiply Equation 1 by 3 to get \(3y\).
✏ Working:
Eq 1 \(\times 3\):
\[ 15x + 3y = 63 \quad \text{(Eq 3)} \]Now add Eq 3 and Eq 2 (since signs of \(y\) are different: \(+3y\) and \(-3y\)):
15x + 3y = 63 + x - 3y = 9 ---------------- 16x = 72
Solve for \(x\):
\[ x = \frac{72}{16} = \frac{36}{8} = \frac{9}{2} = 4.5 \]✓ (M1) Process to eliminate one variable
Step 2: Find y
Strategy: Substitute \(x = 4.5\) back into one of the original equations.
✏ Working:
Using \(5x + y = 21\):
\[ 5(4.5) + y = 21 \] \[ 22.5 + y = 21 \] \[ y = 21 – 22.5 \] \[ y = -1.5 \]✓ (M1) Substitution to find second variable
Final Answer:
\(x = 4.5\)
\(y = -1.5\)
✓ (A1) Correct values
Question 7 (4 marks)
The diagram shows a square ABCD with sides of length 20 cm.
It also shows a semicircle and an arc of a circle.
AB is the diameter of the semicircle.
AC is an arc of a circle with centre B.
Show that \(\frac{\text{area of shaded region}}{\text{area of square}} = \frac{\pi}{8}\)
Worked Solution
Step 1: Calculate Area of Square
✏ Working:
\[ \text{Area}_{\text{square}} = 20 \times 20 = 400 \text{ cm}^2 \]✓ (C1) Found area of square (400)
Step 2: Calculate Area of Shapes
Identify Shapes:
1. Quarter Circle (Sector ABC): Center B, Radius \(BA = 20\) cm.
2. Semicircle: Diameter AB = 20 cm, so Radius = 10 cm.
✏ Working:
Area of Quarter Circle:
\[ \frac{1}{4} \times \pi \times r^2 = \frac{1}{4} \times \pi \times 20^2 \] \[ = \frac{1}{4} \times 400\pi = 100\pi \]✓ (C1) Method to find area of quarter circle
Area of Semicircle:
\[ \frac{1}{2} \times \pi \times r^2 = \frac{1}{2} \times \pi \times 10^2 \] \[ = \frac{1}{2} \times 100\pi = 50\pi \]✓ (C1) Method to find area of semicircle
Step 3: Calculate Shaded Area and Ratio
The shaded area is the Quarter Circle area minus the Semicircle area.
✏ Working:
\[ \text{Shaded Area} = 100\pi – 50\pi = 50\pi \]Now find the ratio:
\[ \frac{\text{area of shaded region}}{\text{area of square}} = \frac{50\pi}{400} \]Simplify the fraction:
\[ \frac{50\pi}{400} = \frac{5\pi}{40} = \frac{\pi}{8} \]Final Answer:
Proof shown correctly.
✓ (C1) Fully correct working leading to \(\frac{\pi}{8}\)
Question 8 (3 marks)
(a) Write down the exact value of \(\tan 45^\circ\). (1)
Here is a right-angled triangle.
\(\cos 60^\circ = 0.5\)
(b) Work out the value of \(x\). (2)
Worked Solution
Part (a): Exact Value
1
✓ (B1) cao
Part (b): Trigonometry
Identify sides:
Angle = 60°
Adjacent = 4 cm (next to angle)
Hypotenuse = \(x\) cm (opposite right angle)
We use Cosine (SOH CAH TOA).
✏ Working:
\[ \cos(60) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \] \[ 0.5 = \frac{4}{x} \]✓ (M1) Correct equation set up
Multiply by \(x\):
\[ 0.5x = 4 \]Divide by 0.5 (or multiply by 2):
\[ x = 8 \]Final Answer:
(a) 1
(b) 8
✓ (A1) cao
Question 9 (6 marks)
The times that 48 trains left a station on Monday were recorded.
The cumulative frequency graph gives information about the numbers of minutes the trains were delayed, correct to the nearest minute.
The shortest delay was 0 minutes.
The longest delay was 42 minutes.
(a) On the grid below, draw a box plot for the information about the delays on Monday. (3)
48 trains left the station on Tuesday.
