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GCSE Mathematics 2020 Paper 1H (Non-Calculator)
📝 Exam Instructions
- Time: 1 hour 30 minutes
- Total Marks: 80
- Type: Non-Calculator – Show all arithmetic working clearly
- Diagrams: Not accurately drawn unless specified
📋 Table of Contents
- Question 1 (Arithmetic Sequence)
- Question 2 (Fraction Multiplication)
- Question 3 (Graphs)
- Question 4 (Congruent Triangles)
- Question 5 (Percentage Profit)
- Question 6 (Geometry Proof)
- Question 7 (Stem & Leaf)
- Question 8 (Prism Pressure)
- Question 9 (Standard Form)
- Question 10 (Ratio)
- Question 11 (Indices)
- Question 12 (Cumulative Frequency)
- Question 13 (Density)
- Question 14 (Probability)
- Question 15 (Perpendicular Lines)
- Question 16 (Capture-Recapture)
- Question 17 (Rearranging Formula)
- Question 18 (Proportion)
- Question 19 (Functions)
- Question 20 (Surds)
- Question 21 (Vector Proof)
- Question 22 (Sector Area)
- Question 23 (Probability with Shapes)
Question 1 (2 marks)
The first five terms of an arithmetic sequence are
1 4 7 10 13
Write down an expression, in terms of \( n \), for the \( n \)th term of this sequence.
Worked Solution
Step 1: Find the common difference
Why: An arithmetic sequence changes by the same amount each time. This amount is the “common difference” and tells us the multiple of \( n \).
How: Look at the gap between terms:
The sequence goes up by 3 each time.
This means the formula starts with \( 3n \).
Step 2: Adjust the formula
Why: The \( 3n \) sequence would be 3, 6, 9, 12… We need to adjust this to match our sequence which starts at 1.
How: Compare the first term of \( 3n \) with our first term.
For \( n=1 \), \( 3n = 3 \times 1 = 3 \).
Our first term is 1.
How do we get from 3 to 1? We subtract 2.
\[ 3 – 2 = 1 \]Final Answer:
\[ 3n – 2 \]✓ (Total: 2 marks)
Question 2 (3 marks)
Show that \( 2\frac{1}{3} \times 3\frac{3}{4} = 8\frac{3}{4} \)
Worked Solution
Step 1: Convert mixed numbers to improper fractions
Why: To multiply mixed numbers, it is much easier to work with improper (top-heavy) fractions first.
How: Multiply the whole number by the denominator and add the numerator.
For \( 2\frac{1}{3} \):
\[ \frac{(2 \times 3) + 1}{3} = \frac{6+1}{3} = \frac{7}{3} \]For \( 3\frac{3}{4} \):
\[ \frac{(3 \times 4) + 3}{4} = \frac{12+3}{4} = \frac{15}{4} \]Step 2: Multiply the fractions
Why: We multiply numerator by numerator and denominator by denominator.
We can simplify before multiplying. 15 and 3 are both divisible by 3:
\[ \frac{7}{1} \times \frac{5}{4} = \frac{35}{4} \](Alternatively: \( \frac{105}{12} \))
Step 3: Convert back to a mixed number
Why: The question asks us to show the result is \( 8\frac{3}{4} \), so we must convert our result.
How: How many times does 4 go into 35?
Final Answer:
Shown correctly as:
\[ \frac{7}{3} \times \frac{15}{4} = \frac{105}{12} = \frac{35}{4} = 8\frac{3}{4} \]✓ (Total: 3 marks)
Question 3 (2 marks)
The diagram shows four graphs.
Each of the equations in the table is the equation of one of the graphs.
Complete the table.
| Equation | Letter of graph |
|---|---|
| \( y = -x^3 \) | |
| \( y = x^3 \) | |
| \( y = x^2 \) | |
| \( y = \frac{1}{x} \) |
Worked Solution
Step 1: Identify the shape of each function
Reciprocal Graph \( y = \frac{1}{x} \): This graph has asymptotes at \(x=0\) and \(y=0\). It is split into two parts. This matches Graph A.
Quadratic Graph \( y = x^2 \): This is a U-shaped parabola. It is always positive (above the x-axis). This matches Graph D.
Cubic Graph \( y = x^3 \): This graph goes from bottom-left to top-right, passing through the origin. It increases as x increases. This matches Graph C.
Negative Cubic Graph \( y = -x^3 \): This graph is the reflection of \( x^3 \). It goes from top-left to bottom-right. This matches Graph B.
