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Edexcel GCSE Mathematics (Higher) – Nov 2017 Paper 3
Mark Scheme Legend
- M: Method mark (for a correct method)
- P: Process mark (for a correct process in problem-solving)
- A: Accuracy mark (dependent on a correct method/process)
- B: Unconditional accuracy mark (no method needed)
- C: Communication mark
- oe: Or equivalent
- cao: Correct answer only
Table of Contents
- Question 1 (Grouped Data & Frequency Polygon)
- Question 2 (Ratio & Currency Conversion)
- Question 3 (Density, Mass, and Volume)
- Question 4 (Ratio Combination)
- Question 5 (Reciprocals & Error Intervals)
- Question 6 (Algebra & Compound Perimeter)
- Question 7 (Standard Form)
- Question 8 (Relative Frequency)
- Question 9 (Depreciation & Reverse Percentage)
- Question 10 (Probability & Least Common Multiples)
- Question 11 (Cumulative Frequency)
- Question 12 (Probability Trees)
- Question 13 (Completing the Square)
- Question 14 (Similar Shapes – Area & Volume)
- Question 15 (Iteration & Roots)
- Question 16 (Upper and Lower Bounds)
- Question 17 (Sine Rule, Cosine Rule & Triangle Area)
- Question 18 (Speed-Time Graph Area under Curve)
- Question 19 (Coordinate Geometry – Circle Tangent)
- Question 20 (Circle Geometry Proof)
- Question 21 (Vectors)
Question 1 (3 marks)
The table shows information about the heights of 80 children.
| Height (\(h\) cm) | Frequency |
|---|---|
| \(130 < h \le 140\) | 4 |
| \(140 < h \le 150\) | 11 |
| \(150 < h \le 160\) | 24 |
| \(160 < h \le 170\) | 22 |
| \(170 < h \le 180\) | 19 |
(a) Find the class interval that contains the median.
(b) Draw a frequency polygon for the information in the table.
Worked Solution
Part (a): Finding the Median Class Interval
What are we being asked to find?
We need to find which height group contains the median child. The median is the middle value when all items are arranged in order of size.
Why we do this:
There are 80 children in total. The median child will be halfway through the data. We can find this position by halving the total frequency.
Working:
Position of median = \(80 \div 2 = 40^{\text{th}}\) child.
We add the frequencies (running total) until we pass the 40th child:
- First group: 4 children (up to 4)
- Second group: \(4 + 11 = 15\) children (up to 15)
- Third group: \(15 + 24 = 39\) children (up to 39)
- Fourth group: \(39 + 22 = 61\) children (up to 61)
Since the 40th child falls in the fourth group, the median class interval is \(160 < h \le 170\).
✓ (B1) for identifying the correct class interval.
Part (b): Drawing the Frequency Polygon
Why we do this:
To plot a frequency polygon for grouped continuous data, we plot the frequency against the midpoint of each class interval, and then join the points with straight lines.
Working:
First, calculate the midpoint of each interval:
- \(130 < h \le 140\) \(\rightarrow\) Midpoint = 135. Plot \((135, 4)\)
- \(140 < h \le 150\) \(\rightarrow\) Midpoint = 145. Plot \((145, 11)\)
- \(150 < h \le 160\) \(\rightarrow\) Midpoint = 155. Plot \((155, 24)\)
- \(160 < h \le 170\) \(\rightarrow\) Midpoint = 165. Plot \((165, 22)\)
- \(170 < h \le 180\) \(\rightarrow\) Midpoint = 175. Plot \((175, 19)\)
✓ (C2) for a fully correct frequency polygon drawn through the correct midpoints. (C1 for points correctly plotted but not joined, or joined but plotted at the interval boundaries instead of midpoints).
Final Answer:
(a) \(160 < h \le 170\)
(b) See correctly drawn frequency polygon above.
Total: 3 marks
Question 2 (3 marks)
In London, 1 litre of petrol costs 108.9p.
In New York, 1 US gallon of petrol costs $2.83.
1 US gallon = 3.785 litres.
£1 = $1.46.
In which city is petrol better value for money, London or New York?
You must show your working.
Worked Solution
Step 1: Understanding the Strategy
What are we being asked to find?
We need to compare two prices given in entirely different units (pence per litre vs dollars per US gallon). To compare them fairly, we must convert them to the same volume unit and the same currency.
Why we do this:
Let’s convert the London price to match New York’s format: Dollars per US gallon.
Step 2: Converting London’s price
Working:
First, turn the London price into pounds (£):
108.9p = £1.089 per litre
Next, find out how much this would cost for a whole US gallon (which is 3.785 litres):
Calculator Steps:
- Cost per gallon = £1.089 \(\times\) 3.785
- = £4.121865 per gallon
Now, convert this cost from Pounds to Dollars so we can compare directly with New York:
- £1 = $1.46
- Cost in Dollars = £4.121865 \(\times\) 1.46
- = $6.0179… per gallon
✓ (P1) for a complete process to give values that can be used for comparison.
What this tells us:
In London, petrol costs roughly $6.02 for a US gallon.
Step 3: Comparing the values
Working:
Cost in London: $6.02 per US gallon.
Cost in New York: $2.83 per US gallon.
Comparing the two: $2.83 is much less than $6.02.
✓ (C1) for New York (supported by correct comparative values).
Final Answer:
New York is better value for money (supported by $2.83 compared to $6.02 per gallon).
Total: 3 marks
Question 3 (3 marks)
A gold bar has a mass of 12.5 kg.
The density of gold is \(19.3 \text{ g/cm}^3\).
Work out the volume of the gold bar.
Give your answer correct to 3 significant figures.
Worked Solution
Step 1: Check Units
Why we do this:
Density is given in grams per cubic centimetre (\(\text{g/cm}^3\)), but the mass is given in kilograms (kg). We must convert the mass into grams so the units match before using the formula.
Working:
1 kg = 1000 g.
Mass = \(12.5 \times 1000 = 12500 \text{ g}\).
Step 2: Calculating Volume
Why we do this:
The relationship between Density, Mass, and Volume is given by the formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We need to rearrange this to find Volume.
Working:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substitute our values into the formula:
\[ \text{Volume} = \frac{12500}{19.3} \]
✓ (M2) for a complete method (e.g., \(12.5 \times 1000 \div 19.3\)).
Calculator Steps:
- Press:
12500÷19.3= - Calculator shows: 647.6683938…
Step 3: Rounding the Answer
Working:
We need to give the answer to 3 significant figures.
Look at the first three important digits: 6, 4, 7.
The next digit is 6, which is 5 or more, so we round the last significant digit up.
647.6… rounds up to 648.
✓ (A1) for an answer in the range 647 to 648.
