2024 AQA GCSE Mathematics 8300 Practice Paper with Answers

First, calculate the brackets: \( 3.2 + 4.8 = 8.0 \). Next, perform the multiplication: \( 1.5 \times 8 = 12 \). Then, calculate the square: \( 2.5^2 = 6.25 \). Finally, subtract: \( 12 - 6.25 = 5.75 \).

M1 for a correct intermediate step, such as finding the bracket sum is 8, or calculating \( 1.5 \times 8 = 12 \), or evaluating \( 2.5^2 = 6.25 \). A0.5 for the correct final answer of 5.75 (or equivalent fraction, e.g. \( \frac{23}{4} \) or \( 5\frac{3}{4} \)).

Actual distance in cm is \( 8.4 \times 25\,000 = 210\,000\text{ cm} \). To convert centimetres to metres, divide by 100: \( 210\,000 \div 100 = 2100\text{ m} \). To convert metres to kilometres, divide by 1000: \( 2100 \div 1000 = 2.1\text{ km} \).

M1 for correctly scaling up the map distance, e.g. \( 8.4 \times 25\,000 \) or finding \( 210\,000\text{ cm} \), or showing a correct method for converting cm to km (dividing by 100,000). A0.5 for 2.1.

To simplify the expression, find the greatest common divisor for each part. For the coefficients, \( \gcd(12, 18) = 6 \), which simplifies \( \frac{12}{18} \) to \( \frac{2}{3} \). For the variable parts, simplify \( \frac{x^2}{x} = x \) and \( \frac{y}{y^3} = \frac{1}{y^2} \). Combining these gives \( \frac{2x}{3y^2} \).

M1 for simplifying the coefficients to \( \frac{2}{3} \) or correctly simplifying at least one of the algebraic terms (e.g., getting \( x \) in the numerator or \( y^2 \) in the denominator). A0.5 for \( \frac{2x}{3y^2} \) or equivalent expression.

The total sum of probabilities must equal 1. Let the probability of landing on Yellow be \( x \). This means the probability of landing on Blue is \( 2x \). Setting up the equation: \( 0.4 + 2x + x = 1 \) which simplifies to \( 0.4 + 3x = 1 \). Subtract 0.4 from both sides to get \( 3x = 0.6 \), giving \( x = 0.2 \). Therefore, the probability of Blue is \( 2 \times 0.2 = 0.4 \).

M1 for setting up a correct equation or finding the sum of the remaining probabilities, e.g., \( 1 - 0.4 = 0.6 \) and writing \( 3x = 0.6 \). A0.5 for 0.4.

Multiply both sides of the equation by 5: \( 2x - 3 = 20 \). Add 3 to both sides: \( 2x = 23 \). Divide by 2: \( x = 11.5 \) or \( \frac{23}{2} \).

M1 for a correct first step to isolate the numerator, e.g., \( 2x - 3 = 20 \). A0.5 for 11.5 (accept \( 11\frac{1}{2} \) or \( \frac{23}{2} \)).

First, write the proportion of students who walk as a fraction: \( \frac{28}{80} \). Simplify this fraction by dividing both the numerator and the denominator by 4: \( \frac{7}{20} \). To turn this into a percentage, multiply by 100: \( \frac{7}{20} \times 100 = 7 \times 5 = 35\% \).

M1 for writing the fraction \( \frac{28}{80} \) and attempting to simplify it or multiply by 100. A0.5 for 35 (accept 35%).

Method 1: A regular octagon has 8 sides. The sum of the exterior angles is \( 360^\circ \), so each exterior angle is \( 360^\circ \div 8 = 45^\circ \). Since the interior and exterior angles sum to \( 180^\circ \), each interior angle is \( 180^\circ - 45^\circ = 135^\circ \). Method 2: Use the formula for the sum of interior angles: \( (8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ \). Divide by the number of angles: \( 1080^\circ \div 8 = 135^\circ \).

M1 for finding the exterior angle \( 360 \div 8 = 45 \), or showing a correct calculation for the sum of interior angles \( (8-2) \times 180 \). A0.5 for 135.

First, find the common difference between consecutive terms: \( 11 - 5 = 6 \). This means the formula contains \( 6n \). Compare the sequence \( 6n \) (which is 6, 12, 18, 24, ...) with the given sequence (5, 11, 17, 23, ...). We can see that each term in our sequence is 1 less than the corresponding term of \( 6n \). Therefore, the \(n\)-th term is \( 6n - 1 \).

M1 for identifying a common difference of 6 or writing an expression of the form \( 6n + c \) where \( c \) is any integer constant. A0.5 for \( 6n - 1 \) (or equivalent).

Let \(x = 0.1\dot{7} = 0.1777...\)
Multiply by 10:
\(10x = 1.777...\)
Multiply by 100:
\(100x = 17.777...\)
Subtracting the two equations:
\(100x - 10x = 17.777... - 1.777...\)
\(90x = 16\)
\(x = \frac{16}{90} = \frac{8}{45}\)

M1 for setting up two appropriate equations, e.g., \(10x = 1.77...\) and \(100x = 17.77...\), and subtracting them to get \(90x = 16\) or equivalent.
A0.5 for \(\frac{8}{45}\) (must be fully simplified).

The difference in parts between Bill and Ted is \(5 - 3 = 2\) parts.
Since Bill receives \(\pounds 24\) more than Ted, these 2 parts represent \(\pounds 24\).
Therefore, 1 part is \(\frac{\pounds 24}{2} = \pounds 12\).
The total number of parts is \(5 + 3 = 8\) parts.
The total amount of money shared is \(8 \times \pounds 12 = \pounds 96\).

M1 for finding the value of one part: \(\frac{24}{5 - 3} = 12\) or setting up an equation such as \(5x - 3x = 24\).
A0.5 for \(96\) (or \(\pounds 96\)).

First, expand the bracket on the left-hand side:
\(5x - 10 > 3x + 7\)
Next, subtract \(3x\) from both sides:
\(2x - 10 > 7\)
Then, add 10 to both sides:
\(2x > 17\)
Finally, divide by 2:
\(x > 8.5\) (or \(x > \frac{17}{2}\))

M1 for correctly expanding the bracket and collecting like terms on both sides to get \(2x > 17\) (or equivalent).
A0.5 for \(x > 8.5\) or \(x > \frac{17}{2}\).

First, simplify the numerator by applying the power of 3 to each term inside the brackets:
\((3x^3 y^2)^3 = 3^3 \times (x^3)^3 \times (y^2)^3 = 27x^9y^6\)
Now, divide this by the denominator:
\(\frac{27x^9y^6}{9x^4y^5} = \frac{27}{9} \times x^{9-4} \times y^{6-5} = 3x^5y\)

M1 for expanding the numerator correctly to \(27x^9y^6\), or for correctly dividing indices with at most one arithmetic slip.
A0.5 for \(3x^5y\) (or \(3x^5y^1\)).

To find the highest common factor (HCF), we identify the common prime factors in both numbers and choose the lowest power for each.
The common prime factors are 2 and 3.
The lowest power of 2 is \(2^2\).
The lowest power of 3 is \(3^1\) (or 3).
Therefore, the \(\text{HCF} = 2^2 \times 3 = 4 \times 3 = 12\).

M1 for identifying the correct common prime factors with their lowest powers, e.g., writing \(2^2 \times 3\) or listing factors of both numbers to find common ones.
A0.5 for \(12\).

