Edexcel GCSE (9-1) Mathematics – June 2017 Paper 1 (Higher)

Working:

Looking at the graph, most of the points form a path sloping upwards from the bottom left to the top right. There is one cross on its own in the top left area. To read its coordinates, we look straight down to the x-axis (hours of sunshine) and straight across to the y-axis (temperature).

The x-value is exactly on the line for \(10\).

The y-value is exactly halfway between \(18\) and \(20\), which is \(19\).

(B1) Correct coordinates

Working:

Ignoring the outlier, the points start low on the left and get higher as we move to the right. As the hours of sunshine increase, the maximum temperature increases.

(C1) Stating “positive correlation”

Working:

1. Draw a straight line of best fit through the main cluster of points.

2. Locate \(16.4\) on the y-axis. (Each small square on the y-axis represents \(0.2^{\circ}\text{C}\), so \(16.4\) is 2 small squares above \(16\)).

3. Draw a horizontal line from \(16.4\) to hit your line of best fit.

4. Draw a vertical line straight down from that point to the x-axis to read the estimated hours of sunshine.

Depending on exactly how you drew your line of best fit, you should hit the x-axis somewhere between \(12\) and \(13\).

(M1) Drawing an appropriate line of best fit OR showing a mark at \((x, 16.4)\).

(A1) Value in the range 12 to 13.

Working:

The graph shows a positive correlation. This means exactly what the weatherman said: as the sunshine hours (x-axis) get higher, the temperature (y-axis) also tends to get higher.

(C1) Valid explanation linking the positive correlation to the statement.

Working:

Start with \(56\). Since it’s even, we know it divides by the prime number \(2\).

\(56 \div 2 = 28\). Circle the \(2\) (it’s prime). Continue with \(28\).

\(28 \div 2 = 14\). Circle the \(2\).

\(14 \div 2 = 7\). Circle the \(2\). Since \(7\) is also a prime number, we circle it and stop.

The prime factors we circled are \(2\), \(2\), \(2\), and \(7\).

(M1) Complete method to find prime factors (e.g., using a complete factor tree with no more than 1 arithmetic error).

Working:

Product = \(2 \times 2 \times 2 \times 7\)

Using index notation: \(2^3 \times 7\)

(A1) Correct final answer. (Accept \(2^3 \times 7\) or \(2 \times 2 \times 2 \times 7\)).

Working:

We will use the column multiplication method to calculate \(546 \times 43\).

    5 4 6
  x   4 3
  -------
  1 6 3 8  (This is 546 × 3.   6×3=18, 4×3=12+1=13, 5×3=15+1=16)
2 1 8 4 0  (This is 546 × 40.  Remember the zero! 6×4=24, 4×4=16+2=18, 5×4=20+1=21)
  -------
2 3 4 7 8  (Add them together: 8+0=8, 3+4=7, 6+8=14, 1+1+1=3, 2)

(M1) Complete method with relative place value correct including addition of all the appropriate elements.

(A1) For digits \(23478\) correctly found.

Working:

Our digits are \(23478\).

We place the decimal point so there are two digits after it: \(234.78\).

(A1) Correct placement of the decimal point into their final answer (following through their digits).

Working:

Look at the diagram. The full length of the top edge (from A to B) is made of two parts: \(3\text{ cm}\) and \(x\text{ cm}\). So the total side length is \((x + 3)\).

Since it’s a square, the height (from A to D) is also \((x + 3)\).

Area = \((x + 3) \times (x + 3)\)

Alternatively, you can sum the four smaller areas inside: \(3\times3=9\), \(3\times x=3x\), \(x\times3=3x\), and \(x\times x=x^2\). Area = \(x^2 + 3x + 3x + 9\).

(M1) Writes the area using algebraic terms, e.g. \((x+3)^2\) or \(x^2+3x+3x+9\).

Working:

Set the expression equal to the given area:

\[(x + 3)(x + 3) = 10\]

Expand the brackets using FOIL (First, Outer, Inner, Last):

\[x^2 + 3x + 3x + 9 = 10\]

Simplify by collecting like terms (\(3x + 3x = 6x\)):

\[x^2 + 6x + 9 = 10\]

(M1) Expands the brackets and includes the given \(10\).

