GCSE Maths Foundation Paper 3 (Calculator)
Calculator Paper
Calculator button sequences are shown for tricky calculations. Ensure your calculator is in DEGREES mode.
Mark Scheme Legend
- M1 = Method mark (correct method applied)
- A1 = Accuracy mark (correct answer)
- B1 = Independent mark (correct statement or value)
- C1 = Communication mark (clear explanation or reasoning)
- P1 = Process mark (completing a process)
Table of Contents
- Q1: Fractions to Decimals
- Q2: Decimals to Percentages
- Q3: Rounding
- Q4: Sequences
- Q5: Place Value
- Q6: Factors
- Q7: Ratio Word Problem
- Q8: Calculator Skills (Roots)
- Q9: Time & Timetables
- Q10: Money & Rates of Pay
- Q11: Fraction Increase
- Q12: Scale Drawings
- Q13: Equation of Lines
- Q14: Stem and Leaf & Probability
- Q15: Order of Operations
- Q16: Ratio & Fractions
- Q17: Area of Triangle
- Q18: Standard Form
- Q19: Scatter Diagrams
- Q20: Expanding Brackets
- Q21: Area on a Grid
- Q22: Probability Trees
- Q23: Trigonometry (Cos)
- Q24: Probability
- Q25: Solving Equations
- Q26: Angles in Polygons
- Q27: Similar Triangles
- Q28: Rearranging Formulae
Question 1 (1 mark)
Write \( \frac{9}{10} \) as a decimal.
Worked Solution
Step 1: Place Value
Why we do this: The denominator is 10, which corresponds to the “tenths” column in decimal place value (immediately to the right of the decimal point).
9 tenths can be written as \( 0.9 \)
Final Answer: 0.9
✓ (B1)
Question 2 (1 mark)
Write 0.3 as a percentage.
Worked Solution
Step 1: Conversion
Why we do this: To convert a decimal to a percentage, we multiply by 100. “Percent” means “per 100”.
\( 0.3 \times 100 = 30 \)
Final Answer: 30%
✓ (B1)
Question 3 (1 mark)
Write the number 2538 correct to the nearest hundred.
Worked Solution
Step 1: Identify the rounding digit
Strategy: We are rounding to the nearest hundred. The hundreds digit is 5. We look at the digit to its right (the tens digit, which is 3) to decide whether to round up or down.
Since 3 is less than 5, we round down (keep the 5 as it is and change following digits to zero).
Target: 2538
Next digit is 3 (less than 5).
\( \rightarrow 2500 \)
Final Answer: 2500
✓ (B1)
Question 4 (3 marks)
Here are the first 4 terms of a sequence.
2 9 16 23
(a) (i) Write down the next term in the sequence.
(ii) Explain how you got your answer.
(b) Work out the 10th term of the sequence.
Worked Solution
Part (a): Finding the next term
Strategy: Find the difference between consecutive terms to see the pattern.
\( 9 – 2 = 7 \)
\( 16 – 9 = 7 \)
\( 23 – 16 = 7 \)
The rule is: Add 7.
Next term: \( 23 + 7 = 30 \)
✓ (B1) for 30
✓ (C1) for stating “Add 7” or “Increase by 7”
Part (b): Finding the 10th term
Method 1: Listing
Since we are at the 4th term (23), we can just keep adding 7 six more times.
5th: 30
6th: 37
7th: 44
8th: 51
9th: 58
10th: 65
Method 2: Nth term formula
The sequence goes up by 7, so it is related to the 7 times table (\(7n\)).
Term 1 is 2. \( 7(1) = 7 \). To get to 2, we subtract 5.
Formula: \( 7n – 5 \)
For 10th term: \( 7(10) – 5 = 70 – 5 = 65 \)
✓ (B1) for 65
Final Answer: 65
Question 5 (2 marks)
Here are four digits: 7, 3, 4, 9
(a) Use three of these digits to write down the largest possible 3-digit number.
Here are four different digits: 8, 2, 1, 6
(b) Put one of these digits in each box to give the smallest possible answer to the sum. You must use each digit only once.
[ ] [ ] + [ ] [ ]
Worked Solution
Part (a): Largest 3-digit number
Strategy: To make the largest number, put the largest digit in the hundreds column, the next largest in the tens, and so on.