Final Answer:
648 cm³
Total: 3 marks
Question 4 (3 marks)
There are only blue pens, green pens and red pens in a box.
The ratio of the number of blue pens to the number of green pens is 2 : 5
The ratio of the number of green pens to the number of red pens is 4 : 1
There are less than 100 pens in the box.
What is the greatest possible number of red pens in the box?
Worked Solution
Step 1: Combining the Ratios
What are we being asked to find?
We are given two separate ratios that share a common element (green pens). To solve the problem, we must combine these into a single 3-part ratio (Blue : Green : Red).
Working:
Blue : Green = 2 : 5
Green : Red = 4 : 1
Notice that ‘Green’ is represented by 5 in the first ratio and 4 in the second ratio. We need to make these numbers the same.
The lowest common multiple of 5 and 4 is 20.
Multiply the first ratio by 4:
Blue : Green = \((2 \times 4)\) : \((5 \times 4)\) = 8 : 20
Multiply the second ratio by 5:
Green : Red = \((4 \times 5)\) : \((1 \times 5)\) = 20 : 5
✓ (P1) strategy to start the problem, e.g., 8:20 and 20:5.
Now we can combine them into a single ratio:
Blue : Green : Red = 8 : 20 : 5
Step 2: Finding Total Parts and Maximum Pens
Why we do this:
The total number of pens must be a multiple of the total parts in our ratio. We need to find the largest multiple that is strictly less than 100.
Working:
Total parts = \(8 + 20 + 5 = 33\) parts.
This means the total number of pens in the box must be a multiple of 33.
Let’s list the multiples of 33:
33, 66, 99, 132…
We are told there are less than 100 pens.
The greatest possible total number of pens is 99 (since \(33 \times 3 = 99\)).
✓ (P1) process to solve the problem (finding multiplier).
Step 3: Calculating Number of Red Pens
Why we do this:
We know the maximum total pens represent exactly 3 sets of our base ratio (since \(99 \div 33 = 3\)). We just multiply the ‘Red’ ratio part by 3.
Working:
Multiplier = 3
Red parts = 5
Number of Red pens = \(5 \times 3 = 15\)
✓ (A1) for 15.
Final Answer:
15
Total: 3 marks
Question 5 (3 marks)
(a) Find the value of the reciprocal of 1.6
Give your answer as a decimal.
Jess rounds a number, \(x\), to one decimal place.
The result is 9.8
(b) Write down the error interval for \(x\).
Worked Solution
Part (a): Finding the Reciprocal
What is a reciprocal?
The reciprocal of a number is 1 divided by that number. For a fraction, you flip the fraction upside down.
Working:
Reciprocal of 1.6 = \(\frac{1}{1.6}\)
Calculator Method:
Press 1 ÷ 1.6 =
Calculator displays: 0.625
(Without a calculator, you could do: \(1.6 = \frac{16}{10} = \frac{8}{5}\). Flipped, it is \(\frac{5}{8}\), which is 0.625.)
✓ (B1) for 0.625.
Part (b): Finding the Error Interval
What is an error interval?
When a number is rounded to 1 decimal place to become 9.8, we need to find the smallest possible value it could have been (lower bound) and the theoretical upper limit before it would round up to 9.9.
Working:
The number was rounded to 1 decimal place (0.1).
To find the bounds, halve the rounding degree: \(0.1 \div 2 = 0.05\).
Lower bound: \(9.8 – 0.05 = 9.75\)
Upper bound: \(9.8 + 0.05 = 9.85\)
The error interval is written as an inequality. It includes the lower bound (as 9.75 rounds up to 9.8) but excludes the upper bound (as 9.85 would round up to 9.9).
Therefore: \(9.75 \le x < 9.85\)
✓ (B2) for \(9.75 \le x < 9.85\). (B1 for identifying 9.75 or 9.85).
Final Answer:
(a) 0.625
(b) \(9.75 \le x < 9.85\)
Total: 3 marks
Question 6 (5 marks)
Here is a rectangle.
The length of the rectangle is 7 cm longer than the width of the rectangle.
4 of these rectangles are used to make this 8-sided shape.
The perimeter of the 8-sided shape is 70 cm.
Work out the area of the 8-sided shape.
Worked Solution
Step 1: Setting up an algebraic expression for the perimeter
What are we being asked to find?
We need to find the total area of the shape. To do this, we first need to find the dimensions of one of the rectangles. We can use algebra to find the unknown width, \(x\).
Why we do this:
We are told the total perimeter is 70 cm. If we define the width as \(x\) and the length as \(x + 7\), we can walk around the outside of the 8-sided shape and add up all the side lengths to create an equation.
Working:
Let width = \(x\)
Let length = \(x + 7\)
✓ (P1) starts process, e.g., uses \(x\) and \(x + 7\).
Tracing around the 8 outer edges of the shape, we have 2 sides that are widths (\(x\)) and 6 sides that are lengths (\(x + 7\)).
Perimeter = \(x + (x + 7) + (x + 7) + (x + 7) + x + (x + 7) + (x + 7) + (x + 7)\)
✓ (P1) starts to work with at least 6 correct sides.
Simplify the expression by collecting like terms:
We have 8 \(x\)’s in total, and six 7’s.
Perimeter = \(8x + (6 \times 7) = 8x + 42\)
✓ (P1) gives a correct expression for the perimeter and equates to 70.
Step 2: Solving the equation for \(x\)
Why we do this:
Now that we have an expression for the perimeter, we can set it equal to the given perimeter (70 cm) and solve for \(x\) to find the physical dimensions of the rectangle.
Working:
\[ 8x + 42 = 70 \]
Subtract 42 from both sides:
\[ 8x = 28 \]
Divide by 8:
\[ x = \frac{28}{8} = 3.5 \]
So, the width is \(3.5\text{ cm}\).
The length is \(x + 7 = 3.5 + 7 = 10.5\text{ cm}\).
✓ (A1) for width = 3.5 and length = 10.5.
Step 3: Calculating the total area
Why we do this:
The total shape is made of 4 identical rectangles. Once we find the area of one rectangle, we multiply it by 4 to get the final answer.
Working:
Area of one rectangle = \(\text{width} \times \text{length}\)
Area = \(3.5 \times 10.5\)
Calculator Steps:
- Press:
3.5×10.5= - Calculator shows: 36.75
There are 4 rectangles in the shape:
Total Area = \(4 \times 36.75 = 147\text{ cm}^2\)
✓ (B1) for correct area using their \(x\).
What this tells us:
The total area of the 8-sided shape is exactly 147 square centimetres.
Final Answer:
147 cm²
Total: 5 marks
Question 7 (2 marks)
Work out \((13.8 \times 10^7) \times (5.4 \times 10^{-12})\)
Give your answer as an ordinary number.