Working:

Subtract \(9\) from both sides of the equation:

\[x^2 + 6x + 9 – 9 = 10 – 9\]

\[x^2 + 6x = 1\]

This matches exactly what we were asked to show.

(A1) Rearranges correctly to give the given equation.

Working:

Pythagoras’ theorem states: \(a^2 + b^2 = c^2\) (where \(c\) is the hypotenuse/diagonal).

Let the diagonal be \(c\).

\[c^2 = 5^2 + 12^2\]

\[c^2 = 25 + 144\]

\[c^2 = 169\]

Square root both sides to find \(c\):

\[c = \sqrt{169} = 13\text{ m}\]

(P1) Starts process of Pythagoras, e.g., \(5^2 + 12^2\).

(P1) Complete process for Pythagoras, e.g., \(\sqrt{169} = 13\).

Working:

The frame has:

• Two horizontal lengths: \(12 + 12 = 24\text{ m}\)
• Two vertical lengths: \(5 + 5 = 10\text{ m}\)
• One diagonal length: \(13\text{ m}\)

Total length = \(12 + 12 + 5 + 5 + 13 = 47\text{ m}\).

(P1) Process of adding all 5 lengths together: \(5+5+12+12+13 = 47\).

Working:

Total weight = \(47 \times 1.5\)

To do this without a calculator, we can think of it as \(47 \times 1\) plus half of \(47\).

\[47 \times 1 = 47\]

Half of \(47 = 23.5\)

\[47 + 23.5 = 70.5\text{ kg}\]

(P1) Process of multiplying lengths by \(1.5\).

(A1) Correct final answer of \(70.5\).

Working:

The equation for \(L_1\) is already in the correct format: \(y = 3x – 2\)

Comparing this to \(y = mx + c\), the number in front of the \(x\) is \(3\).

So, the gradient of \(L_1\) is \(3\).

Working:

Start with: \(3y – 9x + 5 = 0\)

First, move the \(x\) term and the number to the right side by adding \(9x\) and subtracting \(5\):

\[3y = 9x – 5\]

(M1) Starts to manipulate expression, e.g., \(3y = 9x – 6\) or \(3y = 9x – 5\).

Next, we need \(y\) on its own, not \(3y\). So, divide every term by \(3\):

\[y = \frac{9x}{3} – \frac{5}{3}\]

\[y = 3x – \frac{5}{3}\]

Comparing this to \(y = mx + c\), the gradient (the number in front of \(x\)) is \(3\).

(A1) Gives equation which can be used to show that the gradients are the same, e.g., \(y = 3x – \frac{5}{3}\).

Working:

Number of people in the whole class = \(10 \text{ boys} + 20 \text{ girls} = 30 \text{ students}\).

Mean mark for all the class = \(60\).

Total marks for the whole class = \(30 \times 60\)

\[3 \times 6 = 18\text{, so } 30 \times 60 = 1800\]

(P1) For showing the process of \(30 \times 60 = 1800\).

Working:

Number of girls = \(20\).

Mean mark for the girls = \(54\).

Total marks for the girls = \(20 \times 54\)

Without a calculator, do \(2 \times 54 = 108\), and then multiply by \(10\) to get \(1080\).

(P1) For showing the process of \(20 \times 54 = 1080\).

Working:

Total marks for boys = (Total for class) – (Total for girls)

Total marks for boys = \(1800 – 1080\)

  7 10
  8  0  0
- 1  0  8  0
------------
     7  2  0

Total marks for boys = \(720\)

Mean for boys = \(\frac{\text{Total boys marks}}{\text{Number of boys}}\)

Mean for boys = \(\frac{720}{10} = 72\)

(P1) For showing the complete process, e.g., \((1800 – 1080) \div 10\).

(A1) Concluding the answer is \(72\).

Working:

Start with \(7.97\).

