2016 AQA IGCSE Mathematics (9260) Practice Paper with Answers

Since 1 litre is equal to 1000 millilitres, we multiply 4.05 by 1000: \(4.05 \times 1000 = 4050\) millilitres.

B1 for 4050

To convert a percentage to a fraction, write it over 100: \(\frac{12.5}{100} = \frac{125}{1000} = \frac{1}{8}\).

B1 for \(\frac{1}{8}\)

Group the like terms: \(4a + 2a = 6a\) and \(-3b - b = -4b\). Combining them gives \(6a - 4b\).

B1 for \(6a - 4b\)

The range is the difference between the largest number and the smallest number. Range = \(11 - 3 = 8\).

B1 for 8

First, find the flour needed for 1 person: \(300 \div 4 = 75\) grams. Then multiply by 6 for 6 people: \(75 \times 6 = 450\) grams.

B1 for 450g

Add 5 to both sides of the equation: \(3x = 21\). Divide both sides by 3: \(x = 7\).

B1 for x = 7

A rectangle has exactly two lines of symmetry (one vertical and one horizontal). A square has four, an equilateral triangle has three, and a standard parallelogram has zero.

B1 for Rectangle

The total number of counters is \(5 + 3 + 2 = 10\). The number of blue counters is 3. So the probability is \(\frac{3}{10}\).

B1 for \(\frac{3}{10}\)

Find the number of black jackets: \(120 \times \frac{2}{5} = 48\). Find the number of black coats: \(180 \times 0.35 = 63\). Find the total black items: \(48 + 63 = 111\). Find the total outerwear items: \(120 + 180 = 300\). The fraction of black outerwear is \(\frac{111}{300}\), which simplifies to \(\frac{37}{100}\).

M1 for finding the number of black jackets is 48. M1 for finding the number of black coats is 63. M1 for \(\frac{111}{300}\). A1 for \(\frac{37}{100}\) (or equivalent simplified fraction).

The initial number of red counters is \(40 \times \frac{3}{10} = 12\). After removing 5 red counters, there are \(12 - 5 = 7\) red counters left. The new total number of counters in the bag is \(40 - 5 = 35\). The new probability of picking a red counter is \(\frac{7}{35} = \frac{1}{5}\).

M1 for finding the initial number of red counters is 12. M1 for both 7 red counters remaining and 35 total counters remaining. A1 for \(\frac{1}{5}\) (or equivalent, e.g. 0.2).

The total number of ratio parts is \(7 + 3 = 10\). The mass of copper in 15 kg of alloy is \(15 \times \frac{7}{10} = 10.5\text{ kg}\). The mass of zinc is \(15 \times \frac{3}{10} = 4.5\text{ kg}\). Cost of copper: \(10.5 \times 6.40 = \$67.20\). Cost of zinc: \(4.5 \times 4.80 = \$21.60\). Total cost: \(67.20 + 21.60 = \$88.80\).

M1 for dividing the total mass by the sum of ratio parts (\(15 \div 10 = 1.5\)). M1 for finding the masses of copper (10.5 kg) and zinc (4.5 kg). M1 for a complete method to find the total cost: \((10.5 \times 6.40) + (4.5 \times 4.80)\). A1 for $88.80 (accept 88.8).

Expand both sides of the equation: \(8x - 12 = 3x + 15 - 2\). Simplify the right-hand side: \(8x - 12 = 3x + 13\). Subtract \(3x\) from both sides: \(5x - 12 = 13\). Add 12 to both sides: \(5x = 25\). Divide by 5: \(x = 5\).

B1 for correctly expanding at least one of the brackets, i.e., \(8x - 12\) or \(3x + 15\). M1 for isolating terms with x on one side and numbers on the other side, e.g. \(8x - 3x = 13 + 12\). M1dep for \(5x = 25\). A1 for \(5\).

First, use Pythagoras' theorem to find the missing height, \(h\): \(h^2 + 15^2 = 17^2\), which gives \(h^2 + 225 = 289\). Then \(h^2 = 64\), so \(h = \sqrt{64} = 8\text{ cm}\). The perimeter is the sum of all three sides: \(15 + 17 + 8 = 40\text{ cm}\).

