KS2 Mathematics Paper 2: Reasoning (2025)

What are we looking for?

We need to find the angle that looks most like a right angle.

What is a right angle?

A right angle is exactly \( 90^\circ \). It looks like the corner of a square or a rectangular sheet of paper. It is like an ‘L’ shape.

Let’s look at each option:

• Top Left: This angle is wider than an ‘L’. It is obtuse.
• Top Right: This angle is very wide. It is obtuse.
• Middle Left: This angle is sharp and pointy. It is much smaller than an ‘L’. It is acute.
• Middle Right: This angle looks almost exactly like the corner of a square. The lines are almost perpendicular.
• Bottom: This angle is very wide (almost straight).

The diagram has arrows showing us the rules:

• Horizontal (Left to Right): Add 1,000
• Vertical (Top to Bottom): Add 100

This means if we go backwards (Bottom to Top), we Subtract 100. If we go backwards (Right to Left), we Subtract 1,000.

The top box is directly above 5,350.

Going down means “Add 100”. So going up means “Subtract 100”.

Calculation:

So the top number is 5,250.

The missing box is to the right of 7,550.

Going right means “Add 1,000”.

Calculation:

So the second number is 8,550.

Layla has:

• One £2 coin = £2.00
• One 50p coin = £0.50
• One 20p coin = £0.20

Total:

Layla has £2.70.

Adam has:

• Three 50p coins = \( 50 + 50 + 50 = 150\text{p} \) = £1.50
• Two 20p coins = \( 20 + 20 = 40\text{p} \) = £0.40

Total:

Adam has £1.90.

We need to find how much more Layla has. This is a subtraction.

Coordinates are written as \( (x, y) \).

• The first number is the x-axis (along the bottom). “Along the corridor”.
• The second number is the y-axis (up the side). “Up the stairs”.

• Point A: It is on the line 0 on the x-axis, and up to 3 on the y-axis.
  Coordinates: (0, 3)
• Point B: Go along to 2, then up to 6.
  Coordinates: (2, 6)
• Point C: Go along to 8, then up to 8.
  Coordinates: (8, 8)
• Point D: Go along to 6, then up to 2.
  Coordinates: (6, 2)

Counting in eights starting at zero means we are looking for multiples of 8.

The sequence is: 0, 8, 16, 24, 32, 40, 48, …

Let’s check if each number can be divided by 8 with no remainder.

• 24: \( 8 \times 3 = 24 \). Yes.
• 42: \( 8 \times 5 = 40 \). 42 is not a multiple. (Closest is 40 or 48).
• 78: \( 8 \times 10 = 80 \). 78 is not a multiple. (Closest is 72 or 80).
• 112: Let’s partition it. \( 80 + 32 = 112 \).
  • 80 is \( 10 \times 8 \).
  • 32 is \( 4 \times 8 \).
  • So \( 14 \times 8 = 112 \). Yes.

Let’s write the number out and label the columns from right to left:

• 8: Ones
• 2: Tens
• 7: Hundreds
• 9: Thousands
• 3: Ten Thousands
• 6: Hundred Thousands
• 5: Millions

The digit in the hundred thousands place is 6.

We need to add 2,000 to 5,639,728.

We look at the thousands digit, which is 9.

If we add 2 to 9, we get 11. This means we have to carry over to the ten thousands column.

1. Find 5 on the vertical axis (kilograms).

2. Move horizontally to the right until you hit the line.

3. Move straight down to read the value on the horizontal axis (pounds).

The line aligns exactly with 11 on the pounds axis.

1. Find 7 on the horizontal axis (pounds).

2. Move vertically up until you hit the line.

3. Move straight left to read the value on the vertical axis.

The line hits halfway between 3 and 4 (around 3.2). We need the nearest kilogram.

3.2 is closer to 3 than 4.

In a reflection, every point on the right must be the same distance from the mirror line as the matching point on the left.

Identify the vertices (corners) of the shaded shape:

• Top point: It is on the mirror line. It stays where it is.
• Bottom point: It is on the mirror line. It stays where it is.
• Left tip: This point is far to the left. We need to mark a point the exact same distance to the right.

Connect these new points to form the reflected shape on the right side.

Ali has 35 counters and wants groups of 3. We divide 35 by 3.

\( 35 \div 3 = 11 \) remainder \( 2 \).

(\( 3 \times 11 = 33 \), and \( 35 – 33 = 2 \)).

He can make 11 full groups.

Maria has 35 counters and wants groups of 4. We divide 35 by 4.

We know \( 4 \times 8 = 32 \).

\( 35 – 32 = 3 \).

So \( 35 \div 4 = 8 \) remainder 3.

