2022 Edexcel GCSE Mathematics (1MA1) Practice Paper with Answers

In the number 5382, the digit 5 is in the thousands place, the digit 3 is in the hundreds place, the digit 8 is in the tens place, and the digit 2 is in the units place. Therefore, the value of the 3 is 300.

[1] B1 for 300 or three hundred (accept 3 hundreds).

Group the terms in \(x\): \(5x - 3x + x = 2x + x = 3x\).

[1] B1 for 3x (or \(3 \times x\)).

A fair 6-sided dice has six equally likely outcomes: 1, 2, 3, 4, 5, 6. Only one of these outcomes is a 5. Therefore, the probability is \(\frac{1}{6}\).

[1] B1 for 1/6 (accept 0.167 or 16.7%).

To simplify the ratio \(12 : 30\), we divide both terms by their highest common factor, which is 6. \(12 \div 6 = 2\) and \(30 \div 6 = 5\). So the simplified ratio is \(2 : 5\).

[1] B1 for 2:5 (or equivalent simplest ratio, e.g., 2 to 5).

A triangular prism consists of two triangular bases and three rectangular sides. Each triangular base has 3 vertices. Since there are 2 bases, there are \(2 \times 3 = 6\) vertices in total.

[1] B1 for 6.

The mode is the number that appears most frequently in the list. The number 2 appears 1 time, 3 appears 1 time, 5 appears 3 times, and 8 appears 2 times. Since 5 appears the most times, the mode is 5.

[1] B1 for 5.

To convert a fraction to a percentage, we write it with a denominator of 100: \(\frac{3}{10} = \frac{30}{100} = 30\%\).

[1] B1 for 30% or 30 (accept with or without the percentage sign if the context implies percentage).

To solve for \(y\), add 4 to both sides of the equation: \(y = 11 + 4\), which gives \(y = 15\).

[1] B1 for 15 (or \(y = 15\)).

To convert a percentage to a decimal, we divide by 100. This gives \(7 \div 100 = 0.07\).

B1 for 0.07

To simplify the expression, we multiply the coefficients: \(4 \times 3 = 12\). Therefore, \(4 \times 3y = 12y\).

B1 for 12y

To calculate \( 2.34 \times 1.8 \), first calculate \( 234 \times 18 \).
\( 234 \times 10 = 2340 \)
\( 234 \times 8 = 1872 \)
\( 2340 + 1872 = 4212 \).
Since there is a total of three decimal places in the numbers being multiplied (two in 2.34 and one in 1.8), the decimal point is placed three places from the right.
So, \( 2.34 \times 1.8 = 4.212 \).

M1 for a complete method to multiply 234 by 18 (e.g., showing grid method or column multiplication with no more than one arithmetic slip), or for placing the decimal point correctly in a number with the digits 4212 (e.g., 42.12 or 0.4212).
A1 for 4.212 (cao).

Group the like terms together:
\( 4x - x = 3x \)
\( 3y + 2y = 5y \)
Combining these gives \( 3x + 5y \).

M1 for grouping like terms, e.g. showing \( 4x - x \) or \( 3y + 2y \), or having either \( 3x \) or \( 5y \) as part of a two-term expression.
A1 for \( 3x + 5y \) (or equivalent).

First, find the total number of parts:
\( 2 + 7 = 9 \) parts.
Next, find the value of one part:
\( £45 \div 9 = £5 \).
Multiply each part of the ratio by the value of one part:
\( 2 \times £5 = £10 \)
\( 7 \times £5 = £35 \).
So, the shared amounts are £10 and £35.

M1 for \( 45 \div (2 + 7) \) or \( 45 \div 9 \) or showing one of the correct values (£10 or £35).
A1 for £10 and £35 (accept 10 and 35).

The formula for the area of a triangle is:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
Substitute the values into the formula:
\( \text{Area} = \frac{1}{2} \times 7 \times 6 = \frac{42}{2} = 21 \text{ cm}^2 \).

M1 for a correct method to find the area, e.g. \( \frac{1}{2} \times 7 \times 6 \) or \( 7 \times 6 \div 2 \).
A1 for 21.

There are 12 counters in total.
The number of blue counters is 4.
Therefore, the number of counters that are not blue is:
\( 12 - 4 = 8 \).
The probability of taking a counter that is not blue is:
\( \frac{8}{12} = \frac{2}{3} \).

M1 for finding the number of non-blue counters (8) or writing a fraction with a denominator of 12, or for \( 1 - \frac{4}{12} \).
A1 for \( \frac{2}{3} \) (or equivalent fraction, e.g., \( \frac{8}{12} \)).

First, write the numbers in order of size:
\( 0, 1, 1, 1, 2, 2, 3, 4, 5 \)
Since there are 9 numbers, the median is the 5th number in the ordered list.
The 5th number is 2.

M1 for ordering the numbers, or for identifying the middle position (e.g., finding the 5th term).
A1 for 2.

To divide by a fraction, multiply by its reciprocal:
\( \frac{3}{4} \div \frac{2}{5} = \frac{3}{4} \times \frac{5}{2} \)
Multiply the numerators and the denominators:
\( \frac{3 \times 5}{4 \times 2} = \frac{15}{8} \)
Convert the improper fraction to a mixed number:
\( \frac{15}{8} = 1 \frac{7}{8} \).

M1 for \( \frac{3}{4} \times \frac{5}{2} \) or for a correct equivalent improper fraction (e.g. \( \frac{15}{8} \)).
A1 for \( 1 \frac{7}{8} \) (cao).

To find which bag is better value, we can find the cost per apple for each bag.

**Bag A:**
Cost of 1 apple = \(\frac{£1.32}{6} = £0.22\) (or 22p)

**Bag B:**
Cost of 1 apple = \(\frac{£1.68}{8} = £0.21\) (or 21p)

Comparing the two unit costs:
\(21\text{p} < 22\text{p}\)

Therefore, Bag B is better value for money.

M1: for a correct method to find the cost per apple for one bag, e.g., \(1.32 \div 6\) or \(1.68 \div 8\), or for finding the cost of a common number of apples (e.g. 24 apples: Bag A is \(4 \times 1.32 = £5.28\), Bag B is \(3 \times 1.68 = £5.04\)).
M1: for finding both comparative values correctly, e.g., 22p and 21p (or £0.22 and £0.21), or £5.28 and £5.04.
A1: for Bag B with consistent supporting calculations.

To solve the equation \(5x - 3 = 2x + 12\):

Subtract \(2x\) from both sides:
\(5x - 2x - 3 = 12\)
\(3x - 3 = 12\)

Add 3 to both sides:
\(3x = 12 + 3\)
\(3x = 15\)

Divide both sides by 3:
\(x = \frac{15}{3}\)
\(x = 5\)

M1: for a correct first step to collect \(x\) terms on one side or constant terms on the other side, e.g., \(5x - 2x - 3 = 12\) or \(5x = 2x + 15\).
M1: for a correct process to reach \(3x = 15\) or equivalent.
A1: for \(x = 5\) (or just 5).

The ratio of Alice's share to Bob's share is \(3 : 7\).

First, find the difference in the number of parts between Alice and Bob:
\(7 - 3 = 4\text{ parts}\)

We know Bob receives £24 more than Alice, so these 4 parts represent £24.

Next, find the value of 1 part:
\(1\text{ part} = \frac{£24}{4} = £6\)

Alice receives 3 parts, so calculate Alice's share:
\(3 \times £6 = £18\)

M1: for finding the difference in parts, i.e., \(7 - 3 = 4\) parts, or for writing a correct ratio equation like \(7x - 3x = 24\).
M1: for a complete method to find the value of one share, e.g., \(24 \div 4 = 6\), or to find the total sum of money, e.g., \((3+7) \times 6 = 60\).
A1: for £18 (accept 18).