The total number of counters in the bag is \(4 + 6 = 10\).
The probability of drawing a red counter on the first draw is \(\frac{4}{10} = \frac{2}{5}\).
Since the counter is replaced, the probability of drawing a red counter on the second draw is also \(\frac{2}{5}\).
Since these are independent events, the probability of both being red is:
\(\frac{2}{5} \times \frac{2}{5} = \frac{4}{25}\) (or \(0.16\))

M1 for finding the probability of drawing one red counter (\(\frac{4}{10}\) or \(0.4\)) and multiplying it by itself, i.e., \(\frac{4}{10} \times \frac{4}{10}\).
A0.5 for \(\frac{4}{25}\) or equivalent fraction (e.g., \(\frac{16}{100}\)) or decimal (\(0.16\)).

First, we address the negative exponent by taking the reciprocal:
\( 125^{-\frac{2}{3}} = \frac{1}{125^{\frac{2}{3}}} \)

Next, evaluate the cube root (the denominator of the fractional exponent):
\( 125^{\frac{1}{3}} = \sqrt[3]{125} = 5 \)

Then, square the result:
\( 5^2 = 25 \)

Combining these steps gives:
\( 125^{-\frac{2}{3}} = \frac{1}{25} \)

M1 for correctly identifying the cube root of 125 is 5, e.g., showing \( 125^{\frac{1}{3}} = 5 \) or \( (5^3)^{-\frac{2}{3}} \) or writing \( \frac{1}{125^{\frac{2}{3}}} \)
A1 for \( \frac{1}{25} \)

First, solve the quadratic equation \( x^2 - 2x - 15 = 0 \) to find the critical values.
Factorising the quadratic expression:
\( (x - 5)(x + 3) = 0 \)
This gives the critical values:
\( x = 5 \) and \( x = -3 \)

Since we require \( x^2 - 2x - 15 \le 0 \), the solution lies on or between the two critical values on a graph.
Thus, \( -3 \le x \le 5 \).

M1 for factorising to \( (x - 5)(x + 3) \) or finding the critical values \( -3 \) and \( 5 \).
A1 for \( -3 \le x \le 5 \) (allow equivalent interval notation, e.g., \( x \ge -3 \) and \( x \le 5 \)).

Let \( x = 0.4181818... \)
Then, multiply by 10 to align the decimal point before the recurring digits:
\( 10x = 4.181818... \) (Equation 1)

Multiply by 1000 to move one full repeating period past the decimal point:
\( 1000x = 418.181818... \) (Equation 2)

Subtract Equation 1 from Equation 2:
\( 1000x - 10x = 418.1818... - 4.1818... \)
\( 990x = 414 \)

Solve for \( x \):
\( x = \frac{414}{990} \)

Simplify the fraction by dividing the numerator and denominator by 18:
\( \frac{414 \div 18}{990 \div 18} = \frac{23}{55} \)

Hence, \( 0.4\dot{1}\dot{8} = \frac{23}{55} \).

M1 for setting up two appropriate equations that when subtracted eliminate the recurring part (e.g., \( 1000x = 418.\dot{1}\dot{8} \) and \( 10x = 4.\dot{1}\dot{8} \)).
A1 for showing complete steps from \( 990x = 414 \) to the simplified fraction \( \frac{23}{55} \).

Let the ratio of red to blue be \( R : B = 1 : 3 \).
Let the ratio of blue to green be \( B : G = 9 : 8 \).

To combine these ratios, we make the parts for blue (\( B \)) equal. The lowest common multiple of 3 and 9 is 9.
Multiply the first ratio \( R : B \) by 3:
\( R : B = 3 : 9 \)

Now, we can write the combined ratio:
\( R : B : G = 3 : 9 : 8 \)

Calculate the total number of parts:
\( 3 + 9 + 8 = 20 \) parts

The proportion of blue pens is:
\( \frac{9}{20} \)

Convert this to a percentage:
\( \frac{9}{20} \times 100 = 45\% \)

M1 for finding a common scale for the ratios to write a combined ratio, e.g., \( R:B:G = 3:9:8 \) (or equivalent).
A1 for \( 45\% \)

First, factorise the numerator \( 3x^2 - 14x - 5 \):
We need two numbers that multiply to \( 3 \times (-5) = -15 \) and add to \( -14 \). These numbers are \( -15 \) and \( 1 \).
\( 3x^2 - 15x + x - 5 = 3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5) \)

Next, factorise the denominator \( x^2 - 25 \) using the difference of two squares:
\( x^2 - 25 = (x - 5)(x + 5) \)

Now rewrite the fraction with the factorised terms:
\( \frac{(3x + 1)(x - 5)}{(x - 5)(x + 5)} \)

Cancel the common factor of \( (x - 5) \):
\( \frac{3x + 1}{x + 5} \)

M1 for factorising either the numerator to \( (3x + 1)(x - 5) \) or the denominator to \( (x - 5)(x + 5) \).
A1 for \( \frac{3x + 1}{x + 5} \)

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

There are two scenarios where the discs are of different colours:
1. Red then Blue (RB)
2. Blue then Red (BR)

Calculate the probability of Red then Blue:
\( P(\text{Red then Blue}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} \)

Calculate the probability of Blue then Red:
\( P(\text{Blue then Red}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} \)

Sum the two probabilities:
\( P(\text{different colours}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} \)

Simplify the fraction by dividing the numerator and denominator by 6:
\( \frac{48 \div 6}{90 \div 6} = \frac{8}{15} \)

M1 for a correct product for one option, e.g., \( \frac{4}{10} \times \frac{6}{9} \) (or equivalent branch on a tree diagram).
A1 for \( \frac{8}{15} \) (accept equivalent fully simplified fraction).

The perimeter of a sector consists of two straight edges (each equal to the radius) and the curved arc length.

Straight edges:
\( 2 \times r = 2 \times 9 = 18\text{ cm} \)

Arc length:
\( \text{Arc length} = \frac{\theta}{360} \times 2\pi r \)
\( \text{Arc length} = \frac{120}{360} \times 2 \times \pi \times 9 \)
\( \text{Arc length} = \frac{1}{3} \times 18\pi = 6\pi\text{ cm} \)

Perimeter:
\( \text{Perimeter} = 18 + 6\pi \)

M1 for calculating the arc length of the sector, e.g., showing \( \frac{120}{360} \times 2 \times \pi \times 9 \) or obtaining \( 6\pi \).
A1 for \( 18 + 6\pi \) (or equivalent with correct integer values for \( a \) and \( b \)).

Multiply both sides by the denominator \( (x + 2) \):
\( y(x + 2) = 3x - 5 \)

Expand the brackets:
\( xy + 2y = 3x - 5 \)

Rearrange the equation to collect all terms containing \( x \) on one side and terms without \( x \) on the other side:
\( 2y + 5 = 3x - xy \)

Factorise \( x \) from the right-hand side:
\( 2y + 5 = x(3 - y) \)

Divide both sides by \( (3 - y) \) to isolate \( x \):
\( x = \frac{2y + 5}{3 - y} \)

M1 for multiplying by \( (x+2) \) and expanding correctly to get \( xy + 2y = 3x - 5 \), or for isolating terms in \( x \) on one side and factorising to \( x(3 - y) = 2y + 5 \) (or equivalent).
A1 for \( x = \frac{2y + 5}{3 - y} \) (or equivalent, e.g. \( x = \frac{-2y - 5}{y - 3} \)).

Multiply the fraction by 100 to convert it to a percentage:

\(\frac{7}{16} \times 100 = \frac{700}{16}\)

Simplify by dividing the numerator and denominator by 4:

\(\frac{700 \div 4}{16 \div 4} = \frac{175}{4}\)

Convert \(\frac{175}{4}\) to a decimal by division:

\(175 \div 4 = 43.75\)

Therefore, \(\frac{7}{16} = 43.75\%\).