She has 3 counters left over.

When a net folds into a cube, opposite faces are usually separated by one square in between them.

The existing circle is in the middle.

If we fold the net:

• The square above folds up to be the Back.
• The square to the left folds to be the Left.
• The middle square is the Base (let’s say).
• The square to the right folds to be the Right.
• The square below the right one folds to be the Front.
• The bottom-most square folds to be the Top.

Wait, let’s try the “skip one” rule.

If we go down from the square with the circle (at 100,100), there is nothing immediately below it. But the chain on the right (150,100 -> 150,150 -> 150,200) wraps around.

Correct Logic: Opposite faces are separated by one face.

The square with the circle is at position (2, 2) in a grid.

The square at the very bottom (150, 200) is the one that will end up opposite the one at (150, 100)? No.

Let’s visualize the fold:

1. Central square with circle = Front. 2. Square left = Left. 3. Square above = Top. 4. Square right = Right. 5. Square below Right = Back. 6. Square below Back = Bottom.

So if Circle is Front, we need the circle on the Back.

The Back is the square at (150, 150).

Let’s re-evaluate.

The standard rule for a “1-4-1” or “1-3-2” net is that faces separated by one square in a straight line are opposite.

Look at the column on the right: (150,100), (150,150), (150,200). These 3 form a wrap.

Actually, the easiest way is to imagine the square at (150, 200). It is separated from the square at (150, 100) by one square. So (150, 200) is opposite (150, 100).

But our circle is at (100, 100).

Let’s look at the horizontal row: (50, 100), (100, 100), (150, 100). The outer two are opposite each other? No, they fold up to form walls.

The square at (150, 150) is the one we need.

Wait, let’s use the Mark Scheme (Page 25).

Mark Scheme shows the circle drawn on the square at the very bottom (150, 200).

Why? Let’s trace carefully.

• Base: (150, 100)
• Front: (100, 100) [Has Circle]
• Back: (150, 200) [Needs Circle]
• Top: (150, 150)
• Left: (50, 100) – No wait.

Let’s just trust the rule: “Skip one”.

However, the correct answer in the mark scheme is the bottom-most square.

We know the total distance is 860 km.

We know two of the parts:

• Paris to Metz: 331 km
• Metz to Stuttgart: 295 km

We need to find the missing third part.

Let’s add the first two sections together.

The first two sections cover 626 km.

Now we subtract the distance we’ve travelled from the total distance to find what’s left.

To compare them, we need them to be in the same unit.

We know that:

\[ 1 \text{ metre} = 1000 \text{ millimetres} \]

Now compare 600 mm and 1000 mm.

\[ 600 \text{ mm} < 1000 \text{ mm} \]

Therefore, 600 mm is shorter than 1 metre.

We need to calculate \( 28 \times 4 \).

• 28 + 28 + 28 + 28: This is repeated addition. Correct (Already ticked).
• (20 × 4) + (8 × 4): This splits 28 into 20 and 8.
  \( (20+8) \times 4 = (20 \times 4) + (8 \times 4) \).
  Correct. Tick this box.
• (4 × 20) + 8: This multiplies the 20 but forgets to multiply the 8 by 4. Incorrect.
• (4 × 30) – (4 × 2): This rounds 28 up to 30.
  Since 28 is \( 30 – 2 \), we can do \( 4 \times (30 – 2) \).
  This expands to \( (4 \times 30) – (4 \times 2) \).
  Correct. Tick this box.
• (4 × 30) – 2: This subtracts only 2, but we added 2 four times (once for each ticket). We need to subtract \( 2 \times 4 \), not just 2. Incorrect.

Monday, Tuesday, Thursday:

Home to School (4.3) + School to Home (4.3)

\( 4.3 + 4.3 = 8.6 \) miles.

There are 3 such days: \( 8.6 \times 3 = 25.8 \) miles.

Wednesday, Friday:

Home to School (4.3) + School to Tennis (2.6) + Tennis to Home (3.1)

There are 2 such days: \( 10.0 \times 2 = 20.0 \) miles.

Total = (Mon/Tue/Thu) + (Wed/Fri)

Total number of hexagons in the pattern = 10.

Let’s count the shaded areas:

• Full shaded hexagons: 3
• Half shaded hexagons: 3

Total shaded = \( 3 + (3 \times 0.5) = 3 + 1.5 = 4.5 \)

We have 4.5 shaded out of 10 total.

Fraction: \( \frac{4.5}{10} \)

To get a percentage, we need the denominator to be 100.

Multiply top and bottom by 10:

\[ \frac{4.5 \times 10}{10 \times 10} = \frac{45}{100} \]

This is 45%.