To find the area of the L-shape, we can split it into two rectangles.

**Method 1: Split vertically**
- Left rectangle: width = \(4\text{ cm}\), height = \(8\text{ cm}\)
\(\text{Area} = 4 \times 8 = 32\text{ cm}^2\)
- Right rectangle: width = \(10 - 4 = 6\text{ cm}\), height = \(3\text{ cm}\)
\(\text{Area} = 6 \times 3 = 18\text{ cm}^2\)
- Total Area = \(32 + 18 = 50\text{ cm}^2\)

**Method 2: Split horizontally**
- Bottom rectangle: width = \(10\text{ cm}\), height = \(3\text{ cm}\)
\(\text{Area} = 10 \times 3 = 30\text{ cm}^2\)
- Top-left rectangle: width = \(4\text{ cm}\), height = \(8 - 3 = 5\text{ cm}\)
\(\text{Area} = 4 \times 5 = 20\text{ cm}^2\)
- Total Area = \(30 + 20 = 50\text{ cm}^2\)

M1: for splitting the shape into two rectangles and finding a missing dimension, e.g., \(10 - 4 = 6\text{ cm}\) or \(8 - 3 = 5\text{ cm}\).
M1: for a complete method to calculate the area of both rectangles, e.g., \((4 \times 8) + (6 \times 3)\) or \((10 \times 3) + (4 \times 5)\).
A1: for 50 (or \(50\text{ cm}^2\)).

The sum of the probabilities of all possible outcomes is 1.

Let the probability of landing on Blue be \(x\).
Then, the probability of landing on Yellow is \(2x\).

We can write the equation:
\(P(\text{Red}) + P(\text{Blue}) + P(\text{Yellow}) = 1\)
\(0.4 + x + 2x = 1\)

Simplify the equation:
\(0.4 + 3x = 1\)

Subtract \(0.4\) from both sides:
\(3x = 0.6\)

Divide by 3:
\(x = \frac{0.6}{3} = 0.2\)

Therefore, the probability that the spinner lands on Blue is \(0.2\).

M1: for a correct start to find the combined probability of Blue and Yellow, e.g., \(1 - 0.4 = 0.6\).
M1: for a process to find the probability of Blue by dividing by 3, e.g., \(0.6 \div 3\) or setting up \(3x = 0.6\).
A1: for 0.2 (or equivalent fraction, e.g., \(\frac{1}{5}\)).

To find the mean number of goals, we first find the total number of goals scored in all 20 matches:

- Matches with 0 goals: \(0 \times 5 = 0\)
- Matches with 1 goal: \(1 \times 7 = 7\)
- Matches with 2 goals: \(2 \times 4 = 8\)
- Matches with 3 goals: \(3 \times 3 = 9\)
- Matches with 4 goals: \(4 \times 1 = 4\)

Total number of goals = \(0 + 7 + 8 + 9 + 4 = 28\)

Total number of matches = \(20\)

Mean number of goals = \(\frac{\text{Total goals}}{\text{Total matches}} = \frac{28}{20} = 1.4\)

M1: for finding at least 3 correct products of goals scored \(\times\) frequency (can be implied by addition, e.g., \(0 + 7 + 8 + 9 + 4\)).
M1: for a complete method to divide the total number of goals by the total number of matches, e.g., \(28 \div 20\).
A1: for 1.4 (or equivalent fraction, e.g., \(\frac{7}{5}\) or \(1\frac{2}{5}\)).

To find when both buses next leave at the same time, we need to find the Least Common Multiple (LCM) of 12 and 15.

**Method 1: List multiples**
- Multiples of 12: 12, 24, 36, 48, **60**, 72, ...
- Multiples of 15: 15, 30, 45, **60**, 75, ...

The LCM of 12 and 15 is 60.

This means both buses will leave at the same time again in 60 minutes.

60 minutes is equal to 1 hour.

9:00 am + 1 hour = 10:00 am.

M1: for listing at least three multiples of 12 and 15 (or prime factorization of both: \(12 = 2^2 \times 3\) and \(15 = 3 \times 5\)).
M1: for finding the LCM is 60 (minutes).
A1: for 10:00 am (accept 10:00, 10 am, or 10 o'clock).

Let the shares of Amy, Ben and Chloe be \(3x\), \(5x\) and \(8x\) respectively.

Chloe receives \(\pounds 60\) more than Amy, so:
\(8x - 3x = 60\)
\(5x = 60\)
\(x = 12\)

Now we can find the initial amount of money each person has:
- Amy's share: \(3 \times 12 = \pounds 36\)
- Ben's share: \(5 \times 12 = \pounds 60\)
- Chloe's share: \(8 \times 12 = \pounds 96\)

Chloe gives \(25\%\) of her money to Ben:
- Amount Chloe gives: \(25\%\text{ of }\pounds 96 = \frac{1}{4} \times 96 = \pounds 24\)
- Chloe's new total: \(96 - 24 = \pounds 72\)
- Ben's new total: \(60 + 24 = \pounds 84\)
- Amy's total remains: \(\pounds 36\)

The new ratio of Amy's money to Ben's money to Chloe's money is:
\(36 : 84 : 72\)

To simplify, divide all parts of the ratio by their highest common factor, which is 12:
\(36 \div 12 = 3\)
\(84 \div 12 = 7\)
\(72 \div 12 = 6\)

So, the ratio in its simplest form is \(3 : 7 : 6\).

**M1**: For setting up an equation or expression to find the value of one share, e.g. \(8x - 3x = 60\) or \(5\text{ shares} = \pounds 60\) or finding \(1\text{ share} = \pounds 12\).

**M1**: For finding the initial amounts of money for at least two of the people (Amy: \(\pounds 36\), Ben: \(\pounds 60\), Chloe: \(\pounds 96\)).

**M1**: For finding \(25\%\) of Chloe's share (\(\pounds 24\)) and using this to find the new amounts for Ben (\(\pounds 84\)) and Chloe (\(\pounds 72\)).

**A1**: For the final simplified ratio of \(3 : 7 : 6\) (accept in this order only).

First, calculate the cost of the 3 sourdough loaves using Offer A:
- Mr Henderson buys 2 loaves at full price: \(2 \times \pounds 4.20 = \pounds 8.40\)
- The 3rd loaf is half price: \(\pounds 4.20 \div 2 = \pounds 2.10\)
- Total cost for the loaves = \(\pounds 8.40 + \pounds 2.10 = \pounds 10.50\)

Next, calculate the cost of the 8 croissants using Offer B:
- Offer B is "Buy 4 for the price of 3".
- Since Mr Henderson wants 8 croissants, this is exactly two sets of 4 croissants.
- For each set of 4, he only pays for 3. So, he pays for \(3 \times 2 = 6\) croissants in total.
- Cost of 6 croissants = \(6 \times \pounds 1.80 = \pounds 10.80\)

Finally, calculate the total cost:
- Total cost = \(\pounds 10.50 + \pounds 10.80 = \pounds 21.30\)

**M1**: For a method to calculate the cost of 3 sourdough loaves with Offer A, e.g. \(2 \times 4.20 + 2.10\) (or showing \(\pounds 10.50\)).

**M1**: For recognizing that 8 croissants can be bought as two sets of 4 under Offer B, so Mr Henderson only pays for 6 croissants (or showing \(6 \times 1.80\)).

**M1**: For a method to calculate the total cost of 6 croissants, e.g. \(6 \times 1.80 = 10.80\) (or showing \(\pounds 10.80\)).