Move the decimal point \(6\) places left. Add zeros to fill the empty spaces:

\[0.00000797\]

Tip: A quick check for a power of \(-6\) is that there should be \(6\) zeros in total before the first non-zero digit, including the zero before the decimal point.

(B1) Correct answer only.

Working:

The calculation is: \(\frac{2.52 \times 10^5}{4 \times 10^{-3}}\)

Step 1: Divide the number parts

Calculate \(2.52 \div 4\).

    0 . 6  3
   ---------
 4 )2 .25 12

So, \(2.52 \div 4 = 0.63\).

Step 2: Divide the powers of 10

When dividing powers with the same base, subtract the indices: \(a^m \div a^n = a^{m-n}\).

\[10^5 \div 10^{-3} = 10^{5 – (-3)}\]

Subtracting a negative is the same as adding: \(5 – (-3) = 5 + 3 = 8\).

So, \(10^5 \div 10^{-3} = 10^8\).

(M1) For partial calculation involving powers of \(10\), e.g., \(0.63 \times 10^8\).

Step 3: Combine and adjust to standard form

Our combined answer is \(0.63 \times 10^8\).

Standard form must be written as \(A \times 10^n\), where \(A\) is between \(1\) and \(10\) (\(1 \le A < 10\)).

\(0.63\) is too small. We must multiply it by \(10\) to make it \(6.3\).

Because we made the number part \(10\) times bigger, we must make the power part \(10\) times smaller (by subtracting \(1\) from the index) to keep the overall value the same.

\[0.63 \times 10^8 = 6.3 \times 10^7\]

(A1) Correct final answer in standard form.

Working:

Original Price = \(100\%\)

Price with VAT = \(120\%\)

So, we can write an equation: \(120\% = \text{£}600\)

(M1) Recognition of \(1.2\) or \(120\%\) equivalent, e.g., \(120\% = 600\) or \(600 \div 1.2\).

Working:

If \(120\% = 600\)

Divide both sides by \(12\) to find what \(10\%\) is:

\[120\% \div 12 = 10\%\]

\[600 \div 12 = 50\]

So, \(10\% = 50\)

Now, multiply both sides by \(10\) to find \(100\%\):

\[10\% \times 10 = 100\%\]

\[50 \times 10 = 500\]

The original price without VAT is £\(500\).

(A1) Correct answer only.

Working:

Let’s expand the first two brackets: \((x + 1)(x + 2)\)

Using FOIL (First, Outer, Inner, Last):

Firsts: \(x \times x = x^2\)

Outers: \(x \times 2 = 2x\)

Inners: \(1 \times x = 1x\)

Lasts: \(1 \times 2 = 2\)

Expression: \(x^2 + 2x + 1x + 2\)

Simplify by collecting like terms: \(x^2 + 3x + 2\)

(M1) For method to find the product of any two linear expressions (3 correct terms).

Working:

We need to expand: \((x^2 + 3x + 2)(x + 3)\)

Multiply everything in the first bracket by \(x\):

\(x^2 \times x = x^3\)

\(3x \times x = 3x^2\)

\(2 \times x = 2x\)

Multiply everything in the first bracket by \(3\):

\(x^2 \times 3 = 3x^2\)

\(3x \times 3 = 9x\)

\(2 \times 3 = 6\)

Write out the full expanded expression:

\[x^3 + 3x^2 + 2x + 3x^2 + 9x + 6\]

(M1) For method of multiplying out remaining products, half of which are correct.

Working:

Original expansion: \(x^3 + 3x^2 + 2x + 3x^2 + 9x + 6\)

Collect the \(x^2\) terms: \(3x^2 + 3x^2 = 6x^2\)

Collect the \(x\) terms: \(2x + 9x = 11x\)

Put it all together in descending powers of \(x\):

\[x^3 + 6x^2 + 11x + 6\]

(A1) Correct final expression.

Working:

Looking at the graph, the lowest point is exactly on the grid line for \(x = 1\).

Reading across to the y-axis, this point aligns perfectly with \(y = -3\).

So, the coordinates are \((1, -3)\)[cite: 87].

(B1) Correct coordinates.