Worked Solution
Step 1: Grouping similar terms
Why we do this:
When multiplying numbers in standard form, it’s easiest to group the regular numbers together and group the powers of 10 together. This uses the commutative property of multiplication (the order doesn’t matter).
Working:
\[ (13.8 \times 5.4) \times (10^7 \times 10^{-12}) \]
First, multiply the numbers:
\(13.8 \times 5.4 = 74.52\)
✓ (M1) digits 7452 seen.
Next, multiply the powers of 10 by adding the indices:
\(10^7 \times 10^{-12} = 10^{7 + (-12)} = 10^{-5}\)
Combine them back together:
\[ 74.52 \times 10^{-5} \]
Step 2: Converting to an ordinary number
Why we do this:
The question specifically asks for an ordinary number, not standard form. Multiplying by \(10^{-5}\) means we need to move the decimal point 5 places to the left (making the number smaller).
Working:
Start with: \(74.52\)
Move decimal 5 places left:
1 place: \(7.452\)
2 places: \(0.7452\)
3 places: \(0.07452\)
4 places: \(0.007452\)
5 places: \(0.0007452\)
✓ (A1) for correct answer only.
Final Answer:
0.0007452
Total: 2 marks
Question 8 (3 marks)
When a drawing pin is dropped it can land point down or point up.
Lucy, Mel and Tom each dropped the drawing pin a number of times.
The table shows the number of times the drawing pin landed point down and the number of times the drawing pin landed point up for each person.
Rachael is going to drop the drawing pin once.
(a) Whose results will give the best estimate for the probability that the drawing pin will land point up?
Give a reason for your answer.
Stuart is going to drop the drawing pin twice.
(b) Use all the results in the table to work out an estimate for the probability that the drawing pin will land point up the first time and point down the second time.
Worked Solution
Part (a): Evaluating Best Estimates
What are we being asked to find?
We need to decide which person’s individual data is the most reliable for calculating probability, and explain why.
Why we do this:
In experimental probability (relative frequency), the more trials you conduct, the more reliable and accurate your estimate becomes. We just need to total up how many drops each person did.
Working:
- Lucy’s total drops: \(31 + 14 = 45\)
- Mel’s total drops: \(53 + 27 = 80\)
- Tom’s total drops: \(16 + 9 = 25\)
Mel did the most drops (80).
✓ (B1) for choosing Mel with reference to the greatest number of throws.
Part (b): Calculating Combined Probability
Why we do this:
The question asks us to use all the results. We must pool everyone’s data together to get the best overall estimate for the probabilities of landing ‘up’ and ‘down’. Then, we use the “AND” rule in probability (multiplying) for two independent events: landing Up and then landing Down.
Working:
First, find the total combined data:
- Total Point Down = \(31 + 53 + 16 = 100\)
- Total Point Up = \(14 + 27 + 9 = 50\)
- Overall Total Drops = \(100 + 50 = 150\)
Calculate the individual probabilities:
\[ \text{P(Up)} = \frac{50}{150} = \frac{1}{3} \]
\[ \text{P(Down)} = \frac{100}{150} = \frac{2}{3} \]
Now, calculate the combined probability for (Up then Down):
\[ \text{P(Up then Down)} = \text{P(Up)} \times \text{P(Down)} \]
\[ = \frac{50}{150} \times \frac{100}{150} \]
\[ = \frac{1}{3} \times \frac{2}{3} \]
✓ (M1) for selecting overall total and multiplying P(point up) × P(point down).
\[ = \frac{2}{9} \]
✓ (A1) for \(\frac{2}{9}\) or equivalent.
Final Answer:
(a) Mel, because she did the greatest number of throws.
(b) \(\frac{2}{9}\)
Total: 3 marks
Question 9 (5 marks)
Jack bought a new boat for £12 500
The value, \(V\), of Jack’s boat at the end of \(n\) years is given by the formula
\[ V = 12500 \times (0.85)^n \]
(a) At the end of how many years was the value of Jack’s boat first less than 50% of the value of the boat when it was new?
A savings account pays interest at a rate of \(R\%\) per year.
Jack invests £5500 in the account for one year.
At the end of the year, Jack pays tax on the interest at a rate of 40%.
After paying tax, he gets £79.20
(b) Work out the value of \(R\).
Worked Solution
Part (a): Finding the Depreciation Year
What are we being asked to find?
We need to find the smallest whole number \(n\) (years) where the boat’s value drops below half of its original £12,500.
Working:
First, calculate 50% of the new value:
\[ 50\% \text{ of } 12500 = 12500 \div 2 = 6250 \]
Now, use trial and improvement with the formula \(V = 12500 \times (0.85)^n\) until \(V < 6250\):
✓ (M1) for evaluating \(12500 \times (0.85)^n\) for at least one value of \(n\).
- Year 1 (\(n=1\)): \(12500 \times 0.85^1 = 10625\)
- Year 2 (\(n=2\)): \(12500 \times 0.85^2 = 9031.25\)
- Year 3 (\(n=3\)): \(12500 \times 0.85^3 = 7676.56…\)
- Year 4 (\(n=4\)): \(12500 \times 0.85^4 = 6525.07…\)
- Year 5 (\(n=5\)): \(12500 \times 0.85^5 = 5546.31…\)
At the end of Year 5, the value (£5546) is finally less than £6250.
✓ (A1) for 5.
Part (b): Finding the Interest Rate \(R\)
Why we do this:
We are given the net interest (what he gets after 40% tax is removed). To find the original interest rate, we first need to work backwards to find the gross interest (the total interest before tax). Since 40% was taken as tax, the £79.20 represents the remaining 60% of the interest.
Working:
Let Total Interest before tax = \(I\).
He keeps 60% of the interest (since \(100\% – 40\% = 60\%\)).
\[ 0.60 \times I = 79.20 \]
\[ I = \frac{79.20}{0.60} \]
\[ I = \text{£}132 \]
✓ (P1) for process to find the amount of interest before tax (e.g., \(79.20 \div 0.6 = 132\)).
Now, we know Jack earned £132 in interest from his £5500 investment in one year.
To find the interest rate \(R\) as a percentage, we divide the interest by the principal investment and multiply by 100:
\[ R = \left( \frac{132}{5500} \right) \times 100 \]
✓ (P1) for process to find value of R.
Calculator Steps:
- Press:
132÷5500×100= - Calculator shows: 2.4
✓ (A1) for correct answer only.
Final Answer:
(a) 5
(b) 2.4
Total: 5 marks
Question 10 (3 marks)
There are only blue counters, yellow counters, green counters and red counters in a bag.
A counter is taken at random from the bag.
