2024 Key Stage 2 Mathematics – Paper 2: Reasoning

💡 What is reflection?

Reflection acts like a mirror. Every point on the reflected shape must be the same distance from the mirror line as the original point, but on the opposite side.

Let’s look at the three corners (vertices) of the triangle:

• Bottom-Left Corner: It is 2 squares to the right of the mirror line. The reflection must be 2 squares to the left.
• Bottom-Right Corner: It is 5 squares to the right of the mirror line. The reflection must be 5 squares to the left.
• Top Corner: It is 3.5 squares to the right. The reflection must be 3.5 squares to the left.

💡 Why? We know how much Amir paid and how much change he got back. The difference tells us how much he actually spent.

Calculation:

Amir pays: £2.00

Change: £1.50

Cost = \( £2.00 – £1.50 = 50\text{p} \)

We need to find two prices from the table that equal 50p.

Let’s check the combinations:

• banana (12p) + pear (38p) = \( 12 + 38 = 50\text{p} \) ✅
• plum (23p) + apple (32p) = \( 23 + 32 = 55\text{p} \) ❌
• banana (12p) + apple (32p) = \( 12 + 32 = 44\text{p} \) ❌

Let’s see how many points Layla scored in the first three games by reading the height of each bar:

• Game 1: 7 points
• Game 2: 5 points
• Game 3: 10 points

First, find the total for the first 3 games:

\( 7 + 5 + 10 = 22 \)

We know the total after 4 games is 25 points.

So, points for Game 4 = \( 25 – 22 = 3 \)

We need to draw a bar for Game 4 that goes up to 3 on the vertical axis.

3 is halfway between 2 and 4.

💡 Strategy: Look for two numbers that are next to each other to find the jump size.

We have 1 and 7.

Difference = \( 7 – 1 = 6 \)

So, the sequence increases by 6 each time.

First missing number (between -11 and 1):

Check: \( -11 + 6 = -5 \)

Check again: \( -5 + 6 = 1 \) (Matches the next number ✅)

Second missing number (between 7 and 19):

\( 7 + 6 = 13 \)

Check again: \( 13 + 6 = 19 \) (Matches the next number ✅)

We have \( \text{something} \times 3 \) ending in 2.

Let’s think of the 3 times table:

• \( 1 \times 3 = 3 \)
• \( 2 \times 3 = 6 \)
• \( 3 \times 3 = 9 \)
• \( 4 \times 3 = 12 \) (Ends in 2! This works.)

So the top right box is 4. We carry the 1.

Now we multiply the middle digit (5) by 3:

\( 5 \times 3 = 15 \)

Add the carry (1): \( 15 + 1 = 16 \)

The digit 6 goes in the answer box. We carry the 1.

We need \( \text{something} \times 3 + \text{carry (1)} = 7 \).

So, \( \text{something} \times 3 = 6 \).

\( 2 \times 3 = 6 \).

The top left box is 2.

Condition 1 & 2: Greater than 236 and less than 245.

The numbers are: 237, 238, 239, 240, 241, 242, 243, 244.

Condition 3: Has a 3 in the tens’ place.

Let’s look at our list:

• 237 ✅
• 238 ✅
• 239 ✅
• 240 ❌ (4 in tens place)
• 241 ❌
• 242 ❌
• 243 ❌
• 244 ❌

We are left with: 237, 238, 239.

Condition 4: Is an even number.

Even numbers end in 0, 2, 4, 6, or 8.

• 237 (Odd) ❌
• 238 (Even) ✅
• 239 (Odd) ❌

We have 40 packets, and each one has 25 envelopes.

To find the total, we multiply: \( 40 \times 25 \).

Method A: Split the number

\( 40 \times 25 \) is the same as \( 4 \times 10 \times 25 \).

First, do \( 4 \times 25 = 100 \).

Then, multiply by 10: \( 100 \times 10 = 1,000 \).

Method B: Standard Multiplication

We need a number that rounds to 4,000.

When rounding to the nearest 1000, we look at the hundreds digit.

• Smallest possible number: 3,500 (500 rounds up).
• Largest possible number: 4,499 (499 rounds down).

Any whole number between 3,500 and 4,499 is correct.

Example: 4,100

We need a number that rounds to 820,000.

When rounding to the nearest 10,000, we look at the thousands digit.

• Smallest possible number: 815,000 (5000 rounds up).
• Largest possible number: 824,999 (4999 rounds down).

Any whole number between 815,000 and 824,999 is correct.

Example: 821,000

Let’s solve each one:

• \( 4 \div 10 = \mathbf{0.4} \)
• \( 40 \div 10 = \mathbf{4} \)
• \( 4 \div 100 = \mathbf{0.04} \)
• \( 40 \div 100 = \mathbf{0.4} \)

We can see that \( 4 \div 10 \) and \( 40 \div 100 \) both equal 0.4.