M1 for attempt to multiply the fraction by 100 or convert to decimal (e.g. \(7 \div 16\))
A1 for 43.75% (or 43.75)

To compare the ratios directly, make the parts representing blue counters equal in both ratios.

For Red : Blue = \(3 : 5\), multiply both parts by 4:
\(R : B = 12 : 20\)

For Blue : Green = \(4 : 7\), multiply both parts by 5:
\(B : G = 20 : 35\)

Now we can combine them into a single ratio:
\(R : B : G = 12 : 20 : 35\)

Thus, the ratio of red counters to green counters is \(12 : 35\).

M1 for attempting to find a common value for the blue parts (e.g. multiplying the first ratio by 4 and the second by 5, or equivalent)
A1 for 12 : 35 (accept 12 to 35)

Factorise both the numerator and the denominator:

Numerator: \(6x^2 - 18x = 6x(x - 3)\)

Denominator: \(3x^2 - 27 = 3(x^2 - 9) = 3(x - 3)(x + 3)\) (using the difference of two squares)

Substitute these back into the fraction:

\(\frac{6x(x - 3)}{3(x - 3)(x + 3)}\)

Divide the numerator and the denominator by their common factors, \(3\) and \(x - 3\):

\(\frac{2x}{x + 3}\).

M1 for factorising either the numerator as \(6x(x-3)\) or the denominator as \(3(x-3)(x+3)\)
A1 for \(\frac{2x}{x+3}\)

First, write the negative power as a reciprocal:

\(64^{-\frac{2}{3}} = \frac{1}{64^{\frac{2}{3}}}\)

Next, evaluate the fractional power in the denominator. The cube root of 64 is:

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

Then square the result:

\(4^2 = 16\)

So, \(64^{\frac{2}{3}} = 16\).

Therefore, \(64^{-\frac{2}{3}} = \frac{1}{16}\).

M1 for expressing the term with a positive power as \(\frac{1}{64^{2/3}}\), or for correctly finding the cube root of 64 as 4
A1 for \(\frac{1}{16}\)

The sum of all probabilities must equal 1:

\(0.25 + x + 0.45 + 2x = 1\)

Combine like terms:

\(0.70 + 3x = 1\)

Subtract 0.70 from both sides:

\(3x = 0.30\)

Divide by 3:

\(x = 0.10\)

The probability of landing on 4 is \(2x\):

\(P(4) = 2 \times 0.10 = 0.2\) (or \(\frac{1}{5}\)).

M1 for setting up the equation \(0.25 + x + 0.45 + 2x = 1\) and solving to find \(x = 0.1\)
A1 for 0.2 (or equivalent fraction/percentage)

Multiply the first equation by 3 and the second equation by 2 to align the coefficients of \(y\):

\(9x + 6y = 12\)

\(8x - 6y = 22\)

Add the two equations together:

\(17x = 34\)

\(x = 2\)

Substitute \(x = 2\) back into the first equation:

\(3(2) + 2y = 4\)

\(6 + 2y = 4\)

\(2y = -2\)

\(y = -1\)

So the solution is \(x = 2, y = -1\).

M1 for a correct method to eliminate one variable (e.g. multiplying both equations to align coefficients and adding/subtracting)
A1 for both \(x = 2\) and \(y = -1\)

The formula for the total surface area of a cylinder is:

\(A = 2\pi r^2 + 2\pi r h\)

where \(r = 3\text{ cm}\) and \(h = 8\text{ cm}\).

First, find the combined area of the two circular ends:

\(2 \times ̇\pi \times 3^2 = 18\pi\)

Next, find the curved surface area:

\(2 \times \pi \times 3 \times 8 = 48\pi\)

Add these together to find the total surface area:

\(18\pi + 48\pi = 66\pi\)

So, the total surface area is \(66\pi\text{ cm}^2\).

M1 for substituting \(r=3\) and \(h=8\) correctly into either \(2\pi r^2\) (giving \(18\pi\)) or \(2\pi r h\) (giving \(48\pi\))
A1 for \(66\pi\) (accept with units \(66\pi\text{ cm}^2\))

Find the first differences between the terms:

\(11 - 4 = 7\)

\(20 - 11 = 9\)

\(31 - 20 = 11\)

First differences: \(7, 9, 11\)

Find the second differences:

\(9 - 7 = 2\)

\(11 - 9 = 2\)

Since the second difference is \(2\), the coefficient of the \(n^2\) term is \(\frac{2}{2} = 1\).

Subtract \(n^2\) from each term of the original sequence to find the linear part:

For \(n = 1\): \(4 - 1^2 = 3\)

For \(n = 2\): \(11 - 2^2 = 7\)

For \(n = 3\): \(20 - 3^2 = 11\)

For \(n = 4\): \(31 - 4^2 = 15\)

The linear sequence is \(3, 7, 11, 15, \dots\) which has a common difference of \(4\), so its formula is \(4n - 1\).

Combining these parts gives the \(n\)-th term expression:

\(n^2 + 4n - 1\).

M1 for finding the second difference of 2 and identifying the \(n^2\) term, or for attempting to subtract \(n^2\) from the sequence terms
A1 for \(n^2 + 4n - 1\) (or equivalent)

To find the percentage, write the number of sci-fi novels as a fraction of the total books: \(\frac{27}{120}\). Simplify the fraction by dividing the numerator and denominator by 3: \(\frac{27}{120} = \frac{9}{40}\). To convert this fraction to a percentage, multiply by 100: \(\frac{9}{40} \times 100 = \frac{900}{40} = \frac{90}{4} = 22.5\%\).

M1 for setting up the fraction \(\frac{27}{120}\) or simplifying to \(\frac{9}{40}\). A1 for 22.5%.

Multiply each term in the first bracket by each term in the second bracket: \((3x \times 2x) + (3x \times 5) + (-2 \times 2x) + (-2 \times 5) = 6x^2 + 15x - 4x - 10\). Simplify by combining the like terms: \(15x - 4x = 11x\). This gives \(6x^2 + 11x - 10\).

M1 for expansion of at least three terms correctly (e.g. \(6x^2 + 15x - 4x\) or similar). A1 for fully correct simplified expression \(6x^2 + 11x - 10\).

Convert the total mass to grams: \(1.5\text{ kg} = 1500\text{ g}\). Find the total number of parts in the ratio: \(5 + 3 + 2 = 10\text{ parts}\). Calculate the mass of one part: \(1500\text{ g} / 10 = 150\text{ g}\). Sugar corresponds to 3 parts: \(3 \times 150\text{ g} = 450\text{ g}\).

M1 for converting to grams (1500) and dividing by 10, or for finding 3/10 of 1.5kg (0.45kg). A1 for 450 (or 450g).

Subtract \(3x\) from both sides of the inequality: \(2x - 3 > 8\). Add 3 to both sides of the inequality: \(2x > 11\). Divide both sides by 2: \(x > 5.5\) (or \(x > \frac{11}{2}\)).

M1 for rearranging terms correctly to get \(2x > 11\) (or equivalent). A1 for \(x > 5.5\) or \(x > 11/2\).

Multiply the numbers: \(4 \times 8 = 32\). Multiply the powers of 10: \(10^{-5} \times 10^8 = 10^{-5+8} = 10^3\). Combine them: \(32 \times 10^3\). Convert to standard form: \(32 \times 10^3 = 3.2 \times 10^1 \times 10^3 = 3.2 \times 10^4\).