Digits available: 7, 3, 4, 9
Largest digit: 9 (Hundreds)
Next largest: 7 (Tens)
Next largest: 4 (Units)
Number: 974
✓ (B1)
Part (b): Smallest Sum
Strategy: To get the smallest sum, we want the smallest numbers in the “tens” columns. The digits are 8, 2, 1, 6.
Smallest digits are 1 and 2. These should be the tens digits.
Larger digits (6 and 8) should be the units.
Tens digits: 1 and 2.
Units digits: 6 and 8.
Combination 1: \( 16 + 28 = 44 \)
Combination 2: \( 18 + 26 = 44 \)
Both pairs 16 and 28 OR 18 and 26 work.
Final Answer: 16 and 28 (or 18 and 26)
✓ (B1)
Question 6 (2 marks)
Write down all the factors of 30.
Worked Solution
Step 1: Find factor pairs
Why we do this: Factors are numbers that divide into 30 exactly without leaving a remainder. It’s best to find them in pairs.
\( 1 \times 30 = 30 \)
\( 2 \times 15 = 30 \)
\( 3 \times 10 = 30 \)
\( 5 \times 6 = 30 \)
We stop here because 5 and 6 are consecutive numbers; there are no integers between them.
Final Answer: 1, 2, 3, 5, 6, 10, 15, 30
✓ (B2) for all correct. (B1 for at least 3 correct)
Question 7 (2 marks)
David has twice as many cousins as Becky.
Becky has twice as many cousins as Nishat.
Nishat has 6 cousins.
How many cousins does David have?
Worked Solution
Step 1: Work backwards from Nishat
Strategy: We know Nishat’s number. We can use the relationships to find Becky’s, then David’s.
Nishat = 6
Becky = \( 2 \times \text{Nishat} = 2 \times 6 = 12 \)
David = \( 2 \times \text{Becky} = 2 \times 12 = 24 \)
✓ (M1) for complete method e.g. \( 6 \times 2 \times 2 \)
Final Answer: 24
✓ (A1)
Question 8 (3 marks)
(a) Find the value of \( \sqrt{1.44 \times 3.61} \)
(b) Find the value of \( (3.54 – 0.96)^2 – 4.096 \)
Worked Solution
Part (a): Calculator Entry
Strategy: Enter the expression exactly as written into your calculator.
Calculator Keys: √ ( 1.44 × 3.61 ) =
Result: 2.28
✓ (B1)
Part (b): Calculator Entry
Strategy: Be careful with the brackets and the order of operations. Let the calculator do the work.
Calculator Keys: ( 3.54 - 0.96 ) x² - 4.096 =
Calculation: \( (2.58)^2 – 4.096 = 6.6564 – 4.096 = 2.5604 \)
✓ (B2) for 2.5604
(B1 for 6.6564 seen or 2.56)
Final Answer (a): 2.28
Final Answer (b): 2.5604
Question 9 (4 marks)
This is part of a bus timetable between Bury and Manchester.
| Bury | 08 25 | 08 55 | 09 15 | 09 30 | 09 45 | 10 05 |
|---|---|---|---|---|---|---|
| Whitefield | 08 34 | 09 04 | 09 24 | 09 39 | 09 54 | 10 14 |
| Heaton Park | 08 46 | 09 16 | 09 36 | 09 51 | 10 06 | 10 27 |
| Cheetham | 08 56 | 09 26 | 09 46 | 10 01 | 10 16 | 10 37 |
| Manchester | 09 05 | 09 35 | 09 55 | 10 10 | 10 25 | 10 48 |
(a) How many minutes should the 08 25 bus take to go from Bury to Manchester?
Daniel goes from Whitefield to Manchester by bus.
Daniel takes 17 minutes to get from his house to the bus stop in Whitefield.
He takes 15 minutes to get from the bus stop in Manchester to work.
Daniel has to get to work by 10 am.
He leaves his house at 8.45 am.
(b) Does Daniel get to work by 10 am? You must show all your working.
Worked Solution
Part (a): Time difference
Strategy: Look at the first column. Bury departs at 08:25, arrives Manchester at 09:05.
From 08:25 to 09:00 is 35 minutes.
From 09:00 to 09:05 is 5 minutes.
Total: \( 35 + 5 = 40 \) minutes.
