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Key Stage 2 Mathematics Paper 2: Reasoning (2017)

Mark Scheme Legend

  • M1 – Method mark
  • A1 – Accuracy mark
  • (1) – Total marks for question

Question 1 (2 marks)

William asks the children in Year 2 and Year 6 if they walk to school.

This graph shows the results.

0 20 40 60 80 100 120 140 Number of children Year 2 Year 6 walk to school don’t walk to school

1. Altogether, how many children don’t walk to school?

2. How many more Year 6 children than Year 2 children walk to school?

Worked Solution

Step 1: Reading the Graph

What does this graph tell us?

The y-axis goes up in steps of 20 (0, 20, 40…). There is one line between each major mark, representing 10.

  • Year 2 Walk (dark): Between 60 and 80 = 70
  • Year 2 Don’t Walk (light): Between 100 and 120 = 110
  • Year 6 Walk (dark): On the line 120
  • Year 6 Don’t Walk (light): Between 80 and 100 = 90
Part 1: Altogether, how many don’t walk?

Why we do this: We need to add the “don’t walk” (light grey) bars for both years.

\[ 110 + 90 = 200 \]

Answer: 200

Part 2: How many more Year 6 walk?

Why we do this: We compare the “walk” (dark grey) bars. “How many more” means subtraction.

\[ \text{Year 6 Walk} – \text{Year 2 Walk} \] \[ 120 – 70 = 50 \]

Answer: 50

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Question 2 (1 mark)

Circle the number that is 10 times greater than nine hundred and seven.

9,700    907    9,007    970    9,070

Worked Solution

Step 1: Understanding the Operation

Why we do this: “10 times greater” means we multiply the number by 10.

First, write “nine hundred and seven” as a number: \( 907 \).

Now multiply by 10:

\[ 907 \times 10 = 9,070 \]

Answer: 9,070

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Question 3 (1 mark)

Write the missing numbers to make this multiplication grid correct.

× 9 63 54 56 48

Worked Solution

Step 1: Finding the Top Row Numbers

Why we do this: We know that \( 9 \times \text{something} = 63 \) and \( 9 \times \text{something else} = 54 \).

\[ 63 \div 9 = 7 \] \[ 54 \div 9 = 6 \]

So the top row numbers are 7 and 6.

Step 2: Finding the Side Number

Why we do this: Now we look at the bottom row. \( \text{Something} \times 7 = 56 \).

\[ 56 \div 7 = 8 \]

Check with the last number: \( 8 \times 6 = 48 \). This works!

Answers:

Top boxes: 7 and 6

Side box: 8

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Question 4 (2 marks)

This table shows the heights of three mountains.

Mountain Height in metres Mount Everest 8,848 Mount Kilimanjaro 5,895 Ben Nevis 1,344

How much higher is Mount Everest than the combined height of the other two mountains?

Worked Solution

Step 1: Calculate Combined Height

Why we do this: First, we need to find the total height of Mount Kilimanjaro and Ben Nevis.

  5895
+ 1344
──────
  7239
  11  
Step 2: Calculate the Difference

Why we do this: Now subtract this combined height from the height of Mount Everest to find “how much higher” it is.

  8848
- 7239
──────
  1609

Answer: 1,609 m

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Question 5 (2 marks)

Complete this table with the missing numbers. One row has been done for you.

Number 1,000 more 3,500 4,500 85 9,099 15,250

Worked Solution

Step 1: Adding 1,000

Why we do this: For the first missing box, we need to find “1,000 more” than 85.

\[ 85 + 1000 = 1085 \]
Step 2: Working Backwards (Subtracting 1,000)

Why we do this: For the bottom two rows, we are given the “1,000 more” column and need to find the original number. We must subtract 1,000.

For 9,099:

\[ 9099 – 1000 = 8099 \]

For 15,250:

\[ 15250 – 1000 = 14250 \]

Answers:

1,085

8,099

14,250

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Question 6 (1 mark)

Write these numbers in order of size, starting with the smallest.

1.9     0.96     1.253     0.328

Worked Solution

Step 1: Compare Place Value

Why we do this: To compare decimals easily, look at the units column first, then tenths, then hundredths.