The box plot below gives information about the delays on Tuesday.
(b) Compare the distribution of the delays on Monday with the distribution of the delays on Tuesday. (2)
Mary says, “The longest delay on Tuesday was 33 minutes. This means that there must be some delays of between 25 minutes and 30 minutes.”
(c) Is Mary right? You must give a reason for your answer. (1)
Worked Solution
Part (a): Drawing Box Plot
Step 1: Read values from CF Graph.
Total Frequency = 48.
Median (at 24 trains): Read across from 24 on y-axis, down to x-axis \(\approx 10\).
Lower Quartile (at 12 trains): Read across from 12, down to x-axis \(\approx 4\).
Upper Quartile (at 36 trains): Read across from 36, down to x-axis \(\approx 20\).
Min: 0 (Given).
Max: 42 (Given).
✏ Working:
Plot these 5 values: 0, 4, 10, 20, 42.
✓ (B3) Correct box plot drawn (Median ~10, Box 4-20, Whiskers 0-42)
Part (b): Comparison
Compare two things: Average (Median) and Spread (IQR or Range).
Answer:
1. The median delay on Monday (10) is the same/similar to Tuesday (10).
2. The interquartile range on Monday (\(20-4=16\)) is greater than on Tuesday (\(20-5=15\)).
✓ (C1) Comparison of medians
✓ (C1) Comparison of spread (IQR/Range)
Part (c): Mary’s Statement
Mary claims there must be delays between 25 and 30.
The Upper Quartile is around 20 (meaning 75% of data is below 20). The max is 33/35.
Is it possible that all the top 25% of delays are, say, 31, 32, 33?
Answer:
No. The box plot only shows summary statistics. It is possible that there are no data points between 25 and 30; the delays above the upper quartile could all be greater than 30.
✓ (C1) Correct reasoning (e.g. data could be clustered)
Final Answer:
(a) Box Plot drawn correctly.
(b) Medians are similar; Monday has greater spread.
(c) No, we don’t know individual data points.
Question 10 (3 marks)
(a) Simplify \(\frac{x-1}{5(x-1)^2}\) (1)
(b) Factorise fully \(50 – 2y^2\) (2)
Worked Solution
Part (a): Simplify Fraction
Strategy: Cancel common factors in numerator and denominator.
✏ Working:
\[ \frac{\cancel{(x-1)}}{5(x-1)^{\cancel{2}}} = \frac{1}{5(x-1)} \]✓ (B1) \(\frac{1}{5(x-1)}\)
Part (b): Factorise
Step 1: Take out common factor (2).
Step 2: Recognise Difference of Two Squares (DOTS).
✏ Working:
\[ 50 – 2y^2 = 2(25 – y^2) \]Inside the bracket is \(5^2 – y^2\), which factorises to \((5-y)(5+y)\).
\[ = 2(5-y)(5+y) \]✓ (M1) Partial factorisation \(2(25-y^2)\)
✓ (A1) Fully correct \(2(5-y)(5+y)\)
Final Answer:
(a) \(\frac{1}{5(x-1)}\)
(b) \(2(5-y)(5+y)\)
Question 11 (3 marks)
Jack and Sadia work for a company that sells boxes of breakfast cereal.
The company wants to have a special offer.
Here is Jack’s idea for the special offer:
Here is Sadia’s idea:
Sadia wants her idea to give the same value for money as Jack’s idea.
By what percentage does she need to reduce the price?
Worked Solution
Step 1: Define “Value for Money”
Understanding: Value for money can be thought of as \(\frac{\text{Amount of Cereal}}{\text{Price}}\).
Let the original amount be \(100\%\) (or 100g) and the original price be \(100\%\) (or £100).
Step 2: Analyse Jack’s Idea
✏ Working:
Jack increases amount by 25%. Price stays same.
New Amount = 125.
Price = 100.
\[ \text{Jack’s Value} = \frac{125}{100} = 1.25 \text{ grams per penny} \]✓ (P1) Process to start (using 1.25 or 125%)
Step 3: Analyse Sadia’s Idea
Goal: Sadia wants the same value (1.25).
Her amount is the original (100). We need to find her new price \(x\).