Working:

1. On the left side, the curve crosses between \(0\) and \(-1\). It is closer to \(-1\). Each small square represents \(0.1\). It crosses at approximately \(-0.75\).

2. On the right side, the curve crosses between \(2\) and \(3\). It is past the halfway mark, at approximately \(2.75\).

The mark scheme accepts any values between \(-0.7\) to \(-0.8\) and \(2.7\) to \(2.8\)[cite: 87].

(B1) Both estimates correctly identified.

Working:

Find \(1.5\) on the x-axis (exactly halfway between \(1\) and \(2\)).

Move straight down to the curve.

From that point on the curve, move horizontally across to the y-axis.

The value on the y-axis is approximately \(-2.75\).

(B1) Accepts any value from \(-2.7\) to \(-2.8\).

Working:

Step 1 (Negative power): \(81^{-\frac{1}{2}} = \frac{1}{81^{\frac{1}{2}}}\)

Step 2 (Fractional power): \(\frac{1}{81^{\frac{1}{2}}} = \frac{1}{\sqrt{81}}\)

(M1) For showing a method using either reciprocal or square root[cite: 87].

Step 3 (Calculate): The square root of \(81\) is \(9\) (since \(9 \times 9 = 81\)).

\[\frac{1}{9}\]

(A1) Correct final answer of \(\frac{1}{9}\)[cite: 87].

Working:

Expression: \(\left(\frac{64}{125}\right)^{\frac{2}{3}}\)

Step 1: Apply the denominator (Cube Root)

We need to find the cube root of \(64\) and \(125\).

\(\sqrt[3]{64} = 4\) (because \(4 \times 4 \times 4 = 64\))

\(\sqrt[3]{125} = 5\) (because \(5 \times 5 \times 5 = 125\))

So our fraction becomes \(\left(\frac{4}{5}\right)^2\).

(M1) For showing cube root of 64 as 4 and the cube root of 125 as 5[cite: 87].

Step 2: Apply the numerator (Square)

Now we square the result: \(\left(\frac{4}{5}\right)^2 = \frac{4^2}{5^2}\)

\[\frac{16}{25}\]

(A1) Correct answer only.

Working:

Set up the equation: \(y = \frac{k}{x^2}\)

Look at the table for the easiest pair of numbers to use. \(x=3\) and \(y=1\) is perfect.

Substitute \(x=3\) and \(y=1\) into our formula:

\[1 = \frac{k}{3^2}\]

\[1 = \frac{k}{9}\]

(M1) Begins to work with \(y = \frac{k}{x^2}\) e.g., substitutes a pair of numbers into the formula[cite: 87].

Multiply both sides by \(9\) to find \(k\):

\[k = 9\]

Now, rewrite the full equation with our found value of \(k\):

\[y = \frac{9}{x^2}\]

(A1) Correct equation: \(y = \frac{9}{x^2}\)[cite: 87].

Working:

Use our formula: \(y = \frac{9}{x^2}\)

Substitute \(y = 16\):

\[16 = \frac{9}{x^2}\]

(M1) Substitutes \(y=16\) into proportional formula[cite: 87].

Multiply both sides by \(x^2\):

\[16x^2 = 9\]

Divide by \(16\):

\[x^2 = \frac{9}{16}\]

To find \(x\), take the square root of both sides. The square root of a fraction is just the square root of the top over the square root of the bottom:

\[x = \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}\]

The question specifically asks for the positive value, so we don’t need to worry about \(-\frac{3}{4}\).

(A1) Correct answer: \(\frac{3}{4}\)[cite: 87].

Working:

Total White shapes = \(\frac{3}{10}\) of all shapes.

Total Black shapes = \(\frac{7}{10}\) of all shapes.

Working:

White circles = \(\frac{4}{9}\) of the White shapes.

Fraction of all shapes that are White Circles = \(\frac{4}{9} \times \frac{3}{10}\)

\[\frac{4 \times 3}{9 \times 10} = \frac{12}{90}\]

Simplify \(\frac{12}{90}\) by dividing top and bottom by \(6\):

\[\frac{2}{15}\]

(P1) Process to solve the problem, e.g. \(\frac{3}{10} \times \frac{4}{9} = \frac{12}{90}\)[cite: 88].