M1 for \(32 \times 10^3\) or \(32000\). A1 for \(3.2 \times 10^4\).

To estimate the number of times, multiply the number of spins by the probability of landing on Heads: \(300 \times 0.35 = 3 \times 35 = 105\).

M1 for showing multiplication \(300 \times 0.35\). A1 for 105.

Find the common difference between consecutive terms: \(13 - 7 = 6\), \(19 - 13 = 6\). Since the difference is 6, the expression starts with \(6n\). For \(n=1\), \(6(1) = 6\). To get the first term of 7, we must add 1. Therefore, the \(n\)-th term is \(6n + 1\).

M1 for finding a common difference of 6 or writing \(6n\). A1 for \(6n + 1\).

The formula for the area of a trapezium is \(\text{Area} = \frac{1}{2}(a+b)h\), where \(a\) and \(b\) are the parallel sides and \(h\) is the height. Substitute the given values: \(\text{Area} = \frac{1}{2}(8 + 12) \times 5.5 = \frac{1}{2}(20) \times 5.5 = 10 \times 5.5 = 55\text{ cm}^2\).

M1 for substituting correctly into the formula: \(\frac{1}{2}(8+12) \times 5.5\). A1 for 55.

First, convert both mixed numbers to improper fractions: \( 2\frac{1}{4} = \frac{9}{4} \) and \( 1\frac{3}{8} = \frac{11}{8} \). Next, divide the fractions by multiplying by the reciprocal of the second fraction: \( \frac{9}{4} \div \frac{11}{8} = \frac{9}{4} \times \frac{8}{11} = \frac{72}{44} \). Simplify the fraction by dividing the numerator and denominator by 4: \( \frac{18}{11} \). Convert the improper fraction back to a mixed number: \( 1\frac{7}{11} \).

M1: For converting both mixed numbers to improper fractions, \( \frac{9}{4} \) and \( \frac{11}{8} \), or for multiplying by the reciprocal of their second fraction. A1: For \( 1\frac{7}{11} \) (or equivalent mixed number in its simplest form).

Multiply both sides of the equation by 2 to clear the fraction: \( 3x + 1 = 2(8 - x) \). Expand the bracket on the right-hand side: \( 3x + 1 = 16 - 2x \). Add \( 2x \) to both sides to collect the x terms on one side: \( 5x + 1 = 16 \). Subtract 1 from both sides: \( 5x = 15 \). Divide both sides by 5: \( x = 3 \).

M1: For a correct first step of multiplying both sides by 2, e.g. \( 3x + 1 = 2(8 - x) \) or \( 3x + 1 = 16 - 2x \). A1: For \( x = 3 \) (or 3).

The measurement is correct to the nearest millimetre, which is \(0.1\text{ cm}\). The maximum error is half of this unit, which is \(0.05\text{ cm}\). To find the upper bound, we add this error to the measured value: \(12.4 + 0.05 = 12.45\text{ cm}\).

M1: For identifying the error interval boundary of \(0.05\text{ cm}\) (or \(0.5\text{ mm}\)) or for showing \(12.4 + 0.05\).
A1: For \(12.45\) (accept with or without units).

First, calculate the total number of parts in the ratio: \(2 + 3 + 5 = 10\). Next, find the value of one part: \(£340 \div 10 = £34\). The largest share consists of \(5\) parts: \(5 \times £34 = £170\).

M1: For finding the total number of parts (10) and attempting to find the value of one part (\(340 \div 10\)), or for calculating \(340 \times \frac{5}{10}\).
A1: For \(170\) (accept \(£170\)).

The multiplier for a \(12\%\) depreciation is \(1 - 0.12 = 0.88\). The duration from the start of 2021 to the start of 2023 is \(2\) years. The value of the car after \(2\) years is: \(15000 \times 0.88^2 = 15000 \times 0.7744 = 11616\).

M1: For a complete method to find the depreciated value after 2 years, e.g. \(15000 \times 0.88^2\) or finding \(15000 \times 0.88 = 13200\) and then \(13200 \times 0.88\).
A1: For \(11616\) (accept \(£11,616\)).

To estimate the number of successful outcomes, multiply the total number of trials by the probability of landing on Heads: \(300 \times 0.65 = 195\).

M1: For the calculation \(300 \times 0.65\).
A1: For \(195\).

First, add \(7\) to both sides of the inequality: \(4x > 22\). Next, divide both sides by \(4\): \(x > \frac{22}{4}\), which simplifies to \(x > 5.5\).

M1: For a correct first step to isolate the term in \(x\), e.g., \(4x > 22\) or \(4x = 22\).
A1: For \(x > 5.5\) or \(x > \frac{11}{2}\).

The gradient \(m\) is given by the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the coordinates gives: \(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\).

M1: For substituting the values into the gradient formula, e.g., \(\frac{17-5}{6-2}\) or finding the change in \(y\) (12) and the change in \(x\) (4).
A1: For \(3\).

To write \(0.000045\) in standard form, we move the decimal point \(5\) places to the right to obtain a number between 1 and 10, which is \(4.5\). Since the decimal moved to the right, the index is negative: \(4.5 \times 10^{-5}\).

M1: For identifying \(4.5\) and a power of 10, e.g., \(4.5 \times 10^k\) where \(k \neq 0\) or showing division by \(100,000\).
A1: For \(4.5 \times 10^{-5}\).

The area of a sector is given by the formula \(\text{Area} = \frac{\theta}{360} \times \pi r^2\). Substituting the given values: \(\text{Area} = \frac{120}{360} \times \pi \times 6^2 = \frac{1}{3} \times 36\pi = 12\pi\).

M1: For substituting the values into the sector area formula, e.g., \(\frac{120}{360} \times \pi \times 6^2\).
A1: For \(12\pi\) (accept \(12\pi\text{ cm}^2\)).

First year value: \(15000 \times (1 - 0.12) = 15000 \times 0.88 = 13200\)
Second year value: \(13200 \times (1 - 0.08) = 13200 \times 0.92 = 12144\).
Therefore, the value of the car at the end of the second year is \(£12,144\).

M1: For a correct method to find the value after the first year (e.g., \(15000 \times 0.88\) or finding \(13200\)) or for a complete method to find the final value (e.g., \(15000 \times 0.88 \times 0.92\)).
A0.5: For \(12144\).

Using the density formula: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\).
\(\text{Density} = \frac{168}{240} = 0.7\text{ g/cm}^3\).

M1: For \(168 \div 240\).
A0.5: For \(0.7\).

Multiply both sides by \(4\):
\(3x - 5 = 5.5 \times 4\)
\(3x - 5 = 22\)
Add \(5\) to both sides:
\(3x = 27\)
Divide by \(3\):
\(x = 9\).

M1: For a correct first step to isolate the numerator, e.g., \(3x - 5 = 22\) or \(\frac{3x}{4} = 6.75\).
A0.5: For \(9\).

First, find the probability of landing on Tails:
\(P(\text{Tails}) = 1 - 0.65 = 0.35\).
Next, calculate the expected number of Tails:
\(300 \times 0.35 = 105\).

M1: For \((1 - 0.65) \times 300\) or showing \(0.35 \times 300\) or finding \(195\) (expected number of Heads) and subtracting from \(300\).
A0.5: For \(105\).

First year depreciation:
\(18000 \times (1 - 0.12) = 18000 \times 0.88 = £15,840\)

Second and third year depreciation (8% per year for 2 years):
\(15840 \times 0.92^2 = 15840 \times 0.8464 = 13407.024\)

To the nearest pound, the value is £13,407.

