this icon indicates an accompanying worksheet. this icon indicates teacher’s notes in the notes...

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This icon indicates teacher’s notes in the Notes field.

© Boardworks Ltd 20151 of 23

Product informationProduct informationand sample resources


The KS4 Maths Problem Solving product is designed to give your students plenty of practice and guidance with solving problems in maths.

What’s in the product?

● Each presentation provides one problem solving question.

● It explains how to “decode” the question, highlighting key wording and information.

● The solution is worked through in detail.



Each presentation has an accompanying worksheet to allow students to work through the problem alone first.


Choose to find a question or activity based on a particular topic, browse a selection of functional maths problems, or watch a video solving a problem in real time.


There are also starter activities providing a quick problem for use as a plenary in a lesson.



© Boardworks Ltd 20155 of 23 © Boardworks Ltd 2010

Sample question 1Sample question 1Flower bed


Question: Flower bed

A garden designer is planning to make a rectangular flower bed containing rows of red and yellow flowers.

If there are 60 red flowers and 36 yellow flowers, how many flowers will be in each row? How many rows of each colour flower will there be?

● Each row can only contain one colour of flower.

● The rows must each contain the same number of flowers.

● The rows must contain as many flowers as possible.

● All the flowers must be used.


Decoding the question: Flower bed

● Each row can only contain one colour of flower.

The highest common factor (HCF) of 60 and 36 will give us the number of flowers in each row. The number of rows can then be found afterwards.

There are 60 red flowers and 36 yellow flowers.

● The rows must each contain the same number of flowers.

● The rows must contain as many flowers as possible.

● All the flowers must be used.

How many flowers will be in each row? How many rows of each colour flower will there be?

This means the flowers must divide up into rows exactly.


Worked solution: Flower bed

Factors of 36:

Part 1: How many flowers will be in each row?

Factors of 60:

12 is the highest common factor (HCF) of 36 and 60.

There will be 12 flowers in each row.

12346

123456

There are 36 yellow flowers and 60 red flowers.

Look at the numbers that go into 36 and 60 (their factors).

3618129

603020151210


Worked solution: Flower bed

Part 2: How many rows of each colour flower will there be?

Or calculate 36 ÷ 12 = 3

Yellow flowers: Red flowers:

1 row: 12 flowers2 rows: 24 flowers3 rows: 36 flowers

1 row: 12 flowers2 rows: 24 flowers3 rows: 36 flowers4 rows: 48 flowers5 rows: 60 flowers

There are 36 yellow flowers and 60 red flowers.

Or calculate 60 ÷ 12 = 5

There will be 3 rows of yellow flowers and 5 rows of red flowers.




Sample question 2Sample question 2Two lines


Question: Two lines

The line l1 in the diagram has the equation:

3x – 4y + 6 = 0

a) Find an equation of the line l2.

b) Find the length of AB.

A B

The lines l1 and l2 cross the x-axis at the points A and B respectively.

The line l2 is perpendicular tothe line l1 and passes through the point (2, 4).


Decoding the question: Two lines

This question involves the equation of a straight line, as well as the equations of perpendicular lines.

The line l1 in the diagram has the equation: 3x – 4y + 6 = 0

The lines l1 and l2 cross the x-axis at the points A and B respectively.

The line l2 is perpendicular to the line l1 and passes through the point (2, 4).

a) Find an equation of the line l2.

b) Find the length of AB.

A B


34

Worked solution: Two lines

Rearrange into the form y = mx + c. To do this make y the subject:

3x – 4y + 6 = 0

4y = 3x + 6The gradient of y = mx + c is m. y = x + 3

2

34

We are told that lines l1 and l2 are perpendicular.

If a line has a gradient of m, then any linesperpendicular to it will have a gradient of – .1

m

The equation of l1 is 3x – 4y + 6 = 0. Find its gradient.

The gradient of the line l1 is .

Part (a): What is the equation of the line l2 ?


4 = – × 2 + c


4Since l2 is perpendicular to l1, the gradient of l2 is – . 3

34

The gradient of the line l1 is .

We also know that the line l2 goes through the point (2, 4). Substitute these values into y = mx + c to find the value of c:

43

y = mx + c

4 = – + c83

4 + = c83

= c203

y = – x + 43

203

The equation of l2 is:

This could be rewritten by multiplying by 3, and then rearranging to give:

4x + 3y – 20 = 0



Part (b): What is the distance AB?

A B

At the points A and B, y = 0, because they both lie on the x-axis.

To find the length of AB, you first need to know the coordinates of A and B.

l1 is 3x – 4y + 6 = 0 and it intersects the x-axis at A.

l2 is 4x + 3y – 20 = 0 and it intersects the x-axis at B.


A = (–2, 0)


To find out the x-coordinates of A and B, substitute y = 0 into each equation and then solve for x:

l1: 3x – 4y + 6 = 0

3x + 6 = 0

3x = –6

x = –2

l2: 4x + 3y – 20 = 0

4x – 20 = 0

4x = 20

x = 5

B = (5, 0)

If A is (–2, 0) and B is (5, 0), the horizontal distance between them is:

5 – (–2) = 5 + 2 = 7




Sample question 3Sample question 3Faulty goods


Question: Faulty goods

Should the cats be recalled by the company? Show your working.

● Cost of recall per cat = £15

A fault in the programming of a company’s robot cats means that they sometimes bite!

You need to decide whether or not to recall the cats, based on what it will cost the company.

● Probability of cat biting = 0.1%

● Probability of a bite causing injury = 9/10

● Average claim if bite causes injury = £15,000

● Average claim if bite does not cause injury = £4000


Decoding the question: Faulty goods

Should the cats be recalled by the company? Show your working.

The company will need to recall the cats if they think the cost of not recalling them will be higher.

So, we need to compare the cost of recalling with the expected cost of not recalling.

You need to justify your answer by showing your calculations.

This question is about using probability to find expected values. You will need to work out the expected cost per cat if they are not recalled.


Worked solution: Faulty goods

Step 1: find the cost of recalling per cat

● Cost of recall per cat = £15

We are told in the question that the cost of recall per cat is £15.



Step 2: find the cost of not recalling per cat

If the cats are not recalled, then they might bite someone, and claims may be made. We need to usethe rest of the information from the question to find the expected cost of claims.

P(cat biting) expected claim if cat bites ×

The expected cost is:

We know from the question that

P(cat biting) = 0.1% = 0.001

● Probability of cat biting = 0.1%



● Probability of a bite causing injury = 9/10 = 0.9

● Average claim if bite causes injury = £15,000

● Average claim if bite does not cause injury = £4000

Now we need to find the using the rest of the information from the question.

expected claim if a cat bites

expected claim if cat bites = P(bite causes injury) ×

+

average claim ifbite causes injury

average claim if bitedoes not cause injury

= 0.9 × £15,000 + (1 – 0.9) × £4000

P(no injury) ×

= £13,900

If a cat bites, it may or may not cause injury.



So, the expected cost of not recalling the cat is:

P(cat biting) expected claim if cat bites ×

= 0.001 × £13,900 = £13.90

Step 3: compare the two costs

cost of recalling per cat = £15cost of not recalling per cat = £13.90

The cost of not recalling the cats is smaller, so the cats should not be recalled by the company.

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