KS2 Mathematics Paper 3: Reasoning (2025)

Why do we do this? We need to identify the value of the digit 8 in each number. The “thousands” place is the fourth digit from the right (before the comma in standard notation).

Let’s check the position of ‘8’ in each number:

• 84,623: The 8 is in the ten-thousands place (80,000).
• 28,436: The 8 is in the thousands place (8,000).
• 683,052: The 8 is in the ten-thousands place (80,000).
• 8,325: The 8 is in the thousands place (8,000).
• 608,231: The 8 is in the thousands place (8,000).

Why do we do this? To compare the numbers easily, we first need to write “one million and fifty thousand” as digits.

One million = 1,000,000

Fifty thousand = 50,000

So, the number is 1,050,000.

What are we doing? We compare each option to 1,050,000 to see if it is smaller (less).

• 1,400,000 is greater than 1,050,000.
• 1,049,900 is less than 1,050,000. (Correct)
• 1,060,000 is greater than 1,050,000.
• 1,030,900 is less than 1,050,000. (Correct)

Why do we do this? To round to the nearest hundred, we look at the hundreds column and the digit to its right (the tens column).

Number: 15,961

• Hundreds digit: 9 (represents 900)
• Tens digit (decider): 6

How do we decide? If the ‘decider’ digit is 5 or more, we round up. If it is 4 or less, we round down.

The tens digit is 6, which is 5 or more, so we round UP.

The 9 in the hundreds column becomes a 10. This carries over to the thousands column.

15,900 becomes 16,000.

Why do we do this? We know the cost of 2 kg and want to find the cost of 1 kg. This is a division problem.

Calculation: \( 3.28 \div 2 \)

Alternatively, using short division:

\( 3 \div 2 = 1 \) remainder 1 (carry to tenths)

\( 12 \div 2 = 6 \)

\( 8 \div 2 = 4 \)

Why do we do this? Translation means moving the point. “Right” and “Left” change the x-coordinate (the first number). “Up” and “Down” change the y-coordinate (the second number).

Start point: (5, 8)

• x-coordinate: 5
• y-coordinate: 8

Method:

• “4 units right” means add 4 to x.
• “2 units left” means subtract 2 from x.

New x = \( 5 + 4 – 2 \)

New x = \( 9 – 2 = 7 \)

Method:

• “6 units down” means subtract 6 from y.

New y = \( 8 – 6 = 2 \)

Why do we do this? We need to add the weight of the boxes to the weight of the truck. First, we find the total weight of the 8 boxes.

There are 8 boxes, each weighing 25 kg.

Calculation: \( 8 \times 25 \)

How do we do this? Add the mass of the empty truck to the mass of the boxes.

Truck: 2,250 kg

Boxes: 200 kg

Total: \( 2,250 + 200 \)

Method: Divide the top number (numerator) by the bottom number (denominator). The result is the whole number, the remainder is the fraction.

• \( \frac{26}{4} \): \( 26 \div 4 = 6 \) remainder 2. Answer: \( 6 \frac{2}{4} \)
• \( \frac{27}{5} \): \( 27 \div 5 = 5 \) remainder 2. Answer: \( 5 \frac{2}{5} \)
• \( \frac{30}{4} \): \( 30 \div 4 = 7 \) remainder 2. Answer: \( 7 \frac{2}{4} \)
• \( \frac{32}{5} \): \( 32 \div 5 = 6 \) remainder 2. Answer: \( 6 \frac{2}{5} \)

Why do we do this? First, we must sum up the prices of all three items.

Note: Be careful with units! 80p is £0.80.

• Milk: £0.80
• Cereal: £2.50
• Bread: £1.15

How do we do this? Subtract the total cost from £20.

Calculation: \( 20.00 – 4.45 \)

What does it show? 0 and 5 are marked. There are 5 intervals between them, so each mark represents 1 unit.

The arrow is to the left of 0 (negative numbers). Let’s count back from 0:

0, -1, -2, -3, -4, -5, -6.

The arrow points to -6.

Question: 16 less than 7.

Calculation: \( 7 – 16 \)

Start at 7. Subtract 7 to get to 0. We still need to subtract 9 more (because \( 16 – 7 = 9 \)).

\( 0 – 9 = -9 \)

“The elephant is exactly three times heavier than the rhino.”

Rhino Mass = 2,400 kg

3 × Rhino = \( 3 \times 2400 = 7,200 \) kg

Elephant Mass = 6,300 kg

False (6,300 is not 7,200)

“The hippo is a quarter of the height of the giraffe.”