Final Answer:
| \( y = -x^3 \) | B |
| \( y = x^3 \) | C |
| \( y = x^2 \) | D |
| \( y = \frac{1}{x} \) | A |
✓ (Total: 2 marks)
Question 4 (1 mark)
The diagram shows four triangles.
Two of these triangles are congruent.
Write down the letters of these two triangles.
Worked Solution
Step 1: Calculate missing angles
Why: To compare triangles, we need to know as many properties (angles and sides) as possible. The angles in a triangle add up to 180°.
Triangle A:
\[ 180^\circ – (55^\circ + 45^\circ) = 180^\circ – 100^\circ = 80^\circ \]Triangle A has angles 55°, 45°, 80° and a side of 10 cm between the 55° and 45° angles (opposite the 80° angle).
Triangle D:
\[ 180^\circ – (80^\circ + 45^\circ) = 180^\circ – 125^\circ = 55^\circ \]Triangle D has angles 55°, 45°, 80°.
In Triangle D, the 10 cm side is opposite the 80° angle.
Step 2: Compare using Congruence Rules (ASA or AAS)
Check A and D:
- Both have angles 45°, 55°, 80°.
- In Triangle A, the 10cm side is opposite the 80° angle (between 45° and 55°).
- In Triangle D, the 10cm side is opposite the 80° angle.
Since they have two matching angles and a matching corresponding side (AAS), they are congruent.
Final Answer:
Triangle A and Triangle D
✓ (Total: 1 mark)
Question 5 (3 marks)
Sean pays £10 for 24 chocolate bars.
He sells all 24 chocolate bars for 50p each.
Work out Sean’s percentage profit.
Worked Solution
Step 1: Calculate Total Revenue (Selling Price)
Why: We need to know how much money he made in total to calculate profit.
How: Multiply number of bars by price per bar. Be careful with units (£ vs p).
Convert to pounds:
\[ 1200\text{p} = £12.00 \]Step 2: Calculate Profit
Why: Profit is the difference between what he sold them for and what he bought them for.
Cost Price = £10
Selling Price = £12
\[ \text{Profit} = £12 – £10 = £2 \]Step 3: Calculate Percentage Profit
Why: Percentage profit is calculated based on the original cost.
Formula: \( \frac{\text{Profit}}{\text{Cost Price}} \times 100 \)
Convert fraction to percentage:
\[ \frac{2}{10} = \frac{1}{5} = 20\% \]Final Answer:
20%
✓ (Total: 3 marks)
Question 6 (5 marks)
\( ADC \) is a triangle.
\( AED \) and \( ABC \) are straight lines.
\( EB \) is parallel to \( DC \).
Angle \( EBC = 148^\circ \)
Angle \( ADC = 63^\circ \)
Work out the size of angle \( EAB \).
You must give a reason for each stage of your working.
Worked Solution
Step 1: Find Angle \( ABE \)
Why: We know angles on a straight line add up to \( 180^\circ \). Since \( ABC \) is a straight line, we can find the interior angle at \( B \).
Reason: Angles on a straight line add up to \( 180^\circ \).
Step 2: Find Angle \( AEB \)
Why: Since lines \( EB \) and \( DC \) are parallel, we can use parallel line rules. Angle \( AEB \) and angle \( ADC \) are corresponding angles.
Reason: Corresponding angles are equal.
Step 3: Calculate Angle \( EAB \)
Why: The angles in a triangle (triangle \( ABE \)) always add up to \( 180^\circ \).
Reason: Angles in a triangle add up to \( 180^\circ \).
Final Answer:
\[ 85^\circ \]✓ (Total: 5 marks)
Question 7 (3 marks)
The table shows information about the heights, in cm, of a group of Year 9 girls.
| least height | 150 cm |
| median | 165 cm |
| greatest height | 170 cm |
This stem and leaf diagram shows information about the heights, in cm, of a group of 15 Year 9 boys.
Compare the distribution of the heights of the girls with the distribution of the heights of the boys.
Worked Solution
Step 1: Calculate Statistics for Boys
Why: To compare distributions, we need a measure of average (median) and a measure of spread (range).
Median: There are 15 boys. The median position is \( \frac{15+1}{2} = 8 \text{th} \) value.
Counting the leaves:
15: 8, 9, 9 (3 values)
16: 4, 5, 7, 7, 8 (5 values)
The 8th value is \( 168 \text{ cm} \).
