SATs – 2016 – Paper 2

KS2 2016 Maths Paper 2: Reasoning – Interactive Practice

KS2 2016 Maths Paper 2: Reasoning

Mark Scheme Legend

  • M1 = Method mark
  • A1 = Accuracy mark
  • B1 = Independent mark
  • 1m = 1 mark

Question 1 (2 marks)

Ali puts these five numbers in their correct places on a number line.

511    499    502    555    455

Write the number closest to 500.

Write the number furthest from 500.

Worked Solution

Step 1: Find the closest number to 500

Why we do this: To find the closest number, we calculate the difference (distance) between each number and 500. The smallest difference is the closest.

✏ Working:

  • $511 – 500 = 11$
  • $500 – 499 = 1$ (This is very small)
  • $502 – 500 = 2$
  • $555 – 500 = 55$
  • $500 – 455 = 45$

Difference of 1 is the smallest.

Closest: 499 (1m)

Step 2: Find the furthest number from 500

Why we do this: We look for the largest difference calculated above.

✏ Working:

Comparing differences: 11, 1, 2, 55, 45.

55 is the largest difference.

Furthest: 555 (1m)

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Question 2 (1 mark)

Put these houses in order of price starting with the lowest price.

  • A: £135,300
  • B: £119,125
  • C: £130,500
  • D: £131,500
  • E: £91,500

One has been done for you: B is placed second.

Worked Solution

Step 1: Compare the values

Why we do this: To order them, we look at the number of digits and the value of the highest place value columns first.

✏ Working:

  • A: 135,300 (6 digits, starts with 13)
  • B: 119,125 (6 digits, starts with 11) – 2nd lowest
  • C: 130,500 (6 digits, starts with 13)
  • D: 131,500 (6 digits, starts with 13)
  • E: 91,500 (5 digits) – Lowest

Comparing the 130s:

  • C: 130,…
  • D: 131,…
  • A: 135,…

Order: 91,500 (E) → 119,125 (B) → 130,500 (C) → 131,500 (D) → 135,300 (A)

Order: E, B, C, D, A (1m)

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Question 3 (2 marks)

Write the three missing digits to make this addition correct.

 1 5 [ ]
+ 4 [ ] 4
——-
[ ] 1 5

Worked Solution

Step 1: Ones Column

Why we do this: Start from the rightmost column. We need a number that adds to 4 to make 5.

✏ Working:

$? + 4 = 5$

So the top right digit is 1.

Step 2: Tens Column

Why we do this: The result ends in 1. Since $5 + \text{something}$ cannot equal 1, it must equal 11. This means we carry 1.

✏ Working:

$5 + ? = 11$

$? = 6$

So the middle digit is 6.

Don’t forget to carry the 1 to the hundreds column!

Step 3: Hundreds Column

Why we do this: Add the digits we have plus the carried 1.

✏ Working:

$1 + 4 + 1 \text{ (carried)} = 6$

So the bottom left digit is 6.

Top box: 1, Middle box: 6, Bottom box: 6

 1 5 1
+ 4 6 4
——-
6 1 5

(2 marks)

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Question 4 (2 marks)

This table shows the number of people living in various towns in England.

Town Population
Bedford 82,448
Carlton 48,493
Dover 34,087
Formby 24,478
Telford 166,640

a) What is the total of the numbers of people living in Formby and in Telford?

b) What is the difference between the numbers of people living in Bedford and in Dover?

Worked Solution

Part A: Total of Formby and Telford

Why we do this: “Total” means addition. We need to add the population of Formby (24,478) and Telford (166,640).

✏ Working:

  166640
+  24478
——–
  191118

Total: 191,118 (1m)

Part B: Difference between Bedford and Dover

Why we do this: “Difference” means subtraction. We subtract Dover (34,087) from Bedford (82,448).

✏ Working:

  82448
–  34087
——–
  48361

Difference: 48,361 (1m)

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Question 5 (2 marks)

Write each number in its correct place on the diagram.

16   17   18   19

prime numbers square numbers even numbers

Worked Solution

Step 1: Analyze each number

Why we do this: We check the properties of each number to see which sets (Prime, Square, Even) they belong to.