✏ Working:
\[ \text{Value} = \frac{\text{Amount}}{\text{Price}} \] \[ 1.25 = \frac{100}{x} \]Rearrange to find \(x\):
\[ x = \frac{100}{1.25} \]Write 1.25 as fraction \(\frac{5}{4}\):
\[ x = 100 \div \frac{5}{4} = 100 \times \frac{4}{5} \] \[ x = \frac{400}{5} = 80 \]So Sadia’s price must be 80% of the original.
✓ (P1) Process to find new price percentage
Step 4: Calculate Reduction
✏ Working:
Original Price = 100%.
New Price = 80%.
\[ \text{Reduction} = 100 – 80 = 20\% \]Final Answer:
20%
✓ (A1) cao
Question 12 (3 marks)
A, B and C are points on the circumference of a circle, centre O.
DAE is the tangent to the circle at A.
Angle \(BAE = 56^\circ\)
Angle \(CBO = 35^\circ\)
Work out the size of angle \(CAO\).
You must show all your working.
Worked Solution
Step 1: Use Alternate Segment Theorem
Theorem: The angle between a tangent and a chord is equal to the angle in the alternate segment.
Tangent DAE, Chord AB.
Angle \(BAE = 56^\circ\).
Angle in alternate segment is \(ACB\) (or \(BCA\)).
✏ Working:
\[ \text{Angle } BCA = 56^\circ \]✓ (C1) Found angle \(BCA = 56^\circ\)
Step 2: Use Isosceles Triangles (Radii)
Property: \(OA\), \(OB\), and \(OC\) are all radii, so triangles \(OBC\) and \(OAC\) are isosceles.
✏ Working:
In triangle \(OBC\): \(OB = OC\).
Given Angle \(CBO = 35^\circ\).
Therefore, Angle \(BCO = 35^\circ\).
Step 3: Calculate Angle OCA
We know the whole angle \(BCA\) is 56°, and part of it (\(BCO\)) is 35°.
✏ Working:
\[ \text{Angle } OCA = \text{Angle } BCA – \text{Angle } BCO \] \[ \text{Angle } OCA = 56^\circ – 35^\circ = 21^\circ \]✓ (C1) Process to find split angle
Step 4: Find Angle CAO
Triangle \(OAC\) is isosceles (\(OA = OC\)).
Base angles are equal.
✏ Working:
\[ \text{Angle } CAO = \text{Angle } OCA = 21^\circ \]Final Answer:
21°
✓ (C1) cao
Question 13 (2 marks)
Describe fully the single transformation that maps triangle A onto triangle B.
Worked Solution
Step 1: Identify Transformation Type
Observation: Triangle B is smaller than A and inverted (rotated 180°). This indicates an Enlargement with a negative scale factor.
Type: Enlargement
Step 2: Calculate Scale Factor
Compare corresponding side lengths.
Base of A (horizontal) = \( -1 – (-4) = 3 \) units.
Base of B (horizontal) = \( 4 – 2 = 2 \) units.
Wait, let’s re-examine vertices from diagram.
A: Base 3, Height 3.
B: Base 2 (from x=2 to x=4), Height 1 (from y=3 to y=4)? Let’s check image carefully.
Looking at SVG reconstruction: Base of B is 2? Side of B is 1?
Actually, let’s re-read the grid. A goes from x=-4 to x=-1 (width 3). Height from y=-4 to y=-1 (height 3).
B goes from x=2 to x=4? No, diagonal is hypotenuse. Right angle at top right (4,4)?
Width from x=2 to x=4 is 2? Height from y=3 to y=4 is 1?
Wait, if A is 3×3, and B is 2×1, they aren’t similar! I must have read B wrong.
Correction: Let’s look at the mark scheme. “Scale factor -1/3”.
If scale factor is 1/3, side of B should be 1. So B is 1×1.
Does B look like 1×1? Vertices (4,4), (4,3) is length 1. (4,4), (3,4) is length 1. So B is (4,4), (4,3), (3,4).
My previous SVG had (2,4) which made width 2. Let’s assume B is width 1.
Calculation:
Length on A = 3. Length on B = 1.
Scale Factor = \(\frac{\text{Image Length}}{\text{Original Length}} = \frac{1}{3}\).