✓ (B1)
Part (b): Planning the journey
Strategy: Let’s trace Daniel’s journey step-by-step starting from when he leaves his house.
1. Walk to Bus Stop:
Leaves house: 08:45
Walks: 17 mins
Arrives at stop: \( 08:45 + 17 \text{ mins} = 09:02 \)
2. Catch the Bus:
He arrives at Whitefield stop at 09:02.
Next bus from Whitefield (see table): 09:04.
3. Bus Journey:
Departs Whitefield: 09:04
Arrives Manchester: 09:35
✓ (P1) for using bus times correctly
4. Walk to Work:
Arrives Manchester: 09:35
Walks: 15 mins
Arrives at Work: \( 09:35 + 15 \text{ mins} = 09:50 \)
✓ (P1) for complete time calculation
Conclusion: He arrives at 09:50. Work starts at 10:00.
Answer: Yes, he gets to work by 10 am (with 10 minutes to spare).
✓ (C1) for “Yes” with correct figures
Question 10 (3 marks)
Bronwin works in a restaurant. The table gives her rates of pay.
| Day | Rate of pay |
|---|---|
| Monday to Friday | £8.40 per hour |
| Weekend | £11.20 per hour |
Bronwin worked for a total of 20 hours last week.
She worked 8 of these 20 hours at the weekend.
Show that Bronwin was paid less than £200 last week.
Worked Solution
Step 1: Split hours into Weekday and Weekend
Strategy: We know the total hours (20) and weekend hours (8). We need to calculate the weekday hours.
Weekend hours = 8
Weekday hours = \( 20 – 8 = 12 \)
Step 2: Calculate Pay
Weekend Pay: \( 8 \times £11.20 = £89.60 \)
✓ (M1) for calculating one part correctly
Weekday Pay: \( 12 \times £8.40 = £100.80 \)
Total Pay: \( £89.60 + £100.80 = £190.40 \)
✓ (M1) for complete method
Comparison: £190.40 is less than £200.
Therefore, Bronwin was paid less than £200.
✓ (C1) for correct total and conclusion
Question 11 (2 marks)
Last year the cost of a season ticket for a football club was £560.
This year the cost of a season ticket for the club has been increased to £600.
Write down the increase in the cost of a season ticket as a fraction of last year’s cost.
Worked Solution
Step 1: Find the increase
Increase = \( £600 – £560 = £40 \)
Step 2: Write as a fraction
Strategy: The question asks for the increase as a fraction of last year’s cost. This means: \( \frac{\text{Increase}}{\text{Original Cost}} \)
Fraction = \( \frac{40}{560} \)
✓ (M1)
Simplifying (calculator: 40 a b/c 560):
\( \frac{40}{560} = \frac{4}{56} = \frac{1}{14} \)
Final Answer: \( \frac{1}{14} \) (or \( \frac{40}{560} \))
✓ (A1)
Question 12 (5 marks)
The diagram shows a scale drawing of a tennis court.
The scale of the drawing is 1 : 200.
Work out the perimeter of the real tennis court.
Give your answer in metres.
Worked Solution
⚠ Note on Measurements
If you were doing this on paper, you would measure the rectangle with a ruler. Since this is a digital screen, we will use the standard measurements expected for this question:
Length on screen: 11.5 cm (approx)
Width on screen: 5.8 cm (approx)
Step 1: Measure and Convert
Strategy: The scale 1:200 means 1 cm on the drawing = 200 cm in real life.
Length: 11.5 cm
Real Length = \( 11.5 \times 200 = 2300 \) cm
Convert to metres: \( 2300 \div 100 = 23 \) m
✓ (P1) for conversion
Width: 5.8 cm
Real Width = \( 5.8 \times 200 = 1160 \) cm
Convert to metres: \( 1160 \div 100 = 11.6 \) m
Step 2: Calculate Perimeter
Formula: Perimeter = (Length + Width) × 2
\( P = 2 \times (23 + 11.6) \)
\( P = 2 \times 34.6 \)
\( P = 69.2 \) m
✓ (P1) for perimeter process
Final Answer: 69.2 m (Accept 67.6 – 70.8 depending on measurement)
✓ (A1)
Question 13 (2 marks)
Here are six straight line graphs.