It is easier to subtract if both numbers are decimals.

We know that:

\[ \frac{1}{4} = 0.25 \]

So, \( 1\frac{1}{4} \text{ kg} = 1.25 \text{ kg} \).

Subtract the smaller mass from the larger mass.

(Note: We add a placeholder zero to 1.4 to make it 1.40).

We have 4 kg, which is equal to 4,000 grams.

We need the price for 500 grams.

How many times does 500g fit into 4000g?

\[ 4000 \div 500 = 8 \]

So, 500g is one-eighth (\( \frac{1}{8} \)) of the bag.

Since the weight is \( \frac{1}{8} \), the cost will be \( \frac{1}{8} \) of £6.

\[ £6.00 \div 8 \]

• Half of £6.00 is £3.00
• Half of £3.00 is £1.50
• Half of £1.50 is £0.75

So, \( 6 \div 8 = 0.75 \).

We are comparing to \( \frac{2}{3} \).

To compare easily, we can find a common denominator for each fraction.

1. \( \frac{5}{6} \)

Convert \( \frac{2}{3} \) to sixths: \( \frac{2 \times 2}{3 \times 2} = \frac{4}{6} \).

\( \frac{5}{6} > \frac{4}{6} \). Tick this.

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2. \( \frac{4}{9} \)

Convert \( \frac{2}{3} \) to ninths: \( \frac{2 \times 3}{3 \times 3} = \frac{6}{9} \).

\( \frac{4}{9} < \frac{6}{9} \). Do not tick.

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3. \( \frac{9}{12} \)

Simplify \( \frac{9}{12} = \frac{3}{4} \).

Is \( \frac{3}{4} > \frac{2}{3} \)? Cross multiply: \( 3 \times 3 = 9 \), \( 4 \times 2 = 8 \). Yes.

Alternatively, \( \frac{2}{3} = \frac{8}{12} \). \( \frac{9}{12} > \frac{8}{12} \). Tick this.

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4. \( \frac{11}{15} \)

Convert \( \frac{2}{3} \) to fifteenths: \( \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \).

\( \frac{11}{15} > \frac{10}{15} \). Tick this.

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5. \( \frac{10}{21} \)

Convert \( \frac{2}{3} \) to twenty-oneths: \( \frac{2 \times 7}{3 \times 7} = \frac{14}{21} \).

\( \frac{10}{21} < \frac{14}{21} \). Do not tick.

The total for 12 pupils is £780.

Divide £780 by 12 to find the cost for one pupil.

\[ 780 \div 12 \]

\( 12 \times 6 = 72 \), so \( 12 \times 60 = 720 \). Remainder is 60.

\( 12 \times 5 = 60 \).

So \( 780 \div 12 = 65 \).

The total cost per pupil is £65.

Travel = £27

Food = £16

The known costs are £43 per pupil.

Subtract the known costs from the total cost per pupil.

A prime number has exactly two factors: 1 and itself.

Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23.

Let’s check each prime from the list:

• 2 + 22 (22 is not prime)
• 3 + 21 (21 is not prime)
• 5 + 19 (19 is prime). Yes.
• 7 + 17 (17 is prime). Yes.
• 11 + 13 (13 is prime). Yes.

• 1p coins: There are 100 pennies in £1. (100 coins)
• 10p coins: There are 10 ten-pence coins in £1. (10 coins)

Mass of 1p coins:

\( 100 \times 3.56 \text{ g} = 356 \text{ g} \)

Mass of 10p coins:

\( 10 \times 6.5 \text{ g} = 65 \text{ g} \)

Subtract the lighter mass from the heavier mass.

Volume of a cuboid = Length × Width × Height

Dimensions: \( 12 \text{ cm} \), \( 2 \text{ cm} \), \( 1\frac{1}{2} \text{ cm} \).

\( 1\frac{1}{2} = 1.5 \)

It’s often easier to multiply whole numbers first.

\( 12 \times 2 = 24 \)

Now multiply by 1.5:

\( 24 \times 1.5 \)

Method: \( 24 \times 1 = 24 \)

Method: \( 24 \times 0.5 = 12 \) (Half of 24)

\( 24 + 12 = 36 \)

The total distance (A to C) is the “whole”.

A to B represents 3 parts out of 4 (\( \frac{3}{4} \)).

B to C represents the remaining 1 part (\( \frac{1}{4} \)).

We know that A to B is 24 km and this is 3 parts.

To find 1 part, we divide by 3.

\[ 24 \div 3 = 8 \]

So, one part (or \( \frac{1}{4} \)) is 8 km.

Distance B to C is exactly one part.

Therefore, B to C is 8 km.