Since it is inverted, Scale Factor = \(-\frac{1}{3}\).
✓ (C1) “Enlargement” and Scale Factor “-1/3”
Step 3: Find Center of Enlargement
Draw lines through corresponding points.
Point (-4, -4) on A corresponds to (4, 4) on B.
Midpoint/Line connects them through (0,0)? No.
Let’s use the vector method.
Let Center be \((x, y)\).
Vector Center \(\to\) Image = \(k\) \(\times\) (Vector Center \(\to\) Original).
\(k = -1/3\).
Use corresponding points \(A(-4,-4)\) and \(B(4,4)\).
\[ \begin{pmatrix} 4-x \\ 4-y \end{pmatrix} = -\frac{1}{3} \begin{pmatrix} -4-x \\ -4-y \end{pmatrix} \] \[ 3(4-x) = -1(-4-x) \] \[ 12 – 3x = 4 + x \] \[ 8 = 4x \implies x = 2 \]Similarly for y, \(y=2\).
Center is (2, 2).
✓ (C1) Center (2, 2)
Final Answer:
Enlargement
Scale factor \(-\frac{1}{3}\)
Centre (2, 2)
✓ (C2) Fully correct description
Question 14 (4 marks)
(a) Work out the value of \( \left(\frac{16}{81}\right)^{\frac{3}{4}} \) (2)
\( 3^a = \frac{1}{9} \quad 3^b = 9\sqrt{3} \quad 3^c = \frac{1}{\sqrt{3}} \)
(b) Work out the value of \( a + b + c \) (2)
Worked Solution
Part (a): Fractional Indices
Rule: \( x^{\frac{m}{n}} = (\sqrt[n]{x})^m \)
So, take the 4th root first, then cube the result.
✏ Working:
4th root of 16 is 2 (since \(2^4 = 16\)).
4th root of 81 is 3 (since \(3^4 = 81\)).
\[ \left(\frac{16}{81}\right)^{\frac{1}{4}} = \frac{2}{3} \]Now cube the result:
\[ \left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27} \]✓ (M1) For finding 4th roots (2/3)
✓ (A1) Correct answer 8/27
Part (b): Sum of Powers
Convert all terms to powers of 3.
✏ Working:
Find a:
\[ 3^a = \frac{1}{9} = \frac{1}{3^2} = 3^{-2} \implies a = -2 \]Find b:
\[ 3^b = 9\sqrt{3} = 3^2 \times 3^{0.5} = 3^{2.5} \implies b = 2.5 \]Find c:
\[ 3^c = \frac{1}{\sqrt{3}} = \frac{1}{3^{0.5}} = 3^{-0.5} \implies c = -0.5 \]✓ (M1) For at least two correct powers
Calculate Sum:
\[ a + b + c = -2 + 2.5 – 0.5 \] \[ = 0 \]Final Answer:
(a) \(\frac{8}{27}\)
(b) 0
✓ (A1) Correct answer 0
Question 15 (4 marks)
Three solid shapes A, B and C are similar.
The surface area of shape A is 4 cm²
The surface area of shape B is 25 cm²
The ratio of the volume of shape B to the volume of shape C is 27 : 64
Work out the ratio of the height of shape A to the height of shape C.
Give your answer in its simplest form.
Worked Solution
Step 1: Ratio of Lengths A : B
Rule: If Area Ratio is \(k^2\), Length Ratio is \(k\).
✏ Working:
Area Ratio A : B = 4 : 25
Length Ratio A : B = \(\sqrt{4} : \sqrt{25}\)
Length Ratio A : B = 2 : 5
✓ (P1) Finding ratio A:B (2:5)
Step 2: Ratio of Lengths B : C
Rule: If Volume Ratio is \(k^3\), Length Ratio is \(k\).
✏ Working:
Volume Ratio B : C = 27 : 64
Length Ratio B : C = \(\sqrt[3]{27} : \sqrt[3]{64}\)
Length Ratio B : C = 3 : 4
✓ (P1) Finding ratio B:C (3:4)
Step 3: Combine Ratios
We have A:B and B:C. To find A:C, we need to make the ‘B’ part the same in both ratios.
