2019 Key Stage 2 Mathematics Paper 3: reasoning

💡 What do we need to do?

We start with an original price of £8,999. The car is on sale with £1,100 off.

“Off” means we need to subtract the discount from the original price.

We need to calculate: \( 8,999 – 1,100 \)

We can use column subtraction.

🔍 Check:

\( 9 – 0 = 9 \)
\( 9 – 0 = 9 \)
\( 9 – 1 = 8 \)
\( 8 – 1 = 7 \)

💡 Strategy: Let’s label the place value of each digit starting from the right (ones).

9 = Ones
1 = Tens
2 = Hundreds
6 = Thousands
7 = Ten Thousands
5 = Hundred Thousands
3 = Millions

The digit in the ten thousands place is 7.

💡 Strategy:

1. Find the millions digit: 3.

2. Look at the digit to its right (the hundred thousands digit): 5.

3. Rule: If the digit to the right is 5 or more, round up.

Start: Dev had £10.

Action: He gave away \( a \).

Result: “Giving away” means the amount decreases, so we use subtraction.

He starts with 10 and takes away \( a \).

This is written as \( 10 – a \).

Let’s check the others:

• \( 10 + a \) means he got more money.
• \( a – 10 \) means he started with \( a \) and gave away 10.
• \( a \times 10 \) means he has 10 times as much.

💡 Strategy: To compare decimals easily, it helps to give them all the same number of decimal places.

The longest decimal (1.025) has 3 decimal places. Let’s add zeros to the others so they all have 3 decimal places.

Now, look at the numbers as if they were whole numbers (ignoring the decimal for a moment) or compare column by column.

1. 0.009 (smallest, 0 ones and 0 tenths)
2. 0.990 (0 ones but 9 tenths)
3. 1.025 (1 one and 0 tenths)
4. 1.250 (1 one and 2 tenths)

The calculation is: a 3-digit number ending in “2something” plus a 2-digit number ending in 2 equals 200.

Let’s write it vertically:

We know the answer is 200.

We have a number ending in 2. What must we add to a number ending in 2 to get a number ending in 0?

\( 2 + 8 = 10 \). So the units digit of the first number must be 8.

Now we have:

Wait, let’s check the digits provided in the question carefully.

Question format: [ ] 2 [ ] + [ ] 2 = 200

Usually, this implies a 3-digit number added to a 2-digit number.

Let’s use subtraction to find the missing part.

\( 200 – \text{something ending in 2} = \text{something} \)

Alternatively, let’s look at the structure again.

\( [1] 2 [8] + [7] 2 = 200 \)?

\( 128 + 72 = 200 \)

Does this fit the boxes? Yes.

Box 1: 1 (Hundreds)
Box 2: 8 (Units of first number)
Box 3: 7 (Tens of second number)

Why? First, we need to know how much John spent in total.

We add the price of the car (£1.49) and the stickers (£1.64).

Why? Change is the money you get back. We subtract the total cost from the amount paid (£10).

Remember: £10 can be written as £10.00 to make subtraction easier.

1. Identify Heaviest and Lightest:

Heaviest: 410 g

Lightest: 255 g

2. Calculate Difference: Subtract lightest from heaviest.

Let’s sort the masses into the groups:

• 250–299: 255, 275 (Count: 2)
• 300–349: 305, 310, 345 (Count: 3)
• 350–399: 375, 360 (Count: 2)
• 400–449: 410 (Count: 1, already done)

First, find out how many points Ken has now by adding his new score to his old score.

Now, find the difference between his target (6,000) and his current total (4,644).

We use subtraction.

The key tells us that one full circle = 1,000 satellites.

• We have 2 full circles: \( 2 \times 1,000 = 2,000 \)
• We have 1 quarter circle (a pie slice): \( \frac{1}{4} \text{ of } 1,000 \)

To find a quarter of 1,000, we divide by 4:

\( 1,000 \div 4 = 250 \)

Start at the origin (0,0).

Go 1 left (to -1 on x-axis).