**A1**: For \(\pounds 21.30\) (or \(21.30\)).

To find the volume of a prism, we use the formula:

\(\text{Volume} = \text{Area of cross-section} \times \text{length}\)

First, calculate the area of the cross-section, which is made of a rectangle and a right-angled triangle.

1. Area of the rectangle:
\(\text{Area} = \text{length} \times \text{width}\)
\(\text{Area} = 12\text{ cm} \times 6\text{ cm} = 72\text{ cm}^2\)

2. Area of the right-angled triangle:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times 5\text{ cm} \times 6\text{ cm} = 15\text{ cm}^2\)

3. Total area of the cross-section:
\(\text{Total Area} = 72\text{ cm}^2 + 15\text{ cm}^2 = 87\text{ cm}^2\)

Now, calculate the volume of the prism by multiplying the cross-sectional area by the length of the prism:
\(\text{Volume} = 87\text{ cm}^2 \times 20\text{ cm}\)
\(\text{Volume} = 1740\text{ cm}^3\)

**M1**: For finding the area of the rectangle, e.g. \(12 \times 6 = 72\).

**M1**: For finding the area of the triangle, e.g. \(\frac{1}{2} \times 5 \times 6 = 15\).

**M1**: For adding the two areas together to find the cross-sectional area (e.g. \(72 + 15 = 87\)) and multiplying by the length of the prism (e.g. \(87 \times 20\)).

**A1**: For \(1740\) (accept \(1740\text{ cm}^3\)).

To find the number of each type of toy, we first find the total number of parts in the ratio:
\[4 + 5 + 3 = 12 \text{ parts}\]

Next, find the number of toys represented by 1 part:
\[360 \div 12 = 30 \text{ toys}\]

Now, calculate the number of plastic and wooden toys:
- Plastic toys: \(4 \times 30 = 120\)
- Wooden toys: \(5 \times 30 = 150\)
- Metal toys: \(3 \times 30 = 90\)

Next, calculate the number of red toys for each material:
- Red plastic toys: \(25\% \text{ of } 120 = \frac{1}{4} \times 120 = 30\)
- Red wooden toys: \(\frac{2}{5} \text{ of } 150 = \frac{150}{5} \times 2 = 30 \times 2 = 60\)
- Red metal toys: \(0\)

Finally, add these together to get the total number of red toys:
\[30 + 60 = 90\]

**M1**: For finding the total number of parts and dividing the total number of toys by this sum, e.g., \(4 + 5 + 3 = 12\) and \(360 \div 12\) (or 30 seen).

**M1**: For finding the number of plastic toys (120) and wooden toys (150).

**M1**: For a method to find \(25\%\) of their number of plastic toys, e.g., \(120 \div 4\) (or 30 seen).

**M1**: For a method to find \(\frac{2}{5}\) of their number of wooden toys, e.g., \((150 \div 5) \times 2\) (or 60 seen).

**A1**: For the correct final answer of 90.

First, calculate the area of the L-shaped floor.
We can split the shape into two rectangles:
- Left rectangle: \(3\text{ m} \times 5\text{ m} = 15\text{ m}^2\)
- Right rectangle: \((8 - 3)\text{ m} \times 2\text{ m} = 5\text{ m} \times 2\text{ m} = 10\text{ m}^2\)

Total area of the floor:
\[15 + 10 = 25\text{ m}^2\]

Next, convert the dimensions of a tile to metres to find its area:
\[50\text{ cm} = 0.5\text{ m}\]
Area of one tile:
\[0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2\]

Now, calculate the number of tiles needed:
\[25 \div 0.25 = 100\text{ tiles}\]
(Alternatively, each \(1\text{ m}^2\) requires \(4\) tiles, so \(25 \times 4 = 100\) tiles).

Calculate the number of boxes needed:
\[100 \div 8 = 12.5\text{ boxes}\]
Since Jan must buy whole boxes, we round 12.5 up to 13 boxes.

Finally, calculate the total cost:
\[13 \times £15 = £195\]

**M1**: For splitting the L-shape and finding the total area of the floor, e.g., \((3 \times 5) + (5 \times 2) = 25\text{ m}^2\) or \((3 \times 3) + (8 \times 2) = 25\text{ m}^2\).

**M1**: For a method to find the area of one tile in \(\text{m}^2\) (e.g., \(0.5 \times 0.5 = 0.25\)) or to find the number of tiles per \(\text{m}^2\) (e.g., 4 tiles) or to convert floor area to \(\text{cm}^2\) (e.g., \(25 \times 10,000 = 250,000\)).

**M1**: For a method to find the total number of tiles needed, e.g., \(25 \div 0.25\) or \(250,000 \div 2,500\) (giving 100 tiles).

**M1**: For dividing the number of tiles by 8 and rounding up to the next integer, e.g., \(100 \div 8 = 12.5\) so 13 boxes.

**A1**: For £195 (accept 195).

To find the best offer, we first need to determine the number of children going on the trip.

**Step 1: Find the number of children**
The ratio of adults to children is \(1 : 4\).
Total parts = \(1 + 4 = 5\)
Each part represents:
\[150 \div 5 = 30\text{ people}\]

Number of children:
\[4 \times 30 = 120\text{ children}\]

**Step 2: Calculate the cost with Offer A (Buy 2, get 1 free)**
With this offer, for every 3 tickets needed, you only pay for 2.
Number of groups of 3 in 120:
\[120 \div 3 = 40\text{ groups}\]

Total tickets paid for:
\[40 \times 2 = 80\text{ tickets}\]

Total cost for Offer A:
\[80 \times £15 = £1200\]

**Step 3: Calculate the cost with Offer B (15% off every child ticket)**
Normal cost for 120 child tickets without any discount:
\[120 \times £15 = £1800\]

Now find \(15\%\) of \(£1800\):
\[10\%\text{ of } £1800 = £180\]
\[5\%\text{ of } £1800 = £90\]
\[15\%\text{ of } £1800 = £180 + £90 = £270\]

Total cost for Offer B:
\[£1800 - £270 = £1530\]

*(Alternatively, discount per ticket = \(15\%\text{ of } £15 = £2.25\). Price per ticket = \(£15 - £2.25 = £12.75\). Total cost = \(120 \times £12.75 = £1530\).)*

**Step 4: Compare the offers and find the saving**
Comparing the two costs:
\[£1530 \text{ (Offer B)} > £1200 \text{ (Offer A)}\]

So, Offer A is the cheaper offer.

Saving:
\[£1530 - £1200 = £330\]

The leader should choose **Offer A**, saving **\(£330\)**.

**M1:** For a complete method to find the number of children, e.g., \(\frac{150}{1+4} \times 4\) or \(150 \div 5 \times 4\).
**A1:** For \(120\) children.
**M1:** For a method to calculate the total cost using Offer A, e.g., finding that they only pay for \(80\) tickets (\(120 \div 3 \times 2\)) and multiplying by \(£15\).
**A1:** For \(£1200\) (Offer A cost).
**M1:** For a method to calculate the total cost using Offer B, e.g., finding \(15\%\) of \(£1800\) (or \(15\%\) of \(£15\)) and subtracting this discount from the total normal cost.
**A1:** For the correct choice of **Offer A** and the saving of **\(£330\)** (supported by a correct cost of \(£1530\) for Offer B).

To write \( 0.35 \) as a fraction, we place it over 100 because the last digit is in the hundredths place:

\( 0.35 = \frac{35}{100} \)

We can simplify this fraction by dividing both the numerator and the denominator by 5, their highest common factor:

\( 35 \div 5 = 7 \)

\( 100 \div 5 = 20 \)

So, the fraction in its simplest form is \( \frac{7}{20} \).

