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Mathematics Teaching Self-evaluation Tools

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Here you can see a summary of the areas in which you are confident and those in which you are less confident; there are some ideas and suggestions which may help you in your professional learning.

Question £3000 is invested at 2.5% p.a. compound interest. What is the amount after 2 years?

Answer
Compound interest can be found by repeatedly applying the simple interest formula.

A = P + PRT 100, where P = principal investment, R = rate, T = time

Year 1: A= 3000 + 3000 x 2.5 x 1
100

Using P = principal = £3000, R = rate = 2.5%, T = time = 1 year for each year

Now, at the end of Y1 the amount is: 3000 + 75 = £3075

Year 2:

A= 3000 + 3075 x 2.5 x 1
100

Using P = principal = £3075, R = rate = 2.5%, T = time = 1 year for each year

Now, at the end of Y2 the amount is: 3075 + 76.875 = £3075 = £3151.875 = £3151.87 to the nearest penny.

However, it is much more efficient to use the compound interest formula:

Using P = principal = £3000, R = rate = 2.5%, T = time = 2 years

So, A = 3000 × 1.025^{2} = £3151.87 to the nearest penny.

Taking this mathematics further

The Rule of 72, 70 and 69
In finance, the rule of 72, the rule of 70 and the rule of 69 are methods for estimating how long it will take for your investment to double.

The number (72, 70 or 69) is divided by the interest rate to obtain the doubling time. For example using the rule of 72.... at 4% p.a., it will take approximately 18 years (72 ÷ 4) for your investments to double and at 6% p.a., it will take approximately 12 years (72 ÷ 6) for your investments to double.

The choice of number (72, 70 or 69) is mostly a matter of preference but 69 is more accurate, while 72 is more easily divisible.

.

Making connections

If learners are familiar with the idea of working with multipliers they should very quickly be able to identify the multiplier for a compound interest problem and it is then far more efficient to use the compound interest formula.

This principle is applicable to many other types of problem in addition to compound interest; for example: population growth and exponential decay

With simple interest, the amount of interest paid is not reinvested whereas with compound interest, the amount of interest paid is reinvested and earns interest itself.

Simple interest formulae:
I = PRT 100
A = P + PRT 100
where P = principal investment, R = rate, T = time, I = simple interest

Compound interest formulae:

where
P = principal or original investment
R = rate (% per annum)
T = time (in years)
A = total amount

calculate reverse percentages, choosing the correct numbers to take as 100%?(show/hide all)

What this might look like in the classroom

Question 1
A computer is advertised in a sale at £335.75 after a price reduction of 15%. What was the original price of the computer?

Answer 1
£335.75 represents 85% of the original price (100% − 15%)

So 85% of the original price = £335.75
Therefore 1% of the original price = £335.75 ÷ 85 = £3.95
So 100% of the original price = 100 x £3.95 = £395

The original price of the television was £395

Question 2 A gas bill costs £108.10 including VAT at 15%.
What is the cost of the bill without the VAT?

Answer 2 £108.10 represents 115% of the bill (100% + 15%)

115% of the bill = £108.10
Therefore 1% of the bill = £108.10 115 = £0.94
So 100% of the bill = 100 x £0.94 = £94

The gas bill was £94 without the VAT

Taking this mathematics further

Find examples of incorrect uses of percentages. For example, a shop which advertises,

‘WE PAY YOUR VAT – 15% OFF EVERYTHING!’

Making connections

Alternatively, the ideas of proportional reasoning can be used to solve such a problem. Consider the following:

A computer is advertised in a sale at £335.75 after a price reduction of 15%. What was the original price of the computer?

Place the known facts in a table:

?

335.75

x 0.85

So, to ‘undo’, working backwards we need to divide by 0.85 to find the missing value:
335.75 ÷ 0.85 = 395, and the original value of the computer was £395.

divide a quantity into several parts in given ratios?(show/hide all)

What this might look like in the classroom

Question 1 A sum of £50 is to be divided between two children in the ratio 3:2.

How much does each get?

Answer 1
The number of parts is 3 + 2 = 5
The value of each part is £50 ÷ 5 = £10

The two children receive £30 (3 parts at £10 each) and £20 (2 parts at £10 each)

NOTE: It is useful to check that the amounts do add up as a useful check; i.e. £30 + £20 = £50

Question 2
In a fruit cocktail, pineapple juice, orange juice, and apple juice are mixed in the ratio 11:6:3.