M1 for setting up Pythagoras' theorem to find the missing side: \(17^2 - 15^2\). A1 for finding the missing side is 8. M1 for adding all three sides: \(15 + 17 + \text{their } 8\). A1 for 40.

The exterior angle of the regular polygon is \(180^\circ - 144^\circ = 36^\circ\). The sum of all exterior angles of any polygon is \(360^\circ\). The number of sides is \(\frac{360^\circ}{36^\circ} = 10\).

M1 for calculating the exterior angle: \(180 - 144 = 36\). M1 for dividing 360 by their exterior angle: \(360 \div \text{their } 36\). A1 for 10.

Find the highest common factors of both terms. For the coefficients 6 and 15, the HCF is 3. For \(a^2b\) and \(ab^2\), the common variable factor is \(ab\). Factoring out \(3ab\) gives: \(3ab(2a + 5b)\).

M1 for identifying a common factor of at least \(3\), \(a\), or \(b\) outside the bracket. A1 for \(3ab(\dots)\) with one term correct inside the bracket. A1 for \(3ab(2a + 5b)\).

Find the gradient, \(m\): \(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\). Substitute the gradient and one point, e.g. \((2, 5)\), into the equation \(y = mx + c\): \(5 = 3(2) + c\), which simplifies to \(5 = 6 + c\), so \(c = -1\). The equation is \(y = 3x - 1\).

M1 for calculating the gradient \(m = 3\). M1 for substituting their gradient and a point into \(y = mx + c\) to find \(c\). A1 for \(y = 3x - 1\).

Find the number of sides, \(n\):
\(n = \dfrac{360^\circ}{45^\circ} = 8\)

Now find the sum of the interior angles of an 8-sided polygon:
\(\text{Sum} = (n - 2) \times 180^\circ = (8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ\)

- **M1**: for \(360 \div 45\) or 8 seen
- **M1**: for \((\text{their } 8 - 2) \times 180\)
- **A1**: for 1080

First reduction of \(15\%\):
\(\$80 \times (1 - 0.15) = \$80 \times 0.85 = \$68\)

Second reduction of \(10\%\):
\(\$68 \times (1 - 0.10) = \$68 \times 0.90 = \$61.20\)

- **M1**: for finding \(15\%\) of 80 (\(12\)) and subtracting it, or for \(80 \times 0.85\) or 68
- **M1**: for finding \(10\%\) of their 68 (\(6.80\)) and subtracting it, or for \(\text{their } 68 \times 0.90\)
- **A1**: for 61.20 or 61.2

The total probability is 1, so the probability of landing on 4 is:
\(1 - (0.25 + 0.35 + 0.18) = 1 - 0.78 = 0.22\)

Estimate of the number of times it lands on 4:
\(250 \times 0.22 = 55\)

- **M1**: for \(1 - (0.25 + 0.35 + 0.18)\) or 0.22 seen
- **M1**: for \(250 \times \text{their } 0.22\)
- **A1**: for 55

Expand the brackets on both sides:
\(8x - 12 = 3x + 18\)

Rearrange to get \(x\) on one side:
\(8x - 3x = 18 + 12\)
\(5x = 30\)

Divide by 5:
\(x = 6\)

- **B1**: for expanding left hand side correctly: \(8x - 12\)
- **B1**: for expanding right hand side correctly: \(3x + 18\)
- **M1**: for isolating the \(x\) term, e.g. \(8x - 3x = 18 + 12\) or \(5x = 30\) (ft their expansion)
- **A1**: for 6

The difference in parts between Chloe and Anna is:
\(7 - 2 = 5\text{ parts}\)

Since 5 parts correspond to \(\$45\), 1 part is:
\(\$45 \div 5 = \$9\)

The total number of parts shared is:
\(2 + 5 + 7 = 14\text{ parts}\)

The total money shared is:
\(14 \times \$9 = \$126\)

- **M1**: for finding difference in parts: \(7 - 2 = 5\)
- **M1**: for \(45 \div \text{their } 5\) or 9
- **M1**: for \((2 + 5 + 7) \times \text{their } 9\)
- **A1**: for 126

The perimeter of a semi-circle consists of the curved arc plus the straight diameter.
Curved arc length:
\(\text{Arc} = \dfrac{1}{2} \times \pi \times d = \dfrac{1}{2} \times \pi \times 14 = 7\pi \approx 21.991\text{ cm}\)

Perimeter:
\(\text{Perimeter} = \text{Arc} + d = 21.991 + 14 = 35.991\text{ cm}\)

Rounding to 1 decimal place gives \(36.0\text{ cm}\).