B1 for \( \frac{7}{20} \) (or equivalent simplified fraction).

To simplify the expression, we multiply the numbers together and then the variables:

\( 4 \times 3 = 12 \)

\( y \times x = xy \)

Combining these gives \( 12xy \).

B1 for \( 12xy \) or \( 12yx \).

A 7-sided polygon is mathematically known as a heptagon.

B1 for heptagon (accept septagon).

To simplify the ratio \( 15 : 35 \), we find the highest common factor of 15 and 35, which is 5.

Divide both parts of the ratio by 5:

\( 15 \div 5 = 3 \)

\( 35 \div 5 = 7 \)

This gives \( 3 : 7 \).

B1 for \( 3:7 \).

Comparing the temperatures, we can order them from coldest to warmest:

\( -5^\circ\text{C} < -3^\circ\text{C} < 0^\circ\text{C} < 2^\circ\text{C} < 4^\circ\text{C} \)

The lowest temperature is \( -5^\circ\text{C} \).

B1 for \(-5\) (accept \(-5^\circ\text{C}\)).

We need to find a number that, when multiplied by itself three times, equals 512:

\( 8 \times 8 \times 8 = 512 \)

Therefore, \( \sqrt[3]{512} = 8 \).

B1 for 8.

The possible outcomes when rolling a six-sided dice are 1, 2, 3, 4, 5, 6.

The outcomes greater than 4 are 5 and 6 (2 outcomes in total).

The probability is the number of successful outcomes divided by the total number of outcomes:

\( \frac{2}{6} = \frac{1}{3} \).

B1 for \( \frac{1}{3} \) or equivalent fraction, decimal, or percentage.

There are 1000 millilitres in 1 litre.

To convert 3.4 litres into millilitres, we multiply by 1000:

\( 3.4 \times 1000 = 3400 \).

B1 for 3400.

To write the ratio in its simplest form, divide both numbers by their highest common factor (HCF). The factors of 18 are 1, 2, 3, 6, 9, 18 and the factors of 45 are 1, 3, 5, 9, 15, 45. The highest common factor is 9. Divide both parts of the ratio by 9: \(18 \div 9 = 2\) and \(45 \div 9 = 5\). This gives the ratio \(2 : 5\).

B1 for \(2:5\) (or \(2 : 5\))

First, calculate the numerator:
\(3.8^2 + 15.6 = 14.44 + 15.6 = 30.04\)
\(\sqrt{30.04} \approx 5.4808758\)

Next, calculate the denominator:
\(1.4 \times 0.9 = 1.26\)

Now, divide the numerator by the denominator:
\(\frac{5.4808758}{1.26} \approx 4.3499\)

Rounding to 3 significant figures gives \(4.35\).

M1 for working out the numerator as \(\sqrt{30.04}\) (or \(5.48...\)) OR the denominator as \(1.26\) OR for writing down \(4.349...\)
A1 for \(4.35\)

Expand the brackets:
\(8x - 12 = 18\)

Add 12 to both sides of the equation:
\(8x = 30\)

Divide both sides by 8:
\(x = \frac{30}{8} = 3.75\)

Alternatively, divide both sides by 4 first:
\(2x - 3 = 4.5\)
\(2x = 7.5\)
\(x = 3.75\)

M1 for a correct first step, e.g., expanding brackets to get \(8x - 12 = 18\) OR dividing both sides by 4 to get \(2x - 3 = 4.5\)
A1 for \(3.75\) (or equivalent fraction, e.g., \(\frac{15}{4}\) or \(3\frac{3}{4}\))

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

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

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

M1 for \(8.4 \times 25\ 000\) (or \(210\ 000\)) OR for dividing a distance by \(100\ 000\) to convert from cm to km
A1 for \(2.1\)

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

Substitute the given values into the formula:
\(V = \pi \times 3.5^2 \times 12\)
\(V = \pi \times 12.25 \times 12\)
\(V = 147\pi \approx 461.814...\text{ cm}^3\)

Correct to 1 decimal place, the volume is \(461.8\text{ cm}^3\).

M1 for \(\pi \times 3.5^2 \times 12\) (or \(147\pi\) or \(461.8...\))
A1 for \(461.8\) (accept answers in the range \(461.8\) to \(462\) if working is shown)

First, find the sum of all the numbers in the list:
\(12 + 15 + 14 + 17 + 20 + 15 + 18 + 13 = 124\)

There are 8 numbers in the list, so divide the sum by 8:
\(124 \div 8 = 15.5\)

M1 for showing a correct method to find the sum of the numbers (summing to \(124\)) or dividing the sum by \(8\)
A1 for \(15.5\) (or equivalent fraction, e.g., \(\frac{31}{2}\))

To find the value of one share, divide Charlie's share of the money by his ratio part: \(\pounds 120 \div 4 = \pounds 30\). Alice's share is \(3 \times \pounds 30 = \pounds 90\). Bob's share is \(5 \times \pounds 30 = \pounds 150\). The difference between Bob's and Alice's shares is \(\pounds 150 - \pounds 90 = \pounds 60\).

M1: for \(\pounds 120 \div 4\) (or \(30\)). M1: for \((5 - 3) \times 30\) or \((5 \times 30) - (3 \times 30)\). A1: for \(60\) (or \(\pounds 60\)).

Expand the brackets first: \(8x - 12 = 18\). Add \(12\) to both sides of the equation: \(8x = 30\). Divide both sides by \(8\): \(x = \frac{30}{8} = 3.75\).

M1: for expanding brackets to get \(8x - 12 = 18\), or for dividing both sides by \(4\) to get \(2x - 3 = 4.5\). M1: (dep) for isolating the \(x\) term, e.g., \(8x = 30\) or \(2x = 7.5\). A1: for \(3.75\) (or equivalent fraction).

First, calculate the area of the trapezium: \(\text{Area} = \frac{12 + 18}{2} \times 9 = 15 \times 9 = 135\text{ m}^2\). Next, find the number of bags needed by dividing the total area by the coverage of one bag: \(135 \div 25 = 5.4\). Since grass seed must be purchased in whole bags, we round up to the next whole number, which is \(6\) bags.

M1: for a method to find the area of the trapezium, e.g., \(\frac{12+18}{2} \times 9\) (or \(135\)). M1: for dividing their area by \(25\), e.g., \(\text{"135"} \div 25\) (or \(5.4\)). A1: for \(6\) (do not accept \(5.4\)).

Find the total number of goals scored: \((0 \times 3) + (1 \times 6) + (2 \times 5) + (3 \times 4) + (4 \times 2) = 0 + 6 + 10 + 12 + 8 = 36\). Divide this total by the number of matches: \(36 \div 20 = 1.8\).

M1: for finding at least three correct products of goals and frequency (can be implied by addition). M1: (dep) for dividing the sum of their products by \(20\). A1: for \(1.8\) (or equivalent fraction).

Calculate the total value of the investment after \(3\) years: \(\pounds 2400 \times (1.015)^3 = \pounds 2509.6281...\). Subtract the initial investment to find the interest earned: \(\pounds 2509.6281... - \pounds 2400 = \pounds 109.6281...\). Rounded to the nearest penny, this is \(\pounds 109.63\).

M1: for a correct method to find the total value or interest after \(3\) years, e.g., \(2400 \times 1.015^3\) (or \(2509.6281...\)). M1: for subtracting \(2400\) from their total value. A1: for \(109.63\) (accept with or without \(\pounds\) symbol).

The probability of landing on Tails is \(1 - 0.6 = 0.4\). The two outcomes that give exactly one Head are Heads then Tails (HT) and Tails then Heads (TH). Probability of HT is \(0.6 \times 0.4 = 0.24\). Probability of TH is \(0.4 \times 0.6 = 0.24\). Total probability is \(0.24 + 0.24 = 0.48\).