✏ Working:

  • 16: Is it Prime? No. Is it Square? Yes ($4 \times 4$). Is it Even? Yes. -> Intersection of Square and Even.
  • 17: Is it Prime? Yes. Is it Square? No. Is it Even? No. -> Prime only.
  • 18: Is it Prime? No. Is it Square? No. Is it Even? Yes. -> Even only.
  • 19: Is it Prime? Yes. Is it Square? No. Is it Even? No. -> Prime only.
Step 2: Place them in the diagram
prime numbers square numbers even numbers 17, 19 16 18

17, 19 in left circle (not even part).

16 in intersection of Square and Even.

18 in Even region (not prime or square).

(2 marks)

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Question 6 (1 mark)

This diagram shows a shaded shape inside a border of squares.

Draw the reflection of the shape in the mirror line.

mirror line

Worked Solution

Step 1: Reflect each vertex

Why we do this: For a reflection, every point on the new shape must be the same distance from the mirror line as the original point, but on the opposite side.

The reflected shape is drawn on the left side of the mirror line.

(1 mark)

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Question 7 (1 mark)

Write the two missing values to make these equivalent fractions correct.

$\frac{\square}{3} = \frac{8}{12} = \frac{4}{\square}$

Worked Solution

Step 1: Simplify the middle fraction

Why we do this: The middle fraction is $\frac{8}{12}$. We can simplify it or compare it to others.

To get from $\frac{8}{12}$ to $\frac{\square}{3}$:

The denominator goes from 12 to 3. This is dividing by 4 ($12 \div 4 = 3$).

So we must divide the numerator by 4: $8 \div 4 = 2$.

First box: 2

Step 2: Find the third fraction

Why we do this: Now compare $\frac{8}{12}$ to $\frac{4}{\square}$.

The numerator goes from 8 to 4. This is dividing by 2 ($8 \div 2 = 4$).

So we must divide the denominator by 2: $12 \div 2 = 6$.

Second box: 6

$\frac{2}{3} = \frac{8}{12} = \frac{4}{6}$

Missing numbers: 2 and 6 (1m)

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Question 8 (1 mark)

Circle two numbers that add together to equal 0.25

0.05   0.23   0.2   0.5

Worked Solution

Step 1: Understand place value

Why we do this: It helps to write all numbers with the same number of decimal places to avoid confusion.

0.25 is the target.

  • 0.05
  • 0.23
  • 0.20 (added zero)
  • 0.50 (added zero)
Step 2: Test pairs

We need a pair that sums to 0.25.

  • $0.05 + 0.20 = 0.25$ -> YES

0.05 and 0.2 (1m)

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Question 9 (2 marks)

6 pencils cost £1.68

3 pencils and 1 rubber cost £1.09

What is the cost of 1 rubber?

Worked Solution

Step 1: Find the cost of 3 pencils

Why we do this: We know the cost of 6 pencils (£1.68). 3 pencils is half of 6 pencils.

Cost of 3 pencils = $£1.68 \div 2$

$168 \text{p} \div 2 = 84\text{p}$ (or £0.84)

Step 2: Find the cost of the rubber

Why we do this: The second statement says “3 pencils and 1 rubber cost £1.09”. We now know the 3 pencils part is £0.84.

Rubber cost = Total cost – Cost of pencils

Rubber cost = $£1.09 – £0.84$

  1.09
– 0.84
——-
  0.25

Cost of 1 rubber: 25p (or £0.25) (2m)

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Question 10 (2 marks)

Each diagram below is divided into equal sections.

Shade three-quarters of each diagram.

Worked Solution

Step 1: Calculate “Three-Quarters” for each shape

Why we do this: $\frac{3}{4}$ means for every 4 equal parts, we shade 3. Or if there are 8 parts, we shade 6 ($\frac{6}{8} = \frac{3}{4}$).