How much of each juice is required to make a litre of fruit cocktail?

Answer 2
The number of parts is 11+ 6 + 3 = 20
The value of each part is 1000 ÷ 20 = 50
Now, using 1 litre = 1000 ml:

Pineapple juice: 11 parts x 50 = 550 ml
Orange juice: 6 parts x 50 = 300 ml
Apple juice: 3 parts x 50 = 150 ml

NOTE: Again, it is useful to check that the amounts do add up to the original amount; i.e. 550 + 200 + 150 = 1000 ml (=1 litre)

Taking this mathematics further

There are many genuinely useful real−life applications of this process, (for example, in electronics) but they tend to be so technical that they are not always appropriate for a secondary school classroom. As a result, it is easy to rely on lottery wins, number of days worked and inheritance payments! Take time to research real−life applications yourself and select appropriate examples for use in your classroom.

Research the idea of Proportional Representation (PR) in elections where seats in parliament are decided in proportion to the votes cast.

Making connections

To share an amount into proportional parts you should add up the individual parts and divide the amount by this number to find the value of one part.

The checking mechanism suggested here is likely to ensure that the answer is correct, although there is a slight possibility that a wrong solution could still sum to the original total.

‘Example 1’ and ‘Example 2’ show a different way of approaching the problem by finding fractions of an amount. If this is a learner’s preferred approach they will need to understand where the denominator of the chosen fractions comes from (i.e. the sum of the parts in the ratio).

Proportional parts are also used in sampling techniques such as stratified sampling.

use a spreadsheet to check whether two variables are directly proportional to one another?(show/hide all)

What this might look like in the classroom

Question 1 Which of the following are likely to demonstrate direct proportion:

the speed of a car on a motorway and the distance travelled by the car

the number of hours worked and the amount of pay earned

the length of a side of a cube and the volume of the same cube

the distance between two points on a map and the two points on the ground

The time taken to paint a room and the number of painters employed

the number of questions on an examination paper and the time to complete the examination paper

Answer 1

direct proportion (if speed is constant)

direct proportion

not direct proportion (as volume is proportional to the cube of the?length)

direct proportion

not direct proportion (inverse proportion)

direct proportion (if all questions take the same time − which is probably not likely)

Question 2 Frank takes measurements of the circumference and diameter of a number of circular objects as shown

Use a spreadsheet to test whether one set of numbers is directly proportional to the other set

Answer 2
The spreadsheet produces the following results calculating the ratio circumference ÷ diameter

From the evidence, the two sets of numbers are directly proportional although the third reading is rather suspect and should be checked.

NOTE: The diagram might be also be used to draw a graph which should give a straight line if the two variables are directly proportional to one another

Taking this mathematics further

Other relationships such as yx^{2}, y x^{3}, y y

1

x

^{2}, ....etc. might also be tested as well as exponential and logarithmic proportionality.

Two variables are exponentially proportional to one another if if one of the variables is directly proportional to the exponential function of the other. If y is exponentially proportional to x, you can write y ax or y = ka^{x}, where k is a constant.

Two variables are logarithmically proportional to one another if one of the variables is directly proportional to the logarithm of the other. If y is logarithmically proportional to x, you can write y log_{a}x or y = klog_{a}x, where k is a constant.

Making connections

To determine experimentally whether two variables are directly proportional, the variables can be plotted on a graph. If the points lie on or close to a straight line which passes through the origin (0, 0), then the two variables are probably proportional, with the constant of proportionality given by the gradient of the line.

Two variables are directly proportional to one another if if one of the variables is directly proportional to the other (i.e. they are always in the same ratio). If y is proportional to x, you can write yx or y = kx, where k is a constant.

Similarly, two variables are inversely proportional to one another if one of the variables is directly proportional to the reciprocal of the other. If y is inversely proportional to x, you can write y

find the upper and lower bounds of a calculation using real measures?(show/hide all)

What this might look like in the classroom

Question 1
A rectangle measures 10 cm by 5 cm where each measurement is given to the nearest cm. Write down an interval approximation for the area of a rectangle.