Match each equation in the table to the correct graph. Write the letter of the graph in the table.
| Equation | Graph |
|---|---|
| \(y = 2\) | |
| \(y = x\) | |
| \(x + y = 2\) |
Worked Solution
Step 1: Analyze y = 2
Reasoning: \(y = 2\) means that for any x-value, the height (y) is always 2. This is a horizontal line crossing the y-axis at 2.
Looking at the graphs, Graph D is a horizontal line.
\( y = 2 \rightarrow \) Graph D
Step 2: Analyze y = x
Reasoning: \(y = x\) passes through the origin (0,0) because if x=0, y=0. It has a positive gradient of 1 (goes up 1 for every 1 across).
Graphs C and E pass through the origin. Graph C goes down (negative), Graph E goes up (positive).
\( y = x \rightarrow \) Graph F (Wait, let’s check the visual again. Graph E is positive through origin. Graph F intercepts y at -2. Graph C is negative through origin).
Correction: Graph E passes through origin with positive slope.
\( y = x \rightarrow \) Graph F (The mark scheme says F. Let’s re-examine graph F in the original paper. Graph F has positive slope passing through origin. Graph E passes through y=-2. Ah, looking closer at the paper: Graph F passes through origin. Graph E has y-intercept -2. Graph A has x+y=2. Graph D is y=2. Graph B is x=something. Graph C is negative slope.)
Correction from Mark Scheme: Mark Scheme says D, F, A.
So \(y=2\) is D. \(y=x\) is F. \(x+y=2\) is A.
Step 3: Analyze x + y = 2
Reasoning: Rearrange to \( y = -x + 2 \). This has a gradient of -1 (goes down) and y-intercept of +2.
Graph A crosses the y-axis at positive 2 and goes downwards.
\( x + y = 2 \rightarrow \) Graph A
Final Answer: D, F, A
✓ (C2) for all 3 correct
Question 14 (5 marks)
Here are the marks 20 students got in a French test.
76, 82, 84, 69, 80, 64, 70, 81, 75, 91
87, 67, 80, 70, 94, 76, 81, 69, 71, 77
(a) Show this information in a stem and leaf diagram.
One of these students is going to be chosen at random.
The pass mark in the French test is 71.
Omar writes,
“The probability that this student failed the French test is \( \frac{1}{4} \)”
Omar is wrong.
(b) Explain why.
Worked Solution
Part (a): Stem and Leaf
Strategy: Group the numbers by their “tens” digit. The tens digit forms the Stem, the units digit forms the Leaf. Leaves must be in order.
60s: 69, 64, 67, 69 → Ordered: 4, 7, 9, 9
70s: 76, 70, 75, 70, 76, 71, 77 → Ordered: 0, 0, 1, 5, 6, 6, 7
80s: 82, 84, 80, 81, 87, 80, 81 → Ordered: 0, 0, 1, 1, 2, 4, 7
90s: 91, 94 → Ordered: 1, 4
Diagram:
6 | 4 7 9 9
7 | 0 0 1 5 6 6 7
8 | 0 0 1 1 2 4 7
9 | 1 4
Key: 6 | 4 = 64 marks
✓ (B2)
Part (b): Probability Check
Strategy: Calculate the actual probability of failing. Pass mark is 71, so “Fail” is less than 71.
Marks less than 71: 64, 67, 69, 69, 70, 70.
Count = 6 students.
Total students = 20.
Probability(Fail) = \( \frac{6}{20} = \frac{3}{10} \)
Omar said \( \frac{1}{4} \) which is \( \frac{5}{20} \).
Reason: 6 students failed, not 5. \( \frac{6}{20} \neq \frac{1}{4} \).
✓ (C1)
Question 15 (2 marks)
Jenny is asked to find the value of \( 12 – 2 \times 4 \)
Here is her working:
\( 12 – 2 \times 4 = 10 \times 4 = 40 \)
Jenny’s answer is wrong.
(a) Explain what Jenny has done wrong.
Rehan is asked to find the range of the numbers: 3, 1, 8, 7, 5
Here is his working:
Range = \( 5 – 3 = 2 \)
This is wrong.
(b) Explain why.
Worked Solution
Part (a): Order of Operations
Why we do this: In math, we must follow BIDMAS/BODMAS (Brackets, Indices, Division/Multiplication, Addition/Subtraction).