Step 2: Apply to shapes
  • Square: Divided into 4 triangles. Shade 3 triangles.
  • Grid Shape: Divided into squares with diagonals. Total 6 triangles? Or 8? Let’s check: 3 squares x 2 triangles = 6 triangles. $\frac{3}{4}$ of 6 is 4.5? That doesn’t work.
  • Let’s re-examine Shape 2 (L-shape). It’s 4 squares? Top row 2, Middle row 1? No.
  • It looks like 4 squares in an upside down L. 3 across top, 1 down?
  • Let’s count sections. If we shade 3/4, the number of sections must be a multiple of 4. Likely 4, 8, 12.
  • Counting from screenshot p12: Top row has 3 squares. Bottom left has 1 square. Total 4 squares. Each square cut in half? No, only some?
  • Actually, the grid lines show 4 squares total. 3 in top row, 1 in bottom left. Each has a diagonal. Total 8 triangles.
  • Target: $\frac{3}{4}$ of 8 = 6 triangles.
  • Circle: Divided into 8 sectors. $\frac{3}{4}$ of 8 = 6 sectors.

Shape 1: Shade 3 of the 4 sections.

Shape 2: Shade 6 of the 8 small triangles.

Shape 3: Shade 6 of the 8 sectors.

(2 marks)

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Question 11 (2 marks)

A packet contains 1.5 kg of oats.

Every day Maria uses 50 g of oats to make porridge.

How many days does the packet of oats last?

Worked Solution

Step 1: Convert units

Why we do this: We have kg and g. We must work in the same unit. Let’s convert kg to g.

1 kg = 1000 g.

$1.5 \text{ kg} = 1.5 \times 1000 = 1500 \text{ g}$

Step 2: Divide total by daily usage

Why we do this: To find the number of days, divide the total amount by the amount used per day.

$1500 \div 50$

This is the same as $150 \div 5$.

$150 \div 5 = 30$

Answer: 30 days (2m)

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Question 12 (2 marks)

n = 22

What is $2n + 9$?


$2q + 4 = 100$

Work out the value of $q$.

Worked Solution

Part 1: Substitute and Calculate

$n = 22$

$2n$ means $2 \times n = 2 \times 22 = 44$.

$44 + 9 = 53$.

Answer: 53 (1m)

Part 2: Solve for q

Why we do this: We need to isolate $q$. We do the inverse operations in reverse order.

$2q + 4 = 100$

Subtract 4 from both sides:

$2q = 100 – 4$

$2q = 96$

Divide by 2:

$q = 96 \div 2$

$q = 48$

Answer: 48 (1m)

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Question 13 (2 marks)

A stack of 20 identical boxes is 140 cm tall.

Stefan takes three boxes off the top.

How tall is the stack now?

Worked Solution

Step 1: Find the height of one box

Why we do this: We know 20 boxes = 140 cm. Division gives the height of a single box.

$140 \div 20 = 7$

Each box is 7 cm tall.

Step 2: Find the new height

Why we do this: If 3 boxes are removed, there are 17 boxes left. Or we can subtract the height of 3 boxes.

Method A:

Boxes remaining = $20 – 3 = 17$

Height = $17 \times 7$

$10 \times 7 = 70$

$7 \times 7 = 49$

$70 + 49 = 119$


Method B:

Height removed = $3 \times 7 = 21$ cm

New height = $140 – 21 = 119$ cm

Answer: 119 cm (2m)

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Question 14 (1 mark)

Write all the common multiples of 3 and 8 that are less than 50.

Worked Solution

Step 1: List multiples

Why we do this: A common multiple is a number that is in both times tables.

Multiples of 8: 8, 16, 24, 32, 40, 48…

Now check which of these divide by 3:

  • 8? No.
  • 16? No.
  • 24? Yes ($3 \times 8$).
  • 32? No.
  • 40? No.
  • 48? Yes ($3 \times 16$).

Answer: 24 and 48 (1m)

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Question 15 (2 marks)

This thermometer shows temperatures in both °C and °F.

Work out what 25 °C is in °F.

°C 40 30 20 10 0 -10 °F 104 86 68 50 32 14

Worked Solution

Step 1: Identify the pattern

Why we do this: Look at the gap between 20°C and 30°C. 25°C is exactly halfway.

At 20°C, temperature is 68°F.

At 30°C, temperature is 86°F.

Step 2: Find the midpoint in °F

Why we do this: Since 25 is halfway between 20 and 30, the answer will be halfway between 68 and 86.

Difference: $86 – 68 = 18$.