M1: for finding the probability of Tails is \(0.4\). M1: for a method to find the probability of one outcome, e.g., \(0.6 \times 0.4\) (or \(0.24\)), or for adding two correct products, e.g., \((0.6 \times 0.4) + (0.4 \times 0.6)\). A1: for \(0.48\) (or equivalent fraction).

The first part of the journey takes \(1\text{ hour } 15\text{ minutes}\) (which is \(75\text{ minutes}\)). For the second part, time taken is \(\frac{\text{distance}}{\text{speed}} = \frac{85\text{ km}}{100\text{ km/h}} = 0.85\text{ hours}\). Convert \(0.85\text{ hours}\) to minutes: \(0.85 \times 60 = 51\text{ minutes}\). Total time in minutes is \(75 + 51 = 126\text{ minutes}\). Converting \(126\text{ minutes}\) to hours and minutes gives \(2\text{ hours } 6\text{ minutes}\).

M1: for a method to calculate the time for the second stage, e.g., \(85 \div 100\) (or \(0.85\text{ hours}\)). M1: for converting \(0.85\text{ hours}\) to \(51\text{ minutes}\) and adding to \(75\text{ minutes}\) (or adding \(1.25 + 0.85 = 2.1\text{ hours}\)). A1: for \(2\text{ hours } 6\text{ minutes}\) (or \(126\text{ minutes}\)).

Set the expression for the \(n\)th term equal to \(115\): \(4n + 3 = 115\). Subtract \(3\) from both sides: \(4n = 112\). Divide both sides by \(4\): \(n = 28\). Since \(28\) is an integer (whole number), \(115\) is a term in the sequence. It is the \(28\)th term.

M1: for setting up the equation \(4n + 3 = 115\). M1: for solving to find \(n = 28\). A1: for 'Yes' with a correct explanation indicating \(n = 28\) is a whole number.

First, find the value of one share. The difference in parts between Beth and Amir is \(5 - 3 = 2\) parts. Since Beth receives \(\pounds 120\) more than Amir, \(2\) parts = \(\pounds 120\). Therefore, \(1\) part = \(\pounds 120 \div 2 = \pounds 60\). Next, calculate the original amounts of money Beth and Chloe receive. Beth's original share: \(5 \times \pounds 60 = \pounds 300\). Chloe's original share: \(8 \times \pounds 60 = \pounds 480\). Then, calculate the money Beth gives to Chloe. Beth gives \(25\%\) of her share to Chloe: \(0.25 \times \pounds 300 = \pounds 75\). Finally, add this to Chloe's original share to find her new total: \(\pounds 480 + \pounds 75 = \pounds 555\).

M1: for finding the difference in ratio parts (e.g., \(5 - 3 = 2\)) or for writing an equation such as \(5x - 3x = 120\). M1: for finding the value of one part as \(\pounds 60\) or for calculating Beth's original share as \(\pounds 300\) or Chloe's original share as \(\pounds 480\). M1: for a method to find \(25\%\) of Beth's share (e.g., \(0.25 \times 300 = 75\)) and adding it to Chloe's share. A1: for \(555\) (accept \(\pounds 555\)).

First, calculate the area of the trapezium: \(\text{Area} = \frac{12 + 17.5}{2} \times 8 = 14.75 \times 8 = 118\text{ m}^2\). Next, find the area of one tile in square metres. A tile has side length \(50\text{ cm} = 0.5\text{ m}\), so the area of one tile is \(0.5 \times 0.5 = 0.25\text{ m}^2\). Alternatively, convert the trapezium area to square centimetres: \(118 \times 10000 = 1,180,000\text{ cm}^2\) and the tile area is \(50 \times 50 = 2500\text{ cm}^2\). Find the total number of tiles needed: \(118 \div 0.25 = 472\) tiles (or \(1,180,000 \div 2500 = 472\) tiles). Find the number of packs needed: \(472 \div 10 = 47.2\) packs. Since you can only buy whole packs, you need to buy \(48\) packs. Finally, calculate the total cost: \(48 \times \pounds 34.50 = \pounds 1656\).

M1: for a method to find the area of the trapezium, e.g., \(\frac{12 + 17.5}{2} \times 8\) (condone one arithmetic slip), resulting in \(118\). M1: for a method to find the number of tiles needed, e.g., \(118 \div 0.25\) or \(1,180,000 \div 2500\), resulting in \(472\). M1: for dividing the number of tiles by 10 and rounding up to the nearest whole integer to find the number of packs (e.g., \(472 \div 10 = 47.2\) so \(48\) packs). A1: for \(1656\) (accept \(\pounds 1656\)).

First, find the number of male employees: \(35\%\) of \(160 = 0.35 \times 160 = 56\). The number of female employees is \(160 - 56 = 104\) (though this is not strictly needed for the main ratio calculation). Next, find the number of part-time male employees: \(\frac{1}{4}\) of \(56 = 14\). The ratio of part-time male employees to part-time female employees is \(2 : 3\). This means \(2\) parts correspond to \(14\) male employees. Find the value of \(1\) part: \(14 \div 2 = 7\) employees. Find the number of part-time female employees: \(3 \times 7 = 21\). Finally, find the total number of part-time employees: \(14\text{ (male)} + 21\text{ (female)} = 35\).

M1: for finding the number of male employees, e.g., \(0.35 \times 160 = 56\). M1: for finding the number of part-time male employees, e.g., \(56 \div 4 = 14\). M1: for using the ratio to find the number of part-time female employees, e.g., \(14 \div 2 \times 3 = 21\). A1: for \(35\) (with or without working).

First, write down an expression for the perimeter of the rectangle: \(\text{Perimeter of rectangle} = 2 \times ((2x - 3) + (3x + 1)) = 2 \times (5x - 2) = 10x - 4\). Next, write down an expression for the perimeter of the equilateral triangle: \(\text{Perimeter of triangle} = 3 \times (2x + 5) = 6x + 15\). Set the two perimeters equal to each other: \(10x - 4 = 6x + 15\). Solve the equation for \(x\): Subtract \(6x\) from both sides: \(4x - 4 = 15\). Add \(4\) to both sides: \(4x = 19\). Divide by \(4\): \(x = \frac{19}{4} = 4.75\).

M1: for writing a correct expression for the perimeter of the rectangle, e.g., \(2(2x - 3 + 3x + 1)\) or \(10x - 4\). M1: for writing a correct expression for the perimeter of the equilateral triangle, e.g., \(3(2x + 5)\) or \(6x + 15\). M1: for setting the two expressions equal and showing a correct algebraic step to solve, e.g., \(10x - 6x = 15 + 4\) or \(4x = 19\). A1: for \(4.75\) (or equivalent fraction, e.g., \(\frac{19}{4}\)).