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

• 15: Factors are 1, 3, 5, 15 (Not prime)
• 17: Prime ✅
• 19: Prime ✅
• 21: Factors are 1, 3, 7, 21 (Not prime)
• 23: Prime ✅
• 25: Factors are 1, 5, 25 (Not prime)

The prime numbers in the list are 17, 19, and 23.

Let’s check the differences between our primes:

• \( 19 – 17 = 2 \) ✅
• \( 23 – 19 = 4 \) ❌

So the pair is 17 and 19.

The column headers tell us the relationship:

\[ \text{Number of children} \div \text{Number of adults} = \text{Number of children per adult} \]

Let’s check Row 1 to be sure: \( 12 \div 4 = 3 \). Correct.

We have: 20 children, unknown adults, 4 children per adult.

\[ 20 \div ? = 4 \]

What number goes into 20 four times? \( 20 \div 4 = 5 \).

Number of adults = 5

We have: unknown children, 3 adults, 8 children per adult.

\[ ? \div 3 = 8 \]

To find the missing number, we multiply: \( 3 \times 8 = 24 \).

Number of children = 24

It is a 10 by 10 square.

Total squares = \( 10 \times 10 = 100 \).

To find the fraction for each shape, we count its squares and put it over 100.

Shape Y (Smallest): Looks like just 1 square.

Fraction: \( \frac{1}{100} \).

Shape W (Top Left): It is 5 squares wide and 5 squares high.

Area = \( 5 \times 5 = 25 \) squares.

Fraction: \( \frac{25}{100} \). We can simplify this by dividing top and bottom by 25: \( \frac{1}{4} \).

Shape Z (Bottom Left): It is 5 squares wide and 2 squares high.

Area = \( 5 \times 2 = 10 \) squares.

Fraction: \( \frac{10}{100} = \frac{1}{10} \).

Shape X (Right Side): It is 5 squares wide and 8 squares high (rows 1 to 8).

Area = \( 5 \times 8 = 40 \) squares.

Fraction: \( \frac{40}{100} \). Simplify by dividing by 20: \( \frac{2}{5} \).

• Cube: A box shape. 4 corners on top, 4 on bottom. Total = 8.
• Square-based pyramid: 4 corners on the square base, plus 1 point at the top. Total = 5.
• Triangular-based prism: A triangle at the front (3 corners) and a triangle at the back (3 corners). Total = 6.
• Octagonal-based pyramid: An octagon base has 8 corners, plus 1 point at the top. Total = 9.

The whole class represents 1 whole. We are dealing with quarters, so let’s split the class into 4 equal parts.

Sam got 3 parts (three-quarters).

Alex got the rest.

From the diagram, we can see that Alex represents 1 part out of 4.

So, 1 quarter = 7 pupils.

Since the parts are equal, every quarter contains 7 pupils.

Total pupils = \( 7 \times 4 = 28 \).

First, let’s work out the left side of the equals sign:

\[ 3.207 \times 100 \]

To multiply by 100, move the decimal point 2 places to the right:

\( 3.207 \rightarrow 32.07 \rightarrow 320.7 \)

So, the calculation is: \( 320.7 = [ ] \times 10 \)

We need to find a number that, when multiplied by 10, equals 320.7.

\[ [ ] \times 10 = 320.7 \]

To find the missing number, we do the opposite (inverse) of multiplying by 10. We divide by 10.

\[ 320.7 \div 10 \]

To divide by 10, move the decimal point 1 place to the left:

\( 320.7 \rightarrow 32.07 \)

Let’s write the number 9,658,214 into a place value chart:

• 9 : Millions (\( 9,000,000 \))
• 6 : Hundred Thousands (\( 600,000 \))
• 5 : Ten Thousands (\( 50,000 \))
• 8 : Thousands (\( 8,000 \))
• 2 : Hundreds (\( 200 \))
• 1 : Tens (\( 10 \))
• 4 : Ones (\( 4 \))

1. The digit 5 represents 50,000:
   Yes, 5 is in the ten-thousands place. \( 5 \times 10,000 = 50,000 \). ✅ TRUE
2. The value of the digit 9 is nine hundred thousands:
   No, 9 is in the millions place. It is 9 million. ❌
3. The digit 6 represents 6 millions:
   No, 6 is in the hundred-thousands place. It represents 600,000. ❌
4. The value of the digit 2 is twenty tens:
   The digit 2 is in the hundreds place, so it is 200.
   “Twenty tens” means \( 20 \times 10 = 200 \).
   Since 200 = 200, this is correct. ✅ TRUE

Chen paid £10 and got £3.85 back.

So, total cost = \( £10.00 – £3.85 \).

The total spent was £6.15.

Apples: £1.50

Bread: £0.70

Cereal: £1.45

The known items cost £3.65.