Range: \( \text{Highest} – \text{Lowest} \)
\[ 182 – 158 = 24 \text{ cm} \]Step 2: Calculate Statistics for Girls
Why: Use the table provided.
Median: \( 165 \text{ cm} \)
Range: \( 170 – 150 = 20 \text{ cm} \)
Step 3: Write Comparison Statements
Why: A valid comparison needs to mention both the average (median) and the spread (range) in context.
1. On average, boys are taller. The median height for boys (168 cm) is greater than the median height for girls (165 cm).
2. The boys’ heights are more spread out. The range of boys’ heights (24 cm) is greater than the range of girls’ heights (20 cm).
Final Answer:
Comparison of Medians: Boys (168 cm) > Girls (165 cm)
Comparison of Ranges: Boys (24 cm) > Girls (20 cm)
✓ (Total: 3 marks)
Question 8 (3 marks)
The diagram shows a prism placed on a horizontal floor.
The prism has height \( 3 \text{ m} \)
The volume of the prism is \( 18 \text{ m}^3 \)
The pressure on the floor due to the prism is \( 75 \text{ newtons/m}^2 \)
Work out the force exerted by the prism on the floor.
Worked Solution
Step 1: Calculate the Area of the Base
Why: The pressure formula is \( \text{Pressure} = \frac{\text{Force}}{\text{Area}} \). We are given pressure and need to find force, but first we need the area.
How: The volume of a prism is \( \text{Area of Base} \times \text{Height} \).
Step 2: Calculate the Force
Why: Now we can rearrange the pressure formula to find Force.
Formula: \( \text{Force} = \text{Pressure} \times \text{Area} \)
Calculation:
\[ 75 \times 2 = 150 \] \[ 150 \times 3 = 450 \]Alternatively: \( 70 \times 6 = 420 \), \( 5 \times 6 = 30 \), \( 420 + 30 = 450 \).
Final Answer:
\[ 450 \text{ newtons} \]✓ (Total: 3 marks)
Question 9 (2 marks)
Write these numbers in order of size.
Start with the smallest number.
Worked Solution
Step 1: Convert all numbers to the same format
Why: It’s hard to compare standard form with different powers. Let’s convert them all to ordinary numbers or standard form with the same power.
1. \( 6.72 \times 10^5 = 672,000 \)
2. \( 67.2 \times 10^{-4} \): Move decimal 4 places left.
\[ 67.2 \rightarrow 6.72 \rightarrow 0.672 \rightarrow 0.0672 \rightarrow 0.00672 \]3. \( 672 \times 10^4 = 6,720,000 \)
4. \( 0.000672 \)
Step 2: Order the numbers
Comparison:
- \( 0.000672 \) (Smallest)
- \( 0.00672 \) (from \( 67.2 \times 10^{-4} \))
- \( 672,000 \) (from \( 6.72 \times 10^5 \))
- \( 6,720,000 \) (from \( 672 \times 10^4 \)) (Largest)
Final Answer:
\[ 0.000672, \quad 67.2 \times 10^{-4}, \quad 6.72 \times 10^5, \quad 672 \times 10^4 \]✓ (Total: 2 marks)
Question 10 (3 marks)
Given that \( \frac{a}{b} = \frac{2}{5} \) and \( \frac{b}{c} = \frac{3}{4} \)
find \( a : b : c \)
Worked Solution
Step 1: Write as ratios
Why: Fractions like \( \frac{a}{b} = \frac{2}{5} \) represent the ratio \( a : b = 2 : 5 \).
Step 2: Equate the common term (b)
Why: The variable \( b \) appears in both ratios. To combine them, \( b \) must have the same value in both.
How: Find the Lowest Common Multiple (LCM) of 5 and 3, which is 15.
Multiply the first ratio by 3:
\[ a : b = (2 \times 3) : (5 \times 3) = 6 : 15 \]Multiply the second ratio by 5:
\[ b : c = (3 \times 5) : (4 \times 5) = 15 : 20 \]Step 3: Combine into a single ratio
Why: Now that \( b = 15 \) in both, we can join them.
Final Answer:
\[ 6 : 15 : 20 \]✓ (Total: 3 marks)
Question 11 (6 marks)
(a) Find the value of \( \sqrt[4]{81 \times 10^8} \)
………………………………………………. (2)
(b) Find the value of \( 64^{-\frac{1}{2}} \)
………………………………………………. (2)
(c) Write \( \frac{3^n}{9^{n-1}} \) as a power of 3
………………………………………………. (2)
Worked Solution
Part (a): Evaluate \( \sqrt[4]{81 \times 10^8} \)
Why: The 4th root applies to both parts of the product separately. \( \sqrt[4]{ab} = \sqrt[4]{a} \times \sqrt[4]{b} \).