First, find the scaling factor for 40 cakes from 12 cakes:
\(\text{Scale factor} = \frac{40}{12} = \frac{10}{3}\)

Now, calculate the total quantity of each ingredient needed for 40 cakes:
* Flour: \(3\text{ kg} \times \frac{10}{3} = 10\text{ kg}\)
* Sugar: \(1.8\text{ kg} \times \frac{10}{3} = 6\text{ kg}\)
* Eggs: \(24 \times \frac{10}{3} = 80\text{ eggs}\)
* Butter: \(1.5\text{ kg} \times \frac{10}{3} = 5\text{ kg}\)

Convert in-stock quantities into matching units and find the shortfall for each ingredient:
* Flour shortfall: \(10\text{ kg} - 1.2\text{ kg} = 8.8\text{ kg}\)
* Sugar shortfall: \(6\text{ kg} - 0.5\text{ kg} = 5.5\text{ kg}\) (since \(500\text{ g} = 0.5\text{ kg}\))
* Eggs shortfall: \(80 - 8 = 72\text{ eggs}\)
* Butter shortfall: \(5\text{ kg} - 0.3\text{ kg} = 4.7\text{ kg}\) (since \(300\text{ g} = 0.3\text{ kg}\))

Calculate the number of packs/boxes/blocks to buy (always rounding up to the nearest whole pack):
* Flour: \(8.8\text{ kg} \div 1.5\text{ kg} = 5.86... \rightarrow 6\text{ bags}\)
* Sugar: \(5.5\text{ kg} \div 1\text{ kg} = 5.5 \rightarrow 6\text{ bags}\)
* Eggs: \(72 \div 6 = 12\text{ boxes}\)
* Butter: \(4.7\text{ kg} \div 0.5\text{ kg} = 9.4 \rightarrow 10\text{ blocks}\)

Calculate the cost for each ingredient:
* Flour: \(6 \times £1.40 = £8.40\)
* Sugar: \(6 \times £1.10 = £6.60\)
* Eggs: \(12 \times £1.80 = £21.60\)
* Butter: \(10 \times £2.20 = £22.00\)

Calculate the total cost:
\(£8.40 + £6.60 + £21.60 + £22.00 = £58.60\)

**M1**: For a method to scale recipe quantities to 40 cakes (at least two correct quantities: e.g. \(10\text{ kg}\) flour, \(6\text{ kg}\) sugar, \(80\text{ eggs}\), or \(5\text{ kg}\) butter).
**M1**: For converting units correctly (e.g. \(500\text{ g} = 0.5\text{ kg}\) or \(300\text{ g} = 0.3\text{ kg}\)) and calculating the shortfall for at least two ingredients.
**M1**: For a method to find the correct number of packs/bags/boxes needed by dividing the shortfall by pack size and rounding up to the next integer (at least three ingredients correct).
**M1**: For a complete method to calculate the total cost by multiplying the number of packs by their unit costs and summing them.
**A1**: For \(£58.60\) (or \(58.60\)).

First, calculate the cross-sectional area of a single concrete block:
\(\text{Area of trapezium} = \frac{a + b}{2} \times h\)
\(\text{Area} = \frac{12 + 20}{2} \times 8 = 16 \times 8 = 128\text{ cm}^2\)

Next, calculate the volume of a single block:
\(\text{Volume of prism} = \text{cross-sectional area} \times \text{length}\)
\(\text{Volume} = 128 \times 45 = 5760\text{ cm}^3\)

Calculate the total volume for 35 blocks:
\(\text{Total Volume} = 5760 \times 35 = 201,600\text{ cm}^3\)

Convert this total volume into litres:
\(\text{Volume in litres} = \frac{201,600}{1000} = 201.6\text{ litres}\)

Calculate the number of bags of concrete mix needed:
\(\text{Number of bags} = \frac{201.6}{12} = 16.8\text{ bags}\)
Since the builder must buy whole bags, we round up to 17 bags.

Calculate the total cost of these bags:
\(\text{Total Cost} = 17 \times £6.40 = £108.80\)

**M1**: For a method to find the area of the trapezium cross-section, e.g. \(\frac{12 + 20}{2} \times 8\) (or \(128\)).
**M1**: For a method to find the total volume of 35 blocks, e.g. \(\text{their area} \times 45 \times 35\) (or \(201,600\)).
**M1**: For converting the total volume from \(\text{cm}^3\) to litres, e.g. \(\text{their volume} \div 1000\) (or \(201.6\)).
**M1**: For dividing the volume in litres by 12 and rounding up to the next integer to find the number of bags (e.g. \(201.6 \div 12 = 16.8 \rightarrow 17\)).
**A1**: For \(£108.80\) (or \(108.80\)).

The first non-zero digit in \( 0.04073 \) is \( 4 \) in the hundredths place. The second significant figure is \( 0 \) in the thousandths place. The next digit to the right is \( 7 \), which is \( 5 \) or more, so we round up. Therefore, \( 0.04073 \) rounded to 2 significant figures is \( 0.041 \).

B1 for \( 0.041 \) (or equivalent)

We multiply the numbers together: \( 4 \times 3 = 12 \). Then we multiply the variables: \( x \times y = xy \). Combining these gives \( 12xy \).

B1 for \( 12xy \) (or equivalent, e.g., \( 12yx \))

\( 35\% = \frac{35}{100} \). To simplify, we divide both the numerator and denominator by their greatest common divisor, which is \( 5 \). This gives \( \frac{35 \div 5}{100 \div 5} = \frac{7}{20} \).

B1 for \( \frac{7}{20} \) (or equivalent simplest form fraction)

A triangular prism consists of 2 triangular bases at each end and 3 rectangular sides connecting them. This gives a total of \( 2 + 3 = 5 \) faces.

B1 for 5 (or five)

The possible outcomes when rolling a six-sided dice are \( 1, 2, 3, 4, 5, 6 \). The prime numbers in this set are \( 2, 3, \) and \( 5 \). There are 3 prime numbers out of 6 possible outcomes. The probability is \( \frac{3}{6} = \frac{1}{2} \) (or \( 0.5 \)).

B1 for \( \frac{1}{2} \) or \( 0.5 \) or \( 50\% \) (accept \( \frac{3}{6} \))

The highest score is \( 19 \) and the lowest score is \( 11 \). The range is the difference between the highest and lowest scores: \( 19 - 11 = 8 \).

B1 for 8

Average speed is calculated by dividing the total distance by the total time: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{135}{3} = 45 \) miles per hour.

B1 for 45

First, calculate the terms in the numerator:
\(\sqrt{38.44} = 6.2\)
\(4.1^2 = 16.81\)
Add these together:
\(6.2 + 16.81 = 23.01\)
Now, divide by the denominator:
\(\frac{23.01}{0.3} = 76.7\)

M1 for \(\sqrt{38.44} = 6.2\) or \(4.1^2 = 16.81\) or \(23.01\) seen
A1 for \(76.7\) (accept \(\frac{767}{10}\))

Expand the bracket:
\(5y - 10 = 13\)
Add 10 to both sides:
\(5y = 23\)
Divide by 5:
\(y = \frac{23}{5} = 4.6\)

M1 for a correct first step, e.g. expanding to \(5y - 10 = 13\) or dividing both sides by 5 to get \(y - 2 = 2.6\)
A1 for \(4.6\) (or \(\frac{23}{5}\) or \(4\frac{3}{5}\))

Total number of parts = \(2 + 3 + 7 = 12\)
Value of one part = \(450 \div 12 = £37.50\)
Charity A receives \(2 \times 37.50 = £75\)
Charity C receives \(7 \times 37.50 = £262.50\)
Difference = \(262.50 - 75 = £187.50\)
Alternatively, difference in parts = \(7 - 2 = 5\) parts
Difference in money = \(5 \times 37.50 = £187.50\)

M1 for \(450 \div (2 + 3 + 7)\) or \(37.50\) seen, or for finding the share of A (\(75\)) and C (\(262.50\))
A1 for \(187.50\) (accept \(187.5\))

Use the formula for the area of a trapezium:
\(\text{Area} = \frac{1}{2}(a + b)h\)
Substitute the given values:
\(\text{Area} = \frac{1}{2}(8 + 12) \times 5.5\)
\(\text{Area} = \frac{1}{2}(20) \times 5.5 = 10 \times 5.5 = 55\text{ cm}^2\)