- **M1**: for \(\dfrac{1}{2} \times \pi \times 14\) or \(7\pi\) or \([21.9, 22.0]\)
- **M1**: for adding the diameter: \(\text{their arc length} + 14\)
- **A1**: for 36.0 (or 36)

Identify the highest common factor of \(12\) and \(18\), which is \(6\).
Identify the highest common factor of \(x^2y\) and \(xy^2\), which is \(xy\).

Combine these to get the common factor of \(6xy\):
\(12x^2y - 18xy^2 = 6xy(2x - 3y)\)

- **M1**: for extracting any common factor of \(6\), \(x\) or \(y\) (e.g. \(2xy(6x - 9y)\) or \(6x(2xy - 3y^2)\))
- **M1**: for \(6xy(\dots)\) with one term correct inside the bracket
- **A1**: for \(6xy(2x - 3y)\)

First, find the total number of goals scored:
\((0 \times 4) + (1 \times 7) + (2 \times 5) + (3 \times 3) + (4 \times 1) = 0 + 7 + 10 + 9 + 4 = 30\text{ goals}\)

Now, divide by the total number of matches (20):
\(\text{Mean} = \dfrac{30}{20} = 1.5\)

- **M1**: for sum of products \((0 \times 4) + (1 \times 7) + (2 \times 5) + (3 \times 3) + (4 \times 1)\) (allow at least 3 correct products)
- **M1**: for \(\text{their } 30 \div 20\)
- **A1**: for 1.5

First, expand the brackets on both sides of the equation: \(8x - 12 = 2x + 10 - 4\) Simplify the right-hand side: \(8x - 12 = 2x + 6\) Subtract \(2x\) from both sides: \(6x - 12 = 6\) Add \(12\) to both sides: \(6x = 18\) Divide by \(6\): \(x = 3\)

M1 for correct expansion of brackets: \(8x - 12\) or \(2x + 10\) seen. M1 for rearranging to the form \(ax = b\), e.g., \(6x = 18\). A1 for \(3\) (or \(x = 3\)).

The total number of parts in the ratio is \(5 + 3 = 8\). The value of each part is \(1200 \div 8 = 150\) books. The number of history books is \(5 \times 150 = 750\). The number of science books is \(3 \times 150 = 450\). The difference is \(750 - 450 = 300\). Alternatively, the difference in ratio parts is \(5 - 3 = 2\) parts. The difference in books is \(2 \times 150 = 300\).

M1 for finding the total number of parts: \(5 + 3 = 8\). M1 for dividing the total books by the total parts: \(1200 \div 8 = 150\). M1 for calculating the number of books of at least one type (e.g. \(750\) or \(450\)) OR for multiplying the difference in parts by the value of one part (e.g. \(2 \times 150\)). A1 for \(300\).

The sum of probabilities of all possible outcomes is 1. Probability of choosing a yellow counter is \(1 - (0.35 + 0.4) = 1 - 0.75 = 0.25\). Let \(T\) be the total number of counters in the bag. Since the probability of choosing a yellow counter is \(0.25\), we have: \(0.25 \times T = 50\). Therefore, \(T = 50 \div 0.25 = 200\).

M1 for finding the probability of a yellow counter: \(1 - (0.35 + 0.4) = 0.25\). M1 for \(50 \div 0.25\) or an equivalent calculation (e.g. \(50 \times 4\)). A1 for \(200\).