The table shows the probabilities of getting a blue counter or a yellow counter or a green counter.
(a) Work out the probability of getting a red counter.
(b) What is the least possible number of counters in the bag?
You must give a reason for your answer.
Worked Solution
Part (a): Finding the Missing Probability
Why we do this:
The sum of all possible mutually exclusive probabilities in a complete set must always equal exactly 1. We just add up the known probabilities and subtract from 1 to find the missing one.
Working:
Sum of known probabilities = \(0.2 + 0.35 + 0.4 = 0.95\)
P(red) = \(1 – 0.95 = 0.05\)
✓ (B1) for 0.05 or equivalent.
Part (b): Finding the Least Possible Counters
What are we being asked to find?
We need to find the smallest number of counters that allows all these probabilities to happen perfectly. You cannot have a fraction of a counter, so every probability must result in a whole number when multiplied by the total.
Working:
Let’s convert all the decimal probabilities into fractions over 100:
- Blue = \(\frac{20}{100}\)
- Yellow = \(\frac{35}{100}\)
- Green = \(\frac{40}{100}\)
- Red = \(\frac{5}{100}\)
Now, simplify the smallest probability (Red) to its simplest form:
Red = \(\frac{5}{100} = \frac{1}{20}\)
This tells us that exactly 1 out of every 20 counters is red. For this to be physically possible (having at least 1 whole red counter), there must be a minimum of 20 counters in the bag.
We can double-check the other colours with a total of 20:
- Blue: \(0.2 \times 20 = 4\) counters
- Yellow: \(0.35 \times 20 = 7\) counters
- Green: \(0.4 \times 20 = 8\) counters
- Red: \(0.05 \times 20 = 1\) counter
Since all these answers are whole numbers, 20 works perfectly.
✓ (C1) for stating that at least 20 are required.
✓ (C1) for the reason explaining that the number of each colour must be a whole number, OR showing \(0.05 = \frac{1}{20}\) so there must be at least 1 red counter.
Final Answer:
(a) 0.05
(b) 20. Reason: 0.05 is the same as the fraction \(\frac{1}{20}\). Because you must have a whole number of counters, there must be at least 1 red counter, which requires a minimum of 20 total counters in the bag.
Total: 3 marks
Question 11 (4 marks)
The cumulative frequency graph shows information about the weights of 60 potatoes.
(a) Use the graph to find an estimate for the median weight.
Jamil says,
“\(80 – 40 = 40\) so the range of the weights is 40 g.”
(b) Is Jamil correct?
You must give a reason for your answer.
(c) Show that less than 25% of the potatoes have a weight greater than 65 g.
Worked Solution
Part (a): Estimating the Median
What are we being asked to find?
We need to find the weight of the median (middle) potato using the graph.
Working:
There are 60 potatoes in total.
The median is at the \(\frac{60}{2} = 30^{\text{th}}\) value.
Look at the y-axis (Cumulative frequency) and find 30. Draw a horizontal line across to the curve, and then a vertical line straight down to the x-axis (Weight).
Following this path on the graph, the line comes down at 57 g.
✓ (B1) for 57 (accept 56 to 58 depending on reading accuracy).
Part (b): Evaluating the Range
Why we do this:
Range is calculated as the Maximum value minus the Minimum value. We need to look at the graph and see if 80 and 40 are definitely the exact maximum and minimum weights of individual potatoes.
Working:
The curve starts at 40g and ends at 80g. These are the boundaries of the classes (e.g., the grouped intervals).
However, this is grouped data. The lightest potato could be anywhere between the lowest boundary, and the heaviest potato could be anywhere up to 80g, but might just be 78g, for example.
We do not know the exact weights of the individual potatoes.
✓ (C1) Jamil might not be correct. Reason: e.g., the maximum weight could be less than 80, or the minimum weight could be greater than 40.
Part (c): Checking the Top 25%
What are we being asked to find?
We need to prove that the number of potatoes heavier than 65g is less than a quarter (25%) of the total crop.
Working:
First, calculate how many potatoes make up 25% of the total:
\(25\% \text{ of } 60 = \frac{1}{4} \times 60 = 15\) potatoes.
Next, use the graph to find how many potatoes weigh up to 65 g:
Find 65 g on the x-axis, draw a line up to the curve, and across to the y-axis.
The cumulative frequency at 65 g is approximately 49.
✓ (C1) for evidence of reading from the graph at weight 65 (gives value 48 to 49).
This means 49 potatoes weigh 65 g or less. To find how many are greater than 65 g:
\(60 – 49 = 11\) potatoes.
Since 11 potatoes is less than 15 potatoes, it is true that less than 25% have a weight greater than 65 g.
✓ (C1) for clear explanation (e.g., 25% of 60 is 15 but only 11 potatoes have a weight greater than 65g).
Final Answer:
(a) 57 g
(b) No, because we only have grouped data. The exact minimum and maximum weights are unknown.
(c) 25% of 60 = 15. The graph shows ~49 potatoes weigh 65g or less, meaning only 11 weigh more. 11 < 15, so the statement is true.
Total: 4 marks
Question 12 (3 marks)
Alan has two spinners, spinner A and spinner B.
Each spinner can land on only red or white.
The probability that spinner A will land on red is 0.25
The probability that spinner B will land on red is 0.6
The probability tree diagram shows this information.
Alan spins spinner A once and he spins spinner B once.
He does this a number of times.
The number of times both spinners land on red is 24
Work out an estimate for the number of times both spinners land on white.
Worked Solution
Step 1: Finding the Probabilities of Specific Outcomes
What are we being asked to find?
We know how many times “Red AND Red” happened. We want to find out how many times “White AND White” happened. First, we need the theoretical probabilities of these two specific paths along the tree diagram.
Working:
To find the probability of a combined path on a tree diagram, we multiply along the branches.
Probability of (Red AND Red):
\[ \text{P(Red, Red)} = 0.25 \times 0.6 \]
\[ \text{P(Red, Red)} = 0.15 \]
✓ (M1) for \(0.25 \times 0.6 = 0.15\).
Probability of (White AND White):
\[ \text{P(White, White)} = 0.75 \times 0.4 \]
\[ \text{P(White, White)} = 0.3 \]
Step 2: Linking Probability to Real Results
Why we do this:
We are told that a probability of \(0.15\) resulted in exactly 24 occurrences. We can use this to find out the total number of times he spun the spinners, and then use that total to find how many times “White AND White” happened.
Working:
Let \(N\) be the total number of spins.
\[ 0.15 \times N = 24 \]
\[ N = \frac{24}{0.15} = 160 \]
He spun the spinners 160 times.
✓ (M1) for process to find total spins (\(24 \div 0.15 = 160\)) or for calculating P(White, White) = 0.3.
