2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Let \(R\) be selecting a red card and \(B\) be selecting a blue card.\
\
The total number of cards is \(5 + 3 = 8\).\
\
Since the selection is without replacement:\
- The probability of choosing a red card then a blue card is:\
\(P(R \text{ then } B) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)\
\
- The probability of choosing a blue card then a red card is:\
\(P(B \text{ then } R) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)\
\
To find the probability of selecting cards of different colours, we add these two probabilities together:\
\(P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\)

M1 for \(\frac{5}{8} \times \frac{3}{7}\) (or \(\frac{3}{8} \times \frac{5}{7}\))\
M1 for summing both possible orders: \(2 \times \left(\frac{5}{8} \times \frac{3}{7}\right)\)\
A0.5 for the correct simplified fraction \(\frac{15}{28}\)

First, find the vector \(\overrightarrow{AB}\):\
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\)\
\
Since \(AP : PB = 3 : 2\), the point \(P\) lies \(\frac{3}{5}\) of the way along the line segment \(AB\) from \(A\). Thus:\
\(\overrightarrow{AP} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\)\
\
Now find the position vector \(\overrightarrow{OP}\):\
\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\)\
\(\overrightarrow{OP} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\)\
\(\overrightarrow{OP} = \mathbf{a} - \frac{3}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\)\
\(\overrightarrow{OP} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\)

M1 for finding \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or writing \(\overrightarrow{OP} = \overrightarrow{OA} + \frac{3}{5}\overrightarrow{AB}\)\
M1 for substituting and expanding: \(\mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\)\
A0.5 for the correct simplified vector \(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\) or equivalent fractional form such as \(\frac{1}{5}(2\mathbf{a} + 3\mathbf{b})\)

Multiply both sides by the denominator \(2x + 5\) to clear the fraction:\
\(y(2x + 5) = 3x - 4\)\
\
Expand the left-hand side:\
\(2xy + 5y = 3x - 4\)\
\
Rearrange the equation to group all terms containing \(x\) on one side and the other terms on the opposite side:\
\(5y + 4 = 3x - 2xy\)\
\
Factorise \(x\) from the right-hand side:\
\(5y + 4 = x(3 - 2y)\)\
\
Divide both sides by \(3 - 2y\) to solve for \(x\):\
\(x = \frac{5y + 4}{3 - 2y}\)\
\
(Alternatively, if terms are grouped differently, this can also be written as \(x = \frac{-5y - 4}{2y - 3}\).)

M1 for correctly eliminating the fraction: \(y(2x + 5) = 3x - 4\)\
M1 for collecting terms containing \(x\) on one side and factoring out \(x\), e.g. \(x(3 - 2y) = 5y + 4\)\
A0.5 for \(x = \frac{5y + 4}{3 - 2y}\) or any equivalent fully simplified form

To simplify the fraction, we factorize the numerator and the denominator separately.

1. Factorize the numerator \(2x^2 - 5x - 3\):
We look for two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These numbers are \(-6\) and \(1\).
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).

2. Factorize the denominator \(4x^2 - 1\):
This is a difference of two squares: \(4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1)\).

3. Substitute these back into the fraction:
\(\frac{2x^2 - 5x - 3}{4x^2 - 1} = \frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\).

4. Divide out the common factor \((2x + 1)\):
\(\frac{x - 3}{2x - 1}\).

M1: Factorizing \(2x^2 - 5x - 3\) to \((2x+1)(x-3)\) (or equivalent)

M1: Factorizing \(4x^2 - 1\) to \((2x-1)(2x+1)\)

A0.5: Correct final simplified fraction \(\frac{x - 3}{2x - 1}\)

The total number of marbles is \(4 + 6 = 10\).

We want to find the probability of choosing two marbles of different colors. This can happen in two mutually exclusive ways:
1. A Red marble first, then a Blue marble (R, B)
2. A Blue marble first, then a Red marble (B, R)

Since the selection is without replacement:
- The probability of Red then Blue is:
\(P(\text{R, B}) = P(\text{R}) \times P(\text{B | R}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\)

- The probability of Blue then Red is:
\(P(\text{B, R}) = P(\text{B}) \times P(\text{R | B}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\)

- The total probability is:
\(P(\text{different colors}) = P(\text{R, B}) + P(\text{B, R}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90}\)

- Simplifying the fraction by dividing both numerator and denominator by 6:
\(\frac{48}{90} = \frac{8}{15}\).