The area of the inner rectangular garden is \(15 \times 8 = 120\text{ m}^2\). The outer rectangle includes the garden and the path on all sides. The outer length is \(15 + 1 + 1 = 17\text{ m}\). The outer width is \(8 + 1 + 1 = 10\text{ m}\). The area of the outer rectangle is \(17 \times 10 = 170\text{ m}^2\). The area of the path is the difference between the outer area and the inner garden area: \(170 - 120 = 50\text{ m}^2\).

M1 for calculating the area of the garden: \(15 \times 8 = 120\). M1 for finding the correct dimensions of the outer rectangle: \(17\text{ m}\) and \(10\text{ m}\). M1 for calculating the area of the outer rectangle: \(17 \times 10 = 170\). A1 for \(50\).

Grouping the like terms:
\( 4x - x = 3x \)
\( -7 + 9 = 2 \)

So, the simplified expression is \( 3x + 2 \).

B1 for \( 3x + 2 \)

\( 51 = 3 \times 17 \)
\( 57 = 3 \times 19 \)
\( 59 \) has no factors other than 1 and itself, so it is a prime number.
\( 63 = 3 \times 21 \)

B1 for \( 59 \)

\( 0.08 = \frac{8}{100} \)

Simplifying the fraction by dividing both the numerator and the denominator by 4:
\( \frac{8 \div 4}{100 \div 4} = \frac{2}{25} \)

B1 for \( \frac{2}{25} \)

First, convert \( 2.5 \text{ km} \) to centimetres:
\( 2.5 \text{ km} = 2500 \text{ m} = 250\,000 \text{ cm} \)

Using the scale \( 1 : 50\,000 \):
\( \frac{250\,000}{50\,000} = 5 \text{ cm} \)

B1 for \( 5 \)

The number of diagonals of an \( n \)-sided polygon is given by the formula:
\( \frac{n(n - 3)}{2} \)

For a regular hexagon, where \( n = 6 \):
\( \frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9 \)

B1 for \( 9 \)

Rearrange the equation to the gradient-intercept form, \( y = mx + c \), by dividing both sides by 2:
\( y = 3x - 2 \)

The coefficient of \( x \) is the gradient, which is \( 3 \).

B1 for \( 3 \)

The possible outcomes when rolling a six-sided dice are \( \{1, 2, 3, 4, 5, 6\} \).

The multiples of 3 are \( 3 \) and \( 6 \) (2 outcomes).

The probability is:
\( \frac{2}{6} = \frac{1}{3} \)

B1 for \( \frac{1}{3} \)

Add 5 to both sides of the inequality:
\( 3x > 18 \)

Divide both sides by 3:
\( x > 6 \)

B1 for \( x > 6 \)

First, calculate the volume of each type of juice in 40 litres of the mixture.
Total parts in ratio = 3 + 5 = 8 parts.
Volume of orange juice = \(\frac{3}{8} \times 40 = 15\) litres.
Volume of cranberry juice = \(\frac{5}{8} \times 40 = 25\) litres.

Next, calculate the cost of each type of juice.
Cost of orange juice = \(15 \times 1.60 = 24.00\) dollars.
Cost of cranberry juice = \(25 \times 2.40 = 60.00\) dollars.

Total cost = \(24.00 + 60.00 = 84.00\) dollars.

M1 for finding the volume of each juice: 15 and 25 (litres)
M1 for multiplying each volume by its cost per litre: \(15 \times 1.60\) and \(25 \times 2.40\)
M1dep for adding their two costs: \(24 + 60\)
A1 for 84 (accept 84.00)

Convert the fraction of members playing squash to a percentage:
\(\frac{1}{5} = 20\%\)

Add the percentages for tennis and squash:
\(45\% + 20\% = 65\%\)

The remaining percentage for badminton is:
\(100\% - 65\% = 35\%\)

This 35% represents the remaining 63 members.
Total number of members = \(\frac{63}{0.35} = 180\).