M1: For calculating the value after Year 1: \(18000 \times 0.88 = 15840\), or for set up of a correct depreciation calculation such as \(15840 \times 0.92^2\).
A1: For 13407 (or 13407.02).

Expand the bracket:
\(5x - 15 > 2x + 9\)

Subtract \(2x\) from both sides:
\(3x - 15 > 9\)

Add 15 to both sides:
\(3x > 24\)

Divide by 3:
\(x > 8\)

M1: For a correct expansion of the bracket to \(5x - 15\), or for a correct step to collect the \(x\) terms on one side (e.g. \(3x - 15 > 9\)).
A1: For \(x > 8\).

Find the probability of landing on Yellow:
\(P(\text{Yellow}) = 1 - (0.35 + 0.4) = 1 - 0.75 = 0.25\)

Calculate the expected number of spins landing on Yellow:
\(250 \times 0.25 = 62.5\)

M1: For finding the probability of Yellow: \(1 - (0.35 + 0.4) = 0.25\) (or for finding expected Red and Blue: \(250 \times 0.75 = 187.5\)).
A1: For 62.5 (accept 62 or 63).

We need two numbers that multiply to give \(-24\) and add to give \(5\).
These numbers are \(+8\) and \(-3\).
Therefore, the factorised expression is \((x + 8)(x - 3)\).

M1: For writing a product of two linear brackets of the form \((x + a)(x + b)\) where \(a \times b = -24\) or \(a + b = 5\).
A1: For \((x + 8)(x - 3)\) (brackets can be in either order).

Area of a sector is given by:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)

Substitute the values:
\(\text{Area} = \frac{135}{360} \times \pi \times 8^2\)
\(\text{Area} = \frac{3}{8} \times 64 \times \pi = 24\pi \approx 75.398...\)

To 1 decimal place, the area is 75.4 \(\text{cm}^2\).

M1: For substituting the values into the correct sector area formula: \(\frac{135}{360} \times \pi \times 8^2\) (or equivalent).
A1: For 75.4 (accept 75.39 to 75.41).

An octagon has \(n = 8\) sides.
Method 1: Using exterior angles.
Exterior angle = \(360^\circ \div 8 = 45^\circ\)
Interior angle = \(180^\circ - 45^\circ = 135^\circ\)

Method 2: Using the sum of interior angles.
Sum of interior angles = \((8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ\)
One interior angle = \(1080^\circ \div 8 = 135^\circ\).

M1: For a correct method to find either the sum of the interior angles (\((8-2) \times 180\)) or the size of an exterior angle (\(360 \div 8\)).
A1: For 135.

The lower bound is the smallest value that rounds up to 7.4, which is 7.35.
The upper bound is the smallest value that rounds to 7.5, which is 7.45.
Therefore, the error interval is \(7.35 \le y < 7.45\).

M1: For identifying either 7.35 or 7.45.
A1: For \(7.35 \le y < 7.45\) (accept equivalent notation or clear identification of the lower and upper bounds).

Number of students who do not walk to school:
\(1200 - 432 = 768\)

Percentage who do not walk:
\(\frac{768}{1200} \times 100 = 64\%\)

Alternatively, percentage who walk:
\(\frac{432}{1200} \times 100 = 36\%\)
Percentage who do not walk:
\(100\% - 36\% = 64\%\).

M1: For finding the number of students who do not walk (768) or for calculating the percentage who walk (36%).
A1: For 64.

First, evaluate the term inside the square root: \(14.8^2 - 6.33 = 219.04 - 6.33 = 212.71\). Next, calculate the square root: \(\sqrt{212.71} \approx 14.58458\). Then, evaluate the denominator: \(3.1 \times 0.45 = 1.395\). Divide the numerator by the denominator: \(\frac{14.58458}{1.395} \approx 10.4549\). Rounding to 3 significant figures gives \(10.5\).

M1 for showing \(14.58\dots\) or \(1.395\) or an unrounded answer of \(10.45\dots\) or \(\frac{\sqrt{212.71}}{1.395}\)
A1 for \(10.5\)

The sale price of \(\pounds 468\) represents \(100\% - 28\% = 72\%\) of the original price. To find the original price, divide the sale price by \(0.72\): \(468 \div 0.72 = 650\). Therefore, the original price was \(\pounds 650\).

M1 for \(468 \div 0.72\) or equating \(72\% = 468\)
A1 for \(650\)

The total number of shares in the ratio is \(5 + 3 + 2 = 10\). One share is equivalent to \(14.5 \div 10 = 1.45\text{ kg}\). Zinc has \(3\) shares, so the mass of zinc is \(3 \times 1.45 = 4.35\text{ kg}\).

M1 for \(14.5 \div 10\) or \(14.5 \times \frac{3}{10}\)
A1 for \(4.35\)

Multiply both sides by 2 to get \(2y = 3x - 5\). Add 5 to both sides to get \(2y + 5 = 3x\). Divide both sides by 3 to isolate \(x\), giving \(x = \frac{2y + 5}{3}\).

M1 for correctly multiplying by 2 (\(2y = 3x - 5\)) or adding 5 to both sides divided by 2 (\(y + 2.5 = 1.5x\))
A1 for \(x = \frac{2y + 5}{3}\) or equivalent correct expression

Using the FOIL method or grid method: \(2x \times x = 2x^2\), \(2x \times 4 = 8x\), \(-3 \times x = -3x\), and \(-3 \times 4 = -12\). Combining these terms gives \(2x^2 + 8x - 3x - 12\). Simplifying the middle terms gives \(2x^2 + 5x - 12\).

M1 for expansion with at least three correct terms from \(2x^2, 8x, -3x, -12\)
A1 for \(2x^2 + 5x - 12\)

The sum of all probabilities is 1. The remaining probability for C and D combined is \(1 - (0.25 + 0.35) = 1 - 0.60 = 0.40\). Let the probability of D be \(x\). Then the probability of C is \(3x\). This means \(3x + x = 0.40\), so \(4x = 0.40\), which gives \(x = 0.10\). Thus, the probability of landing on C is \(3 \times 0.10 = 0.30\).

M1 for setting up \(1 - (0.25 + 0.35) = 0.4\) and dividing by 4 to find \(x = 0.1\) or equating \(3x + x = 0.4\)
A1 for \(0.3\) or \(0.30\)

The sum of the exterior angles of any regular polygon is always \(360^\circ\). To find the number of sides, divide the total sum of the exterior angles by the size of one exterior angle: \(360^\circ \div 15^\circ = 24\). Therefore, the polygon has 24 sides.

M1 for \(360 \div 15\)
A1 for \(24\)

The area of a sector is given by the formula \(A = \frac{\theta}{360} \times \pi r^2\). Substituting the given values: \(A = \frac{135}{360} \times \pi \times 8^2 = \frac{3}{8} \times 64\pi = 24\pi \approx 75.3982\text{ cm}^2\). Rounding to 3 significant figures gives \(75.4\text{ cm}^2\).

M1 for \(\frac{135}{360} \times \pi \times 8^2\) or \(24\pi\) or \(75.39...\)
A1 for \(75.4\)

The multiplier for a \(12.5\%\) decrease is \(1 - 0.125 = 0.875\). After \(3\) years, the value is \(24000 \times 0.875^3 = 16078.125\). To the nearest pound, this is \(16078\).

M1 for \(24000 \times 0.875^3\) or for working step-by-step for \(3\) years. A1 for \(16078\).