How:
- Find the 4th root of 81 (what number multiplied by itself 4 times makes 81?).
- Find the 4th root of \( 10^8 \) (divide the power by 4).
Answer: 300
Part (b): Evaluate \( 64^{-\frac{1}{2}} \)
Why: A negative power means the reciprocal (\( x^{-a} = \frac{1}{x^a} \)). A fractional power of \( \frac{1}{2} \) means the square root.
First, deal with the negative sign:
\[ 64^{-\frac{1}{2}} = \frac{1}{64^{\frac{1}{2}}} \]Next, evaluate the fractional power (square root):
\[ 64^{\frac{1}{2}} = \sqrt{64} = 8 \]So:
\[ \frac{1}{8} \]Answer: \( \frac{1}{8} \)
Part (c): Simplify \( \frac{3^n}{9^{n-1}} \)
Why: To use index laws, the bases must be the same. We need to change the base 9 into base 3.
How: We know \( 9 = 3^2 \). Replace 9 with \( 3^2 \) in the expression.
Multiply the powers in the denominator:
\[ (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n-2} \]Now apply the division law (\( \frac{a^x}{a^y} = a^{x-y} \)):
\[ \frac{3^n}{3^{2n-2}} = 3^{n – (2n-2)} \]Be careful with the double negative:
\[ n – (2n – 2) = n – 2n + 2 = -n + 2 \] \[ = 3^{2-n} \]Answer: \( 3^{2-n} \)
Final Answers:
(a) 300
(b) \( \frac{1}{8} \)
(c) \( 3^{2-n} \)
✓ (Total: 6 marks)
Question 12 (6 marks)
The table gives information about the weekly wages of 80 people.
| Wage (£w) | Frequency |
|---|---|
| \( 200 < w \le 250 \) | 5 |
| \( 250 < w \le 300 \) | 10 |
| \( 300 < w \le 350 \) | 20 |
| \( 350 < w \le 400 \) | 20 |
| \( 400 < w \le 450 \) | 15 |
| \( 450 < w \le 500 \) | 10 |
(a) Complete the cumulative frequency table.
(b) On the grid opposite, draw a cumulative frequency graph for your completed table.
Juan says
“60% of this group of people have a weekly wage of £360 or less.”
(c) Is Juan correct?
You must show how you get your answer.
Worked Solution
Part (a): Cumulative Frequency Table
Why: Cumulative frequency is the running total of frequencies.
| \( 200 < w \le 250 \) | 5 |
| \( 200 < w \le 300 \) | \( 5+10 = \) 15 |
| \( 200 < w \le 350 \) | \( 15+20 = \) 35 |
| \( 200 < w \le 400 \) | \( 35+20 = \) 55 |
| \( 200 < w \le 450 \) | \( 55+15 = \) 70 |
| \( 200 < w \le 500 \) | \( 70+10 = \) 80 |
Part (b): Draw the Graph
Why: Plot cumulative frequency against the upper bound of each interval. Draw a smooth curve starting from the lower bound of the first interval (200).
Points to plot: (250, 5), (300, 15), (350, 35), (400, 55), (450, 70), (500, 80).
Part (c): Evaluate Juan’s Statement
Step 1: Find 60% of the people.
Total people = 80.
\[ 60\% \text{ of } 80 = 0.6 \times 80 = 48 \]Juan claims that 48 people earn £360 or less.
Step 2: Check the graph.
Read the cumulative frequency at Wage = £360.
Look at the blue lines on the graph above.
- At £350, CF is 35.
- At £400, CF is 55.
- At £360, the value is approximately 40 (or between 38 and 42).
Conclusion:
The graph shows approx 40 people earn £360 or less, not 48.
Alternatively, find the wage for 48 people (60%). Reading across from 48 on the y-axis gives a wage of approximately £380-£390.
Graph reading at £360 gives a frequency of approx 40.
60% of the group is 48 people.
40 is not equal to 48.
Final Answer:
No, Juan is incorrect.
60% of 80 is 48 people.
Reading from the graph at wage £360 gives a cumulative frequency of approximately 40, which is less than 48.
✓ (Total: 6 marks)
Question 13 (3 marks)
Liquid A and liquid B are mixed to make liquid C.
