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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
Table of Contents
- Question 1 (Bar Chart)
- Question 2 (Number Sense)
- Question 3 (Multiplication Grid)
- Question 4 (Height Difference)
- Question 5 (Number Patterns)
- Question 6 (Ordering Decimals)
- Question 7 (Time Conversion)
- Question 8 (Word Problem)
- Question 9 (Fractions of Shapes)
- Question 10 (Rounding)
- Question 11 (Formula)
- Question 12 (3D Shapes)
- Question 13 (Money)
- Question 14 (Ratio)
- Question 15 (Roman Numerals)
- Question 16 (Rotation)
- Question 17 (Coordinates)
- Question 18 (Percentages)
- Question 19 (Division/Money)
- Question 20 (Fractions Explanation)
- Question 21 (Map Scale)
- Question 22 (Similar Triangles)
- Question 23 (Circle Fractions)
Question 1 (2 marks)
William asks the children in Year 2 and Year 6 if they walk to school.
This graph shows the results.
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.
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.
Answer: 50
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
Question 3 (1 mark)
Write the missing numbers to make this multiplication grid correct.
Worked Solution
Step 1: Finding the Top Row Numbers
Why we do this: We know that \( 9 \times \text{something} = 63 \) and \( 9 \times \text{something else} = 54 \).
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. \( \text{Something} \times 7 = 56 \).
Check with the last number: \( 8 \times 6 = 48 \). This works!
Answers:
Top boxes: 7 and 6
Side box: 8
Question 4 (2 marks)
This table shows the heights of three mountains.
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
Question 5 (2 marks)
Complete this table with the missing numbers. One row has been done for you.
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.
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
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
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.
Step 2: Hours to Days
Fact: There are 24 hours in 1 day.
Step 3: Days to Weeks
Fact: There are 7 days in 1 week.
Answers: 5 years, 3 days, 12 weeks
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
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.
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
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?
minutes2. What is the mass of a chicken that takes 100 minutes to cook?
kgWorked 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
Question 12 (2 marks)
Here are diagrams of some 3-D shapes.
Tick each shape that has the same number of faces as vertices.
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.
Question 13 (2 marks)
Ally and Jack buy some stickers.
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 \( 12 \times 1p \).
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
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.
There are 6 groups.
Step 2: Calculate Planted Seeds
Why we do this: In each of those 6 groups, she planted 3 seeds.
Answer: 18 seeds
Question 15 (1 mark)
At the end of a film, the year is given in Roman numerals.
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
Answer: 2006
Question 16 (1 mark)
Layla completes one-and-a-half somersaults in a dive.
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°
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.
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.
Answer: Join the points to form a kite shape as shown above.
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.
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.
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%
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.
1.68
──────
3 | 5.04
2 2
Answer: £1.68
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.
Question 21 (2 marks)
On a map, 1cm represents 20km.
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.
Step 2: Perform the Division
Simplify by dividing by 10:
\[ 25 \div 2 \]Half of 25 is 12.5.
Answer: 12.5 cm
Question 22 (1 mark)
Here are two similar right-angled triangles.
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
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.
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.
Answer: \( \frac{7}{12} \)