M1: Finding the probability of one combination, e.g., \(\frac{4}{10} \times \frac{6}{9}\) or \(\frac{6}{10} \times \frac{4}{9}\)

M1: Adding the two possible combinations, e.g., \(2 \times \left(\frac{4}{10} \times \frac{6}{9}\right)\) or \(\frac{24}{90} + \frac{24}{90}\)

A0.5: Correctly simplified fraction \(\frac{8}{15}\)

First, express the vector \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).

Since the point \(P\) divides the line segment \(AB\) in the ratio \(3 : 2\), the fraction of \(AB\) from \(A\) to \(P\) is:
\(\frac{3}{3 + 2} = \frac{3}{5}\).

Therefore:
\(\overrightarrow{AP} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\).

Now find \(\overrightarrow{OP}\):
\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\).

Expand and simplify:
\(\overrightarrow{OP} = \mathbf{a} + \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a} = \left(1 - \frac{3}{5}\right)\mathbf{a} + \frac{3}{5}\mathbf{b} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\).

Alternatively, this can be written as \(\frac{2\mathbf{a} + 3\mathbf{b}}{5}\).

M1: Finding \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or equivalent

M1: Writing a correct vector equation for \(\overrightarrow{OP}\), e.g., \(\overrightarrow{OP} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\) or \(\overrightarrow{OP} = \mathbf{b} - \frac{2}{5}(\mathbf{b} - \mathbf{a})\)

A0.5: Fully simplified expression \(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\) or \(\frac{2\mathbf{a} + 3\mathbf{b}}{5}\)

To simplify the algebraic fraction, we factorise both the numerator and the denominator.

1. Factorise the quadratic expression in the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

2. Factorise the difference of two squares in the denominator:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

3. Divide out the common linear factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

M1: Correctly factorises the numerator into two linear brackets, e.g., \((2x + 1)(x - 3)\)
M1: Correctly factorises the denominator as a difference of two squares, e.g., \((2x - 1)(2x + 1)\)
A0.5: Correct fully simplified fraction \(\frac{x - 3}{2x - 1}\)

The total number of counters in the box is \(5 + 3 = 8\).

There are two mutually exclusive ways to choose counters of different colors:
1. Choosing a Yellow counter first, then a Green counter:
\(\text{P(Y, G)} = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

2. Choosing a Green counter first, then a Yellow counter:
\(\text{P(G, Y)} = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Add the probabilities of these two scenarios together:
\(\text{P(different colors)} = \frac{15}{56} + \frac{15}{56} = \frac{30}{56}\)

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, 2:
\(\frac{30}{56} = \frac{15}{28}\)

M1: Writes a correct product representing one of the sequential paths with descending denominator, i.e., \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\)
M1: Sums both of the valid scenarios together: \(\frac{15}{56} + \frac{15}{56}\) or equivalent calculation
A0.5: Correct fully simplified fraction \(\frac{15}{28}\) (accept equivalent simplified fraction, reject unsimplified \(\frac{30}{56}\))

We analyze the sequence to determine its order:
- Terms: \(4, \quad 7, \quad 14, \quad 25\)
- First differences: \(3, \quad 7, \quad 11\)
- Second differences: \(4, \quad 4\)

Since the second difference is constant and equal to 4, the sequence is quadratic and has an \(n^2\) coefficient of \(\frac{4}{2} = 2\).

Subtract the quadratic term \(2n^2\) from the original sequence to find the remaining linear part:
- For \(n = 1\): \(4 - 2(1)^2 = 2\)
- For \(n = 2\): \(7 - 2(2)^2 = -1\)
- For \(n = 3\): \(14 - 2(3)^2 = -4\)
- For \(n = 4\): \(25 - 2(4)^2 = -7\)

The linear sequence is \(2, -1, -4, -7, \dots\).
This is an arithmetic progression with a first term of 2 and a common difference of \(-3\).
Its general term is \(2 + (n - 1)(-3) = -3n + 5\).