Liquid A has a density of \( 70 \text{ kg/m}^3 \)
Liquid A has a mass of \( 1400 \text{ kg} \)
Liquid B has a density of \( 280 \text{ kg/m}^3 \)
Liquid B has a volume of \( 30 \text{ m}^3 \)
Work out the density of liquid C.
Worked Solution
Step 1: Understand the Goal
Why: To find the density of a mixture (Liquid C), we calculate the Total Mass and divide it by the Total Volume.
Formula: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
Warning: You CANNOT just average the densities.
Step 2: Calculate Missing Values for Liquid A
We know Mass (1400) and Density (70). We need Volume.
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]Step 3: Calculate Missing Values for Liquid B
We know Density (280) and Volume (30). We need Mass.
\[ \text{Mass} = \text{Density} \times \text{Volume} \]Step 4: Calculate Density of Liquid C
Find totals and divide.
Total Mass: \( 1400 + 8400 = 9800 \text{ kg} \)
Total Volume: \( 20 + 30 = 50 \text{ m}^3 \)
\[ \text{Density}_C = \frac{9800}{50} = \frac{980}{5} \]Calculation:
\[ 980 \div 5 = (1000 \div 5) – (20 \div 5) = 200 – 4 = 196 \]Final Answer:
\[ 196 \text{ kg/m}^3 \]✓ (Total: 3 marks)
Question 14 (3 marks)
Sally plays two games against Martin.
In each game, Sally could win, draw or lose.
In each game they play,
- the probability that Sally will win against Martin is 0.3
- the probability that Sally will draw against Martin is 0.1
Work out the probability that Sally will win exactly one of the two games against Martin.
Worked Solution
Step 1: Identify all probabilities
Why: Probabilities sum to 1. We need the probability of losing.
\( P(\text{Win}) = 0.3 \)
\( P(\text{Draw}) = 0.1 \)
\( P(\text{Lose}) = 1 – (0.3 + 0.1) = 0.6 \)
However, for this question, we only care if she Wins or “Does Not Win”.
\( P(\text{Not Win}) = 0.7 \)
Step 2: Identify winning scenarios
Why: “Exactly one win” can happen in two ways over two games:
- Game 1: Win, Game 2: Not Win
- Game 1: Not Win, Game 2: Win
Scenario 1 (Win, Not Win):
\[ 0.3 \times 0.7 = 0.21 \]Scenario 2 (Not Win, Win):
\[ 0.7 \times 0.3 = 0.21 \]Step 3: Add the probabilities
Why: Since these are mutually exclusive events (both satisfy the condition), we add them.
Alternative Method (listing all outcomes):
Win, Draw: \( 0.3 \times 0.1 = 0.03 \)
Win, Lose: \( 0.3 \times 0.6 = 0.18 \)
Draw, Win: \( 0.1 \times 0.3 = 0.03 \)
Lose, Win: \( 0.6 \times 0.3 = 0.18 \)
Total: \( 0.03 + 0.18 + 0.03 + 0.18 = 0.42 \)
Final Answer:
\[ 0.42 \]✓ (Total: 3 marks)
Question 15 (3 marks)
The straight line \( L_1 \) has equation \( y = 3x – 4 \)
The straight line \( L_2 \) is perpendicular to \( L_1 \) and passes through the point \( (9, 5) \)
Find an equation of line \( L_2 \)
Worked Solution
Step 1: Find the gradient of \( L_1 \)
Why: The equation is in the form \( y = mx + c \), where \( m \) is the gradient.
Gradient \( m_1 = 3 \)
Step 2: Find the gradient of \( L_2 \)
Why: Perpendicular gradients multiply to give -1. This means the new gradient is the “negative reciprocal”.
Step 3: Use the point to find the equation
Why: We have the gradient \( m = -\frac{1}{3} \) and a point \( (x, y) = (9, 5) \). We can use \( y = mx + c \) to find \( c \).
Substitute \( x = 9, y = 5 \):
\[ 5 = -\frac{1}{3}(9) + c \] \[ 5 = -3 + c \] \[ c = 8 \]Final Answer:
\[ y = -\frac{1}{3}x + 8 \](Also accept: \( 3y + x = 24 \) or \( y – 5 = -\frac{1}{3}(x – 9) \))
✓ (Total: 3 marks)
Question 16 (4 marks)
Shirley wants to find an estimate for the number of bees in her hive.
On Monday she catches 90 of the bees.
She puts a mark on each bee and returns them to her hive.
On Tuesday she catches 120 of the bees.
She finds that 20 of these bees have been marked.