Let’s line them up:

  • 1.900 (1 unit)
  • 0.960 (0 units, 9 tenths)
  • 1.253 (1 unit)
  • 0.328 (0 units, 3 tenths)
Step 2: Order them

Smallest (0 units): 0.328 is smaller than 0.960.

Next (1 unit): 1.253 is smaller than 1.900.

Order: 0.328, 0.96, 1.253, 1.9

Answer: 0.328, 0.96, 1.253, 1.9

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Question 7 (2 marks)

Write the missing numbers.

60 months = [     ] years

72 hours = [     ] days

84 days = [     ] weeks

Worked Solution

Step 1: Months to Years

Fact: There are 12 months in 1 year.

\[ 60 \div 12 = 5 \]
Step 2: Hours to Days

Fact: There are 24 hours in 1 day.

\[ 72 \div 24 = 3 \]
Step 3: Days to Weeks

Fact: There are 7 days in 1 week.

\[ 84 \div 7 = 12 \]

Answers: 5 years, 3 days, 12 weeks

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Question 8 (2 marks)

At the start of June, there were 1,793 toy cars in the shop.

During June,

  • 8,728 more toy cars were delivered
  • 9,473 toy cars were sold.

How many toy cars were left in the shop at the end of June?

Worked Solution

Step 1: Add the Delivery

Why we do this: First, add the new cars to the starting amount.

  1793
+ 8728
──────
 10521
 11 1 
Step 2: Subtract the Sold Cars

Why we do this: Now take away the cars that were sold from the total.

 10521
- 9473
──────
  1048

Answer: 1,048 toy cars

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Question 9 (1 mark)

Tick two shapes that have \( \frac{3}{4} \) shaded.

Worked Solution

Step 1: Check Each Shape

Why we do this: We need to find the shapes where the shaded fraction equals \( \frac{3}{4} \). Note that \( \frac{3}{4} \) is equivalent to \( \frac{6}{8} \) or \( \frac{12}{16} \).

  • Top Left: 8 total rectangles. 6 are shaded. \( \frac{6}{8} = \frac{3}{4} \). YES.
  • Top Right: 8 total rectangles. 4 are shaded. \( \frac{4}{8} = \frac{1}{2} \). NO.
  • Bottom Left: 9 total squares. 8 are shaded. \( \frac{8}{9} \). NO.
  • Bottom Right: 16 total squares. 12 are shaded (4+4+2+2). \( \frac{12}{16} = \frac{3}{4} \). YES.

Answer: The Top Left and Bottom Right shapes.

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Question 10 (2 marks)

Round 84,516

to the nearest 10: [        ]

to the nearest 100: [        ]

to the nearest 1,000: [        ]

Worked Solution

Step 1: Nearest 10

Number: 84,516

The unit digit is 6 (5 or more rounds up).

84,516 → 84,520

Step 2: Nearest 100

Number: 84,516

The tens digit is 1 (less than 5 rounds down).

84,516 → 84,500

Step 3: Nearest 1,000

Number: 84,516

The hundreds digit is 5 (5 or more rounds up).

84,516 → 85,000

Answers: 84,520, 84,500, 85,000

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Question 11 (2 marks)

Here is a rule for the time it takes to cook a chicken.

Cooking time = 20 minutes plus an extra

40 minutes for each kilogram

1. How many minutes will it take to cook a 3 kg chicken?

minutes

2. What is the mass of a chicken that takes 100 minutes to cook?

kg

Worked Solution

Part 1: Cooking a 3 kg Chicken

Why we do this: The rule says we need 40 minutes for each kilogram, plus 20 minutes extra.

1. Calculate time for the mass:

\[ 3 \times 40 = 120 \text{ minutes} \]

2. Add the extra 20 minutes:

\[ 120 + 20 = 140 \text{ minutes} \]

Answer: 140 minutes

Part 2: Finding Mass from Time (100 minutes)

Why we do this: We need to work backwards. First, remove the “extra 20 minutes”, then divide by 40 to find the kilograms.