We can write \(y = kx^2\). Substituting \(x = 4\) and \(y = 72\) gives \(72 = k \times 4^2\), so \(16k = 72\) which means \(k = 4.5\). When \(x = 5\), \(y = 4.5 \times 5^2 = 4.5 \times 25 = 112.5\).

M1 for writing \(y = kx^2\) and finding \(k = 4.5\). A1 for \(112.5\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = 8\), and \(c = -5\): \(x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-5)}}{2(3)} = \frac{-8 \pm \sqrt{124}}{6}\). This gives \(x \approx 0.52\) and \(x \approx -3.19\).

M1 for correct substitution into the quadratic formula, e.g. \(\frac{-8 \pm \sqrt{8^2 - 4 \times 3 \times (-5)}}{2 \times 3}\). A1 for both \(0.52\) and \(-3.19\).

The sum of the first four numbers is \(4 \times 14 = 56\). The sum of all five numbers is \(5 \times 15.5 = 77.5\). The fifth number is \(77.5 - 56 = 21.5\).

M1 for calculating either \(4 \times 14 = 56\) or \(5 \times 15.5 = 77.5\). A1 for \(21.5\).

The area of a sector is \(\frac{\theta}{360} \times \pi r^2\). Substituting the values: \(\text{Area} = \frac{135}{360} \times \pi \times 8^2 = 24\pi \approx 75.398\text{ cm}^2\). To \(1\) decimal place, this is \(75.4\).

M1 for \(\frac{135}{360} \times \pi \times 8^2\). A1 for \(75.4\).

Dividing the coefficients: \(4.2 \div 1.5 = 2.8\). Subtracting the indices: \(7 - (-3) = 10\). This gives \(2.8 \times 10^{10}\).

M1 for dividing the coefficients to get \(2.8\) or working out the power of \(10\) as \(10^{10}\). A1 for \(2.8 \times 10^{10}\) (accept \(2.8 * 10^{10}\)).

Let \(P(4) = x\), then \(P(3) = 2x\). Since the sum of probabilities is \(1\): \(0.3 + 0.4 + 2x + x = 1 \Rightarrow 0.7 + 3x = 1 \Rightarrow 3x = 0.3 \Rightarrow x = 0.1\). Therefore, \(P(3) = 2 \times 0.1 = 0.2\).

M1 for setting up a correct equation, e.g. \(0.3 + 0.4 + 3x = 1\) or finding that the remaining probability for \(3\) and \(4\) is \(0.3\). A1 for \(0.2\).

Subtract \(u^2\) from both sides: \(v^2 - u^2 = 2as\). Divide both sides by \(2s\): \(a = \frac{v^2 - u^2}{2s}\).

M1 for isolating the \(a\) term, e.g. \(v^2 - u^2 = 2as\). A1 for \(a = \frac{v^2 - u^2}{2s}\) (or equivalent correct rearrangement).

Value after 1 year: \(18500 \times 0.88 = 16280\). Value after 3 years: \(16280 \times 0.94^2 = 14385.008\). Rounding to the nearest pound gives £14385.

M1 for \(18500 \times 0.88 \times 0.94^2\) oe, A1 for 14385 (accept 14385.01)

Area of a sector = \((\theta / 360) \times \pi \times r^2\). Here, \((72 / 360) \times \pi \times 8.5^2 = 0.2 \times \pi \times 72.25 = 14.45\pi \approx 45.396...\). To 3 significant figures, this is 45.4.

M1 for \((72 / 360) \times \pi \times 8.5^2\) oe, A1 for 45.4 (accept answers in range 45.39 to 45.41)

\((2x - 3)(3x + 5) = 6x^2 + 10x - 9x - 15 = 6x^2 + x - 15\). Expanding the second part: \(-x(x - 4) = -x^2 + 4x\). Combining these terms: \(6x^2 + x - 15 - x^2 + 4x = 5x^2 + 5x - 15\).

M1 for expanding either part with at least 3 terms correct in \((2x - 3)(3x + 5)\) (e.g., \(6x^2 + x - 15\)) or expanding \(-x(x - 4)\) to \(-x^2 + 4x\), A1 for \(5x^2 + 5x - 15\)

The probability of landing on tails is \(1 - 0.35 = 0.65\). The expected number of times it lands on tails is \(140 \times 0.65 = 91\).

M1 for \(1 - 0.35\) or \(140 \times 0.35\) oe, A1 for 91

First, convert mass to grams: \(1.17 \text{ kg} = 1170 \text{ g}\). Density = mass / volume = \(1170 / 150 = 7.8 \text{ g/cm}^3\).

M1 for \(1.17 \times 1000\) or correct division formula e.g. \(1.17 / 150\), A1 for 7.8

The sum of the probabilities is 1.

The probability of landing on Yellow is:
\(1 - (0.34 + 0.42) = 1 - 0.76 = 0.24\)

The expected number of times the spinner lands on Yellow is:
\(250 \times 0.24 = 60\)

M1 for finding the probability of Yellow: \(1 - (0.34 + 0.42) = 0.24\)
A1 for multiplying by 250 to get 60

First, calculate the actual distance in centimetres:
\(8.4 \text{ cm} \times 25\ 000 = 210\ 000 \text{ cm}\)

Convert centimetres to metres by dividing by 100:
\(210\ 000 \div 100 = 2100 \text{ m}\)

Convert metres to kilometres by dividing by 1000:
\(2100 \div 1000 = 2.1 \text{ km}\)

M1 for \(8.4 \times 25\ 000\) or showing division by 100 000
A1 for 2.1

The interest rate is 2.3% per annum, so the multiplier is \(1 + 0.023 = 1.023\).

The value of the investment after 3 years is:
\(\text{Value} = 4200 \times 1.023^3\)
\(\text{Value} = 4200 \times 1.070599167 \approx 4496.5165...\)

To the nearest penny, this is £4496.52.

M1 for setting up the calculation \(4200 \times 1.023^3\)
A1 for 4496.52

First, expand the brackets:
\(5x - 7 > 3x + 6\)

Subtract \(3x\) from both sides:
\(2x - 7 > 6\)

Add 7 to both sides:
\(2x > 13\)

Divide by 2:
\(x > 6.5\)

M1 for expanding brackets correctly to \(3x + 6\) and gathering terms on one side, e.g. \(2x > 13\)
A1 for \(x > 6.5\) or \(x > \frac{13}{2}\)

The formula for the volume of a cylinder is:
\(V = \pi r^2 h\)

Substitute the given values:
\(V = \pi \times 4.5^2 \times 12\)
\(V = \pi \times 20.25 \times 12\)
\(V = 243\pi \approx 763.407... \text{ cm}^3\)

To 1 decimal place, the volume is 763.4 cm\(^3\).

M1 for \(\pi \times 4.5^2 \times 12\)
A1 for 763.4

Multiply each term in the first bracket by each term in the second bracket:
\((2x - 3)(3x + 5) = 2x(3x) + 2x(5) - 3(3x) - 3(5)\)
\(= 6x^2 + 10x - 9x - 15\)
\(= 6x^2 + x - 15\)

M1 for at least 3 correct terms of the quadratic expansion: e.g. \(6x^2\), \(10x\), \(-9x\), \(-15\)
A1 for \(6x^2 + x - 15\)

The measurement is given to the nearest 0.1 kg.

The degree of accuracy is 0.1 kg, so the boundary value is \(0.1 \div 2 = 0.05 \text{ kg}\).