M1 for a correct substitution into the trapezium area formula, e.g. \(\frac{1}{2}(8 + 12) \times 5.5\) or \(10 \times 5.5\)
A1 for 55

The sum of probabilities of all possible mutually exclusive outcomes is 1.
Let \(P(\text{Yellow})\) be the probability of choosing a yellow counter.
\(P(\text{Yellow}) = 1 - (0.35 + 0.4)\)
\(P(\text{Yellow}) = 1 - 0.75 = 0.25\)

M1 for \(0.35 + 0.4\) or \(0.75\) seen, or \(1 - 0.35 - 0.4\)
A1 for \(0.25\) (or equivalent fraction, e.g. \(\frac{1}{4}\))

Multiply each number of goals by its frequency:
\(0 \times 5 = 0\)
\(1 \times 8 = 8\)
\(2 \times 4 = 8\)
\(3 \times 3 = 9\)
Sum the products:
\(0 + 8 + 8 + 9 = 25\)

M1 for at least two correct products shown (e.g. \(1 \times 8\) and \(2 \times 4\)) or for \(0 + 8 + 8 + 9\)
A1 for 25

Calculate the interest for one year:
\(2.5\%\text{ of } 2400 = 0.025 \times 2400 = 60\)
Calculate the total interest for 3 years:
\(60 \times 3 = 180\)

M1 for a correct method to find \(2.5\%\) of \(2400\) (e.g. \(0.025 \times 2400\) or \(60\)) or for a correct formula used (e.g. \(\frac{2400 \times 2.5 \times 3}{100}\))
A1 for 180 (accept £180)

Find the common difference between consecutive terms:
\(9 - 5 = 4\)
\(13 - 9 = 4\)
Since the difference is \(4\), the expression will contain \(4n\).
Compare the \(4n\) sequence to the given sequence:
For \(n = 1\), \(4(1) = 4\). The first term is \(5\), which is \(4 + 1\).
For \(n = 2\), \(4(2) = 8\). The second term is \(9\), which is \(8 + 1\).
Thus, the \(n\)-th term is \(4n + 1\).

M1 for \(4n + c\) (where \(c\) is an integer) or \(4n\) seen, or for identifying a common difference of 4
A1 for \(4n + 1\) (or equivalent, e.g. \(1 + 4n\))

The ratio of red marbles to blue marbles is \(3 : 5\).
We are given that there are 40 blue marbles, which represent 5 parts of the ratio.
First, find the value of 1 part:
\(40 \div 5 = 8\)

The total number of parts in the ratio is:
\(3 + 5 = 8\) parts

Now, multiply the total parts by the value of 1 part:
\(8 \times 8 = 64\)

M1 for a correct first step to find the value of 1 part, e.g., \(40 \div 5\) (or 8), or writing \(3 \times 8 = 24\) to find the number of red marbles.
A1 for 64 (cao).

Angles on a straight line add up to \(180^\circ\).
We can write the equation:
\(3x + 2x + 55 = 180\)

Simplify the equation:
\(5x + 55 = 180\)

Subtract 55 from both sides:
\(5x = 125\)

Divide by 5:
\(x = 25\)

M1 for setting up a correct equation, e.g., \(3x + 2x + 55 = 180\) or \(5x = 180 - 55\) (or 125).
A1 for 25.

First, evaluate the terms in the numerator:
\(\sqrt{18.49} = 4.3\)
\(3.2^2 = 10.24\)

Add these values together:
\(4.3 + 10.24 = 14.54\)

Now, divide the numerator by the denominator:
\(\frac{14.54}{0.5} = 29.08\)

M1 for evaluating the numerator correctly as 14.54, or showing \(4.3\) and \(10.24\) in the working.
A1 for 29.08 (or \(\frac{727}{25}\)).

To find the mean, first calculate the total number of goals scored by multiplying each number of goals by its corresponding frequency:

\(0 \times 5 = 0\)
\(1 \times 8 = 8\)
\(2 \times 6 = 12\)
\(3 \times 4 = 12\)
\(4 \times 2 = 8\)

Sum of all goals scored:
\(0 + 8 + 12 + 12 + 8 = 40\)

Total number of matches:
\(25\)

Calculate the mean:
\(\text{Mean} = \frac{40}{25} = 1.6\)

M1: for finding at least three correct products of goals and frequency (can be implied by a sum of 40)
M1: for division of their total sum of goals by 25
A1: for 1.6 (or equivalent fraction)

First, convert the mass of the silver bar from kilograms (kg) to grams (g):
\(1.575 \text{ kg} \times 1000 = 1575 \text{ g}\)

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

Substitute the values into the formula:
\(\text{Density} = \frac{1575 \text{ g}}{150 \text{ cm}^3} = 10.5 \text{ g/cm}^3\)

M1: for converting the mass to grams: \(1.575 \times 1000 = 1575\) (or dividing final density by 1000 if converted later)
M1: for \(\text{their mass in grams} \div 150\)
A1: for 10.5

First, find the area of the rectangle:
\(\text{Area of rectangle} = 8 \text{ m} \times 3 \text{ m} = 24 \text{ m}^2\)

Next, find the radius of the semicircle:
\(\text{Radius} = 3 \text{ m} \div 2 = 1.5 \text{ m}\)

Now, calculate the area of the semicircle:
\(\text{Area of semicircle} = \frac{1}{2} \times \pi \times 1.5^2 \approx 3.53429 \text{ m}^2\)

Find the total area of the flowerbed:
\(\text{Total Area} = 24 + 3.53429 = 27.53429 \text{ m}^2\)

Round to 3 significant figures:
\(27.5 \text{ m}^2\)

M1: for finding the area of the rectangle: \(8 \times 3 = 24\) or for finding the area of a full circle with radius 1.5: \(\pi \times 1.5^2 \approx 7.07\)
M1: for a complete method to find the total area, e.g., \(24 + 0.5 \times \pi \times 1.5^2\)
A1: for 27.5 (accept answers in the range 27.5 to 27.54)

First, find the total number of parts in the ratio:
\(2 + 3 + 5 = 10\) parts.

Next, find the weight of one part:
\(450 \div 10 = 45\text{ kg}\).

Now, calculate the weight and cost of each ingredient:
- Weight of cement: \(2 \times 45 = 90\text{ kg}\)
Cost of cement: \(90 \times £0.45 = £40.50\)
- Weight of sand: \(3 \times 45 = 135\text{ kg}\)
Cost of sand: \(135 \times £0.15 = £20.25\)
- Weight of gravel: \(5 \times 45 = 225\text{ kg}\)
Cost of gravel: \(225 \times £0.08 = £18.00\)

Finally, add the costs of all ingredients to find the total cost:
\(£40.50 + £20.25 + £18.00 = £78.75\).

M1: For finding the weight of one part, e.g., \(450 \div (2+3+5) = 45\) (or finding the correct weight of at least one ingredient: \(90\text{ kg}\), \(135\text{ kg}\), or \(225\text{ kg}\)).
M1: For a method to calculate the cost of at least two ingredients (e.g., \(90 \times 0.45 = 40.5\) and \(135 \times 0.15 = 20.25\)).
M1: For a complete method to find the total cost, i.e., adding all three calculated ingredient costs together.
A1: \(78.75\) (or \(£78.75\)).

First, calculate the total area of the rectangular garden:
\(\text{Area of rectangle} = 12 \times 8 = 96\text{ m}^2\).

Next, calculate the area of the circular pond:
\(\text{Area of circle} = \pi \times 2.5^2 = 6.25\pi \approx 19.635\text{ m}^2\).

Now, subtract the area of the pond from the total area to find the area to be covered with grass seed:
\(\text{Grass area} = 96 - 19.635 = 76.365\text{ m}^2\).