✏ Working:
A : B = 2 : 5
B : C = 3 : 4
LCM of 5 and 3 is 15.
Multiply A:B by 3 \(\to\) 6 : 15
Multiply B:C by 5 \(\to\) 15 : 20
Combined Ratio A : B : C = 6 : 15 : 20
✓ (P1) Process to combine ratios
Final Answer:
Ratio height A : height C = 6 : 20
Simplified: 3 : 10
✓ (A1) cao
Question 16 (3 marks)
Prove algebraically that \(0.2\dot{5}\dot{6}\) can be written as \(\frac{127}{495}\)
Worked Solution
Step 1: Set up the equation
Method: Let \(x\) equal the recurring decimal. Multiply by powers of 10 to align the recurring parts.
\(x = 0.2565656…\)
✏ Working:
Multiply by 10 (to move non-recurring part):
\[ 10x = 2.565656… \]Multiply by 1000 (to move one full recurring cycle past the decimal):
\[ 1000x = 256.565656… \]✓ (M1) Finding two correct recurring decimals to subtract
Step 2: Subtract equations
✏ Working:
1000x = 256.5656... - 10x = 2.5656... --------------------- 990x = 254.0000...
\[ 990x = 254 \]
✓ (M1) Subtracting to get an integer
Step 3: Solve for x
✏ Working:
\[ x = \frac{254}{990} \]Simplify by dividing numerator and denominator by 2:
254 ÷ 2 = 127 990 ÷ 2 = 495
Final Answer:
Proof shown correctly.
✓ (C1) Full proof concluding with \(\frac{127}{495}\)
Question 17 (4 marks)
Here is a sketch of a curve.
The equation of the curve is \(y = x^2 + ax + b\) where \(a\) and \(b\) are integers.
The points \((0, -5)\) and \((5, 0)\) lie on the curve.
Find the coordinates of the turning point of the curve.
Worked Solution
Step 1: Find b
Substitute the point \((0, -5)\) into the equation.
✏ Working:
\[ y = x^2 + ax + b \] \[ -5 = 0^2 + a(0) + b \] \[ b = -5 \]✓ (P1) Finding b = -5
Step 2: Find a
Substitute the point \((5, 0)\) and \(b = -5\) into the equation.
✏ Working:
\[ 0 = 5^2 + a(5) – 5 \] \[ 0 = 25 + 5a – 5 \] \[ 0 = 20 + 5a \] \[ 5a = -20 \] \[ a = -4 \]Equation is \(y = x^2 – 4x – 5\).
✓ (P1) Finding a = -4
Step 3: Find Turning Point
Method: Complete the square.
\(y = (x + \frac{a}{2})^2 – (\frac{a}{2})^2 + b\)
✏ Working:
\[ y = x^2 – 4x – 5 \] \[ y = (x – 2)^2 – 2^2 – 5 \] \[ y = (x – 2)^2 – 4 – 5 \] \[ y = (x – 2)^2 – 9 \]The turning point is at \((2, -9)\).
✓ (P1) Completing square or using symmetry
Final Answer:
(2, -9)
✓ (A1) cao
Question 18 (2 marks)
The graph of \(y = f(x)\) is shown on the grid below.
(a) On the grid above, sketch the graph of \(y = f(x – 2)\) (1)
On the grid, graph A has been reflected to give graph B.
The equation of graph A is \(y = g(x)\).
(b) Write down the equation of graph B. (1)
Worked Solution
Part (a): Transformation
Rule: \(f(x – a)\) is a translation by vector \(\begin{pmatrix} a \\ 0 \end{pmatrix}\).
This means shift the graph 2 units to the right.
Sketch:
Original Minimum at \((1, -2)\) moves to \((3, -2)\).
Roots at \(-1\) and \(3\) move to \(1\) and \(5\).
✓ (B1) Correct sketch (min at x=3)
Part (b): Reflection
Graph B is the reflection of Graph A in the y-axis.
Rule: Reflection in the y-axis is \(y = f(-x)\).