Butter = Total Cost – Cost of known items

Butter = \( £6.15 – £3.65 \)

The scale is in kilograms, but the question asks for 1350 grams.

We know \( 1000 \text{ grams} = 1 \text{ kg} \).

So, \( 1350 \div 1000 = 1.35 \text{ kg} \).

There are 5 gaps between 1 and 1.5.

Distance = \( 1.5 – 1 = 0.5 \).

Each gap = \( 0.5 \div 5 = 0.1 \).

The lines represent: 1.1, 1.2, 1.3, 1.4.

We need to show 1.35.

This is exactly halfway between 1.3 (the 3rd mark) and 1.4 (the 4th mark).

We know the total capacity is 1,250.

We know there are 40 empty seats at 8 pm.

So, Filled seats = Total – Empty

\( 1250 – 40 = 1210 \)

We want to know how many more seats were filled between 7 pm and 8 pm.

Seats at 8 pm: 1210

Seats at 7 pm: 880

Difference = \( 1210 – 880 \)

Morning Break: 10:15 to 10:30 is 15 minutes.

Lunchtime: 12:40 to 1:30.

• 12:40 to 1:00 = 20 minutes
• 1:00 to 1:30 = 30 minutes
• Total lunch = 50 minutes

Total per day = \( 15 + 50 = 65 \) minutes.

We need the total for a 5-day week.

\( 65 \text{ minutes} \times 5 \)

\( 60 \times 5 = 300 \)

\( 5 \times 5 = 25 \)

Total = 325 minutes.

There are 60 minutes in an hour.

\( 325 \div 60 \)

\( 60 \times 5 = 300 \) (5 hours)

Remainder = \( 325 – 300 = 25 \) minutes.

• Parallel: Lines that never meet (like train tracks).
• Perpendicular: Lines that meet at a right angle (90°).

• AB is parallel to CD: AB is horizontal, CD is vertical. They meet at 90°. They are perpendicular, not parallel. ❌
• GH is parallel to AB: Both lines are horizontal and run in the same direction without meeting. ✅
• CD is perpendicular to GH: CD is vertical, GH is horizontal. The square symbol at D confirms they meet at 90°. ✅
• EF is perpendicular to CD: Both lines are vertical. They run in the same direction. They are parallel, not perpendicular. ❌

To find the mean, we:

1. Add up all the values (Total).
2. Divide by the number of values (Count).

\[ 1.8 + 2.4 + 3.2 + 1.6 + 4.5 \]

Let’s pair them up to make it easier:

• \( 2.4 + 1.6 = 4.0 \)
• \( 1.8 + 3.2 = 5.0 \)

So far: \( 4.0 + 5.0 = 9.0 \)

Now add Dev’s distance: \( 9.0 + 4.5 = \mathbf{13.5} \)

There are 5 friends, so we divide by 5.

\[ 13.5 \div 5 \]

We need to find out how many full trays we have and what is left over.

\[ 940 \div 12 \]

She fills 78 trays, and there are 4 seeds remaining.

The last tray has 4 seeds in it.

A full tray holds 12 seeds.

The fraction is \( \frac{4}{12} \).

We can simplify this by dividing top and bottom by 4:

\[ \frac{4 \div 4}{12 \div 4} = \frac{1}{3} \]

Let’s look at the properties of the available numbers (40, 60, 64, 100):

• Square numbers: \( 8 \times 8 = 64 \), \( 10 \times 10 = 100 \). (Could be 64 or 100)
• Cube number: \( 4 \times 4 \times 4 = 64 \). (Must be 64)

Since 64 must be the cube number, then 100 must be the square number.

We have used 64 and 100. We have 40 and 60 left.

Statement: “…is a common multiple of 4 and 5”

Multiples of 4: 4, 8… 20, 40, 60…

Multiples of 5: 5, 10… 20, 40, 60…

Both 40 and 60 fit this description.

Statement: “…is a common factor of 80 and 120”

Factors divide exactly into the number.

Does 60 go into 80? No. (80 ÷ 60 is not a whole number).

Does 40 go into 80? Yes (\( 80 \div 40 = 2 \)).

Does 40 go into 120? Yes (\( 120 \div 40 = 3 \)).

So, 40 must be the common factor.

• Cube: 64
• Square: 100 (since 64 is taken)
• Common Factor: 40
• Common Multiple: 60 (the only one left)

The formula is \( 3 \times b – a = 2 \).

Substitute \( b = 2 \):

\[ 3 \times 2 – a = 2 \]

\[ 6 – a = 2 \]

What number subtracted from 6 gives 2?

\( 6 – 2 = 4 \).

So, a = 4.

The formula is \( 3 \times b – a = 2 \).