1. Subtract the extra time:

\[ 100 – 20 = 80 \text{ minutes} \]

2. Divide by time per kilogram:

\[ 80 \div 40 = 2 \text{ kg} \]

Answer: 2 kg

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Question 12 (2 marks)

Here are diagrams of some 3-D shapes.

Tick each shape that has the same number of faces as vertices.

Cube Square-based pyramid Triangular prism Triangular-based pyramid

Worked Solution

Step 1: Count Faces and Vertices for Each Shape

Why we do this: We need to check the properties of each shape individually.

  • Cube: 6 Faces, 8 Vertices (Corners). No match.
  • Square-based pyramid: 5 Faces (1 base + 4 sides), 5 Vertices (4 base + 1 top). Match!
  • Triangular prism: 5 Faces (2 ends + 3 sides), 6 Vertices. No match.
  • Triangular-based pyramid (Tetrahedron): 4 Faces (1 base + 3 sides), 4 Vertices (3 base + 1 top). Match!

Answer: You should tick the Square-based pyramid and the Triangular-based pyramid.

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Question 13 (2 marks)

Ally and Jack buy some stickers.

PACK Pack of 12 stickers £10.49 12 stickers 99p each

Ally buys a pack of 12 stickers for £10.49.

Jack buys 12 single stickers for 99p each.

How much more does Jack pay than Ally?

Worked Solution

Step 1: Calculate Jack’s Cost

Why we do this: Jack buys 12 stickers at 99p each. We need to find his total cost.

Strategy: \( 12 \times 99p \). It is easier to do \( 12 \times £1 \) and subtract \( 12 \times 1p \).

\[ 12 \times £1.00 = £12.00 \] \[ 12 \times 1p = 12p = £0.12 \] \[ £12.00 – £0.12 = £11.88 \]

So Jack pays £11.88.

Step 2: Calculate the Difference

Why we do this: Compare Jack’s cost (£11.88) with Ally’s cost (£10.49).

  11.88
- 10.49
───────
   1.39

Answer: £1.39

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Question 14 (1 mark)

Amina planted some seeds.

For every 3 seeds Amina planted, only 2 seeds grew.

Altogether, 12 seeds grew.

How many seeds did Amina plant?

Worked Solution

Step 1: Find the Scale Factor

Why we do this: We know 2 seeds grew out of every small group. We need to find how many groups of 2 fit into the total 12 seeds.

\[ 12 \div 2 = 6 \]

There are 6 groups.

Step 2: Calculate Planted Seeds

Why we do this: In each of those 6 groups, she planted 3 seeds.

\[ 6 \times 3 = 18 \]

Answer: 18 seeds

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Question 15 (1 mark)

At the end of a film, the year is given in Roman numerals.

The End MMVI

Write the year MMVI in figures.

Worked Solution

Step 1: Break Down the Numerals

Why we do this: Convert each part of the Roman numeral into numbers.

  • M = 1000
  • M = 1000
  • V = 5
  • I = 1
Step 2: Add Them Up
\[ 1000 + 1000 + 5 + 1 = 2006 \]

Answer: 2006

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Question 16 (1 mark)

Layla completes one-and-a-half somersaults in a dive.

Start

How many degrees does Layla turn through in her dive?

Worked Solution

Step 1: Understand a Full Turn

Fact: A full circle (one somersault) is 360 degrees.

Step 2: Calculate 1.5 Turns

One somersault = 360°

Half a somersault = \( 360 \div 2 = 180^\circ \)

Total = \( 360 + 180 \)

\[ 360 + 180 = 540 \]

Answer: 540°

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Question 17 (1 mark)

The vertices of a quadrilateral have these coordinates:

(1, 5)   (5, 4)   (1, -3)   (-3, 4)

One side of the quadrilateral has been drawn on the grid. Complete the quadrilateral.

x y

Worked Solution

Step 1: Plotting the Missing Points

Why we do this: We need to find the locations of the remaining two coordinates on the grid.

  • (1, -3): Go Right 1, Down 3.
  • (-3, 4): Go Left 3, Up 4.
(1,5) (5,4) (1,-3) (-3,4)

Answer: Join the points to form a kite shape as shown above.

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Question 18 (1 mark)

A cat sleeps for 12 hours each day. 50% of its life is spent asleep.