The interval approximation is 42.75 cm^{2} to 57.75 cm^{2}

Question 2
The value of p is 225 and the value of q is 5, with both figures being given to the nearest whole number. Calculate the maximum and minimum values of:

p + q, p − q, p x q, p ÷ q

Answer 2 Using p_{min} = 224.5, p_{max} = 225.5 and q_{min} = 4.5, q_{max} = 5.5

p + q
maximum = 225.5 + 5.5 = 231
minimum = 224.5 + 4.5 = 229

p − q
maximum = 225.5 − 4.5 = 221
minimum = 224.5 − 5.5 = 219

p x q
maximum = 225.5 x 5.5 = 1240.25
minimum = 224.5 x 4.5 = 1010.25

p ÷ q
maximum = 225.5 ÷ 4.5 = 50.111111....
minimum = 224.5 ÷ 5.5 = 40.818181....

NOTE: To get the maximum value of p − q you work out
p_{max} − q_{min} and to get the minimum value of p − q you work out p_{min} − q_{max}

NOTE: To get the maximum value of p ÷ q you work out
p_{max} ÷ q_{min} and to get the minimum value of p ÷ q you work out p_{min} ÷ q_{max}

Taking this mathematics further

Consider the use of bounds (and limits) in statistics

Making connections

If a length is given as 10 cm to the nearest cm, then the actual length will lie in the interval 9.5 cm to 10.499999.... cm as all values in this interval will be rounded off to 10 cm to the nearest cm.

The length 10.499999....cm is usually written as 10.5 cm, although it is accepted that 10.5 cm would usually be rounded off to 11 cm (to the nearest cm).

The value 9.5 cm is called the lower bound as it is the boundary point lowest value which would be rounded to 10 cm The value 10.5 cm is called the upper bound as it is the boundary point of the highest value which would be rounded to 10 cm

Bounds are particularly useful when undertaking measurements since continuous data will always be rounded. Some consideration of this accuracy should be made when using calculations involving measurement.

Bounds are also useful when graphing points as consideration should be given to the most appropriate point to plot the point. For example in cumulative frequency graphs the point should always be plotted at the upper bound.

use a calculator or spreadsheet to explore exponential growth and decay?(show/hide all)

What this might look like in the classroom

Question 1
The number of bacteria on a table top doubles each hour.
Starting with one bacterium, how many bacteria will there be after 24 hours?

Answer 1
A calculator or a spreadsheet is useful to demonstrate the answer. For exponential growth, the formula is

In this case a = 1 (the number of bacteria), r = 100 (the percentage increase per hour) and t = 24 (the number of hours in question)

So Þ

This can be shown on a spreadsheet as follows

After 24 hours there will be 16777216 bacteria

Question 2 The population of a town is expected to grow at a rate of 10% per annum. The population in 2008 was 18000.
In which year will the population have doubled in size?

Answer 2
For exponential growth, the formula is

In this case, a = 18000, r = 10 and t = ?

So Þ y = 18000(0.1)^{t}

This can be shown on a spreadsheet as follows

Time

0

1

2

3

4

5

6

7

8

9

10

Year

=B1+2000

=C1+2000

=D1+2000

=E1+2000

=F1+2000

=G1+2000

=H1+2000

=I1+2000

=J1+2000

=K1+2000

=L1+2000

Population

=18000*1.1^B1

=18000*1.1^C1

=18000*1.1^D1

=18000*1.1^E1

=18000*1.1^F1

=18000*1.1^G1

=18000*1.1^H1

=18000*1.1^I1

=18000*1.1^J1

=18000*1.1^K1

=18000*1.1^L1

Time

0

1

2

3

4

5

6

7

8

9

10

Year

2000

2001

2002

2003

2004

2005

2006

2007

2008

2009

2010

Population

18000

19800

21780

23958

26353.8

28989.18

31888.1

35076.91

38584.6

42443.06

46687.36

From the spreadsheet, it can be seen that the population will double in 2008 (after 8 years)

Taking this mathematics further

A useful oversight about the exponential function is offered in the video here.
A development of this work might be to look at the exponential function which is further discussed here.

Making connections

Any quantity that grows or decays by a fixed percent at regular intervals is said to possess exponential growth or exponential decay. The growth/decay can be represented mathematically by the formulae

Exponential growth:

Exponential decay:

Where a = initial amount
r = rate expressed as a %
t = time

Exponential growth and decay are prevalent across a number of subject areas including biology (consideration of the growth of microorganisms or the spread of a virus), physics (nuclear reactions), computers (processing power and internet growth) and finance (such as the growth rate of investments).

Exponential growth is also identified in the rice on a chessboard problem (see this link) or else the idea of repeatedly folding a piece of paper in half (see here)

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