The upper bound is:
\(3.4 + 0.05 = 3.45 \text{ kg}\)

M1 for identifying the half-unit interval of 0.05
A1 for 3.45

Find the midpoint (\(x\)) for each interval:
- For \(0 < h \le 10\), midpoint = 5
- For \(10 < h \le 20\), midpoint = 15
- For \(20 < h \le 30\), midpoint = 25

Multiply midpoints by frequencies (\(f \times x\)):
- \(4 \times 5 = 20\)
- \(7 \times 15 = 105\)
- \(9 \times 25 = 225\)

Sum of products = \(20 + 105 + 225 = 350\)
Total frequency = \(4 + 7 + 9 = 20\)

Estimate of the mean = \(350 \div 20 = 17.5\)

M1 for finding at least two correct midpoints and multiplying by their frequencies, and dividing the sum by 20.
A1 for 17.5

Let the original value of the car be \(V\).

The multiplier for an 8% depreciation is \(1 - 0.08 = 0.92\).

After 3 years, the value of the car is given by:
\(V \times 0.92^3 = 11,679.84\)

Calculate \(0.92^3\):
\(0.92^3 = 0.778688\)

Now, solve for \(V\):
\(V = \frac{11,679.84}{0.778688} = 15,000\)

Therefore, the original value of the car was \(£15,000\).

M1: For writing a correct equation involving the depreciation multiplier, e.g. \(V \times 0.92^3 = 11,679.84\), or for evaluating \(0.92^3 = 0.778688\).
A1: For the correct original value of 15000.

The formula for the total surface area of a closed cylinder is:
\(A = 2\pi r^2 + 2\pi r h\)

Substitute the given values \(A = 150\pi\) and \(r = 5\):
\(150\pi = 2\pi(5)^2 + 2\pi(5)h\)
\(150\pi = 50\pi + 10\pi h\)

Subtract \(50\pi\) from both sides:
\(100\pi = 10\pi h\)

Divide by \(10\pi\):
\(h = 10\text{ cm}\)

Now, calculate the volume \(V\) of the cylinder using \(V = \pi r^2 h\):
\(V = \pi \times 5^2 \times 10\)
\(V = 250\pi\text{ cm}^3\)

M1: For setting up a correct equation for the surface area to find the height, e.g. \(150\pi = 50\pi + 10\pi h\), and solving to find \(h = 10\).
A1: For the correct volume of \(250\pi\).

We have the system of equations:
1) \(2x + 3.5y = 15.6\)
2) \(4.5x - 2y = -0.45\)

To eliminate \(y\), multiply equation (1) by 2 and equation (2) by 3.5:
\(4x + 7y = 31.2\)
\(15.75x - 7y = -1.575\)

Add the two equations together:
\(19.75x = 29.625\)

Solve for \(x\):
\(x = \frac{29.625}{19.75} = 1.5\)

Substitute \(x = 1.5\) back into equation (1):
\(2(1.5) + 3.5y = 15.6\)
\(3 + 3.5y = 15.6\)
\(3.5y = 12.6\)
\(y = 3.6\)

Now find \(x + y\):
\(x + y = 1.5 + 3.6 = 5.1\)

M1: For a complete algebraic method to eliminate one variable to find either \(x = 1.5\) or \(y = 3.6\).
A1: For the correct value of \(5.1\).

The sum of the exterior angles of any polygon is \(360^\circ\).

For a regular polygon with \(n\) sides, each exterior angle is \(\frac{360^\circ}{n}\).

So, \(n = \frac{360}{15} = 24\).

M1 for \(360 \div 15\) or \((n-2) \times 180 = (180 - 15)n\)

A1 for 24

Density is calculated by dividing mass by volume:

\(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)

\(\text{Density} = \frac{180}{240} = 0.75\text{ g/cm}^3\).

M1 for \(180 \div 240\) or \(0.75\)

A1 for \(0.75\text{ g/cm}^3\) (or equivalent correct units)

The sale price represents \(100\% - 15\% = 85\%\) of the original price.

Let the original price be \(P\).

\(0.85 \times P = 68\)

\(P = \frac{68}{0.85} = 80\).

So the original price of the coat was \(\text{£}80\).

M1 for \(68 \div 0.85\) or finding \(1\% = 68 \div 85\) (which is \(0.8\))

A1 for \(\text{£}80\) (accept \(80\))

First, collect the terms in \(x\) on one side by subtracting \(3x\) from both sides:

\(2x - 3 > 8\)

Next, add \(3\) to both sides:

\(2x > 11\)

Finally, divide both sides by \(2\):

\(x > 5.5\) (or \(x > \frac{11}{2}\))

M1 for collecting terms in \(x\) on one side and constant terms on the other (e.g. \(2x > 11\) or \(2x = 11\))

A1 for \(x > 5.5\) or \(x > \frac{11}{2}\)

First find the number of times the coin lands on Tails:

\(150 - 93 = 57\)

Then calculate the relative frequency of Tails:

\(\text{Relative frequency} = \frac{57}{150} = 0.38\)

M1 for \(150 - 93\) (or \(57\)) or \(1 - \frac{93}{150}\)

A1 for \(0.38\)

Total score of first \(5\) students \(= 5 \times 12 = 60\).

Total score of all \(6\) students \(= 6 \times 13 = 78\).

Score of the sixth student \(= 78 - 60 = 18\).

M1 for finding the total of either group, e.g. \(5 \times 12 = 60\) or \(6 \times 13 = 78\)

A1 for 18

The common difference between terms is \(4\).

So the sequence is related to the \(4n\) times table:

\(4, 8, 12, 16, \dots\)

Each term in our sequence is \(3\) more than the corresponding term of \(4n\).

Therefore, the \(n\)th term is \(4n + 3\).

M1 for finding a common difference of \(4\) (e.g. \(4n + c\) or showing \(+4\) repeatedly)

A1 for \(4n + 3\) (or equivalent, e.g. \(3 + 4n\))

Multiply each term in the first bracket by each term in the second bracket:

\((x + 3)(x - 5) = x^2 - 5x + 3x - 15\)

Combine the like terms (\(-5x + 3x = -2x\)):

\(= x^2 - 2x - 15\)

M1 for expanding at least 3 terms of the quadratic correctly (e.g. \(x^2 - 5x + 3x\) or \(x^2 - 2x + c\))

A1 for \(x^2 - 2x - 15\)

First, convert the mass from kilograms to grams:
\(4.2 \text{ kg} = 4.2 \times 1000 = 4200 \text{ g}\)

Next, use the formula for density:
\(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)

\(\text{Density} = \frac{4200}{350} = 12 \text{ g/cm}^3\)

M1 for converting 4.2 kg to 4200 g or for writing \(\frac{4200}{350}\)
A1 for 12

The formula for the volume of a cylinder is \(V = \pi r^2 h\).

Substitute the given values into the formula:
\(V = \pi \times 4.5^2 \times 12\)
\(V = \pi \times 20.25 \times 12\)
\(V = 243\pi \approx 763.407... \text{ cm}^3\)

Rounding to 3 significant figures gives 763.

M1 for \(\pi \times 4.5^2 \times 12\) or \(243\pi\) or \(763.4...\)
A1 for 763

The total amount in the account after 3 years is:
\(3500 \times (1.024)^3 = 3500 \times 1.073741824 = £3758.096384\)

The interest earned is:
\(£3758.096384 - £3500 = £258.096384\)

To the nearest penny, this is £258.10.