Now, apply the “White AND White” probability to the total spins:
Expected (White, White) = \(\text{P(White, White)} \times \text{Total Spins}\)
Expected (White, White) = \(0.3 \times 160 = 48\)
(Alternative faster method: Notice that the probability of White, White (0.3) is exactly double the probability of Red, Red (0.15). Therefore, the outcome should be exactly double: \(24 \times 2 = 48\)).
✓ (A1) for 48.
Final Answer:
48
Total: 3 marks
Question 13 (2 marks)
Write \(x^2 + 6x – 7\) in the form \((x + a)^2 + b\) where \(a\) and \(b\) are integers.
Worked Solution
Step 1: Completing the Square
Why we do this:
The form \((x+a)^2 + b\) is achieved by “completing the square”. The rule is to halve the coefficient of the \(x\) term to find \(a\), put it in the bracket, and then subtract the square of \(a\) on the outside to keep the equation balanced.
Working:
Expression: \(x^2 + 6x – 7\)
Look at the \(x\) term, which is \(+6x\). Halve the 6 to get \(+3\).
Set up the bracket: \((x + 3)^2\)
✓ (M1) for \((x + 3)^2\).
If we expanded \((x + 3)^2\), we would get \(x^2 + 6x + 9\). But our original expression doesn’t have a \(+9\), it has a \(-7\).
To fix this, we subtract the \(3^2\) (which is 9) right after the bracket, and then include the original \(-7\):
\[ (x + 3)^2 – 9 – 7 \]
Simplify the numbers on the outside:
\[ -9 – 7 = -16 \]
Final expression: \[ (x + 3)^2 – 16 \]
✓ (A1) for the correct final answer.
Final Answer:
\((x + 3)^2 – 16\)
Total: 2 marks
Question 14 (3 marks)
Cone A and cone B are mathematically similar.
The ratio of the volume of cone A to the volume of cone B is 27 : 8
The surface area of cone A is \(297\text{ cm}^2\)
Show that the surface area of cone B is \(132\text{ cm}^2\)
Worked Solution
Step 1: Finding the Length Scale Factor
Why we do this:
For mathematically similar shapes, there is a strict relationship between lengths, areas, and volumes.
- If the length ratio is \(k\),
- The area ratio is \(k^2\),
- The volume ratio is \(k^3\).
We are given the volume ratio, so we must cube root it to find the basic length ratio first.
Working:
Volume Ratio (A : B) = 27 : 8
Length Ratio = \(\sqrt[3]{27} : \sqrt[3]{8}\)
Length Ratio = 3 : 2
✓ (M1) for \(\sqrt[3]{\frac{27}{8}}\) or identifying the 3:2 ratio.
Step 2: Finding the Area Scale Factor
Why we do this:
The question asks about surface area. Now that we have the length ratio, we square it to find the area ratio.
Working:
Area Ratio = \(3^2 : 2^2\)
Area Ratio = 9 : 4
✓ (M1) for squaring the length ratio to get 9:4.
Step 3: Calculating Area of Cone B
Why we do this:
We know the area of Cone A is 297, which corresponds to “9 parts” in our ratio. We need to find the value of “4 parts” for Cone B.
Working:
Area of A : Area of B = 9 : 4
297 : Area of B = 9 : 4
Find the multiplier by dividing 297 by 9:
\[ 297 \div 9 = 33 \]
Multiply this by 4 to find Cone B’s area:
\[ 33 \times 4 = 132 \]
✓ (A1) for 132 from correct arithmetic showing the proof is complete.
Final Answer:
Area ratio is 9:4. \(297 \div 9 \times 4 = 132\text{ cm}^2\). Shown.
Total: 3 marks
Question 15 (9 marks)
(a) Show that the equation \(x^3 + 7x – 5 = 0\) has a solution between \(x = 0\) and \(x = 1\)
(b) Show that the equation \(x^3 + 7x – 5 = 0\) can be arranged to give
\[ x = \frac{5}{x^2 + 7} \]
(c) Starting with \(x_0 = 1\), use the iteration formula
\[ x_{n+1} = \frac{5}{x_n^2 + 7} \]
three times to find an estimate for the solution of \(x^3 + 7x – 5 = 0\)
(d) By substituting your answer to part (c) into \(x^3 + 7x – 5\),
comment on the accuracy of your estimate for the solution to \(x^3 + 7x – 5 = 0\)
Worked Solution
Part (a): Checking for a solution
Why we do this:
If an equation crosses the x-axis (equals 0) between two x-values, the result of substituting those values into the expression will change from negative to positive (or vice versa). This is called the “change of sign” rule.
Working:
Let \(f(x) = x^3 + 7x – 5\)
Substitute \(x = 0\):
\[ f(0) = (0)^3 + 7(0) – 5 = -5 \]
✓ (M1) for substituting 0 and 1.
Substitute \(x = 1\):
\[ f(1) = (1)^3 + 7(1) – 5 = 1 + 7 – 5 = 3 \]
Because \(f(0) = -5\) (negative) and \(f(1) = 3\) (positive), there is a sign change. Since the function is continuous, it must cross zero between \(x = 0\) and \(x = 1\).
✓ (C1) for stating the correct values and deducing that a sign change implies a root.
Part (b): Rearranging the equation
Why we do this:
We are required to manipulate \(x^3 + 7x – 5 = 0\) using algebra until it looks exactly like \(x = \frac{5}{x^2 + 7}\).
Working:
\[ x^3 + 7x – 5 = 0 \]
Add 5 to both sides to isolate the \(x\) terms:
\[ x^3 + 7x = 5 \]
✓ (C1) for a correct first step.
Factorise the left hand side by pulling out \(x\):
\[ x(x^2 + 7) = 5 \]
Divide both sides by \((x^2 + 7)\) to isolate \(x\):
\[ x = \frac{5}{x^2 + 7} \]
✓ (C1) for clear and correct steps showing complete rearrangement.
Part (c): Using the iteration formula
Why we do this:
Iteration is a method of finding a more accurate answer by repeatedly putting the previous answer back into the formula. We need to do this three times (\(x_1\), \(x_2\), \(x_3\)) starting with \(x_0 = 1\).
Working:
Start with \(x_0 = 1\).
First iteration (\(x_1\)):
\[ x_1 = \frac{5}{(1)^2 + 7} = \frac{5}{1 + 7} = \frac{5}{8} = 0.625 \]
✓ (M1) for substitution of 1 into the formula.
Second iteration (\(x_2\)):
Substitute 0.625 back into the formula:
\[ x_2 = \frac{5}{(0.625)^2 + 7} = 0.6765327696… \]
✓ (M1) for substitution of \(x_1\) and \(x_2\).