Combining the quadratic and linear parts, the overall \(n\)-th term is:
\(2n^2 - 3n + 5\)

M1: Identifies the second difference as 4 and establishes that the \(n^2\) term is \(2n^2\)
M1: Subtracts \(2n^2\) to form the linear sequence \(2, -1, -4, -7, \dots\) and correctly finds its common difference of \(-3\)
A0.5: Correct final expression \(2n^2 - 3n + 5\)

Multiply both sides by \(2 + p\): \(q(2 + p) = 4p - 3\). Expand the bracket: \(2q + pq = 4p - 3\). Rearrange to get all terms with \(p\) on one side and other terms on the other side: \(2q + 3 = 4p - pq\). Factorise the right-hand side: \(2q + 3 = p(4 - q)\). Divide by \(4 - q\) to make \(p\) the subject: \(p = \frac{2q + 3}{4 - q}\). This can also be written as \(p = \frac{-2q - 3}{q - 4}\).

M1 for clearing the fraction to obtain \(q(2 + p) = 4p - 3\) or equivalent. M1 for isolating terms in \(p\) on one side, e.g., \(4p - pq = 2q + 3\). A0.5 for the final correct expression \(p = \frac{2q + 3}{4 - q}\) or equivalent.

Calculate the first differences: 7, 11, 15, 19. Calculate the second differences: 4, 4, 4. Since the second differences are constant and equal to 4, the sequence contains a term of \(an^2\) where \(2a = 4\), so \(a = 2\). Subtract \(2n^2\) from each term of the sequence: For \(n=1\): \(3 - 2(1)^2 = 1\). For \(n=2\): \(10 - 2(2)^2 = 2\). For \(n=3\): \(21 - 2(3)^2 = 3\). This leaves the sequence 1, 2, 3, ..., which has the \(n\)-th term of \(n\). Therefore, the overall \(n\)-th term is \(2n^2 + n\).

M1 for finding the second difference of 4, or for stating the coefficient of \(n^2\) is 2. M1 for subtracting \(2n^2\) from the terms and finding the linear sequence 1, 2, 3, ... has the rule \(n\). A0.5 for the final correct expression \(2n^2 + n\) or equivalent.

First, evaluate the first part: \(\left( \frac{8}{27} \right)^{-\frac{2}{3}} = \left( \frac{27}{8} \right)^{\frac{2}{3}} = \left( \sqrt[3]{\frac{27}{8}} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}\). Next, evaluate the second part: \(16^{\frac{3}{4}} = (\sqrt[4]{16})^3 = 2^3 = 8\). Finally, multiply the two results: \(\frac{9}{4} \times 8 = 9 \times 2 = 18\).

M1 for finding \(\left( \frac{8}{27} \right)^{-\frac{2}{3}} = \frac{9}{4}\) (or showing the intermediate step \((\frac{3}{2})^2\)). M1 for finding \(16^{\frac{3}{4}} = 8\) (or showing the intermediate step \(2^3\)). A0.5 for the correct final answer of 18.

First, factorise the numerator and the denominator. The numerator is a quadratic expression: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). The denominator is a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Rewriting the fraction gives \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\). Cancelling the common factor of \(2x + 1\) yields the final simplified form \(\frac{x - 3}{2x - 1}\).

M1 for attempting to factorise the quadratic numerator to get \((2x + 1)(x - 3)\). M1 for factorising the denominator to get \((2x - 1)(2x + 1)\). A0.5 for the correct simplified fraction \(\frac{x - 3}{2x - 1}\).

To find the \(n\)-th term of the sequence \(3, 10, 21, 36, 55, \dots\), we first look at the differences between successive terms. The first differences are \(7, 11, 15, 19, \dots\). The second differences are constant at \(4\). This indicates a quadratic sequence of the form \(an^2 + bn + c\) where \(2a = 4\), so \(a = 2\). Subtracting \(2n^2\) from each term of the original sequence: for \(n = 1\), \(3 - 2(1) = 1\); for \(n = 2\), \(10 - 8 = 2\); for \(n = 3\), \(21 - 18 = 3\). This leaves the sequence \(1, 2, 3, 4, \dots\), which is simply \(n\). Combining these parts, the \(n\)-th term is \(2n^2 + n\).