Working:

Black circles = \(\frac{2}{7}\) of the Black shapes.

Fraction of all shapes that are Black Circles = \(\frac{2}{7} \times \frac{7}{10}\)

\[\frac{2 \times 7}{7 \times 10} = \frac{14}{70}\]

Simplify \(\frac{14}{70}\) by dividing top and bottom by \(14\) (or just cancel the \(7\)s initially):

\[\frac{2}{10} = \frac{1}{5}\]

(P1) Second step of process, e.g. \(\frac{7}{10} \times \frac{2}{7} = \frac{14}{70}\)[cite: 88].

Working:

Total Circles = (White Circles) + (Black Circles)

Total Circles = \(\frac{2}{15} + \frac{1}{5}\)

To add fractions, we need a common denominator. The lowest common multiple of \(15\) and \(5\) is \(15\).

Multiply top and bottom of \(\frac{1}{5}\) by \(3\): \(\frac{3}{15}\).

\[\frac{2}{15} + \frac{3}{15} = \frac{5}{15}\]

(P1) Complete process to find total circles, e.g. \(\frac{2}{15} + \frac{1}{5} = \frac{5}{15}\)[cite: 88].

Simplify the final fraction by dividing top and bottom by \(5\):

\[\frac{1}{3}\]

(A1) Correct answer: \(\frac{1}{3}\)[cite: 88].

Working:

First, round the given values to \(1\) significant figure:

Volume (\(V\)): \(98 \approx 100\)

Radius (\(r\)): \(5.13 \approx 5\)

\(\pi\): \(\pi \approx 3\)

Now substitute these rounded numbers into the volume formula:

\[V = \frac{1}{3}\pi r^2 h\]

\[100 \approx \frac{1}{3} \times 3 \times 5^2 \times h\]

(M1) Substitution into formula of chosen rounded values for r and V[cite: 88].

Simplify the right side:

The \(\frac{1}{3}\) and the \(3\) cancel each other out (\(\frac{1}{3} \times 3 = 1\)).

\(5^2 = 25\)

So the equation becomes:

\[100 \approx 25 \times h\]

(M1) Uses estimates in calculation and starts rearrangement[cite: 88].

Divide both sides by \(25\):

\[h \approx \frac{100}{25}\]

\[h \approx 4\]

(A1) Arrives at a single value from estimate in the range \(3.5\) to \(4.5\)[cite: 88].

Working:

Look at how we rounded the values:

• Volume (numerator) went UP (from \(98\) to \(100\)).
• Radius and \(\pi\) (denominator) went DOWN (from \(5.13\) to \(5\), and \(3.14…\) to \(3\)).

In a fraction, if you make the top bigger, the overall answer gets bigger. If you make the bottom smaller, the overall answer also gets bigger.

Because both our rounding choices push the fraction to be larger, our estimate must be more than the true answer.

(C1) States “more” with a valid reason: e.g. “more since number in numerator goes up; numbers in denominator go down”[cite: 88].

Working:

Expand \((n – 2)(n – 2)\) using FOIL:

\(n \times n = n^2\)

\(n \times -2 = -2n\)

\(-2 \times n = -2n\)

\(-2 \times -2 = 4\)

Combine like terms:

\[n^2 – 4n + 4\]

(C1) Correct expansion of brackets to give at least 3 correct terms from \(n^2 – 2n – 2n + 4\).

Working:

Original expression: \(n^2 – 2 – (n – 2)^2\)

Substitute our expansion:

\[n^2 – 2 – (n^2 – 4n + 4)\]

Distribute the negative sign:

\[n^2 – 2 – n^2 + 4n – 4\]

(C1) Arrives at \(n^2 – 2 – n^2 + 4n – 4\).

Collect like terms:

The \(n^2\) and \(-n^2\) cancel out to \(0\).

\(-2 – 4 = -6\).

We are left with:

\[4n – 6\]

(C1) Reduces to \(4n – 6\) or \(2(2n – 3)\).