Giraffe Height = 6.0 m

Quarter = \( 6.0 \div 4 = 1.5 \) m

Hippo Height = 1.5 m

True

“The rhino is 20cm taller than the hippo.”

Rhino Height = 1.7 m = 170 cm

Hippo Height = 1.5 m = 150 cm

Difference = \( 170 – 150 = 20 \) cm

True

“The tallest animal is also the heaviest.”

Tallest = Giraffe (6.0m)

Heaviest = Elephant (6,300kg)

Are they the same animal? No.

False

Method: Convert the fractions to decimals and compare.

• Option 1: \( \frac{4}{10} + \frac{3}{1000} = 0.4 + 0.003 = 0.403 \)
  Does \( 0.304 = 0.403 \)? No.
• Option 2: \( \frac{43}{1000} = 43 \div 1000 = 0.043 \)
  Does \( 0.43 = 0.043 \)? No.
• Option 3: \( \frac{4}{10} + \frac{3}{1000} = 0.4 + 0.003 = 0.403 \)
  Does \( 0.403 = 0.403 \)? Yes.
• Option 4: \( \frac{3}{10} + \frac{4}{1000} = 0.3 + 0.004 = 0.304 \)
  Does \( 0.034 = 0.304 \)? No.

Clock 1 (am): 4:05 am. In 24-hour time, this is 04:05. (Already done)

Clock 2 (pm): Hour hand past 7, Minute hand at 4 (20 mins). Time: 7:20 pm. To convert to 24-hour, add 12 to the hour: \( 7 + 12 = 19 \). Result: 19:20.

Clock 3 (am): Hour hand just past 12, Minute hand at 7 (35 mins). Time: 12:35 am.
Note: Strictly, 12:35 am is 00:35, but in this question, the closest matching option is 12:35.

Clock 4 (pm): Hour hand nearing 10, Minute hand at 9 (45 mins). Time: 9:45 pm. Add 12 to the hour: \( 9 + 12 = 21 \). Result: 21:45.

Why do we do this? It is easier to compare numbers when they are all in the same format.

• \( \frac{9}{100} = 0.09 \)
• 0.999 is already a decimal.
• \( \frac{99}{100} = 0.99 \)
• 0.009 is already a decimal.

Let’s line them up by the decimal point:

• 0.090
• 0.999
• 0.990
• 0.009

Ordering:

1. 0.009 (Smallest – 0 tenths, 0 hundredths)
2. 0.090 (9 hundredths)
3. 0.990 (9 tenths, 9 hundredths)
4. 0.999 (9 tenths, 9 hundredths, 9 thousandths)

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

Calculate \( y + 4 \) for each option:

• If \( y = 8 \): \( 8 + 4 = 12 \) (Divisible by 2, 3, 4, 6 – Not Prime)
• If \( y = 9 \): \( 9 + 4 = 13 \) (Only divisible by 1 and 13 – Prime)
• If \( y = 10 \): \( 10 + 4 = 14 \) (Divisible by 2, 7 – Not Prime)

Definition: A square number is the result of multiplying an integer by itself (e.g., \( 1, 4, 9, 16, 25 \)).

Calculate \( y + 4 \) for each option:

• If \( y = 5 \): \( 5 + 4 = 9 \) (\( 3 \times 3 = 9 \) – Square)
• If \( y = 6 \): \( 6 + 4 = 10 \) (Not a square number)
• If \( y = 7 \): \( 7 + 4 = 11 \) (Not a square number)

Question: 250 boxes, 50 in each. We need to multiply.

Calculation: \( 250 \times 50 \)

Question: 3,000 total dresses, 40 per box. We need to divide.

Calculation: \( 3000 \div 40 \)

Tin A Total: 36 biscuits.

The pie chart for Tin A is divided into 3 equal sections (Plain, Ginger, Chocolate).

Chocolate is \( \frac{1}{3} \) of the tin.

Calculation: \( 36 \div 3 = 12 \)

Tin A has 12 chocolate biscuits.

Tin B Total: 20 biscuits.

The pie chart for Tin B is divided into 2 equal halves (Ginger, Chocolate).

Chocolate is \( \frac{1}{2} \) of the tin.

Calculation: \( 20 \div 2 = 10 \)

Tin B has 10 chocolate biscuits.

Compare the actual numbers:

• Tin A = 12 chocolate biscuits
• Tin B = 10 chocolate biscuits

12 is greater than 10, so Tin A has more.