Third iteration (\(x_3\)):
Substitute 0.67653… back into the formula:
\[ x_3 = \frac{5}{(0.67653…)^2 + 7} = 0.6704483001… \]
✓ (A1) for 0.6704 (or more precise).
Part (d): Commenting on Accuracy
Why we do this:
The true solution makes the equation equal exactly 0. By plugging our estimate back in, we can see how close to 0 the result is. The closer to 0, the more accurate the estimate.
Working:
Substitute \(x = 0.670448…\) into \(x^3 + 7x – 5\):
\[ (0.670448…)^3 + 7(0.670448…) – 5 \]
\[ \approx 0.30136… + 4.6931… – 5 \]
\[ \approx -0.00549… \]
✓ (M1) for substituting the answer to (c) into expression.
Comment: The result is very close to 0 (it is -0.005…), so this shows that the estimate \(x_3 \approx 0.6704\) is an accurate solution.
✓ (C1) appropriate comment, e.g., accurate as answer is close to 0.
Final Answer:
(a) \(f(0) = -5\), \(f(1) = 3\). Sign change proves root exists.
(b) Proof shown.
(c) 0.6704
(d) It evaluates to \(-0.005…\), which is very close to 0, so the estimate is accurate.
Total: 9 marks
Question 16 (3 marks)
The petrol consumption of a car, in litres per 100 kilometres, is given by the formula
\[ \text{Petrol consumption} = \frac{100 \times \text{Number of litres of petrol used}}{\text{Number of kilometres travelled}} \]
Nathan’s car travelled 148 kilometres, correct to 3 significant figures.
The car used 11.8 litres of petrol, correct to 3 significant figures.
Nathan says,
“My car used less than 8 litres of petrol per 100 kilometres.”
Could Nathan be wrong?
You must show how you get your answer.
Worked Solution
Step 1: Finding the Error Bounds
What are we being asked to find?
We need to determine if it’s mathematically possible for the petrol consumption to be 8 or more. If it is possible, Nathan could be wrong.
Why we do this:
Because the distance and fuel were rounded, their true values could be slightly higher or lower. We need to find the Upper and Lower Bounds for both values to see the maximum possible consumption.
Working:
Distance (\(d\)) = 148 km (to 3 s.f.). The rounding degree is 1 km.
Half of 1 is 0.5.
Lower Bound for distance = \(148 – 0.5 = 147.5\)
Upper Bound for distance = \(148 + 0.5 = 148.5\)
Petrol (\(p\)) = 11.8 L (to 3 s.f.). The rounding degree is 0.1 L.
Half of 0.1 is 0.05.
Lower Bound for petrol = \(11.8 – 0.05 = 11.75\)
Upper Bound for petrol = \(11.8 + 0.05 = 11.85\)
✓ (B1) for finding 147.5, 148.5, 11.75, or 11.85.
Step 2: Calculating Maximum Possible Consumption
Why we do this:
To find the maximum possible petrol consumption, we must make the numerator (top of the fraction) as large as possible and the denominator (bottom of the fraction) as small as possible. So, we use the Upper Bound for petrol used, and the Lower Bound for distance travelled.
Working:
\[ \text{Max Consumption} = \frac{100 \times \text{Upper Bound (Petrol)}}{\text{Lower Bound (Distance)}} \]
\[ \text{Max Consumption} = \frac{100 \times 11.85}{147.5} \]
✓ (P1) for substituting \(11.85\) and \(147.5\) into the formula.
Calculator Steps:
- Press:
100×11.85÷147.5= - Calculator shows: 8.033898305…
What this tells us:
The calculation shows that the petrol consumption could be as high as ~8.03 litres per 100 km. Because 8.03 is strictly greater than 8, Nathan’s claim that it is definitely “less than 8” might be wrong.
Final Answer:
Yes, Nathan could be wrong. The maximum possible consumption is 8.03… litres per 100 km, which is greater than 8.
Total: 3 marks
Question 17 (5 marks)
\(ABC\) and \(ADC\) are triangles.
The area of triangle \(ADC\) is \(56\text{ m}^2\)
Work out the length of \(AB\).
Give your answer correct to 1 decimal place.
Worked Solution
Step 1: Using the Area to find CD
What are we being asked to find?
Our final goal is length \(AB\) in the top triangle. To get there, we need to know the shared middle side, \(AC\). First, we use the given area of the bottom triangle to find its missing side, \(CD\).
Working:
The formula for the area of a triangle is: \(\text{Area} = \frac{1}{2}ab\sin(C)\)
For triangle \(ADC\):
\[ 56 = 0.5 \times 11 \times CD \times \sin(105^\circ) \]
✓ (P1) for start of process, setting up the area equation.
Rearrange to solve for \(CD\):
\[ CD = \frac{56}{0.5 \times 11 \times \sin(105^\circ)} \]
\[ CD = \frac{56}{5.5 \times \sin(105^\circ)} \]
\[ CD = 10.5401… \text{ m} \]
✓ (P1) for complete process to find \(CD\).
Step 2: Using the Cosine Rule to find AC
Why we do this:
Now we have two sides (\(AD\) and \(CD\)) and the angle between them (\(105^\circ\)) in triangle \(ADC\). We can use the Cosine Rule to find the side opposite the angle, which is the shared line \(AC\).
Working:
Cosine Rule: \(a^2 = b^2 + c^2 – 2bc\cos(A)\)
\[ AC^2 = 11^2 + (10.5401…)^2 – 2 \times 11 \times (10.5401…) \times \cos(105^\circ) \]
✓ (P1) for process to find \(AC\).
Calculator Steps:
- Calculate: \(121 + 111.093… – (231.882… \times -0.2588…)\)
- \(AC^2 = 232.093… + 60.015… = 292.109…\)
- \(AC = \sqrt{292.109…} = 17.091… \text{ m}\)
Step 3: Using the Sine Rule to find AB
Why we do this:
Now focus on the top triangle, \(ABC\). We know \(AC = 17.091…\text{ m}\), angle \(B = 118^\circ\), and angle \(C = 48^\circ\). We want to find side \(AB\), which is opposite angle \(C\). The known side \(AC\) is opposite angle \(B\). This perfect pairing means we use the Sine Rule.
Working:
Sine Rule: \(\frac{a}{\sin(A)} = \frac{b}{\sin(B)}\)
\[ \frac{AB}{\sin(48^\circ)} = \frac{AC}{\sin(118^\circ)} \]
\[ \frac{AB}{\sin(48^\circ)} = \frac{17.091…}{\sin(118^\circ)} \]
✓ (P1) for process to find \(AB\).