Substitute \( a = 13 \):

\[ 3 \times b – 13 = 2 \]

We need to find \( 3 \times b \). Let’s work backwards by adding 13 to the answer.

\[ 3 \times b = 2 + 13 \]

\[ 3 \times b = 15 \]

What number times 3 equals 15?

\( 15 \div 3 = 5 \).

So, b = 5.

• Parallel: Lines that never meet (like train tracks).
• Perpendicular: Lines that meet at a right angle (90°).

• AB is parallel to CD: AB is horizontal, CD is vertical. They meet at 90°. They are perpendicular, not parallel. ❌
• GH is parallel to AB: Both lines are horizontal and run in the same direction without meeting. ✅
• CD is perpendicular to GH: CD is vertical, GH is horizontal. The square symbol at D confirms they meet at 90°. ✅
• EF is perpendicular to CD: Both lines are vertical. They run in the same direction. They are parallel, not perpendicular. ❌

To find the mean, we:

1. Add up all the values (Total).
2. Divide by the number of values (Count).

\[ 1.8 + 2.4 + 3.2 + 1.6 + 4.5 \]

Let’s pair them up to make it easier:

• \( 2.4 + 1.6 = 4.0 \)
• \( 1.8 + 3.2 = 5.0 \)

So far: \( 4.0 + 5.0 = 9.0 \)

Now add Dev’s distance: \( 9.0 + 4.5 = \mathbf{13.5} \)

There are 5 friends, so we divide by 5.

\[ 13.5 \div 5 \]

We need to find out how many full trays we have and what is left over.

\[ 940 \div 12 \]

She fills 78 trays, and there are 4 seeds remaining.

The last tray has 4 seeds in it.

A full tray holds 12 seeds.

The fraction is \( \frac{4}{12} \).

We can simplify this by dividing top and bottom by 4:

\[ \frac{4 \div 4}{12 \div 4} = \frac{1}{3} \]

Let’s look at the properties of the available numbers (40, 60, 64, 100):

• Square numbers: \( 8 \times 8 = 64 \), \( 10 \times 10 = 100 \). (Could be 64 or 100)
• Cube number: \( 4 \times 4 \times 4 = 64 \). (Must be 64)

Since 64 must be the cube number, then 100 must be the square number.

We have used 64 and 100. We have 40 and 60 left.

Statement: “…is a common multiple of 4 and 5”

Multiples of 4: 4, 8… 20, 40, 60…

Multiples of 5: 5, 10… 20, 40, 60…

Both 40 and 60 fit this description.

Statement: “…is a common factor of 80 and 120”

Factors divide exactly into the number.

Does 60 go into 80? No. (80 ÷ 60 is not a whole number).

Does 40 go into 80? Yes (\( 80 \div 40 = 2 \)).

Does 40 go into 120? Yes (\( 120 \div 40 = 3 \)).

So, 40 must be the common factor.

• Cube: 64
• Square: 100 (since 64 is taken)
• Common Factor: 40
• Common Multiple: 60 (the only one left)

The formula is \( 3 \times b – a = 2 \).

Substitute \( b = 2 \):

\[ 3 \times 2 – a = 2 \]

\[ 6 – a = 2 \]

What number subtracted from 6 gives 2?

\( 6 – 2 = 4 \).

So, a = 4.

The formula is \( 3 \times b – a = 2 \).

Substitute \( a = 13 \):

\[ 3 \times b – 13 = 2 \]

We need to find \( 3 \times b \). Let’s work backwards by adding 13 to the answer.

\[ 3 \times b = 2 + 13 \]

\[ 3 \times b = 15 \]

What number times 3 equals 15?

\( 15 \div 3 = 5 \).

So, b = 5.

Moving left means the x-coordinate (the first number) gets smaller.

“Four units left” means we subtract 4 from the x-coordinate.

\[ x \rightarrow x – 4 \]

The y-coordinate (height) should stay the same.

• (0, 2) to (4, 2): \( 0 \rightarrow 4 \). This is \( +4 \). This is Right 4. ❌
• (6, 8) to (2, 8): \( 6 \rightarrow 2 \). This is \( 6 – 4 = 2 \). This is Left 4. ✅
• (–3, 5) to (–7, 5): \( -3 \rightarrow -7 \). Think of a number line: going from -3 down to -7 is a jump of 4 to the left. This is Left 4. ✅

💡 Strategy: Let’s pick a number for Jar B and see what happens.

Let’s say Jar B has 100 beads.

Since Jar A is double Jar B, Jar A has 200 beads.

Olivia says:

• 50% of Jar B: 50% is a half. Half of 100 is 50.
• 25% of Jar A: 25% is a quarter. A quarter of 200 is \( 200 \div 4 = \mathbf{50} \).

50 = 50. She is correct!

We need to explain the relationship generally.

Jar A is double the size of Jar B. 25% is a quarter, and 50% is a half.

Finding a quarter of an amount that is twice as big is the same as finding half of the original amount.