KS2 Mathematics Paper 3 2018 (Reasoning)

Why we do this: To complete the sequence, we first need to know how much the numbers increase by each step. We look at the two numbers that are next to each other.

What this tells us: The sequence increases by 7 each time.

How we do this: We subtract 7 to find numbers to the left, and add 7 to find numbers to the right.

Strategy: We need to check which pairs are already found and which are missing. Let’s list all possibilities systematically starting with each colour.

The condition says: 4 in the tens column.

In a four-digit number (Thousands, Hundreds, Tens, Ones), the tens column is the second from the right.

The condition says: 2 so that it has a higher value than any of the other digits.

For a digit to have the highest value, it must be in the column with the highest place value. In a four-digit number, this is the Thousands column.

The condition says: the remaining two digits so that 7 has the higher value.

We have digits 3 and 7 left.

The empty spots are Hundreds and Ones. Hundreds have a higher value than Ones. So 7 must go in the Hundreds column.

Look at the far right: \( 9 + \square = 6 \).

Since 9 is larger than 6, the result must be 16, not 6. We are carrying 1.

What is a factor? A number that divides another number exactly (with no remainder).

We need numbers that appear in both lists and are in the options provided.

Total cats = 20.

The vertical line splits the circle in half. This means the Right side (Tigers + Lions) is half of 20.

The square symbol on the Left side shows the Cheetah sector is a right angle (90°), which is one-quarter of the circle.

1. "More cheetahs than jaguars": 5 is not more than 5. (False)

2. "Total lions and tigers is 10": We calculated this as 10. (True)

3. "One-quarter are cheetahs": We identified the right angle. (True)

4. "More than 5 jaguars": We calculated exactly 5. (False)

Why: We need to find how many groups of 6 fit into 980.

Method: We can find 10% first, then 30%, then 5%. Or calculate directly.

10% of 400 = 40.

Real Height: 20 m

Model Height: 40 cm

What is the relationship? \( 20 \times 2 = 40 \). The number doubles (ignoring units for the scale factor logic, or seeing that 20m -> 40cm means 1m -> 2cm).

Real Width: 54 m

We must do the same operation: Double the number.

Let's look at the x-coordinates of the known points: -3, 1, 5.

Let's look at the y-coordinates: 1, 5, 1.

The top point is at x=1. The bottom point must also be at x=1 for it to be a square tilted this way.

The left point is at y=1. The right point is at y=1.

Distance from center (1, 1) to top (1, 5) is 4 units.

Distance from center to bottom must be 4 units.

Jug 1: The liquid is exactly halfway between 100 and 200.
The midpoint is 150 ml.

Jug 2: There are three small marks between 100 and 200. This splits the gap into 4 parts.
The gap is 100ml. \( 100 \div 4 = 25 \) ml per mark.
The level is at the third mark above 100.
\( 100 + 25 + 25 + 25 = 175 \) ml.

Add the two amounts to find total water removed from the bottle. Then subtract from the original 600 ml.

To find "how many times larger", we divide the larger number by the smaller number.

Trap: The box is in kg but usage is in grams.

We must convert kg to grams first. 1 kg = 1000 g.

Divide total grams by grams per wash.

Key Fact: In an equilateral triangle, all three sides are the same length AND all three angles are equal.

Since angles in a triangle add up to 180°, each angle in an equilateral triangle must be \( 180 \div 3 = 60^\circ \).

The question gives angles of 70° and 40°.

Neither of these is 60°.

Also, the angles are different, so it cannot be equilateral.

Formula: \( \text{price} = 60\text{p} \times \text{colours} + £1.25 \)

First, substitute 3 for the number of colours.

Strategy: Work backwards.

Subtract the fixed cost (£1.25) from the total money (£5.00) to see how much is left for the colours.

Calculate \(\frac{1}{3}\) of 276. We divide 276 by 3.

Subtract the pages read from the total pages.

Rule: Opposite faces sum to 7.

• Opposite 6 must be 1.
• Opposite 5 must be 2.
• Opposite 4 must be 3.

Let's use the center square as the Back.

• The Top square (6) is above the center. So Center is opposite Top? No.
• If Center is Back, Top (6) folds to be Top. Opposite of Top is Bottom.
• The square Below the center is Bottom. So Bottom = 1 (Opposite 6).
• The square to the Left (4) folds to be Left. Opposite is Right.
• The square to the Right (Empty) is Right. So Right = 3 (Opposite 4).
• The square attached to the 6 (5) folds around to be the Front.
• Opposite of Front (5) is Back (Center). So Center = 2.

Goal: Find the total fraction used by potatoes and cabbages.

We need to add \(\frac{2}{3}\) and \(\frac{1}{4}\).

Common Denominator: 12.

The whole garden is 1 (or \(\frac{12}{12}\)).

Subtract the planted part to find the carrots.

Compare with \( 354 \times 95 = 33,630 \).

9.5 is 10 times smaller than 95.

So the answer must be 10 times smaller.

Compare with \( 354 \times 95 = 33,630 \).

3,540 is 10 times bigger than 354.

So the answer must be 10 times bigger.

This involves division. We know \( 33,630 \div 95 = 354 \).

We are dividing 3,363, which is 10 times smaller than 33,630.

So the answer must be 10 times smaller than 354.

We know how many eggs he collected in the first 20 days. We need to find out how many days are left in March.

To get the greatest total, we must assume he collects the maximum possible number of eggs (4) on every single one of the remaining 11 days.

Add the eggs from the first 20 days to the maximum eggs from the last 11 days.

Why: "After" means we add the time. Ally took longer.

Method: Add minutes to minutes, and seconds to seconds.

Why: "Before" means she was faster. We subtract the time.

Method: Subtract minutes and seconds. We may need to "borrow" a minute if the seconds don't calculate directly.