M1 for finding the constant second differences are 4, implying a \(2n^2\) term. M1 for subtracting \(2n^2\) from the terms to find the linear difference sequence. A0.5 for the final term \(2n^2 + n\).

The total number of counters in the bag is \(5 + 3 = 8\). The two outcomes where the counters are of different colours are (Red then Blue) and (Blue then Red). The probability of drawing a Red then a Blue is \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). The probability of drawing a Blue then a Red is \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). Adding these two mutually exclusive probabilities together gives the total probability of different colours: \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56}\), which simplifies to \(\frac{15}{28}\).

M1 for writing down one correct probability product, e.g., \(\frac{5}{8} \times \frac{3}{7}\). M1 for adding the two different order probabilities. A0.5 for the correct simplified fraction \(\frac{15}{28}\) (or equivalent).

(a) Factorise the numerator and denominator:
Numerator: \(2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)\)
Denominator: \(x^2 + x - 6 = (x+3)(x-2)\)

Divide both by the common factor \((x-2)\):
\(\frac{2(x-2)(x+2)}{(x+3)(x-2)} = \frac{2(x+2)}{x+3}\) (or \(\frac{2x+4}{x+3}\)).

(b) Find a common denominator, which is \((x-2)(x+1)\):
\(\frac{3(x+1) - 2(x-2)}{(x-2)(x+1)} = \frac{3x + 3 - 2x + 4}{(x-2)(x+1)} = \frac{x+7}{(x-2)(x+1)}\).

(a)
- M1 for factorising the numerator: \(2(x-2)(x+2)\) or \(2(x^2 - 4)\)
- M1 for factorising the denominator: \((x+3)(x-2)\)
- A1 for the final simplified fraction: \(\frac{2(x+2)}{x+3}\) or \(\frac{2x+4}{x+3}\)

(b)
- M1 for writing with a common denominator: \(\frac{3(x+1) - 2(x-2)}{(x-2)(x+1)}\) oe
- A1 for final fraction: \(\frac{x+7}{(x-2)(x+1)}\) or \(\frac{x+7}{x^2-x-2}\)

Let \(R\) be red and \(B\) be blue. The total number of marbles is \(5 + 3 = 8\).

(a) The probability that the first marble is blue is \(\frac{3}{8}\).
Since it is without replacement, there are now 7 marbles left, including 2 blue marbles.
The probability that the second marble is blue is \(\frac{2}{7}\).
\(P(\text{both blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}\).

(b) The event 'at least one marble is red' is the complement of the event 'both marbles are blue'.
Therefore:
\(P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - \frac{3}{28} = \frac{25}{28}\).

Alternatively, sum the probabilities of the outcomes containing at least one red:
\(P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\)
\(P(RB) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
\(P(BR) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)
Total probability = \(\frac{20 + 15 + 15}{56} = \frac{50}{56} = \frac{25}{28}\).

(a)
- M1 for \(\frac{3}{8} \times \frac{2}{7}\)
- A1 for \(\frac{3}{28}\) oe

(b)
- M1 for using the complement: \(1 - \text{their } P(BB)\), or summing the three correct cases
- M1 for a correct numerical expression: \(1 - \frac{3}{28}\) or \(\frac{20}{56} + \frac{15}{56} + \frac{15}{56}\) oe
- A1 for \(\frac{25}{28}\) oe

(a) The radius, perpendicular height \(h\), and slant height form a right-angled triangle.
Using Pythagoras' theorem:
\(3^2 + h^2 = 5^2\)
\(9 + h^2 = 25\)
\(h^2 = 16\)
\(h = 4\text{ cm}\).

(b) The total surface area \(A\) of a cone is given by \(A = \pi r^2 + \pi r l\), where \(r = 3\) and \(l = 5\):
\(A = \pi(3)^2 + \pi(3)(5) = 9\pi + 15\pi = 24\pi\text{ cm}^2\).