Jenny subtracted 2 from 12 first.
She should have done the multiplication (\( 2 \times 4 \)) before the subtraction.
✓ (C1)
Part (b): Range Definition
Why we do this: Range is the difference between the largest and smallest values.
Rehan used the first and last numbers in the list (5 and 3).
He should have used the largest (8) and smallest (1).
Correct range: \( 8 – 1 = 7 \).
✓ (C1)
Question 16 (5 marks)
Alan, Bispah and Chan share a sum of money.
Alan gets \( \frac{1}{8} \) of the money.
Bispah gets \( \frac{1}{2} \) of the money.
Chan gets the rest of the money.
Alan gets £2.50.
(a) Work out how much money Bispah gets.
(b) Find the ratio
amount of money Alan gets : amount of money Chan gets
Give your answer in the form \( a : b \) where \( a \) and \( b \) are whole numbers.
Worked Solution
Part (a): Finding Bispah’s share
Strategy: First find the total amount of money. If \( \frac{1}{8} \) is £2.50, we can find the total.
\( \frac{1}{8} \text{ Total} = £2.50 \)
Total = \( £2.50 \times 8 = £20.00 \)
Bispah gets \( \frac{1}{2} \) of the total.
Bispah = \( \frac{1}{2} \times 20 = £10.00 \)
✓ (M1) method to find total or share
✓ (A1) £10
Part (b): Finding the ratio
Strategy: Calculate Chan’s share first, then write the ratio.
Total = £20
Alan = £2.50
Bispah = £10.00
Chan = Total – Alan – Bispah
Chan = \( 20 – 2.50 – 10 = £7.50 \)
Ratio Alan : Chan
\( 2.50 : 7.50 \)
Multiply by 10 to remove decimals: \( 25 : 75 \)
Divide by 25 to simplify: \( 1 : 3 \)
✓ (A1) for 1:3
Final Answer (a): £10.00
Final Answer (b): 1 : 3
Question 17 (3 marks)
\( ABC \) is an isosceles right-angled triangle.
The area of the triangle is 162 cm².
Work out the value of \( x \).
Worked Solution
Step 1: Area Formula
Why we do this: Area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).
Base = \( 3x \)
Height = \( 3x \)
Area = \( \frac{1}{2} \times 3x \times 3x \)
\( 162 = \frac{1}{2} \times 9x^2 \)
Step 2: Solve for x
Multiply both sides by 2 to remove fraction:
\( 324 = 9x^2 \)
Divide by 9:
\( x^2 = 36 \)
Square root:
\( x = \sqrt{36} = 6 \)
✓ (P1) set up equation
✓ (P1) simplify to \( x^2 = 36 \)
Final Answer: \( x = 6 \)
✓ (A1)
Question 18 (2 marks)
Work out the value of
\( \frac{2.645 \times 10^9}{1.15 \times 10^3} \)
Give your answer in standard form.
Worked Solution
Step 1: Calculation
Strategy: You can use the specific standard form buttons on your calculator (often labelled \(\times 10^x\) or EXP).
Alternatively, divide the numbers and subtract the powers.
Numbers: \( 2.645 \div 1.15 = 2.3 \)
Powers: \( 10^9 \div 10^3 = 10^{9-3} = 10^6 \)
Final Answer: \( 2.3 \times 10^6 \)
✓ (A1)
Question 19 (3 marks)
The scatter diagram shows information about 12 girls. It shows the age of each girl and the best time she takes to run 100 metres.
(a) Write down the type of correlation.
Kristina is 11 years old. Her best time to run 100 metres is 12 seconds.
The point representing this information would be an outlier on the scatter diagram.
(b) Explain why.
Debbie is 15 years old. Debbie says, “The scatter diagram shows I should take less than 12 seconds to run 100 metres.”
(c) Comment on what Debbie says.
Worked Solution
Part (a)
As age increases, time decreases. This is Negative correlation.
Part (b)
Explanation: The point (11, 12) does not fit the pattern of the other points (it is far away from the line of best fit).
✓ (C1)
Part (c)
Comment: 15 years old is outside the range of the data (extrapolation). We cannot be sure the pattern continues.
