KS2 SATs Mathematics Paper 2: Reasoning (2023)
✏️ Non-Calculator Paper
You have 40 minutes. Calculators are NOT allowed.
Mark Scheme Legend
- 1m = 1 Mark
- 2m = 2 Marks (Show your working!)
Table of Questions
- Q1: Time (Clocks)
- Q2: Negative Numbers
- Q3: Coordinates
- Q4: Pictogram
- Q5: Logical Reasoning
- Q6: Money Change
- Q7: Number Sequence
- Q8: Rounding
- Q9: Circle Parts
- Q10: Reverse Operations
- Q11: Place Value
- Q12: Symmetry
- Q13: Fractions of Amounts
- Q14: Angles
- Q15: Fractions Number Line
- Q16: Rate Problems
- Q17: Multi-step Division
- Q18: Fraction Addition
- Q19: Missing Number Div
- Q20: Missing Digits Mult
- Q21: Unit Conversion
- Q22: Area Fraction
- Q23: Percentages
- Q24: Pie Charts
- Q25: Ratio & Mass
- Q26: Algebra Formula
Question 1 (1 mark)
Circle the clock that shows 5 minutes past 11.
A
B
C
D
Answer: Clock B
Why?
- The hour hand (short hand) should be just past 11.
- The minute hand (long hand) should be pointing at 1 (which represents 5 minutes).
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Question 2 (1 mark)
Write these temperatures in order, starting with the lowest.
6°C –4°C 1°C –10°C 3°C
💡 Strategy: Think of a thermometer. The lower the number (more negative), the colder it is.
- –10 is the coldest.
- –4 is next.
- Then the positive numbers: 1, 3, 6.
Lowest: –10°C, –4°C, 1°C, 3°C, 6°C Highest
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Question 3 (1 mark)
ABC is a triangle.
What are the coordinates of point C?
💡 How to read coordinates:
Remember: “Along the corridor, then up the stairs.”
- Find point C.
- Go down to the x-axis (bottom) to find the first number: 6.
- Go across to the y-axis (left) to find the second number: 2.
Answer: (6, 2)
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Question 4 (1 mark)
Some children choose their favourite zoo animal. The pictogram shows the results.
Key: 🔵 stands for 2 children
How many more children choose tiger than elephant?
Step 1: Work out the values
- One circle 🔵 = 2 children
- Half a circle ◑ = 1 child
Tiger: 6 circles = \(6 \times 2 = 12\) children.
Elephant: 2 full circles + 1 half = \(2 + 2 + 1 = 5\) children.
Step 2: Find the difference
\(12 – 5 = 7\)
Answer: 7
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Question 5 (1 mark)
Cars and motorbikes are parked in a street.
- Car: 4 wheels
- Motorbike: 2 wheels
Stefan counts 3 motorbikes and 5 cars.
He counts 28 wheels altogether.
Explain why Stefan cannot be correct.
💡 Check Stefan’s maths:
We need to calculate the total number of wheels that 5 cars and 3 motorbikes would actually have.
Cars: \(5 \times 4 = 20\) wheels
Motorbikes: \(3 \times 2 = 6\) wheels
Total: \(20 + 6 = 26\) wheels
Stefan counted 28, but there are only 26 wheels. Therefore, he is incorrect.
Answer: 5 cars have 20 wheels and 3 motorbikes have 6 wheels, which makes 26 wheels in total, not 28.
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Question 6 (1 mark)
Kirsty buys 1 litre of apple juice for £1.39.
She pays with a £5 note.
How much change does Kirsty get?
✏️ Subtraction:
£5.00 – £1.39
\[ \begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c} & {}^{4}\bcancel{5} & . & {}^{9}\bcancel{0} & {}^{1}0 \\ – & 1 & . & 3 & 9 \\ \hline & 3 & . & 6 & 1 \\ \end{array} \]Answer: £3.61
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Question 7 (1 mark)
Here is a number sequence.
75 50 25
Write the next two numbers in the sequence.
💡 Find the rule:
75 to 50 is taking away 25.
50 to 25 is taking away 25.
The rule is subtract 25.
\(25 – 25 = 0\)
\(0 – 25 = -25\)
Answer: 0, –25
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Question 8 (1 mark)
In 2012, there were 24,372 schools in the United Kingdom.
Round the number of schools to the nearest hundred.
💡 Strategy:
Look at the hundreds column: 24,372.
Look at the digit to the right (the tens): 7.
If it is 5 or more, round up. Since 7 is more than 5, we round up.
300 becomes 400.
Answer: 24,400
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Question 9 (1 mark)
Here are some diagrams showing parts of a circle. Match each diagram to the name of the dashed line.
A
B
C
Options: Circumference, Diameter, Radius
- A (Line all the way across): Diameter
- B (Line from center to edge): Radius
- C (The outside edge): Circumference
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Question 10 (1 mark)
Ken thinks of a number.