M1 for \(3500 \times 1.024^3\) or \(3758.10\) or \(258.10\)
A1 for 258.10

Subtract \(3x\) from both sides:
\(2x - 3 < 8\)

Add 3 to both sides:
\(2x < 11\)

Divide by 2:
\(x < 5.5\) or \(x < \frac{11}{2}\)

M1 for isolating the x terms on one side, e.g. \(2x < 11\) or \(2x - 3 < 8\) or \(2x = 11\)
A1 for \(x < 5.5\) or \(x < \frac{11}{2}\)

Multiply each term in the first bracket by each term in the second bracket:
\((2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15\)

Combine like terms:
\(2x^2 + 7x - 15\)

M1 for expanding to get at least 3 correct terms out of 4 (e.g., \(2x^2 + 10x - 3x - 15\))
A1 for \(2x^2 + 7x - 15\)

Find the total score of the girls:
\(12 \times 74 = 888\)

Find the total score of the boys:
\(8 \times 69 = 552\)

Find the total score of all students:
\(888 + 552 = 1440\)

Divide by the total number of students:
\(\frac{1440}{20} = 72\)

M1 for \(12 \times 74\) or \(8 \times 69\) or \(888 + 552\) or \(1440\)
A1 for 72

The sum of all probabilities must be 1:
\(P(\text{Red}) + P(\text{Blue}) + P(\text{Yellow}) = 1\)

Let \(P(\text{Yellow}) = x\), then \(P(\text{Blue}) = 2x\).
\(0.4 + 2x + x = 1\)
\(0.4 + 3x = 1\)
\(3x = 0.6\)
\(x = 0.2\)

Therefore, \(P(\text{Blue}) = 2 \times 0.2 = 0.4\).

M1 for writing \(0.4 + 2x + x = 1\) or \(3x = 0.6\) or finding \(P(\text{Yellow}) = 0.2\)
A1 for 0.4

The formula for the sum of the interior angles of an \(n\)-sided polygon is:
\(\text{Sum} = (n - 2) \times 180^\circ\)

Set this equal to \(1440^\circ\):
\((n - 2) \times 180 = 1440\)
\(n - 2 = \frac{1440}{180}\)
\(n - 2 = 8\)
\(n = 10\)

M1 for \((n - 2) \times 180 = 1440\) or \(\frac{1440}{180} + 2\)
A1 for 10

Rate per minute: \(320 \div 8 = 40\) bottles. Bottles produced in 25 minutes: \(40 \times 25 = 1000\).

M1 for \(320 \div 8\) or showing a correct method to find the rate. A1 for 1000.

Calculate 15% of \(580\): \(580 \times 0.15 = 87\). Subtract this from the original price: \(580 - 87 = 493\). Alternatively, \(580 \times 0.85 = 493\).

M1 for a complete method to find the sale price, e.g. \(580 \times 0.85\) or \(580 - (580 \times 0.15)\). A1 for 493.

Convert mass to grams: \(1.2\text{ kg} = 1200\text{ g}\). Calculate density: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1200}{150} = 8\text{ g/cm}^3\).

M1 for converting the mass to grams (1200 g) or writing a correct density division with consistent units. A1 for 8.

Expand the terms: \(3(2x - 5) = 6x - 15\) and \(-2(x - 4) = -2x + 8\). Simplify: \(6x - 15 - 2x + 8 = 4x - 7\).

M1 for expanding at least one bracket correctly to get \(6x - 15\) or \(-2x + 8\). A1 for \(4x - 7\).

Use the formula for the area of a circle: \(\text{Area} = \pi r^2 = \pi \times 6.5^2\). \(\text{Area} = 42.25\pi \approx 132.73\text{ cm}^2\). To 3 significant figures, this rounds to \(133\text{ cm}^2\).

M1 for substituting \(6.5\) into the area formula: \(\pi \times 6.5^2\) or \(42.25\pi\). A1 for 133 (accept answers in the range 132.7 to 133).

Multiply the numbers: \(4 \times 8 = 32\). Multiply the powers of 10: \(10^5 \times 10^{-2} = 10^3\). Combine to get \(32 \times 10^3\). Convert to standard form: \(3.2 \times 10^4\).

M1 for \(32 \times 10^3\) or \(32000\) or achieving \(3.2 \times 10^n\) where \(n\) is an integer. A1 for \(3.2 \times 10^4\).

Estimate = \(200 \times 0.35 = 70\).

M1 for showing the product \(200 \times 0.35\). A1 for 70.

Subtract \(2x\) from both sides: \(3x - 7 = 11\). Add \(7\) to both sides: \(3x = 18\). Divide by \(3\): \(x = 6\).

M1 for a correct algebraic step to isolate terms in \(x\) on one side and numerical terms on the other, e.g. \(3x - 7 = 11\) or \(5x = 2x + 18\). A1 for 6.

First, calculate the price after the 15% increase:
\(\text{New Price} = 92 \times 1.15 = 105.80\) pounds.

Next, calculate the price after the 20% reduction:
\(\text{Final Price} = 105.80 \times 0.80 = 84.64\) pounds.

M1 for finding the price after the 15% increase: \(92 \times 1.15 = 105.80\) (or finding the 15% increase of 13.80), or for a complete multiplier method: \(92 \times 1.15 \times 0.80\)
A1 for 84.64 (or £84.64)

The formula for the volume of a cylinder is:
\(V = \pi r^2 h\)

Substitute the given values into the formula:
\(350 = \pi \times 4.5^2 \times h\)
\(350 = 20.25\pi \times h\)

Rearrange to solve for \(h\):
\(h = \frac{350}{20.25\pi} \approx 5.50165\text{ cm}\)

To 3 significant figures, the height is 5.50 cm.

M1 for a correct substitution into the volume formula, e.g. \(350 = \pi \times 4.5^2 \times h\) or rearranging to make h the subject: \(h = \frac{350}{\pi \times 4.5^2}\)
A1 for 5.50 (accept 5.5)

First, expand the brackets on both sides of the equation:
\(10x - 15 = 2x + 9\)

Subtract \(2x\) from both sides:
\(8x - 15 = 9\)

Add 15 to both sides:
\(8x = 24\)

Divide by 8:
\(x = 3\)

M1 for correctly expanding at least one side, e.g. \(10x - 15\) or \(2x + 9\), or for a correct step in rearranging once expanded (e.g. \(8x = 24\))
A1 for 3

First, find the total number of goals scored by multiplying each number of goals by its frequency:
\((0 \times 6) + (1 \times 8) + (2 \times 7) + (3 \times 3) + (4 \times 1) = 0 + 8 + 14 + 9 + 4 = 35\)

Next, divide the total number of goals by the total number of matches (25):
\(\text{Mean} = \frac{35}{25} = 1.4\)

M1 for finding the sum of the goals: \(0 \times 6 + 1 \times 8 + 2 \times 7 + 3 \times 3 + 4 \times 1\) or 35 seen
A1 for 1.4

The probability of choosing a blue counter is:
\(1 - 0.35 = 0.65\)

Let the total number of counters in the bag be \(N\).
Since the probability of choosing a blue counter is 0.65, we can set up the equation:
\(0.65 \times N = 26\)
\(N = \frac{26}{0.65} = 40\)

M1 for finding the probability of choosing a blue counter: \(1 - 0.35 = 0.65\) or for setting up the equation \(\frac{26}{N} = 0.65\)
A1 for 40

First, calculate the actual distance in centimetres:
\(8.4 \times 25\\,000 = 210\\,000\text{ cm}\)

Next, convert the distance from centimetres to metres (divide by 100):
\(210\\,000 \div 100 = 2100\text{ m}\)

Finally, convert the distance from metres to kilometres (divide by 1000):
\(2100 \div 1000 = 2.1\text{ km}\)

M1 for \(8.4 \times 25\\,000\) or 210,000 seen, or for a correct conversion method (e.g., dividing by 100,000)
A1 for 2.1