(c) The volume \(V\) of a cone is given by \(V = \frac{1}{3}\pi r^2 h\), where \(r = 3\) and \(h = 4\):
\(V = \frac{1}{3}\pi(3)^2(4) = \frac{1}{3}\pi(9)(4) = 12\pi\text{ cm}^3\).

(a)
- M1 for writing \(3^2 + h^2 = 5^2\) or \(h = \sqrt{5^2 - 3^2}\) showing detailed evaluation to get \(4\).

(b)
- M1 for using the formula \(\pi r^2 + \pi r l\) with correct values: \(\pi \times 3^2 + \pi \times 3 \times 5\)
- A1 for \(24\pi\) (accept equivalent exact values, e.g. \(24\pi\text{ cm}^2\))

(c)
- M1 for using the formula \(\frac{1}{3}\pi r^2 h\) with correct values: \(\frac{1}{3} \times \pi \times 3^2 \times 4\)
- A1 for \(12\pi\) (accept equivalent exact values, e.g. \(12\pi\text{ cm}^3\))

(a) First find \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).

Since \(AP : PB = 2 : 3\), the point \(P\) is \(\frac{2}{5}\) of the way along the vector \(\overrightarrow{AB}\) from \(A\):
\(\overrightarrow{AP} = \frac{2}{5}\overrightarrow{AB} = \frac{2}{5}(\mathbf{b} - \mathbf{a})\).

(b) Find \(\overrightarrow{QP}\) using a vector path, for example:
\(\overrightarrow{QP} = \overrightarrow{QO} + \overrightarrow{OA} + \overrightarrow{AP}\).

Since \(Q\) is the midpoint of \(OB\), we have \(\overrightarrow{QO} = -\frac{1}{2}\mathbf{b}\).

Substituting the known vectors:
\(\overrightarrow{QP} = -\frac{1}{2}\mathbf{b} + \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a})\)

Simplify the expression:
\(\overrightarrow{QP} = \mathbf{a} - \frac{2}{5}\mathbf{a} - \frac{1}{2}\mathbf{b} + \frac{2}{5}\mathbf{b}\)

\(\overrightarrow{QP} = \frac{3}{5}\mathbf{a} + \left(-\frac{5}{10} + \frac{4}{10}\right)\mathbf{b}\)

\(\overrightarrow{QP} = \frac{3}{5}\mathbf{a} - \frac{1}{10}\mathbf{b}\).

(a)
M1 for finding \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or recognizing \(\overrightarrow{AP} = \frac{2}{5}\overrightarrow{AB}\)
A1 for \(\frac{2}{5}(\mathbf{b} - \mathbf{a})\) or equivalent, e.g. \(-\frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\)

(b)
M1 for a correct vector path to find \(\overrightarrow{QP}\), e.g. \(\overrightarrow{QO} + \overrightarrow{OA} + \overrightarrow{AP}\) or \(\overrightarrow{QB} + \overrightarrow{BP}\)
M1 for substituting their vector from part (a) correctly into a valid path, e.g. \(-\frac{1}{2}\mathbf{b} + \mathbf{a} + \text{their } \overrightarrow{AP}\)
A1 for \(\frac{3}{5}\mathbf{a} - \frac{1}{10}\mathbf{b}\) or equivalent simplified expression

(a) The total number of balls initially is \(5 + 3 = 8\).

The probability that the first ball is red is \(\frac{5}{8}\).

Since the selection is without replacement, there are now 7 balls left, of which 4 are red.

The probability that the second ball is red is \(\frac{4}{7}\).

The probability that both balls are red is:
\(\text{P(Red, Red)} = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\).

(b) The two balls are of different colours if we choose a red ball then a blue ball, or a blue ball then a red ball:
\(\text{P(Red, Blue)} = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

\(\text{P(Blue, Red)} = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Combining these probabilities:
\(\text{P(different colours)} = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

(a)
M1 for \(\frac{5}{8} \times \frac{4}{7}\)
A1 for \(\frac{5}{14}\) or any equivalent fraction (e.g., \(\frac{20}{56}\))

(b)
M1 for calculating the probability of one combination of different colours, e.g. \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\)
M1 for adding the two different scenarios, e.g. \(2 \times \left(\frac{5}{8} \times \frac{3}{7}\right)\)
A1 for \(\frac{15}{28}\) or any equivalent fraction (e.g., \(\frac{30}{56}\))

(a) Find the differences between consecutive terms:
First differences: 7, 11, 15
Second differences: 4, 4

Since the second differences are constant, the sequence is quadratic and has the general form \(an^2 + bn + c\).