M1 for converting \(\frac{1}{5}\) to 20% or 0.2
M1 for finding that badminton represents 35% (or 0.35) of the members
A1 for 180

Expand the brackets on both sides of the equation:
\(8x - 12 = 3x + 15 - 2\)

Simplify the right side:
\(8x - 12 = 3x + 13\)

Subtract \(3x\) from both sides:
\(5x - 12 = 13\)

Add 12 to both sides:
\(5x = 25\)

Divide by 5:
\(x = 5\)

B1 for \(8x - 12\) correctly expanded
B1 for \(3x + 15\) correctly expanded
M1 for correctly rearranging to the form \(ax = b\) (e.g. \(5x = 25\))
A1 for 5

First, calculate the volume of the cylindrical tank:
\(\text{Volume} = \pi r^2 h = \pi \times 30^2 \times 80 = 72000\pi \approx 226194.67\text{ cm}^3\)

Convert this volume to litres:
\(\text{Volume in litres} = \frac{226194.67}{1000} \approx 226.19\text{ litres}\)

Calculate the time required to fill the tank at a rate of 0.5 litres per second:
\(\text{Time in seconds} = \frac{226.19}{0.5} \approx 452.39\text{ seconds}\)

Convert 8 minutes to seconds to make a comparison:
\(8\text{ minutes} = 8 \times 60 = 480\text{ seconds}\)

Since \(452.39\text{ seconds} < 480\text{ seconds}\), it takes less than 8 minutes.

M1 for calculating the volume of the cylinder: \(\pi \times 30^2 \times 80\) (values in range [226080, 226200])
M1 for converting the volume to litres (dividing by 1000 to get value in range [226, 226.2])
M1dep for dividing by 0.5 to find the time in seconds (approx 452 seconds)
A1 for "Yes" with fully correct supporting calculations showing time is less than 480 seconds (or 8 minutes)

Set up simultaneous equations using the given information:
\(a + 3b = 110\)
\(a + 8b = 210\)

Subtract the first equation from the second:
\(5b = 100\)
\(b = 20\)

Substitute \(b = 20\) back into the first equation:
\(a + 3(20) = 110\)
\(a + 60 = 110\)
\(a = 50\)

M1 for setting up at least one correct equation or calculating the gradient (rate) as \(\frac{210 - 110}{8 - 3}\)
A1 for \(b = 20\)
A1 for \(a = 50\)

First, find the number of students who play basketball only:
\(15\text{ (total basketball)} - 5\text{ (both)} = 10\text{ students}\)

Since there are 30 students in total, the probability is:
\(\frac{10}{30} = \frac{1}{3}\)

M1 for subtracting to find basketball-only students: \(15 - 5 = 10\)
M1 for putting their value over 30: \(\frac{\text{their } 10}{30}\)
A1 for \(\frac{1}{3}\) (or equivalent fraction, decimal \(0.333...\), or percentage \(33.3\%\))

First, expand the brackets:
\(T = 3v - 12 + 2w\)

Next, isolate the term with \(v\):
\(T + 12 - 2w = 3v\)

Finally, divide by 3:
\(v = \frac{T + 12 - 2w}{3}\)

M1 for correct expansion of brackets: \(T = 3v - 12 + 2w\) (or isolating the bracket term: \(3(v - 4) = T - 2w\))
M1 for isolating the term containing \(v\): \(3v = T + 12 - 2w\) (or \(v - 4 = \frac{T - 2w}{3}\))
A1 for \(v = \frac{T + 12 - 2w}{3}\) (or equivalent expression)

Using Pythagoras' theorem for a right-angled triangle where the ladder is the hypotenuse:
\(a^2 + b^2 = c^2\)
Let \(h\) be the height up the wall:
\(h^2 + 5^2 = 13^2\)
\(h^2 + 25 = 169\)
\(h^2 = 169 - 25\)
\(h^2 = 144\)
\(h = \sqrt{144} = 12\text{ m}\)

M1 for writing down Pythagoras' theorem correctly applied to the context: \(h^2 + 5^2 = 13^2\) or \(13^2 - 5^2\)
M1 for calculating \(\sqrt{169 - 25}\) or \(\sqrt{144}\)
A1 for 12

First, find the total parts in the ratio:
\(2 + 7 = 9\) parts.

Calculate the volume per part:
\(18 \div 9 = 2\) litres per part.