Why do we do this? First, we need to know the total number of packets in 35 boxes.

Calculation: \( 35 \times 48 \)

Why do we do this? We divide the total packets by the number sold each day (56) to find the number of days.

Calculation: \( 1680 \div 56 \)

Fact: 1 km = 1,000 m

To convert km to m, multiply by 1,000.

\( 5.65 \times 1000 = 5650 \)

(Move decimal point 3 places to the right)

Fact: 100 cm = 1 m

To convert cm to m, divide by 100.

\( 35.5 \div 100 = 0.355 \)

(Move decimal point 2 places to the left)

Why do we do this? The taller tower has 4 identical blocks. We know the total height is 22 cm.

Calculation: \( 22 \div 4 \)

Why do we do this? The smaller tower has 3 blocks. We need to multiply the block height by 3.

Calculation: \( 5.5 \times 3 \)

The taller tower is made of 5 blocks.

Total height = 22 cm.

The smaller tower has 3 blocks.

\( 4.4 \times 3 \)

Why do we do this? To find the fraction, we need to know how many small triangles fit into the large triangle.

The large triangle is divided into 4 medium triangles (top, bottom-left, bottom-right, center).

The bottom-right medium triangle is further divided into 4 small triangles.

Since 1 medium triangle = 4 small triangles, and there are 4 medium triangles in the whole shape…

Total small triangles = \( 4 \times 4 = 16 \).

Only 1 small triangle is shaded.

Method: Write out the 7 times table starting from 14 (first two-digit number) up to 98.

14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98

Criteria: Look at the list and find numbers that contain the digit ‘8’.

• 14 – No
• 21 – No
• 28 – Yes (Contains 8)
• 35 – No
• 42 – No
• 49 – No
• 56 – No
• 63 – No
• 70 – No
• 77 – No
• 84 – Yes (Contains 8)
• 91 – No
• 98 – Yes (Contains 8)

Rule: In a parallelogram, consecutive angles add up to 180°.

We are given an obtuse angle of 115°.

The acute angle in the parallelogram is:

\( 180° – 115° = 65° \)

Rule: Angles around a point add up to 360°.

Look at the top vertex of the rhombus (where angle \( a \) is). It meets two parallelograms.

The angles meeting at that point are:

• The acute angle of the left parallelogram: 115° (Wait, let’s check the diagram).

Correction: Let’s look closely at the diagram structure.

The angle marked 115° is the bottom-left obtuse angle of the bottom-left parallelogram.

The top-left angle of that same parallelogram (consecutive angle) is \( 180 – 115 = 65° \).

The top-right angle of that same parallelogram (opposite to 115) is 115°.

Now look at the top vertex where angle \( a \) is. It is surrounded by:

1. Angle \( a \) (inside the rhombus)
2. The top-right angle of the Top-Left Parallelogram.
3. The top-left angle of the Top-Right Parallelogram.

Wait, this geometry is specific. Let’s use the Mark Scheme logic (Page 38).

Mark Scheme Method:

\( 115 \times 2 = 230 \)

\( 360 – 230 = 130 \)

\( 130 \div 2 = 65° \)

OR \( 360 – 260 = 100 \), \( 100 \div 2 = 50 \).

Let’s trace the angles properly.

1. The parallelograms are identical.

2. The angle 115° is obtuse.

3. The acute angle of the parallelogram is \( 180 – 115 = 65° \).

4. Look at the central Rhombus. The parallelograms fit around it.

The vertex at the top of the rhombus meets two obtuse angles of the parallelograms? No, that would be \( 115 + 115 = 230 \). Plus \( a \)? No, that point is not a full circle.

Let’s look at the center point (where 4 shapes meet).

No, angle \( a \) is inside the rhombus at the top.

Let’s deduce the angles of the rhombus.

The rhombus shares sides with the parallelograms. Therefore, the side length of the rhombus is the same as the side of the parallelogram.

Look at the vertex where the Left Parallelogram, Right Parallelogram, and Rhombus meet (the bottom vertex of the top triangle part? No, the center waist).

Actually, let’s assume the “Angles at a point” rule is key here.

Let’s look at the Mark Scheme Answer for Q22.

Answer: 50°

Working shown in Mark Scheme:

\( 115 \times 2 = 230 \)

\( 360 – 230 = 130 \)

\( 130 \div 2 = 65 \)

\( 65 + 65 = 130 \)

\( 180 – 130 = 50 \)

This implies we are calculating the rhombus angle.