(a) Work out an estimate for the total number of bees in her hive.
………………………………………………. (3)
Shirley assumes that none of the marks had rubbed off between Monday and Tuesday.
(b) If Shirley’s assumption is wrong, explain what effect this would have on your answer to part (a).
………………………………………………. (1)
Worked Solution
Part (a): Estimate Total Population
Why: We assume the proportion of marked bees in the sample is the same as the proportion of marked bees in the whole hive.
Formula: \( \frac{\text{Marked in Sample}}{\text{Total in Sample}} = \frac{\text{Marked in Total}}{\text{Total Population}} \)
Let \( N \) be the total population.
Simplify the fraction on the left:
\[ \frac{20}{120} = \frac{1}{6} \]So:
\[ \frac{1}{6} = \frac{90}{N} \]Multiply both sides by \( N \) and by 6:
\[ N = 90 \times 6 \] \[ N = 540 \]Answer: 540 bees
Part (b): Effect of Assumption
Why: If marks rub off, we count fewer marked bees than we should.
Analysis: Look at the formula: \( N = \frac{\text{Total Caught} \times \text{Original Marked}}{\text{Marked Recaptured}} \)
If “Marked Recaptured” (the denominator) is smaller than it should be, the result \( N \) becomes larger.
If marks rub off, the number of marked bees found (20) would be lower.
Dividing by a smaller number gives a larger result.
Therefore, the estimated population would be an overestimate.
Final Answer:
(a) 540
(b) It would lead to an overestimate.
✓ (Total: 4 marks)
Question 17 (4 marks)
Make \( f \) the subject of the formula \( d = \frac{3(1 – f)}{f – 4} \)
Worked Solution
Step 1: Clear the fraction
Why: We can’t rearrange easily when \( f \) is in the denominator. Multiply both sides by \( (f – 4) \).
Step 2: Expand the brackets
Why: We need to “release” the \( f \) terms so we can group them together.
Step 3: Group \( f \) terms on one side
Why: Move all terms with \( f \) to the left, and all non-\( f \) terms to the right.
Add \( 3f \) to both sides. Add \( 4d \) to both sides.
Step 4: Factorise and Divide
Why: Factor out \( f \) to get it by itself.
Divide by \( (d + 3) \):
\[ f = \frac{3 + 4d}{d + 3} \]Final Answer:
\[ f = \frac{4d + 3}{d + 3} \]✓ (Total: 4 marks)
Question 18 (3 marks)
\( x \) is proportional to \( \sqrt{y} \) where \( y > 0 \)
\( y \) is increased by 44%
Work out the percentage increase in \( x \).
Worked Solution
Step 1: Write the proportionality equation
Why: Turn the sentence “x is proportional to root y” into algebra.
Step 2: Apply the increase to \( y \)
Why: An increase of 44% means the multiplier is \( 1 + 0.44 = 1.44 \).
Let the new value of \( y \) be \( y_{new} = 1.44y \).
Substitute the new \( y \) into the formula to find the new \( x \):
\[ x_{new} = k\sqrt{1.44y} \]Step 3: Simplify and compare
Why: We can separate the square root of the number from the variable. \( \sqrt{ab} = \sqrt{a}\sqrt{b} \).
Rearranging to see the original \( x \):
\[ x_{new} = 1.2 \times (k\sqrt{y}) \] \[ x_{new} = 1.2x \]Step 4: Convert multiplier to percentage increase
Why: A multiplier of 1.2 represents an increase.
Final Answer:
\[ 20\% \]✓ (Total: 3 marks)
Question 19 (5 marks)
\( f \) and \( g \) are functions such that
\( f(x) = \frac{12}{\sqrt{x}} \) and \( g(x) = 3(2x + 1) \)
(a) Find \( g(5) \)
………………………………………………. (1)
(b) Find \( gf(9) \)
………………………………………………. (2)
(c) Find \( g^{-1}(6) \)
………………………………………………. (2)
Worked Solution
Part (a): Find \( g(5) \)
Why: Substitute \( x = 5 \) into the function \( g(x) \).
Part (b): Find \( gf(9) \)
Why: This is a composite function. Work from the inside out (right to left). First find \( f(9) \), then put that result into \( g(x) \).
Step 1: Find \( f(9) \)
\[ f(9) = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4 \]Step 2: Find \( g(4) \)
\[ g(4) = 3(2(4) + 1) \] \[ = 3(8 + 1) \] \[ = 3(9) = 27 \]Part (c): Find \( g^{-1}(6) \)
Why: The inverse function reverses the operation. We can set the output to 6 and solve for \( x \).