Working:

Take out a common factor of \(2\) from \(4n – 6\):

\[2(2n – 3)\]

Because the expression has a factor of \(2\) (it is \(2\) multiplied by an integer), it must be a multiple of \(2\), which means it is always an even number.

(C1) For conclusion, e.g. “\(2(2n – 3)\) is always even since it is a multiple of \(2\)”.

Working:

Total counters at start = \(9\)

Path 1: Green then Blue

Probability of Green first = \(\frac{7}{9}\)

Once she takes a Green counter, there are only \(8\) counters left in the bag. The number of Blue counters is still \(2\).

Probability of Blue second = \(\frac{2}{8}\)

Multiply along the path: P(Green, Blue) = \(\frac{7}{9} \times \frac{2}{8}\)

(P1) For finding \(\frac{7}{8}\) or \(\frac{2}{8}\) or \(\frac{6}{8}\) seen in a calculation (the “without replacement” denominator).

Working:

Path 2: Blue then Green

Probability of Blue first = \(\frac{2}{9}\)

Once she takes a Blue counter, there are \(8\) counters left. The number of Green counters is still \(7\).

Probability of Green second = \(\frac{7}{8}\)

Multiply along the path: P(Blue, Green) = \(\frac{2}{9} \times \frac{7}{8}\)

(P1) For \(\frac{7}{9} \times \frac{2}{8}\) or \(\frac{2}{9} \times \frac{7}{8}\).

Working:

P(One of each) = P(Green, Blue) + P(Blue, Green)

\[= \left(\frac{7}{9} \times \frac{2}{8}\right) + \left(\frac{2}{9} \times \frac{7}{8}\right)\]

Calculate the multiplications (multiply tops, multiply bottoms):

\[= \frac{14}{72} + \frac{14}{72}\]

(P1) For \(\left(\frac{7}{9} \times \frac{2}{8}\right) + \left(\frac{2}{9} \times \frac{7}{8}\right)\).

Add the fractions (they already have a common denominator):

\[= \frac{28}{72}\]

Simplify the fraction by dividing top and bottom by \(4\):

\[= \frac{7}{18}\]

(A1) Correct equivalent fraction, e.g. \(\frac{28}{72}\) or \(\frac{7}{18}\).

Working:

The equation of \(DB\) is \(y = \frac{1}{2}x + 6\).

The gradient (\(m_1\)) of \(DB\) is \(\frac{1}{2}\).

Because \(AC\) and \(DB\) are perpendicular, their gradients multiply to make \(-1\):

\[m_1 \times m_2 = -1\]

(P1) Shows evidence of understanding that \(AC\) is perpendicular to \(DB\), or states the gradient of \(DB\) is \(0.5\).

Working:

Gradient of \(DB\) = \(\frac{1}{2}\)

Flip \(\frac{1}{2}\) to get \(\frac{2}{1}\) (which is \(2\)). Change the sign to get \(-2\).

Gradient of \(AC = -2\).

(P1) Shows a process to find the gradient of a perpendicular line (e.g. use of \(-\frac{1}{m}\)).

So, the equation for \(AC\) starts like this: \(y = -2x + c\)

Working:

Substitute \(x = 5\) and \(y = 11\) into our equation:

\[11 = -2(5) + c\]

(P1) Substitutes \(x = 5, y = 11\) into \(y = mx + c\) using their found perpendicular gradient.

\[11 = -10 + c\]

Add \(10\) to both sides:

\[c = 11 + 10 = 21\]

Now write the complete equation by putting our \(c\) value back in:

\[y = -2x + 21\]

(A1) Correct equation: \(y = -2x + 21\).

Working:

To get to \(X\) from \(O\), we travel from \(O\) to \(A\), and then halfway along \(AC\).