Calculate the volume of each component:
Volume of concentrate = \(2 \times 2 = 4\) litres.
Volume of water = \(7 \times 2 = 14\) litres.

Calculate the cost of each component:
Cost of concentrate = \(4 \times \$1.80 = \$7.20\).
Cost of water = \(14 \times \$0.05 = \$0.70\).

Total cost:
\(\$7.20 + \$0.70 = \$7.90\).

M1: \(18 \div (2 + 7)\) or \(2\) (litres per part)
M1: \(4 \times 1.80\) (= \(7.20\)) or \(14 \times 0.05\) (= \(0.70\))
M1dep: their \(7.20\) + their \(0.70\)
A1: \(7.90\) (allow \(7.9\))

Multiply the second equation by \(2\):
\(2x + 2y = 14\)

Subtract this from the first equation:
\((3x + 2y) - (2x + 2y) = 19 - 14\)
\(x = 5\)

Substitute \(x = 5\) back into the second equation:
\(5 + y = 7\)
\(y = 2\)

So the solutions are \(x = 5\) and \(y = 2\).

M1: Equates coefficients or expresses one variable in terms of the other (e.g., \(2x + 2y = 14\) or \(y = 7 - x\))
A1: \(x = 5\) or \(y = 2\)
A1: \(x = 5\) and \(y = 2\)

Calculate the volume of the cylindrical tank:
\(V = \pi r^2 h = \pi \times 6^2 \times 15 = 540\pi \approx 1696.46\text{ cm}^3\).

Calculate the time required to empty the tank in seconds:
\(1696.46 \div 12 \approx 141.37\text{ seconds}\).

Convert the time to minutes:
\(141.37 \div 60 \approx 2.36\text{ minutes}\).

Since \(2.36\text{ minutes} > 2\text{ minutes}\), yes, it takes longer.

M1: \(\pi \times 6^2 \times 15\) or \(540\pi\) or \([1695, 1697]\)
M1: their Volume \(\div 12\) to find seconds \((\approx 141.37)\)
M1dep: their seconds \(\div 60\) to convert to minutes
A1: \([2.35, 2.36]\) and Yes

Percentage of adults = \(55\%\).
Percentage of children = \(100\% - 55\% = 45\%\).

Adult males as a percentage of total audience:
\(0.40 \times 55\% = 22\%\).

Child males as a percentage of total audience:
\(0.30 \times 45\% = 13.5\%\).

Total percentage of males:
\(22\% + 13.5\% = 35.5\%\).

M1: \(55 \times 0.40\) (= \(22\)) or implied by \(22\%\)
M1: \((100 - 55) \times 0.30\) (= \(13.5\)) or implied by \(13.5\%\)
A1: \(35.5\%\) (or \(35.5\))

The total probability must sum to 1.
Probability of picking a gold card:
\(P(\text{gold}) = 1 - 0.62 - 0.28 = 0.10\).

Now, multiply by the total number of cards to find the quantity of gold cards:
\(150 \times 0.10 = 15\).

M1: \(1 - 0.62 - 0.28\) (= \(0.10\))
M1dep: \(150 \times\) their \(0.10\)
A1: \(15\)

Case 1: The given \(50^\circ\) angle is the vertex angle (between the two equal sides).
The other two angles are equal:
\((180^\circ - 50^\circ) \div 2 = 130^\circ \div 2 = 65^\circ\).
So the angles are \(65^\circ\) and \(65^\circ\).

Case 2: The given \(50^\circ\) angle is one of the base angles.
The other base angle must also be \(50^\circ\).
The vertex angle is:
\(180^\circ - 50^\circ - 50^\circ = 80^\circ\).
So the angles are \(50^\circ\) and \(80^\circ\).

M1: \((180 - 50) \div 2\) (= \(65\))
M1: \(180 - 50 - 50\) (= \(80\))
A1: \(65^\circ, 65^\circ\) and \(50^\circ, 80^\circ\) (clearly indicated for both distinct possible triangles)

Calculate the deposit paid:
\(20\% \text{ of } 12\,500 = 0.20 \times 12\,500 = \$2500\).