Go 2 up (to 2 on y-axis).

Mark the point and join it to the top and bottom points to close the shape (forming a kite/arrowhead).

Move every corner point 4 squares to the right.

• (-2, 3) → (-2 + 4, 3) = (2, 3)
• (1, 2) → (1 + 4, 2) = (5, 2)
• (-2, -1) → (-2 + 4, -1) = (2, -1)
• (-1, 2) → (-1 + 4, 2) = (3, 2)

Plot these new points and join them.

We need to check the remaining numbers: 3, 4, 5, 6.

• Prime number: Has exactly two factors (1 and itself).
• Factor of 12: Numbers that divide into 12 exactly (1, 2, 3, 4, 6, 12).
• Factor of 15: Numbers that divide into 15 exactly (1, 3, 5, 15).

A “one-tenth” scale model means the model is 10 times smaller than the real object.

We need to divide both measurements by 10.

• Acute angle: Less than 90°.
• Obtuse angle: Between 90° and 180°.

Kirsty thinks: \( \text{Acute} \times 2 = \text{Obtuse} \)

Can we find an acute angle that, when doubled, is still acute (less than 90°)?

Remember the rhyme:

“30 days hath September,
April, June and November.
All the rest have 31…”

• September: 30 days
• October: 31 days
• November: 30 days

We are given the conversion: 5 miles = 8 km.

We need to convert 250 miles to km.

We need to find out how many “blocks of 5 miles” fit into 250 miles.

For every group of 5 miles, we have 8 km.

We have 50 groups, so we calculate \( 50 \times 8 \).

Potatoes cost £1.50 per kg.

Jack buys \( 1\frac{1}{2} \) kg. This is 1 kg plus half a kg.

Carrots cost £1.80 per kg.

Jack buys \( \frac{1}{2} \) kg.

Total cost = Potatoes + Carrots

Change = £5.00 – Total Cost

We know that \( x + 2y = 20 \).

Also, \( x \) and \( y \) must be whole numbers less than 10.

Let’s pick a value for \( y \) and see if we can find \( x \).

We are comparing to \( \frac{5}{8} \).

It’s easier if all fractions have the same denominator. Let’s use 32, as it works for 2, 4, 8, 16, and 32.

Target Fraction: \( \frac{5}{8} = \frac{20}{32} \)

We need fractions smaller than \( \frac{20}{32} \).

We need to divide the total sheets by the sheets per booklet.

\( 1,000 \div 34 \)

Let’s use chunking or estimation.

• \( 10 \times 34 = 340 \)
• \( 20 \times 34 = 680 \)
• \( 30 \times 34 = 1,020 \) (Too high!)

So the answer is slightly less than 30.

Point B is at the top right.

It shares the same y-coordinate (height) as Point A. So, \( y = 30 \).

To find the x-coordinate, we use the centre point C.

The horizontal distance from A (25) to C (40) is \( 40 – 25 = 15 \).

By symmetry, the distance from C to the right edge (B) is also 15.

Point D is at the bottom right.

It shares the same x-coordinate as Point B. So, \( x = 55 \).

To find the y-coordinate, we look at the vertical distance.

The vertical distance from A (30) to C (22) is \( 30 – 22 = 8 \).

By symmetry, the distance from C down to the bottom edge is also 8.

The rectangle on the right is made of 2 identical right-angled triangles.

The height of the shape is 7 cm. This corresponds to the long leg of the triangle.

So, Long Leg = 7 cm.

The left section is made of 4 identical triangles.

Look at the vertical line where they meet. The total height (7 cm) is made up of two short legs of the triangles stacked on top of each other.

The total length is the width of the left section plus the width of the right section.

• Left Section Width: This corresponds to the Long Leg = 7 cm.
• Right Section Width: This corresponds to the Short Leg = 3.5 cm.

Let the distance from Q to R be 1 part.

The distance from P to Q is 4 times that, so it is 4 parts.

Total distance (P to R) = \( 4 \text{ parts} + 1 \text{ part} = 5 \text{ parts} \).