A koala sleeps for 18 hours each day.

Cat 12h (50%) Koala 18h (?%)

What percentage of a koala’s life is spent asleep?

Worked Solution

Step 1: Write as a Fraction

Why we do this: There are 24 hours in a full day. The koala sleeps for 18 of them.

\[ \frac{18}{24} \]
Step 2: Simplify and Convert to Percentage

Why we do this: Simplify the fraction to make it easier to see the percentage.

Divide top and bottom by 6:

\[ \frac{18 \div 6}{24 \div 6} = \frac{3}{4} \]

We know that:

\[ \frac{3}{4} = 75\% \]

Answer: 75%

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Question 19 (2 marks)

Amina posts three large letters.

The postage costs the same for each letter.

She pays with a £20 note.

Her change is £14.96.

What is the cost of posting one letter?

Worked Solution

Step 1: Calculate Total Cost

Why we do this: Find out how much she actually spent by subtracting the change from the money given.

  20.00
- 14.96
───────
   5.04

She spent £5.04 in total.

Step 2: Calculate Cost per Letter

Why we do this: Divide the total cost by 3 to find the cost of a single letter.

\[ 5.04 \div 3 \]
    1.68
  ──────
3 | 5.04
    2 2 

Answer: £1.68

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Question 20 (1 mark)

Adam says,

0.25 is smaller than \( \frac{2}{5} \)

Explain why he is correct.

Worked Solution

Step 1: Convert to the Same Format

Why we do this: To compare them, both need to be decimals or fractions.

Method 1: Decimals

We know \( \frac{1}{5} = 0.2 \).

So \( \frac{2}{5} = 0.4 \).

Compare: 0.25 is smaller than 0.40.


Method 2: Fractions

\( 0.25 = \frac{1}{4} = \frac{5}{20} \)

\( \frac{2}{5} = \frac{8}{20} \)

\( \frac{5}{20} \) is smaller than \( \frac{8}{20} \).

Answer: Because \( \frac{2}{5} \) equals 0.4, which is greater than 0.25.

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Question 21 (2 marks)

On a map, 1cm represents 20km.

0 20 40 60 80 100

The distance between two cities is 250km.

On the map, what is the distance between the two cities?

Worked Solution

Step 1: Determine the Calculation

Why we do this: Every 20km in real life corresponds to 1cm on the map. We need to find how many “20kms” are in 250km.

\[ 250 \div 20 \]
Step 2: Perform the Division

Simplify by dividing by 10:

\[ 25 \div 2 \]

Half of 25 is 12.5.

Answer: 12.5 cm

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Question 22 (1 mark)

Here are two similar right-angled triangles.

a b

Write the ratio of side \( a \) to side \( b \).

\( a : b = \) [   ] : [   ]

Worked Solution

Step 1: Count the Squares

Why we do this: Since the triangles are on a grid and “similar”, we can compare their corresponding lengths.

  • Triangle A: The base is 3 squares wide. The height is 2 squares high.
  • Triangle B: The base is 6 squares wide. The height is 4 squares high.
Step 2: Determine the Ratio

Why we do this: Compare the dimensions.

Length A = 3, Length B = 6.

Ratio is \( 3 : 6 \).

Simplifying (dividing both by 3):

\[ 1 : 2 \]

Answer: 1 : 2

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Question 23 (2 marks)

In this circle, \( \frac{1}{4} \) and \( \frac{1}{6} \) are shaded.

What fraction of the whole circle is not shaded?

Worked Solution

Step 1: Add the Shaded Fractions

Why we do this: Find the total amount shaded first. We need a common denominator for 4 and 6, which is 12.

\[ \frac{1}{4} = \frac{3}{12} \] \[ \frac{1}{6} = \frac{2}{12} \] \[ \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \]
Step 2: Subtract from the Whole

Why we do this: The whole circle is 1 (or \( \frac{12}{12} \)). We subtract the shaded part to find the unshaded part.

\[ 1 – \frac{5}{12} \] \[ \frac{12}{12} – \frac{5}{12} = \frac{7}{12} \]

Answer: \( \frac{7}{12} \)

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