Calculate the number of boxes of grass seed required:
\(76.365 \div 15 \approx 5.091\) boxes.
Since only whole boxes can be bought, \(6\) boxes are needed.

Finally, calculate the total cost:
\(6 \times £6.40 = £38.40\).

M1: For finding the area of the rectangle (\(12 \times 8 = 96\)) or the area of the circle (\(\pi \times 2.5^2 = 6.25\pi \approx 19.635\)).
M1: For a complete method to find the remaining grass area, i.e., subtracting the area of the circle from \(96\) (giving an answer in the range \(76.3\) to \(76.4\)).
M1: For dividing their grass area by 15 and rounding up to the next integer (e.g., \(5.09 \rightarrow 6\)).
A1: \(38.40\) (or \(£38.40\), accept \(38.4\)).

First, find the missing frequency \(x\).
The total frequency is \(20\), so:
\(3 + x + 6 + 4 + 2 = 20\)
\(15 + x = 20\)
\(x = 5\).

Next, calculate the total number of books read by all students by multiplying each number of books by its frequency:
- For 0 books: \(0 \times 3 = 0\)
- For 1 book: \(1 \times 5 = 5\)
- For 2 books: \(2 \times 6 = 12\)
- For 3 books: \(3 \times 4 = 12\)
- For 4 books: \(4 \times 2 = 8\)

Total number of books = \(0 + 5 + 12 + 12 + 8 = 37\).

Finally, calculate the mean:
\(\text{Mean} = 37 \div 20 = 1.85\).

M1: For finding the missing frequency \(x = 5\).
M1: For a method to find the total number of books by calculating at least three correct products of \(\text{number of books} \times \text{frequency}\).
M1: For dividing the sum of their products by 20.
A1: \(1.85\).

At the start, the car is worth \(£16,000\).

At the end of the first year, after \(12\%\) depreciation:
\(£16,000 \times (1 - 0.12) = £16,000 \times 0.88 = £14,080\).

At the end of the second year, after \(8\%\) depreciation:
\(£14,080 \times (1 - 0.08) = £14,080 \times 0.92 = £12,953.60\).

At the end of the third year, after \(5\%\) depreciation:
\(£12,953.60 \times (1 - 0.05) = £12,953.60 \times 0.95 = £12,305.92\).

M1: For a correct method to find the value after the first year, e.g., \(16000 \times 0.88\) (giving \(14080\)) or finding \(12\%\) and subtracting.
M1: For a correct method to find the value after the second year using their first-year value, e.g., \(14080 \times 0.92\) (giving \(12953.60\)).
M1: For a correct method to find the value after the third year using their second-year value, e.g., \(12953.60 \times 0.95\).
A1: \(12305.92\) (or \(£12305.92\)).

An isosceles triangle has two equal sides, so the sides are \((3x - 2)\text{ cm}\), \((3x - 2)\text{ cm}\), and \((2x + 5)\text{ cm}\).

Write an equation for the perimeter:
\((3x - 2) + (3x - 2) + (2x + 5) = 49\)

Simplify the equation:
\(8x + 1 = 49\)

Solve for \(x\):
\(8x = 48\)
\(x = 6\).

Now, calculate the lengths of the sides:
- The two equal sides: \(3(6) - 2 = 18 - 2 = 16\text{ cm}\).
- The third side: \(2(6) + 5 = 12 + 5 = 17\text{ cm}\).

The shortest side is \(16\text{ cm}\).

M1: For setting up a correct equation for the perimeter, e.g., \(2(3x - 2) + 2x + 5 = 49\).
M1: For simplifying to \(8x + 1 = 49\) or solving to find \(x = 6\).
M1: For substituting \(x = 6\) into at least one of the side expressions (e.g., \(3(6) - 2\) or \(2(6) + 5\)).
A1: \(16\) (or \(16\text{ cm}\)).

The sum of the probabilities is \(1\), so:
\(0.35 + 0.2 + x + 2x = 1\)

Simplify the equation:
\(0.55 + 3x = 1\)
\(3x = 0.45\)
\(x = 0.15\).

Now, find the probability of landing on Yellow:
\(\text{Probability(Yellow)} = 2x = 2 \times 0.15 = 0.3\).

Estimate the number of times it lands on Yellow by multiplying the probability by the total number of spins:
\(\text{Estimate} = 150 \times 0.3 = 45\).

M1: For a correct equation showing that the sum of probabilities is 1, e.g., \(0.35 + 0.2 + x + 2x = 1\) (or \(0.55 + 3x = 1\)).
M1: For a correct method to find \(x\), e.g., \(3x = 0.45 \Rightarrow x = 0.15\).
M1: For finding the probability of Yellow, e.g., \(2 \times 0.15 = 0.3\) (or using their value of \(x\) to find \(2x\)).
A1: \(45\).

First, calculate the number of gallons needed: \( 315 \div 42 = 7.5 \) gallons. Next, convert this to litres: \( 7.5 \times 4.5 = 33.75 \) litres. Then, calculate the total cost of the petrol: \( 33.75 \times 1.40 = £47.25 \). Finally, compare the cost to her budget: \( £47.25 - £45 = £2.25 \). Since the petrol costs £47.25, she does not have enough money. She is £2.25 short.

M1: for finding the number of gallons needed: \( 315 \div 42 \) (= 7.5). M1: for converting gallons to litres: 'their 7.5' \( \times 4.5 \) (= 33.75). M1: for finding the total cost: 'their 33.75' \( \times 1.40 \) (= 47.25). A1: for £47.25 (or £2.25 short). C1: for concluding 'No' with £47.25 or £2.25 short seen.

First, find the area of the trapezium: \( \frac{1}{2} \times (4 + 7) \times 5 = 27.5 \text{ m}^2 \). Next, find the area of one paving slab. Since 50 cm = 0.5 m, the area of one slab is \( 0.5 \times 0.5 = 0.25 \text{ m}^2 \). Then, find the number of slabs needed: \( 27.5 \div 0.25 = 110 \) slabs. Now, find the number of packs needed: \( 110 \div 8 = 13.75 \) packs. Since we must buy whole packs, this rounds up to 14 packs. Finally, calculate the total cost: \( 14 \times £32 = £448 \).

M1: for finding the area of the trapezium: \( 0.5 \times (4 + 7) \times 5 \) (= 27.5). M1: for finding the area of one slab in \( \text{m}^2 \) (\( 0.5 \times 0.5 = 0.25 \)) or converting units consistently. M1: for finding the total number of slabs needed: 'their 27.5' \( \div \) 'their 0.25' (= 110). M1: for dividing 'their 110' by 8 (= 13.75) and rounding up to 14. A1: £448.

First, calculate the number of boys and girls in the class. Total ratio parts = \( 2 + 3 = 5 \). Number of boys = \( \frac{2}{5} \times 30 = 12 \). Number of girls = \( \frac{3}{5} \times 30 = 18 \). Next, find the total score for the entire class: \( 30 \times 68 = 2040 \). Then, find the total score for the boys: \( 12 \times 62 = 744 \). Subtract the boys' total score from the class total score to find the girls' total score: \( 2040 - 744 = 1296 \). Finally, find the mean score for the girls: \( 1296 \div 18 = 72 \).

M1: for finding the number of boys and girls: \( 30 \div 5 \times 2 = 12 \) boys and \( 30 \div 5 \times 3 = 18 \) girls. M1: for finding the total score for the class: \( 30 \times 68 \) (= 2040). M1: for finding the total score for the boys: 'their 12' \( \times 62 \) (= 744). M1: for finding the total score for the girls: 'their 2040' \( - \) 'their 744' (= 1296), and dividing by 'their 18'. A1: 72.