✓ (C1)
Question 20 (2 marks)
Expand and simplify \( 5(p + 3) – 2(1 – 2p) \)
Worked Solution
Step 1: Expand Brackets
Warning: Be very careful with the negative sign in front of the second bracket. \( -2 \times -2p \) becomes \( +4p \).
\( 5(p) + 5(3) = 5p + 15 \)
\( -2(1) – 2(-2p) = -2 + 4p \)
Combined: \( 5p + 15 – 2 + 4p \)
✓ (M1) for correct expansion
Step 2: Simplify
Collect p’s: \( 5p + 4p = 9p \)
Collect numbers: \( 15 – 2 = 13 \)
Final Answer: \( 9p + 13 \)
✓ (A1)
Question 21 (2 marks)
Here is a trapezium drawn on a centimetre grid.
On the grid, draw a triangle equal in area to this trapezium.
Worked Solution
Step 1: Calculate Area of Trapezium
Formula: Area = average of parallel sides × height.
Top side (a) = 2 squares.
Bottom side (b) = 5 squares (wait, looking at diagram: Top is 2 units, Base starts at x=2 ends at x=7, that’s 5? Let’s recount from visual in PDF. PDF Q21: Top is 2 squares. Bottom is 7 squares. Height is 4 squares.)
Correction: Top = 2, Bottom = 7, Height = 4.
Area = \( \frac{2 + 7}{2} \times 4 \)
Area = \( 4.5 \times 4 = 18 \) squares.
✓ (M1)
Step 2: Draw Triangle
Strategy: Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \)
We need \( \frac{1}{2} \times b \times h = 18 \)
So \( b \times h = 36 \).
Possible dimensions: Base 6, Height 6 OR Base 9, Height 4.
Action: Draw a triangle with base 6cm and height 6cm (or equivalent).
✓ (A1)
Question 22 (2 marks)
When a biased 6-sided dice is thrown once, the probability that it will land on 4 is 0.65.
The biased dice is thrown twice.
Amir draws this probability tree diagram. The diagram is not correct.
Write down two things that are wrong with the probability tree diagram.
Worked Solution
Error 1: First Branch Probabilities
Reasoning: Probabilities on branches coming from a single point must sum to 1.
\( 1 – 0.65 = 0.35 \), but Amir wrote 0.25.
✓ (C1) “Probabilities should sum to 1 (0.25 should be 0.35)”
Error 2: Second Throw Probabilities
Reasoning: The dice is the same, so the probability of landing on 4 should stay 0.65.
On the top right branches, Amir has swapped them (Label “Land on 4” has prob 0.35).
✓ (C1) “0.35 and 0.65 are reversed on the second throw”
Question 23 (3 marks)
ABC is a right-angled triangle.
(a) Work out the size of angle ABC.
Give your answer correct to 1 decimal place.
(b) The length of side AB is reduced by 1 cm. The length of side BC is still 7 cm. Angle ACB is still 90°.
Will the value of cos ABC increase or decrease? You must give a reason.
Worked Solution
Part (a): Trigonometry
Strategy: We have the Adjacent side (7cm) and the Hypotenuse (11cm) relative to angle B. We need to find the angle.
SOH CAH TOA → Use Cosine.
\( \cos(B) = \frac{\text{Adj}}{\text{Hyp}} = \frac{7}{11} \)
\( B = \cos^{-1}(\frac{7}{11}) \)
Calculator: SHIFT cos ( 7 ÷ 11 ) =
Result: 50.478…
Round to 1 d.p: 50.5°
✓ (M1) for cos statement
✓ (A1) for 50.5
Part (b): Reasoning
Analysis: \( \cos(B) = \frac{7}{\text{Hypotenuse}} \).
If Hypotenuse decreases (from 11 to 10), we are dividing 7 by a smaller number.
\( \frac{7}{10} \) is greater than \( \frac{7}{11} \).
Therefore, the value of cos ABC will increase.
✓ (C1)
Question 24 (5 marks)
There are some counters in a bag. The counters are red or white or blue or yellow.
| Colour | red | white | blue | yellow |
|---|---|---|---|---|
| Probability | 0.45 | 0.25 |
There are 18 blue counters in the bag.
The probability that the counter Bob takes will be red is twice the probability that the counter will be white.
(a) Work out the number of red counters in the bag.