Half of the difference: $18 \div 2 = 9$.

Add 9 to the lower value: $68 + 9 = 77$.

Answer: 77 °F (2m)

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Question 16 (2 marks)

Write the number that is five less than ten million.

Write the number that is one hundred thousand less than six million.

Worked Solution

Part 1: Five less than ten million

Ten million = 10,000,000

Subtract 5:

9,999,995 (1m)

Part 2: 100,000 less than 6 million

Six million = 6,000,000

One hundred thousand = 100,000

  6,000,000
–   100,000
———–
  5,900,000

Answer: 5,900,000 (1m)

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Question 17 (2 marks)

Calculate the size of angles $a$ and $b$ in this diagram.

a 160° b

Not to scale

$a = \dots\dots\dots\dots\dots ^\circ$

$b = \dots\dots\dots\dots\dots ^\circ$

Worked Solution

Step 1: Calculate angle a

Why we do this: The angle marked $160^\circ$ and angle $a$ are vertically opposite angles. Vertically opposite angles are equal.

$a = 160$

a = 160° (1m)

Step 2: Calculate angle b

Why we do this: The shape formed is a quadrilateral (4-sided shape). The sum of angles in a quadrilateral is $360^\circ$.

The angles are:

  • Two right angles ($90^\circ$ each)
  • Angle $a$ ($160^\circ$)
  • Angle $b$

Sum = $90 + 90 + 160 + b = 360$

$180 + 160 + b = 360$

$340 + b = 360$

$b = 360 – 340$

$b = 20$

b = 20° (1m)

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Question 18 (1 mark)

Write the missing number.

$70 \div \square = 3.5$

Worked Solution

Step 1: Rearrange the calculation

Why we do this: $70 \div \text{something} = 3.5$ is the same as $70 \div 3.5 = \text{something}$.

$70 \div 3.5$

Multiply both by 10 to remove the decimal:

$700 \div 35$

Step 2: Solve

We know $35 \times 2 = 70$.

So $35 \times 20 = 700$.

Answer: 20 (1m)

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Question 19 (3 marks)

Miss Mills is making jam to sell at the school fair.

  • Strawberries cost £7.50 per kg.
  • Sugar costs 79p per kg.
  • 10 glass jars cost £6.90.

She uses 12 kg of strawberries and 10 kg of sugar to make 20 jars full of jam.

Calculate the total cost to make 20 jars full of jam.

Worked Solution

Step 1: Cost of Strawberries

12 kg at £7.50 per kg.

$12 \times 7.50$

$12 \times 7 = 84$

$12 \times 0.50 = 6$

$84 + 6 = £90.00$

Step 2: Cost of Sugar

10 kg at 79p per kg.

$10 \times 79\text{p} = 790\text{p}$

Convert to pounds: £7.90

Step 3: Cost of Jars

Note: 10 jars cost £6.90. She makes 20 jars, so she needs 2 sets of jars.

$£6.90 \times 2 = £13.80$

Step 4: Total Cost
  90.00
   7.90
+ 13.80
——-
 111.70

Answer: £111.70 (3m)

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Question 20 (1 mark)

Here are two triangles drawn on coordinate axes.

Triangle B is a reflection of triangle A in the x-axis.

Two of the new vertices of triangle B are $(10, -10)$ and $(20, -30)$.

What are the coordinates of the third vertex of triangle B?

x y 0 A (-10, 40) (20, 30) (10, 10) B

Worked Solution

Step 1: Identify the third vertex of A

Why we do this: The question gives the coordinates of all three vertices of A in the diagram: $(-10, 40)$, $(20, 30)$, and $(10, 10)$.

The question says B has new vertices $(10, -10)$ and $(20, -30)$.

These correspond to reflecting $(10, 10)$ and $(20, 30)$.

We need to reflect the third vertex: $(-10, 40)$.

Step 2: Reflect in x-axis

Rule: To reflect in the x-axis, the x-coordinate stays the same, and the y-coordinate changes sign (positive becomes negative).

$(x, y) \rightarrow (x, -y)$

Vertex: $(-10, 40)$

Reflected: $(-10, -40)$

Answer: (-10, -40) (1m)

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