He divides it by 3.
The answer is 72.
What number was Ken thinking of?
💡 Strategy: Work backwards (Inverse Operations).
The opposite of dividing by 3 is multiplying by 3.
\(72 \times 3\)
\(70 \times 3 = 210\)
\(2 \times 3 = 6\)
\(210 + 6 = 216\)
Answer: 216
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Question 11 (1 mark)
a) Write the number that is one thousand more than 19,039.
b) Write the number that is one hundred less than 19,039.
Part a:
Add 1 to the thousands column.
19,039 + 1,000 = 20,039
Part b:
Subtract 1 from the hundreds column. Since there is a 0 in the hundreds column, we need to look at the thousands too.
190 hundreds – 1 hundred = 189 hundreds.
Answer: 18,939
a) 20,039
b) 18,939
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Question 12 (1 mark)
Draw all the lines of symmetry on this shape.
The shape has one vertical line of symmetry and one horizontal line of symmetry.
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Question 13 (1 mark)
\(\frac{1}{5}\) of a number is 22.
What is the number?
💡 Strategy:
If one part is 22, and there are 5 parts in the whole, we need to multiply 22 by 5.
\(22 \times 5 = 110\)
Answer: 110
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Question 14 (1 mark)
Measure angle a.
(Note: On this screen, you cannot use a physical protractor. The answer is based on the original exam paper.)
Using a protractor, you would align the baseline with the horizontal line and the center point with the vertex.
The angle is obtuse (greater than 90°).
Answer: 128° to 132° is accepted.
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Question 15 (1 mark)
Here are four fractions:
\(\frac{1}{3} \quad \frac{1}{6} \quad \frac{1}{4} \quad \frac{1}{2}\)
Write the fractions in the correct place on the number line.
💡 Compare the fractions:
When the numerator is 1, the larger the denominator, the smaller the fraction.
- Smallest: \(\frac{1}{6}\)
- Next: \(\frac{1}{4}\)
- Next: \(\frac{1}{3}\)
- Largest: \(\frac{1}{2}\)
Order from left to right: \(\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}\)
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Question 16 (1 mark)
One day last year, the rate of rainfall from 6:30 am until 9:00 am was 2 millimetres per hour.
What was the total rainfall from 6:30 am until 9:00 am?
Step 1: Calculate the time
6:30 am to 9:00 am is 2 hours and 30 minutes (or 2.5 hours).
Step 2: Calculate rainfall
\(2 \text{ mm} \times 2.5 \text{ hours}\)
\(2 \times 2 = 4\)
\(2 \times 0.5 = 1\)
\(4 + 1 = 5\)
Answer: 5 mm
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Question 17 (2 marks)
The manager of a flower shop orders 4 boxes of red roses.
There are 50 roses in each box.
The manager makes bunches with 6 roses in each bunch.
What is the greatest number of bunches that can be made?
Step 1: Total number of roses
\(4 \text{ boxes} \times 50 \text{ roses} = 200 \text{ roses}\)
Step 2: Divide by 6
\(200 \div 6\)
\(180 \div 6 = 30\)
\(200 – 180 = 20\)
\(20 \div 6 = 3\) remainder 2
Total: 33 remainder 2.
The remainder represents 2 roses left over. We cannot make a full bunch with these.
Answer: 33 bunches
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Question 18 (2 marks)
A cinema sells tickets at three different prices.
- \(\frac{1}{20}\) of the tickets are price A.
- \(\frac{3}{5}\) of the tickets are price B.
- The rest of the tickets are price C.
What fraction of the tickets are price C?
Step 1: Add the known fractions
We need a common denominator. Use 20.
\(\frac{3}{5} = \frac{12}{20}\) (Multiply top and bottom by 4)
\(\frac{1}{20} + \frac{12}{20} = \frac{13}{20}\)
Step 2: Find the rest
The whole is \(\frac{20}{20}\).
\(\frac{20}{20} – \frac{13}{20} = \frac{7}{20}\)
Answer: \(\frac{7}{20}\)
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Question 19 (1 mark)
Write the missing number to make this division correct.
\[ 15,000 \div \square = 75 \]
💡 Strategy: We can swap the division.
\(15,000 \div 75 = \square\)
First, ignore the zeros: \(150 \div 75 = 2\).
Now add the two zeros back: \(200\).
Answer: 200
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Question 20 (2 marks)
Write the two missing digits to make this long multiplication correct.
x 5
———
9 7 0 5
1 6 1 7 5 0
———
1 7 1 4 5 5
Step 1: Find the top digit
Look at the second row of working: \(161750\).
This comes from multiplying the top number by the 50 (the 5 in the tens column).
\(\square 23 \times 5 = 1615\).