The coefficient \(a\) is half of the second difference:
\(a = \frac{4}{2} = 2\).

Subtract \(2n^2\) from each term of the original sequence:
For \(n = 1\): \(4 - 2(1)^2 = 2\)
For \(n = 2\): \(11 - 2(2)^2 = 3\)
For \(n = 3\): \(22 - 2(3)^2 = 4\)
For \(n = 4\): \(37 - 2(4)^2 = 5\)

The remaining linear sequence is 2, 3, 4, 5, ... which has the \(n\)th term formula \(n + 1\).

Combining these parts, the \(n\)th term of the sequence is:
\(2n^2 + n + 1\).

(b) Substitute \(n = 5\) into the given formula for the second sequence:
\(5 \times 2^{5-1} = 5 \times 2^4\)

Calculate \(2^4 = 16\):
\(5 \times 16 = 80\).

(a)
M1 for finding the second differences are 4, indicating a \(2n^2\) term
M1 for subtracting \(2n^2\) from the sequence to attempt to find the linear part \(bn + c\)
A1 for \(2n^2 + n + 1\)

(b)
M1 for substituting \(n = 5\) into the formula, e.g. \(5 \times 2^4\) or \(5 \times 16\)
A1 for 80

(a) We know that
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\).
Since \(AD : DB = 1 : 2\), the point \(D\) lies \(\frac{1}{3}\) of the way along \(AB\) from \(A\).
Thus,
\(\overrightarrow{AD} = \frac{1}{3}\overrightarrow{AB} = \frac{1}{3}(\mathbf{b} - \mathbf{a})\).

Now we find \(\overrightarrow{OD}\):
\(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a}) = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\).

(b) Since \(C\) is the midpoint of \(OB\), we have:
\(\overrightarrow{OC} = \frac{1}{2}\mathbf{b}\).

Let us find \(\overrightarrow{CD}\):
\(\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = \left(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\right) - \frac{1}{2}\mathbf{b} = \frac{2}{3}\mathbf{a} - \frac{1}{6}\mathbf{b} = \frac{1}{6}(4\mathbf{a} - \mathbf{b})\).

Now let us find \(\overrightarrow{CE}\):
\(\overrightarrow{CE} = \overrightarrow{OE} - \overrightarrow{OC} = \left(4\mathbf{a} - \frac{1}{2}\mathbf{b}\right) - \frac{1}{2}\mathbf{b} = 4\mathbf{a} - \mathbf{b}\).

Comparing \(\overrightarrow{CD}\) and \(\overrightarrow{CE}\):
\(\overrightarrow{CE} = 6\overrightarrow{CD}\).

Since \(\overrightarrow{CE}\) is a scalar multiple of \(\overrightarrow{CD}\), the vectors are parallel. Because they share the common point \(C\), the points \(C\), \(D\), and \(E\) must lie on a straight line.

(a)
* **M1**: For a correct vector path for \(\overrightarrow{OD}\), e.g., \(\overrightarrow{OA} + \frac{1}{3}\overrightarrow{AB}\) or \(\overrightarrow{OB} + \frac{2}{3}\overrightarrow{BA}\).
* **A1**: For \(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\) or equivalent simplest form.

(b)
* **M1**: For finding a correct expression for \(\overrightarrow{CD}\) (e.g., \(\frac{2}{3}\mathbf{a} - \frac{1}{6}\mathbf{b}\)) or \(\overrightarrow{DE}\) (e.g., \(\frac{10}{3}\mathbf{a} - \frac{5}{6}\mathbf{b}\)).
* **M1**: For finding a correct expression for \(\overrightarrow{CE}\) (e.g., \(4\mathbf{a} - \mathbf{b}\)).
* **A1**: For showing one vector is a scalar multiple of another (e.g., \(\overrightarrow{CE} = 6\overrightarrow{CD}\) or \(\overrightarrow{DE} = 5\overrightarrow{CD}\)) and concluding that they lie on a straight line due to sharing the common point \(C\) (or \(D\)).