Alternatively, find the inverse function first.
Set \( g(x) = 6 \):
\[ 3(2x + 1) = 6 \]Divide by 3:
\[ 2x + 1 = 2 \]Subtract 1:
\[ 2x = 1 \]Divide by 2:
\[ x = 0.5 \text{ (or } \frac{1}{2}) \]Final Answers:
(a) 33
(b) 27
(c) \( \frac{1}{2} \)
✓ (Total: 5 marks)
Question 20 (4 marks)
Show that \( \frac{\sqrt{180} – 2\sqrt{5}}{5\sqrt{5} – 5} \) can be written in the form \( a + \frac{\sqrt{5}}{b} \) where \( a \) and \( b \) are integers.
Worked Solution
Step 1: Simplify the Numerator
Why: The surd \( \sqrt{180} \) can be simplified because 180 has a square factor (36).
Now substitute this back into the numerator:
\[ \text{Numerator} = 6\sqrt{5} – 2\sqrt{5} = 4\sqrt{5} \]So the expression becomes:
\[ \frac{4\sqrt{5}}{5\sqrt{5} – 5} \]Step 2: Rationalise the Denominator
Why: We need to remove the surd from the bottom. Multiply top and bottom by the conjugate \( 5\sqrt{5} + 5 \).
Alternatively, notice we can factor out 5 from the bottom first to make numbers smaller.
Factorising the bottom:
\[ \frac{4\sqrt{5}}{5(\sqrt{5} – 1)} \]Now multiply top and bottom by \( (\sqrt{5} + 1) \):
\[ \frac{4\sqrt{5}(\sqrt{5} + 1)}{5(\sqrt{5} – 1)(\sqrt{5} + 1)} \]Expand the top:
\[ 4\sqrt{5} \times \sqrt{5} + 4\sqrt{5} \times 1 = 4(5) + 4\sqrt{5} = 20 + 4\sqrt{5} \]Expand the bottom (Difference of Two Squares):
\[ 5((\sqrt{5})^2 – 1^2) = 5(5 – 1) = 5(4) = 20 \]Step 3: Simplify the Fraction
Why: Divide both terms in the numerator by the denominator.
Final Answer:
The expression simplifies to:
\[ 1 + \frac{\sqrt{5}}{5} \]Where \( a = 1 \) and \( b = 5 \).
✓ (Total: 4 marks)
Question 21 (4 marks)
\( DEF \) is a triangle.
\( P \) is the midpoint of \( FD \).
\( Q \) is the midpoint of \( DE \).
\( \overrightarrow{FD} = \mathbf{a} \) and \( \overrightarrow{FE} = \mathbf{b} \)
Use a vector method to prove that \( PQ \) is parallel to \( FE \).
Worked Solution
Step 1: Define vectors for the path P to Q
Why: To find vector \( \overrightarrow{PQ} \), we can go from \( P \) to \( D \), then from \( D \) to \( Q \).
Path: \( \overrightarrow{PQ} = \overrightarrow{PD} + \overrightarrow{DQ} \)
Step 2: Express component vectors in terms of \( \mathbf{a} \) and \( \mathbf{b} \)
Vector \( \overrightarrow{PD} \):
We know \( \overrightarrow{FD} = \mathbf{a} \). Since \( P \) is the midpoint, \( \overrightarrow{PD} = \frac{1}{2}\mathbf{a} \).
Vector \( \overrightarrow{DQ} \):
We need \( \overrightarrow{DE} \) first. Using the triangle law: \( \overrightarrow{DE} = \overrightarrow{DF} + \overrightarrow{FE} \).
\( \overrightarrow{DF} = -\mathbf{a} \) (reverse of \( \mathbf{a} \)).
\( \overrightarrow{FE} = \mathbf{b} \).
So, \( \overrightarrow{DE} = -\mathbf{a} + \mathbf{b} = \mathbf{b} – \mathbf{a} \).
Since \( Q \) is the midpoint, \( \overrightarrow{DQ} = \frac{1}{2}(\mathbf{b} – \mathbf{a}) \).
Step 3: Combine to find \( \overrightarrow{PQ} \)
Add the vectors together.
Expand brackets:
\[ \overrightarrow{PQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} – \frac{1}{2}\mathbf{a} \]Simplify:
\[ \overrightarrow{PQ} = \frac{1}{2}\mathbf{b} \]Step 4: Conclusion
Why: Two vectors are parallel if one is a scalar multiple of the other.