First, find the full diagonal vector \(\vec{AC}\):

\[\vec{AC} = \vec{AO} + \vec{OC} = -\mathbf{a} + \mathbf{c}\]

Since \(X\) is the midpoint, \(\vec{AX}\) is half of \(\vec{AC}\):

\[\vec{AX} = \frac{1}{2}(-\mathbf{a} + \mathbf{c}) = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\]

Now, find \(\vec{OX}\):

\[\vec{OX} = \vec{OA} + \vec{AX}\]

\[\vec{OX} = \mathbf{a} + \left(-\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\right) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\]

(P1) For first step to solve problem, e.g., \(\vec{OX} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\).

Working:

Using our path rule to get to \(D\):

\[\vec{OD} = \vec{OX} + \vec{XD}\]

We know \(\vec{OX} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\), and we are given \(\vec{XD} = 3\mathbf{c} – \frac{1}{2}\mathbf{a}\).

\[\vec{OD} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\right) + \left(3\mathbf{c} – \frac{1}{2}\mathbf{a}\right)\]

The \(\frac{1}{2}\mathbf{a}\) and \(-\frac{1}{2}\mathbf{a}\) cancel each other out:

\[\vec{OD} = 3.5\mathbf{c}\]

Now, find \(\vec{CD}\). Since \(O, C, D\) is a straight line, \(\vec{CD}\) is just the part of the journey from \(C\) to \(D\):

\[\vec{CD} = \vec{OD} – \vec{OC}\]

\[\vec{CD} = 3.5\mathbf{c} – 1\mathbf{c} = 2.5\mathbf{c}\]

(P1) For a correct vector statement using \(\vec{CD}\), e.g., \(\vec{CD} = 2.5\mathbf{c}\).

Working:

We have \(\vec{OC} = 1\mathbf{c}\) and \(\vec{CD} = 2.5\mathbf{c}\).

The ratio of their lengths is:

\[OC : CD\]

\[1 : 2.5\]

The question requires the ratio to be in the form \(k : 1\). This means the right side must be exactly \(1\).

To turn \(2.5\) into \(1\), we must divide both sides of the ratio by \(2.5\).

\[\frac{1}{2.5} : \frac{2.5}{2.5}\]

\[\frac{1}{2.5} : 1\]

(P1) For a correct equation or ratio using \(k\), e.g., \(1 : 2.5 = k : 1\).

So, \(k = \frac{1}{2.5}\).

To simplify \(\frac{1}{2.5}\), multiply top and bottom by \(10\): \(\frac{10}{25}\), which simplifies to \(\frac{2}{5}\) (or \(0.4\)).

(A1) Correct answer: \(\frac{2}{5}\) or \(0.4\).

Working:

Rearrange the linear equation: \(y – 3x = 13\)

Add \(3x\) to both sides: \(y = 3x + 13\)

Now, substitute this expression for \(y\) into the quadratic equation \(x^2 + y^2 = 25\):

\[x^2 + (3x + 13)^2 = 25\]

(M1) For substitution of a rearrangement of the linear equation into the quadratic equation.

Working:

Expand \((3x + 13)^2 = (3x + 13)(3x + 13)\)

\[(3x \times 3x) + (3x \times 13) + (13 \times 3x) + (13 \times 13)\]

\[9x^2 + 39x + 39x + 169\]

\[9x^2 + 78x + 169\]

(M1) For expansion of bracket after substitution.

Substitute this back into our main equation:

\[x^2 + 9x^2 + 78x + 169 = 25\]

\[10x^2 + 78x + 169 = 25\]

Subtract \(25\) from both sides to make it equal zero:

\[10x^2 + 78x + 144 = 0\]

(M1) For forming quadratic ready for solving.

Working:

Divide by \(2\):

\[5x^2 + 39x + 72 = 0\]

We need to factorise this. We are looking for factors of \(5 \times 72 = 360\) that add up to \(39\).

Factors of \(360\): … \(10 \times 36\) (adds to \(46\)), \(15 \times 24\) (adds to \(39\)!).

Split the middle term using \(15\) and \(24\):

\[5x^2 + 15x + 24x + 72 = 0\]

Factorise the first two terms and the last two terms:

\[5x(x + 3) + 24(x + 3) = 0\]

\[(5x + 24)(x + 3) = 0\]

(M1) For factorising, e.g., \((5x + 24)(x + 3) = 0\).