(a) The total number of beads in the bag is \(x + 4\).
Since the beads are taken without replacement:
* The probability that the first bead is red is \(\frac{4}{x+4}\).
* The probability that the second bead is red is \(\frac{3}{x+3}\).

Therefore, the probability that both beads are red is:
\(\text{P(Red, Red)} = \frac{4}{x+4} \times \frac{3}{x+3} = \frac{12}{(x+4)(x+3)}\) (or \(\frac{12}{x^2 + 7x + 12}\)).

(b) We are given that this probability is equal to \(\frac{1}{6}\):
\(\frac{12}{(x+4)(x+3)} = \frac{1}{6}\)

Cross-multiplying gives:
\(12 \times 6 = (x+4)(x+3)\)
\(72 = x^2 + 7x + 12\)

Subtracting 72 from both sides to form a quadratic equation:
\(x^2 + 7x - 60 = 0\) (as required).

To solve this quadratic equation, we factorise it. We need two numbers that multiply to \(-60\) and add to \(7\). These are \(12\) and \(-5\):
\((x+12)(x-5) = 0\)

This gives \(x = -12\) or \(x = 5\).
Since the number of blue beads \(x\) must be positive, we reject \(x = -12\).
Thus, \(x = 5\).

(a)
* **M1**: For the product of two correct fractions, i.e., \(\frac{4}{x+4} \times \frac{3}{x+3}\).
* **A1**: For \(\frac{12}{(x+4)(x+3)}\) or \(\frac{12}{x^2 + 7x + 12}\).

(b)
* **M1**: For equating their expression from part (a) to \(\frac{1}{6}\) and cross-multiplying, e.g., \(72 = (x+4)(x+3)\).
* **M1**: For factorising the quadratic equation to \((x+12)(x-5) = 0\) (or using the quadratic formula correctly).
* **A1**: For finding \(x = 5\) and clearly rejecting \(x = -12\).

(a) To find the intersections with the axes: On the \(y\)-axis, \(x = 0 \implies y = 0^3 - 3(0)^2 = 0\), so the point is \((0, 0)\). On the \(x\)-axis, \(y = 0 \implies x^3 - 3x^2 = 0 \implies x^2(x - 3) = 0\), which gives \(x = 0\) or \(x = 3\). So, the points where the curve meets the coordinate axes are \((0, 0)\) and \((3, 0)\). (b) To find the stationary points, differentiate \(y = x^3 - 3x^2\): \(\frac{dy}{dx} = 3x^2 - 6x\). Set \(\frac{dy}{dx} = 0\): \(3x^2 - 6x = 0 \implies 3x(x - 2) = 0\). This gives \(x = 0\) or \(x = 2\). For \(x = 0\), \(y = 0^3 - 3(0)^2 = 0\). The stationary point is \((0, 0)\). For \(x = 2\), \(y = 2^3 - 3(2)^2 = 8 - 12 = -4\). The stationary point is \((2, -4)\). So, the stationary points are \((0, 0)\) and \((2, -4)\). (c) We can determine the nature of these points using the second derivative: \(\frac{d^2y}{dx^2} = 6x - 6\). At \(x = 0\), \(\frac{d^2y}{dx^2} = -6 < 0\), which is a local maximum. At \(x = 2\), \(\frac{d^2y}{dx^2} = 6(2) - 6 = 6 > 0\), which is a local minimum. Thus, the local minimum point is \((2, -4)\).

(a) [2.5 marks] M1 for setting \(y = 0\) and attempting to factorise: \(x^2(x - 3) = 0\). A1.5 for both \((0, 0)\) and \((3, 0)\) correct. Award 0.5 marks if only one correct pair is found. (b) [3 marks] M1 for correct differentiation to obtain \(\frac{dy}{dx} = 3x^2 - 6x\). M1 for setting \(\frac{dy}{dx} = 0\) and solving to find \(x = 0\) and \(x = 2\). A1 for both correct coordinates \((0, 0)\) and \((2, -4)\). (c) [2 marks] M1 for testing the nature of the stationary points using the second derivative or analyzing the gradient of the curve. A1 for identifying \((2, -4)\) as the local minimum point.