Calculate the remaining balance to be paid:
\(12\,500 - 2500 = \$10\,000\).

Divide the balance by the number of monthly instalments:
\(10\,000 \div 48 = \$208.33\) (rounded to the nearest cent).

M1: \(12500 \times 0.80\) (= \(10000\)) or \(12500 \times 0.20\) (= \(2500\))
M1dep: \((12500 -\text{ their } 2500) \div 48\)
A1: \(208.33\) (accept \(208.33\) or \(208.34\), condone \(208.3\))

Substitute the values into the formula:
\(s = (12)(4) + \frac{1}{2}(-9.8)(4)^2\)

Calculate each term:
\((12)(4) = 48\)
\(\frac{1}{2}(-9.8)(16) = -4.9 \times 16 = -78.4\)

Calculate the final result:
\(s = 48 - 78.4 = -30.4\).

M1: Correct substitution of all values: \(12(4) + 0.5(-9.8)(4)^2\)
M1: \(48 - 78.4\) or \(48 - 4.9(16)\) seen as part of simplification
A1: \(-30.4\)

Ratio is \( 3 : 7 \).

Total parts = \( 3 + 7 = 10 \).

Volume of apple juice = \( \frac{3}{10} \times 50 = 15 \) litres.

Volume of sparkling water = \( \frac{7}{10} \times 50 = 35 \) litres.

Cost of apple juice = \( 15 \times 1.80 = 27 \) dollars.

Cost of sparkling water = \( 35 \times 0.40 = 14 \) dollars.

Total cost = \( 27 + 14 = 41 \) dollars.

M1 for \( 50 \div (3 + 7) \) or 5

M1dep for \( \text{their } 5 \times 3 \times 1.80 \) or 27, OR \( \text{their } 5 \times 7 \times 0.40 \) or 14

M1dep for \( \text{their } 27 + \text{their } 14 \)

A1 for 41 or 41.00

First, add the fractions for tennis and badminton:
\( \frac{4}{9} + \frac{1}{3} = \frac{4}{9} + \frac{3}{9} = \frac{7}{9} \)

The remaining fraction of members play squash:
\( 1 - \frac{7}{9} = \frac{2}{9} \)

Since \( \frac{2}{9} \) of the members represents 48:
Total members = \( 48 \div \frac{2}{9} = 48 \times \frac{9}{2} = 216 \)

Number of members playing tennis = \( \frac{4}{9} \times 216 = 96 \).

M1 for \( \frac{4}{9} + \frac{1}{3} = \frac{7}{9} \) or equivalent

M1dep for \( 1 - \text{their } \frac{7}{9} = \frac{2}{9} \)

M1dep for \( 48 \div \text{their } \frac{2}{9} \) or 216

A1 for 96

Volume of the cuboid = \( 80 \times 60 \times 50 = 240,000\text{ cm}^3 \).

Convert to litres: \( \frac{240,000}{1000} = 240 \) litres.

Time to fill = \( \frac{240}{1.5} = 160 \) minutes.

Convert 2.5 hours to minutes: \( 2.5 \times 60 = 150 \) minutes.

Since 160 minutes is more than 150 minutes, yes, it takes more than 2.5 hours.

M1 for \( 80 \times 60 \times 50 \) or 240,000

M1 for \( \text{their } 240,000 \div 1000 \) or 240

M1 for \( \text{their } 240 \div 1.5 \) or 160

A1 for 160 (minutes) and Yes (or equivalent concluding statement)

Factorise the quadratic expression \( x^2 + 2x - 15 \).

We need two numbers that multiply to -15 and add to 2.

These numbers are 5 and -3.

So, \( (x + 5)(x - 3) = 0 \).

This gives \( x + 5 = 0 \) or \( x - 3 = 0 \).

Thus, \( x = -5 \) or \( x = 3 \).

M1 for attempting to factorise, e.g., \( (x \pm 5)(x \pm 3) \)

A1 for correct factorisation \( (x + 5)(x - 3) = 0 \)

A1 for \( -5 \) and \( 3 \) (either order)