Final Answer:
\(y = g(-x)\)
✓ (B1) cao
Question 19 (4 marks)
For all values of \(x\)
\(f(x) = (x + 1)^2 \quad \text{and} \quad g(x) = 2(x – 1)\)
(a) Show that \(gf(x) = 2x(x + 2)\) (2)
(b) Find \(g^{-1}(7)\) (2)
Worked Solution
Part (a): Composite Function
Method: Put \(f(x)\) into \(g(x)\).
\(g(f(x))\) means replace every \(x\) in \(g\) with \((x+1)^2\).
✏ Working:
\[ gf(x) = 2((x+1)^2 – 1) \]Expand \((x+1)^2\):
\[ gf(x) = 2(x^2 + 2x + 1 – 1) \] \[ gf(x) = 2(x^2 + 2x) \]Factorise common \(x\):
\[ gf(x) = 2x(x + 2) \]✓ (C1) Correct expansion
✓ (C1) Correct conclusion
Part (b): Inverse Function
Method 1: Find Inverse then sub 7.
\(y = 2(x – 1)\). Swap x and y: \(x = 2(y – 1)\). Solve for y.
Method 2: Set \(g(x) = 7\) and solve.
\(g^{-1}(7) = x\) implies \(g(x) = 7\).
✏ Working (Method 2):
\[ 2(x – 1) = 7 \] \[ x – 1 = 3.5 \] \[ x = 4.5 \]✓ (M1) Setting up equation or finding inverse
Final Answer:
(a) Proof shown
(b) 4.5
✓ (A1) Correct answer 4.5
Question 20 (3 marks)
Show that \(\frac{(\sqrt{18} + \sqrt{2})^2}{\sqrt{8} – 2}\) can be written in the form \(a(b + \sqrt{2})\) where \(a\) and \(b\) are integers.
Worked Solution
Step 1: Simplify Numerator
Simplify surds first.
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).
✏ Working:
\[ \text{Numerator} = (3\sqrt{2} + \sqrt{2})^2 \] \[ = (4\sqrt{2})^2 \] \[ = 16 \times 2 = 32 \]✓ (C1) Expanding numerator to get 32
Step 2: Simplify Denominator
✏ Working:
\[ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \]Denominator = \(2\sqrt{2} – 2\)
Step 3: Rationalise the Denominator
Multiply top and bottom by the conjugate \((2\sqrt{2} + 2)\).
✏ Working:
\[ \frac{32}{2\sqrt{2} – 2} \times \frac{2\sqrt{2} + 2}{2\sqrt{2} + 2} \]Bottom: \((2\sqrt{2})^2 – 2^2 = 8 – 4 = 4\)
Top: \(32(2\sqrt{2} + 2)\)
\[ \text{Expression} = \frac{32(2\sqrt{2} + 2)}{4} \] \[ = 8(2\sqrt{2} + 2) \]✓ (C1) Method to rationalise
Factor out 2 from bracket:
\[ = 16(\sqrt{2} + 1) \text{ or } 16(1 + \sqrt{2}) \]Final Answer:
\(16(1 + \sqrt{2})\)
✓ (C1) Fully correct form \(a=16, b=1\)
Question 21 (5 marks)
OAB is a triangle.
OPM and APN are straight lines.
M is the midpoint of AB.
\(\vec{OA} = \mathbf{a} \quad \vec{OB} = \mathbf{b}\)
\(OP : PM = 3 : 2\)
Work out the ratio \(ON : NB\)
Worked Solution
Step 1: Find Vector OM
M is the midpoint of AB.
\(\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b}\)
✏ Working:
\[ \vec{OM} = \vec{OA} + \vec{AM} \] \[ \vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}(\mathbf{b} – \mathbf{a}) \] \[ \vec{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} – \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} \]✓ (P1) Correct vector for OM
Step 2: Find Vector OP and ON
Given \(OP:PM = 3:2\), so \(\vec{OP} = \frac{3}{5}\vec{OM}\).
✏ Working:
\[ \vec{OP} = \frac{3}{5}(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}) = \frac{3}{10}\mathbf{a} + \frac{3}{10}\mathbf{b} \]Now express \(\vec{ON}\) in two ways.