Multiply both sides by \(\sin(48^\circ)\):
\[ AB = \frac{17.091… \times \sin(48^\circ)}{\sin(118^\circ)} \]
\[ AB = \frac{12.701…}{0.8829…} \]
\[ AB = 14.384… \text{ m} \]
Rounding to 1 decimal place:
\(AB = 14.4 \text{ m}\)
✓ (A1) for answer in range 14.3 to 14.4.
Final Answer:
14.4 m
Total: 5 marks
Question 18 (4 marks)
Here is a speed-time graph for a train.
(a) Work out an estimate for the distance the train travelled in the first 20 seconds.
Use 4 strips of equal width.
(b) Is your answer to (a) an underestimate or an overestimate of the actual distance the train travelled?
Give a reason for your answer.
Worked Solution
Part (a): Estimating Distance from Speed-Time Graph
What are we being asked to find?
The distance travelled on a speed-time graph is represented by the area under the curve. Because it’s a curve, we estimate this area by splitting it into 4 vertical strips (shapes), which we treat as trapeziums.
Working:
We need 4 strips between \(t = 0\) and \(t = 20\). This means each strip has a width of \(20 \div 4 = 5\) seconds.
Read the speeds (y-values) at each interval from the graph:
- At \(t = 0\), speed = 0 m/s
- At \(t = 5\), speed = 2 m/s
- At \(t = 10\), speed = 5 m/s
- At \(t = 15\), speed = 10 m/s
- At \(t = 20\), speed = 18 m/s
Now calculate the area of each trapezium. The formula is \(\text{Area} = \frac{1}{2}(a + b)h\), where \(a\) and \(b\) are the parallel sides (the speeds) and \(h\) is the width (5).
Strip 1 (0 to 5s): Area = \(0.5 \times 5 \times (0 + 2) = 5\)
✓ (M1) for starting to find area under the curve.
Strip 2 (5 to 10s): Area = \(0.5 \times 5 \times (2 + 5) = 17.5\)
Strip 3 (10 to 15s): Area = \(0.5 \times 5 \times (5 + 10) = 37.5\)
Strip 4 (15 to 20s): Area = \(0.5 \times 5 \times (10 + 18) = 70\)
✓ (M1) for a complete method using 4 strips.
Total Estimated Distance = \(5 + 17.5 + 37.5 + 70 = 130\text{ m}\).
✓ (A1) for 130 (or answer in range 130.1 to 132 from accurate reading).
Part (b): Underestimate or Overestimate?
Why we do this:
Look at the shape of the curve. It is curving upwards (accelerating). When we draw a straight line connecting the top corners of our trapeziums, that straight line will sit slightly above the actual curve.
Working:
Because the straight lines of the trapeziums pass above the actual curve, the trapeziums include some extra space that isn’t actually under the curve.
Therefore, it is an overestimate.
✓ (C1) for stating “overestimate” and giving an appropriate reason linked to the area between trapeziums and the curve.
Final Answer:
(a) 130 m
(b) Overestimate. The straight lines at the top of the trapeziums sit above the curve, including extra area.
Total: 4 marks
Question 19 (5 marks)
Prove algebraically that the straight line with equation \(x – 2y = 10\) is a tangent to the circle with equation \(x^2 + y^2 = 20\)
Worked Solution
Step 1: Setting up Simultaneous Equations
What are we being asked to find?
A tangent is a line that touches a circle at exactly one point. To prove this algebraically, we solve the equations simultaneously. If it is a tangent, we will find exactly one solution (one point of intersection).
Working:
Line: \(x – 2y = 10\)
Circle: \(x^2 + y^2 = 20\)
First, rearrange the linear equation to make \(x\) the subject:
\[ x = 2y + 10 \]
Substitute this into the circle equation to eliminate \(x\):
\[ (2y + 10)^2 + y^2 = 20 \]
✓ (M1) starts process by substituting.
Step 2: Expanding and Simplifying
Why we do this:
We need to expand the brackets and collect all terms to one side to form a quadratic equation equal to zero.
Working:
Expand \((2y + 10)^2\):
\[ (2y + 10)(2y + 10) = 4y^2 + 20y + 20y + 100 \]
\[ = 4y^2 + 40y + 100 \]
✓ (M1) for expanding.
Put this back into the main equation:
\[ 4y^2 + 40y + 100 + y^2 = 20 \]
\[ 5y^2 + 40y + 100 = 20 \]
Subtract 20 from both sides:
\[ 5y^2 + 40y + 80 = 0 \]
✓ (M1) for 3-term quadratic equation ready for solving.
Step 3: Solving the Quadratic
Why we do this:
We solve this quadratic to find the \(y\)-coordinate(s) of intersection.
Working:
Divide the whole equation by 5 to make it simpler:
\[ y^2 + 8y + 16 = 0 \]
✓ (M1) for method to solve the equation (e.g., by factorising).
Factorise the quadratic:
We need two numbers that multiply to 16 and add to 8. The numbers are 4 and 4.
\[ (y + 4)(y + 4) = 0 \]
\[ (y + 4)^2 = 0 \]
This gives only one solution:
\[ y = -4 \]
What this tells us:
Because there is only one repeated root (one value for \(y\)), the line intersects the circle at exactly one point. A line that intersects a circle at exactly one point is, by definition, a tangent.
Final Answer:
Substituting gives \((y+4)^2 = 0\), yielding only one solution (\(y = -4\)). Because there is only one point of intersection, the line is a tangent to the circle. Proof complete.
Total: 5 marks
✓ (C1) fully correct method leading to one solution and concluding statement.
Question 20 (4 marks)
\(A\), \(B\) and \(C\) are points on the circumference of a circle, centre \(O\).
\(AOB\) is a diameter of the circle.
Prove that angle \(ACB\) is \(90^\circ\)
You must not use any circle theorems in your proof.
Worked Solution
Step 1: Constructing Internal Triangles
Why we do this:
Since we are forbidden from simply stating the circle theorem (“angle in a semicircle is a right angle”), we must prove it using basic triangle properties. The best way to do this is to draw a line from the center \(O\) to the point \(C\). This splits the large triangle into two smaller isosceles triangles.
Working:
Draw a straight line connecting point \(O\) to point \(C\).
The lengths \(OA\), \(OB\), and \(OC\) are all radii of the same circle. Therefore, they are all equal in length.
This creates two isosceles triangles: \(\Delta AOC\) and \(\Delta BOC\).
✓ (C1) draws OC and considers angles in an isosceles triangle.
Step 2: Defining the Angles
Why we do this:
In an isosceles triangle, the base angles are equal. We can assign algebra letters (\(x\) and \(y\)) to represent these angles.
Working:
In isosceles triangle \(AOC\), the base angles are equal:
Let angle \(OAC = x\). Then angle \(OCA = x\).