We found \( \overrightarrow{PQ} = \frac{1}{2}\mathbf{b} \).
We know \( \overrightarrow{FE} = \mathbf{b} \).
Since \( \overrightarrow{PQ} \) is a multiple of \( \overrightarrow{FE} \) (specifically half), the lines are parallel.
Final Proof:
\[ \overrightarrow{PQ} = \frac{1}{2}\mathbf{b} \]Since \( \overrightarrow{PQ} \) is a scalar multiple of \( \overrightarrow{FE} \) (\( \frac{1}{2}\overrightarrow{FE} \)), they are parallel.
✓ (Total: 4 marks)
Question 22 (5 marks)
The diagram shows two shaded shapes, A and B.
Shape A is formed by removing a sector of a circle with radius \( (3x – 1) \text{ cm} \) from a sector of the circle with radius \( (5x – 1) \text{ cm} \).
Shape B is a circle of diameter \( (2 – 2x) \text{ cm} \).
The area of shape A is equal to the area of shape B.
Find the value of \( x \).
You must show all your working.
Worked Solution
Step 1: Calculate Area of Shape A
Why: Shape A is the large sector minus the small sector. The angle is 45°, so the fraction of the circle is \( \frac{45}{360} = \frac{1}{8} \).
Large radius \( R = 5x – 1 \). Small radius \( r = 3x – 1 \).
Note: The diagram labels the extra width as \( 2x \), so large radius = \( (3x-1) + 2x = 5x-1 \). This confirms the text.
Factor out \( \frac{1}{8}\pi \):
\[ \text{Area}_A = \frac{\pi}{8} [ (5x-1)^2 – (3x-1)^2 ] \]Expand the brackets:
\[ (5x-1)^2 = 25x^2 – 10x + 1 \] \[ (3x-1)^2 = 9x^2 – 6x + 1 \] \[ \text{Difference} = (25x^2 – 10x + 1) – (9x^2 – 6x + 1) \] \[ = 16x^2 – 4x \]So Area A:
\[ \text{Area}_A = \frac{\pi}{8}(16x^2 – 4x) = \pi(2x^2 – 0.5x) \]Step 2: Calculate Area of Shape B
Why: Area of a circle is \( \pi r^2 \). We are given diameter \( d = 2 – 2x \), so radius \( r = 1 – x \).
Step 3: Equate and Solve
Why: The question states Area A = Area B.
Divide by \( \pi \) and simplify:
\[ 2x^2 – 0.5x = x^2 – 2x + 1 \]Rearrange to form a quadratic equation (=0):
\[ x^2 + 1.5x – 1 = 0 \]Multiply by 2 to remove decimals:
\[ 2x^2 + 3x – 2 = 0 \]Factorise:
\[ (2x – 1)(x + 2) = 0 \]Solutions: \( x = 0.5 \) or \( x = -2 \).
Step 4: Check Validity
Why: Lengths must be positive.
If \( x = -2 \), radius \( 3x – 1 = -7 \) (Impossible).
If \( x = 0.5 \), radius \( 3(0.5) – 1 = 0.5 \) (Valid).
So \( x = 0.5 \).
Final Answer:
\[ x = 0.5 \]✓ (Total: 5 marks)
Question 23 (3 marks)
There are four types of cards in a game.
Each card has a black circle or a white circle or a black triangle or a white triangle.
number of cards with a black shape : number of cards with a white shape = 3 : 5
number of cards with a circle : number of cards with a triangle = 2 : 7
Express the total number of cards with a black shape as a fraction of the total number of cards with a triangle.
Worked Solution
Step 1: Determine a common total
Why: The ratios imply different “total parts”.
- Black : White = 3 : 5 (Total parts = 8)
- Circle : Triangle = 2 : 7 (Total parts = 9)
To compare them, let’s assume a total number of cards that is divisible by both 8 and 9.
LCM(8, 9) = 72.
Step 2: Calculate specific quantities
Assume Total Cards = 72.
Black Cards:
\[ \frac{3}{8} \times 72 = 3 \times 9 = 27 \]Triangle Cards:
\[ \frac{7}{9} \times 72 = 7 \times 8 = 56 \]Step 3: Write as a fraction
Goal: “Black shape as a fraction of Triangle”.
Numerator = Black Cards (27)
Denominator = Triangle Cards (56)
Final Answer:
\[ \frac{27}{56} \]✓ (Total: 3 marks)