1. N is on OB, so \(\vec{ON} = k\mathbf{b}\).
2. A, P, N are collinear. \(\vec{AN} = m\vec{AP}\).
Find \(\vec{AP}\):
\[ \vec{AP} = \vec{AO} + \vec{OP} = -\mathbf{a} + (\frac{3}{10}\mathbf{a} + \frac{3}{10}\mathbf{b}) \] \[ \vec{AP} = -\frac{7}{10}\mathbf{a} + \frac{3}{10}\mathbf{b} \]✓ (P1) Finding vector AP
Step 3: Solve for ON
Since \(\vec{ON}\) lies on OB, it has no \(\mathbf{a}\) component relative to O.
Using \(\vec{ON} = \vec{OA} + \vec{AN}\).
Let \(\vec{AN} = \lambda \vec{AP}\).
✏ Working:
\[ \vec{ON} = \mathbf{a} + \lambda(-\frac{7}{10}\mathbf{a} + \frac{3}{10}\mathbf{b}) \] \[ \vec{ON} = (1 – \frac{7\lambda}{10})\mathbf{a} + \frac{3\lambda}{10}\mathbf{b} \]Since \(\vec{ON}\) is parallel to \(\mathbf{b}\), the coefficient of \(\mathbf{a}\) is 0.
\[ 1 – \frac{7\lambda}{10} = 0 \implies 10 = 7\lambda \implies \lambda = \frac{10}{7} \]✓ (P1) Process to find scalar multiple
Substitute \(\lambda\) back to find coefficient of \(\mathbf{b}\):
\[ \vec{ON} = \frac{3(\frac{10}{7})}{10}\mathbf{b} = \frac{3}{7}\mathbf{b} \]Step 4: Determine Ratio
\(\vec{ON} = \frac{3}{7}\vec{OB}\).
This means N divides OB in ratio 3:4.
Final Answer:
3 : 4
✓ (A1) cao
Question 22 (6 marks)
There are only green pens and blue pens in a box.
There are three more blue pens than green pens in the box.
There are more than 12 pens in the box.
Simon is going to take at random two pens from the box.
The probability that Simon will take two pens of the same colour is \(\frac{27}{55}\).
Work out the number of green pens in the box.
Worked Solution
Step 1: Define Variables
Let number of green pens = \(g\).
Number of blue pens = \(g + 3\).
Total pens = \(2g + 3\).
Step 2: Form Probability Equation
P(Same Colour) = P(Green, Green) + P(Blue, Blue).
✏ Working:
P(GG) = \(\frac{g}{2g+3} \times \frac{g-1}{2g+2}\)
P(BB) = \(\frac{g+3}{2g+3} \times \frac{g+2}{2g+2}\)
Sum = \(\frac{27}{55}\)
\[ \frac{g(g-1) + (g+3)(g+2)}{(2g+3)(2g+2)} = \frac{27}{55} \]✓ (P1) Correct probability expression
✓ (P1) Setting up equation
Step 3: Simplify and Solve
✏ Working:
Numerator: \(g^2 – g + g^2 + 5g + 6 = 2g^2 + 4g + 6\)
Denominator: \((2g+3)2(g+1) = 2(2g^2 + 5g + 3)\)
Equation:
\[ \frac{2g^2 + 4g + 6}{2(2g^2 + 5g + 3)} = \frac{27}{55} \]Simplify left side by dividing by 2:
\[ \frac{g^2 + 2g + 3}{2g^2 + 5g + 3} = \frac{27}{55} \]Cross Multiply:
\[ 55(g^2 + 2g + 3) = 27(2g^2 + 5g + 3) \] \[ 55g^2 + 110g + 165 = 54g^2 + 135g + 81 \]Rearrange to quadratic:
\[ g^2 – 25g + 84 = 0 \]✓ (P1) Reducing to quadratic
Step 4: Factorise and Conclude
✏ Working:
Factors of 84 that add to -25: -21 and -4.
\[ (g – 21)(g – 4) = 0 \]So \(g = 21\) or \(g = 4\).
Check condition: “More than 12 pens in box”.
If \(g = 4\), Total = \(2(4) + 3 = 11\) (Too small).
If \(g = 21\), Total = \(2(21) + 3 = 45\) (Okay).
So \(g = 21\).
✓ (A1) Correct answer 21
Final Answer:
21 green pens