In isosceles triangle \(BOC\), the base angles are equal:
Let angle \(OBC = y\). Then angle \(OCB = y\).
Looking at the whole triangle \(ABC\), the angle at \(C\) (angle \(ACB\)) is made up of these two parts put together:
Angle \(ACB = x + y\)
Step 3: Calculating the Sum of Angles
Why we do this:
The sum of the interior angles of any triangle is exactly \(180^\circ\). We can add up all the corner angles of the big triangle \(ABC\) and set them equal to 180.
Working:
The three angles of the big triangle \(ABC\) are:
- Angle at \(A = x\)
- Angle at \(B = y\)
- Angle at \(C = x + y\)
Summing them up:
\[ x + y + (x + y) = 180^\circ \]
✓ (C1) finds sum of angles in triangle ABC.
Simplify the equation:
\[ 2x + 2y = 180^\circ \]
Divide the entire equation by 2:
\[ x + y = 90^\circ \]
✓ (C1) complete method leading to \(ACB = 90^\circ\).
What this tells us:
We established earlier that angle \(ACB\) is exactly equal to \((x + y)\). Since we just proved that \(x + y = 90^\circ\), then angle \(ACB\) must be \(90^\circ\).
Final Answer:
1. Draw \(OC\).
2. \(OA = OC = OB\) (radii).
3. Base angles of isosceles triangles are equal. Let \(\angle OAC = \angle OCA = x\), and \(\angle OBC = \angle OCB = y\).
4. Angles in triangle \(ABC\) add up to \(180^\circ\), so \(x + y + (x + y) = 180^\circ\).
5. \(2x + 2y = 180^\circ \rightarrow x + y = 90^\circ\).
6. Since \(\angle ACB = x + y\), \(\angle ACB = 90^\circ\).
Total: 4 marks
✓ (C1) for complete proof with all reasons explicitly given.
Question 21 (5 marks)
\(OAN\), \(OMB\) and \(APB\) are straight lines[cite: 464].
\(AN = 2OA\)[cite: 464].
\(M\) is the midpoint of \(OB\)[cite: 466].
\(\vec{OA} = \mathbf{a}\) [cite: 467]
\(\vec{OB} = \mathbf{b}\) [cite: 468]
\(\vec{AP} = k\vec{AB}\) where \(k\) is a scalar quantity[cite: 469].
Given that \(MPN\) is a straight line, find the value of \(k\)[cite: 470].
Worked Solution
Step 1: Finding the fundamental vectors
What are we being asked to find?
We need to find \(k\), which describes how far along the line \(AB\) the point \(P\) lies. Since \(MPN\) forms a straight line, we can solve this by expressing vectors \(\vec{MN}\) and \(\vec{MP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), and then using the fact that they must be parallel (scalar multiples of each other)[cite: 470, 99].
Working:
First, find the vectors for the full lengths of the lines in the diagram:
\(\vec{OA} = \mathbf{a}\) and \(\vec{OB} = \mathbf{b}\)[cite: 467, 468].
Because \(M\) is the midpoint of \(OB\), we know:
\[ \vec{OM} = \frac{1}{2}\mathbf{b} \] [cite: 99, 466]
Because \(AN = 2OA\), and they are in a straight line extending from \(O\):
\[ \vec{ON} = \vec{OA} + \vec{AN} = \mathbf{a} + 2\mathbf{a} = 3\mathbf{a} \] [cite: 464, 99]
The vector \(\vec{AB}\) goes backwards along \(\mathbf{a}\) and forwards along \(\mathbf{b}\):
\[ \vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} – \mathbf{a} \] [cite: 99]
✓ (P1) for process to find \(\vec{AB}\)[cite: 99].
Step 2: Defining vectors \(\vec{MN}\) and \(\vec{MP}\)
Why we do this:
To use the straight-line property of \(MPN\), we must express two vectors originating from the same point (like \(M\)) that lie on this line[cite: 470, 99].
Working:
Let’s find the full vector \(\vec{MN}\):
\[ \vec{MN} = \vec{MO} + \vec{ON} \]
\[ \vec{MN} = -\frac{1}{2}\mathbf{b} + 3\mathbf{a} \]
✓ (P1) for process to find \(\vec{MN}\)[cite: 99].
Now let’s find vector \(\vec{MP}\). We must walk from \(M \rightarrow O \rightarrow A \rightarrow P\):
\[ \vec{MP} = \vec{MO} + \vec{OA} + \vec{AP} \]
Substitute \(\vec{AP} = k\vec{AB} = k(\mathbf{b} – \mathbf{a})\):
\[ \vec{MP} = -\frac{1}{2}\mathbf{b} + \mathbf{a} + k(\mathbf{b} – \mathbf{a}) \]
✓ (P1) for process to find \(\vec{MP}\)[cite: 99].
Group the \(\mathbf{a}\) and \(\mathbf{b}\) components together:
\[ \vec{MP} = \mathbf{a}(1 – k) + \mathbf{b}(k – 0.5) \]
Step 3: Using the straight line property to solve for \(k\)
Why we do this:
Since \(M, P, N\) form a straight line, \(\vec{MP}\) must be a scaled-down version of \(\vec{MN}\). This means their components (\(\mathbf{a}\) and \(\mathbf{b}\)) must share the exact same ratio[cite: 470, 99].
Working:
We know \(\vec{MP} = c \times \vec{MN}\), where \(c\) is a constant multiplier.
\[ \mathbf{a}(1 – k) + \mathbf{b}(k – 0.5) = c(3\mathbf{a} – 0.5\mathbf{b}) \]
✓ (P1) for process to find \(k\) using \(\vec{MN}\) as a multiple of \(\vec{MP}\)[cite: 99].
Equate the coefficients of \(\mathbf{a}\):
\[ 1 – k = 3c \]
\[ c = \frac{1 – k}{3} \]
Equate the coefficients of \(\mathbf{b}\):
\[ k – 0.5 = -0.5c \]
Substitute our expression for \(c\) into the \(\mathbf{b}\) equation:
\[ k – 0.5 = -0.5 \left( \frac{1 – k}{3} \right) \]
Multiply both sides by 3 to remove the fraction:
\[ 3(k – 0.5) = -0.5(1 – k) \]
\[ 3k – 1.5 = -0.5 + 0.5k \]
Subtract \(0.5k\) from both sides:
\[ 2.5k – 1.5 = -0.5 \]
Add 1.5 to both sides:
\[ 2.5k = 1.0 \]
\[ k = \frac{1}{2.5} = \frac{2}{5} \]
✓ (A1) for \(\frac{2}{5}\) or equivalent[cite: 99].
Final Answer:
\(k = \frac{2}{5}\) [cite: 99]
Total: 5 marks