(a) To find the vertices of the region \(R\), we find the intersection points of the boundary lines taken two at a time: Line 1 and Line 2: \(y = 1\) and \(y = 2x + 1\). Substituting \(y = 1\) into \(y = 2x + 1\) gives: \(1 = 2x + 1 \implies 2x = 0 \implies x = 0\). So, the first vertex is \(A(0, 1)\). Line 1 and Line 3: \(y = 1\) and \(x + y = 4\). Substituting \(y = 1\) into \(x + y = 4\) gives: \(x + 1 = 4 \implies x = 3\). So, the second vertex is \(B(3, 1)\). Line 2 and Line 3: \(y = 2x + 1\) and \(x + y = 4\). Substituting \(y = 2x + 1\) into \(x + y = 4\) gives: \(x + (2x + 1) = 4 \implies 3x + 1 = 4 \implies 3x = 3 \implies x = 1\). Since \(x = 1\), \(y = 2(1) + 1 = 3\). So, the third vertex is \(C(1, 3)\). The three vertices of the region \(R\) are \((0, 1)\), \((3, 1)\), and \((1, 3)\). (b) We evaluate \(2x + 3y\) at each of the three vertices: At \(A(0, 1)\): \(2(0) + 3(1) = 3\). At \(B(3, 1)\): \(2(3) + 3(1) = 9\). At \(C(1, 3)\): \(2(1) + 3(3) = 11\). Comparing these values, the maximum value of \(2x + 3y\) is 11.

(a) [4.5 marks] M1 for attempting to find the intersection of \(y = 1\) and \(y = 2x + 1\). A0.5 for obtaining \((0, 1)\). M1 for attempting to find the intersection of \(y = 1\) and \(x + y = 4\). A0.5 for obtaining \((3, 1)\). M1 for attempting to solve the simultaneous equations \(y = 2x + 1\) and \(x + y = 4\). A0.5 for obtaining \((1, 3)\). (b) [3 marks] M1 for substituting at least one of their vertices from part (a) into the expression \(2x + 3y\). M1 for evaluating and comparing the expression \(2x + 3y\) at all three of their vertices. A1 for the correct maximum value of 11.

(a) A rotation of \(90^\circ\) anticlockwise about the origin \((0,0)\) maps any point \((x, y)\) to \((-y, x)\). Applying this mapping: \(P(1, 2) \to P'(-2, 1)\), \(Q(3, 2) \to Q'(-2, 3)\), and \(R(1, 5) \to R'(-5, 1)\). (b) An enlargement with center \(C(1, 1)\) and scale factor \(k = -2\) maps a point \(X\) to \(X''\) such that \(\vec{CX''} = -2 \vec{CX}\). For \(P(1, 2)\): \(\vec{CP} = \begin{pmatrix} 0 \\\\ 1 \end{pmatrix}\), so \(\vec{CP''} = \begin{pmatrix} 0 \\\\ -2 \end{pmatrix}\), giving \(P''(1, -1)\). For \(Q(3, 2)\): \(\vec{CQ} = \begin{pmatrix} 2 \\\\ 1 \end{pmatrix}\), so \(\vec{CQ''} = \begin{pmatrix} -4 \\\\ -2 \end{pmatrix}\), giving \(Q''(-3, -1)\). For \(R(1, 5)\): \(\vec{CR} = \begin{pmatrix} 0 \\\\ 4 \end{pmatrix}\), so \(\vec{CR''} = \begin{pmatrix} 0 \\\\ -8 \end{pmatrix}\), giving \(R''(1, -7)\).

(a) [3 marks] B1 for each correct vertex coordinates: \(P'(-2, 1)\), \(Q'(-2, 3)\), and \(R'(-5, 1)\). (b) [4.5 marks] For each vertex: M1 for applying the scale factor \(-2\) from the center \((1, 1)\); A0.5 for the correct coordinates (\(P''\) is \((1, -1)\), \(Q''\) is \((-3, -1)\), \(R''\) is \((1, -7)\)).