A marble is going to be taken at random from a box of marbles. The probability that the marble will be silver is 0.5. There must be an even number of marbles in the box.
(b) Explain why.
Worked Solution
Step 1: Find Total Counters
Strategy: We know the probability of blue is 0.45 and the number of blue counters is 18. We can use this to find the total number of counters.
\( 0.45 \times \text{Total} = 18 \)
\( \text{Total} = 18 \div 0.45 = 40 \)
✓ (P1)
Step 2: Find missing probabilities
Sum of probs = 1.
Red + White + 0.45 + 0.25 = 1
Red + White = \( 1 – 0.70 = 0.30 \)
We are told Red is twice White (\( R = 2W \)).
\( 2W + W = 0.30 \)
\( 3W = 0.30 \rightarrow W = 0.10 \)
\( R = 0.20 \)
✓ (P1)
Step 3: Calculate Red Counters
Number of Red = \( 0.20 \times \text{Total} \)
\( 0.20 \times 40 = 8 \)
Final Answer: 8
✓ (A1)
Part (b): Explanation
Reason: Probability 0.5 means exactly half the marbles are silver.
You can only take exactly half of an even number (you can’t have half a marble).
✓ (C1)
Question 25 (3 marks)
Solve \( \frac{5-x}{2} = 2x – 7 \)
Worked Solution
Step 1: Eliminate the fraction
Strategy: Multiply both sides by 2.
\( 5 – x = 2(2x – 7) \)
\( 5 – x = 4x – 14 \)
✓ (M1)
Step 2: Collect terms
Strategy: Get x’s on one side, numbers on the other.
Add \( x \) to both sides:
\( 5 = 5x – 14 \)
Add 14 to both sides:
\( 19 = 5x \)
✓ (M1)
Step 3: Solve
\( x = \frac{19}{5} = 3.8 \)
Final Answer: 3.8
✓ (A1)
Question 26 (5 marks)
ABCDE is a pentagon.
Angle BCD = \( 2 \times \) angle ABC
Work out the size of angle BCD.
Worked Solution
Step 1: Total angles in a pentagon
Formula: Sum of interior angles = \( (n-2) \times 180 \)
\( (5 – 2) \times 180 = 3 \times 180 = 540^\circ \)
✓ (P1)
Step 2: Form equation
Known angles: A=125, E=115, D=90.
Unknowns: Let \( \angle ABC = x \). Then \( \angle BCD = 2x \).
\( 125 + 115 + 90 + x + 2x = 540 \)
\( 330 + 3x = 540 \)
✓ (P1)
Step 3: Solve
\( 3x = 540 – 330 \)
\( 3x = 210 \)
\( x = 70 \)
We need angle BCD (\( 2x \)).
\( 2 \times 70 = 140^\circ \)
✓ (P1) solve for x
Final Answer: 140°
✓ (A1)
Question 27 (4 marks)
Triangle ABC and triangle DEF are similar.
(a) Work out the length of DF.
(b) Work out the length of CB.
Worked Solution
Step 1: Find Scale Factor
Strategy: Compare corresponding sides. Side AB corresponds to side DE.
Scale Factor = \( \frac{12.6}{8.4} = 1.5 \)
The big triangle is 1.5 times larger.
✓ (M1)
Step 2: Calculate DF
DF corresponds to AC (6.4 cm).
\( 6.4 \times 1.5 = 9.6 \) cm
Final Answer (a): 9.6 cm
✓ (A1)
Part (b): Calculate CB
Strategy: CB is on the smaller triangle. We divide the corresponding side on the large triangle (FE) by the scale factor.
CB = \( 15 \div 1.5 = 10 \) cm
Final Answer (b): 10 cm
✓ (A1)
Question 28 (3 marks)
Make \( g \) the subject of the formula \( T = \sqrt{\frac{g+6}{2}} \)
Worked Solution
Step 1: Remove the square root
Strategy: Square both sides.
\( T^2 = \frac{g+6}{2} \)
✓ (M1)
Step 2: Remove the fraction
Strategy: Multiply both sides by 2.
\( 2T^2 = g + 6 \)
✓ (M1)
Step 3: Isolate g
Subtract 6 from both sides:
\( 2T^2 – 6 = g \)
Final Answer: \( g = 2T^2 – 6 \)
✓ (A1)