This gives two possible values for \(x\):

Either \(x + 3 = 0 \Rightarrow x = -3\)

Or \(5x + 24 = 0 \Rightarrow 5x = -24 \Rightarrow x = -\frac{24}{5}\)

Working:

When \(x = -3\):

\[y = 3(-3) + 13 = -9 + 13 = 4\]

First solution pair: \(x = -3, y = 4\)

When \(x = -\frac{24}{5}\):

\[y = 3\left(-\frac{24}{5}\right) + 13\]

\[y = -\frac{72}{5} + \frac{65}{5}\] (Note: \(13 = \frac{65}{5}\))

\[y = -\frac{7}{5}\]

Second solution pair: \(x = -\frac{24}{5}, y = -\frac{7}{5}\)

(A1) Correct set of four values paired correctly.

Working:

We will construct (imagine or draw) the diagonals \(AC\) and \(BD\), creating two overlapping triangles: triangle \(ABC\) and triangle \(DCB\).

We will use the SAS (Side-Angle-Side) rule to prove they are congruent.

Working:

1. Side: \(AB = CD\) (This is explicitly given in the question).

(C1) States 1st link: \(AB = CD\).

2. Angle: Angle \(ABC\) = Angle \(DCB\) (This is also explicitly given in the question).

(C1) States 2nd link: angle \(ABC\) = angle \(BCD\).

3. Side: \(BC = CB\) (This side is shared by both triangles, so it is a common side).

(C1) States 3rd link with reason: \(BC = BC\) (common).

Working:

Because we have established Side-Angle-Side matching, triangle \(ABC\) is congruent to triangle \(DCB\).

Since the triangles are congruent, all their corresponding lengths must be equal.

Therefore, \(AC = BD\).

(C1) Concludes proof by stating triangle \(ABC\) is congruent to triangle \(DCB\) with reason SAS, and therefore \(AC = BD\).

Working:

In triangle \(ABC\):

\(AB = x\)

\(BC = x\)

Angle \(ABC = 30^\circ\)

Using the Cosine Rule: \(a^2 = b^2 + c^2 – 2bc\cos A\)

\[AC^2 = x^2 + x^2 – 2(x)(x)\cos(30^\circ)\]

We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\).

(B1) For stating \(\cos 30^\circ = \frac{\sqrt{3}}{2}\).

Substitute this into the equation:

\[AC^2 = 2x^2 – 2x^2\left(\frac{\sqrt{3}}{2}\right)\]

\[AC^2 = 2x^2 – x^2\sqrt{3}\]

Factorising out \(x^2\):

\[AC^2 = x^2(2 – \sqrt{3})\]

Working:

Since \(PQ = AC\), we know \(PQ^2 = x^2(2 – \sqrt{3})\).

(M1) For \(PQ^2 = 10^2 + 10^2 – 2(10)(10)\cos PBQ\) or finding \(AC^2\).

In triangle \(PBQ\):

\(BP = 10\)

\(BQ = 10\)

Rearranged Cosine Rule: \(\cos A = \frac{b^2 + c^2 – a^2}{2bc}\)

\[\cos PBQ = \frac{10^2 + 10^2 – PQ^2}{2(10)(10)}\]

(M1) For \(\cos PBQ = \frac{10^2 + 10^2 – PQ^2}{2(10)(10)}\).

Working:

Substitute \(PQ^2 = x^2(2 – \sqrt{3})\):

\[\cos PBQ = \frac{100 + 100 – x^2(2 – \sqrt{3})}{200}\]

(M1) For substituting the expression for \(PQ^2\) into the equation.

\[\cos PBQ = \frac{200 – x^2(2 – \sqrt{3})}{200}\]

Split the fraction into two separate parts over the common denominator of \(200\):

\[\cos PBQ = \frac{200}{200} – \frac{x^2(2 – \sqrt{3})}{200}\]

\[\cos PBQ = 1 – \frac{(2 – \sqrt{3})}{200}x^2\]

(A1) Conclusion of proof with all working seen.