\(323 \times 5 = 1615\). So the top box is 3.
Step 2: Find the bottom digit
Look at the first row of working: \(9705\).
\(323 \times \square = 9705\).
We know the last digit is 3. \(3 \times \text{what}\) ends in 5? It must be 5.
Check: \(323 \times 5 = 1615\)… wait. Let’s check the calculation again.
\(323 \times \square = 969\) (if box is 3). No, the first row is 9705.
Wait, let’s look at the provided working in the question again carefully.
First row: 9705. Top number: 3235? No, 3 digits? Ah, looking at the Mark Scheme (Page 27), the top number is 3235. It’s a 4-digit number multiplied by a 2-digit number.
Let’s re-read the question layout. It is: [ ] 2 3 5 multiplied by 5 [ ].
Top number: 3235.
Bottom number: 53.
\(3235 \times 3 = 9705\).
\(3235 \times 50 = 161750\).
Top Box: 3
Bottom Box: 3
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Question 21 (2 marks)
The height of the tallest person in history is 8 feet 11 inches.
Conversion table:
One foot = 30 centimetres
One inch = 2.5 centimetres
Calculate the height of the tallest person, in centimetres.
Step 1: Convert feet
\(8 \times 30 = 240 \text{ cm}\)
Step 2: Convert inches
\(11 \times 2.5\)
\(11 \times 2 = 22\)
\(11 \times 0.5 = 5.5\)
\(22 + 5.5 = 27.5 \text{ cm}\)
Step 3: Total
\(240 + 27.5 = 267.5 \text{ cm}\)
Answer: 267.5 cm
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Question 22 (1 mark)
Here is a regular hexagon.
The area of the large shaded triangle is double the area of the small shaded triangle.
What fraction of the whole hexagon is the shaded area?
💡 Visualise:
A regular hexagon can be divided into 6 equal equilateral triangles (imagine lines from the center to each corner).
The large shaded triangle covers exactly 1 of these 6 sections? Actually, let’s look at the problem logic.
Let the small triangle = 1 unit.
The large triangle = 2 units.
Total shaded = 3 units.
A regular hexagon is made of 12 “small triangle” units (if you divide every equilateral triangle in half).
So the fraction is \(\frac{3}{12}\), which simplifies to \(\frac{1}{4}\).
Answer: \(\frac{1}{4}\) (or \(\frac{3}{12}\))
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Question 23 (2 marks)
A small box contains 650 grams of cereal.
A large box contains 20% more cereal.
One portion of cereal is 40 grams.
How many full portions are in a large box?
Step 1: Find the weight of the large box
10% of 650g = 65g.
20% of 650g = 130g.
Large box = \(650 + 130 = 780 \text{ g}\).
Step 2: Divide by portion size
\(780 \div 40\)
\(78 \div 4\)
\(78 \div 2 = 39\)
\(39 \div 2 = 19.5\)
The question asks for full portions.
Answer: 19 portions
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Question 24 (1 mark)
1,200 pupils were asked this question: “How important is it to have a break when using a screen?”
Not important: 12%
Quite important: 41%
Very important: ?
How many pupils answered ‘Very important’?
Step 1: Find percentage for ‘Very important’
Total = 100%
\(100\% – 12\% – 41\% = 47\%\)
Step 2: Calculate 47% of 1,200
1% of 1,200 = 12.
\(47 \times 12\)
\(47 \times 10 = 470\)
\(47 \times 2 = 94\)
\(470 + 94 = 564\)
Answer: 564 pupils
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Question 25 (2 marks)
There are 25 sheets of paper in a small pack.
There are 500 sheets in a large pack.
a) How many small packs make one large pack?
The mass of the paper in the large pack is 2.4 kilograms.
b) What is the mass of one sheet of paper, in grams?
Part a:
\(500 \div 25\)
\(500 \div 100 = 5\), so \(500 \div 25 = 20\). (Because 25 is 4 times smaller than 100).
Part b:
Convert mass to grams: 2.4 kg = 2,400 g.
Divide total mass by number of sheets (500).
\(2400 \div 500 = 24 \div 5\).
\(24 \div 5 = 4.8\).
a) 20 packs
b) 4.8 g
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Question 26 (2 marks)
This formula is used to estimate the mass (in kilograms) of young children:
mass = 2 × (age in years + 5)
a) Stefan’s sister is 4 years of age. Use the formula to estimate her mass.
b) The mass of Megan’s brother is 16 kilograms. Use the formula to estimate his age.
Part a:
Age = 4.
Mass = \(2 \times (4 + 5)\)
Mass = \(2 \times 9 = 18\).
Part b:
Mass = 16.
\(16 = 2 \times (\text{age} + 5)\)
Divide by 2: \(8 = \text{age} + 5\)
Subtract 5: \(\text{age} = 3\).
a) 18 kg
b) 3 years
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