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Proportionality

Proportionality

Proportions

Proportions are used in mathematics to show a relationship between two things. This is very similar to a ratio and gives us information that can be used to work out certain values. You can use proportionality to work out what items in a shop are the best value or to compare a number of other attributes such as weight, volume or size.

Example

In a shop a bag of washing powder costs £4.50 and is 2kg. Another costs £8.00 and is 3.5kg. Work out which is the better value.

Here we can use the proportionality of price to amount of washing powder. So, with the first bag, £4.50 gets us 2kg of powder so that is equal to £2.25 per kilogram. The second bag then gives us 3.5kg for £8. This means that each kilogram costs 8 \div 3.5 which is equal to £2.29 to the nearest penny. This tells us that the best value powder is in the first bag.

When looking at things that are proportional we use the symbol \propto. This shows that two different values are proportional to one another and tells us that two things are only proportional – it does not give us any numerical value or tell us a ratio. For example, if we had t\propto R we would only read this as ‘t is proportional to R’ and would not have any idea what relationship the two have numerically – we simply know that the two are related in some way.

We can use the idea of proportionality when setting up equations. This is done through our knowledge that certain values are linked in some way. For example, if we know that x is proportional to y and that when x is 30, y is 40, we can do the following:

x\propto y

x=ky (where is an unknown number)

When x=30, y=40 so 30=40k

This means that k=30 \div40=\frac{3}{4}

So we are left with x=\frac{3}{4}y

In the above example we have taken the idea that x and y are proportional and used only one point where x is 30 and y is 40. This shows that we can take only small amounts of information and then create an equation by simply knowing that two values are proportional. Don’t worry if you do not understand the notation used here – we will cover algebra later in this course.

Inverse proportions

Not only can values be proportional to one another, they can also be inversely proportional. This means that when one value increases, the other will decrease. For example, we could use inverse proportionality to show the relationship between the amount of people that are working on a task and the time that it takes to complete this.

To show that something is inversely proportional we use the same symbol as previously but put one value as the denominator of a fraction:

r\propto \frac{1}{t}

The expression above would be read out as ‘r is inversely proportional to t’. In this case, when we have a constant that is introduced, it will show as the numerator of the fraction:

r\propto \frac{9}{t} 

Example

A stone wall takes 2 men 5 weeks to build. Show this as a relationship between the amount of men and the time it would take to construct the wall.

Here we see that the time taken to create the wall (t) is inversely proportional to the number of men (which we can show as m) that are helping. Therefore, we can show this relationship using the following:

t\propto \frac{1}{m}

When we now look at the fact that it takes 2 men 5 weeks to build the wall we can adapt the above expression to show this where time is in weeks:

t\propto \frac{1}{m}

t\propto \frac{k}{m}

When t=5, m=2 so 5= \frac{k}{2}

This means that k=10

So we are left with t\propto \frac{10}{m}

Finding the value of a constant

Obviously for a situation we will need to find the value of the constant k when given a real-life set of numbers. To do this we must be told the proportionality of the two variables and be given at least one set of numbers. From this we can then put the numbers given into the equation for the situation and then find the value of k.

Example

Find the value of the constant in the following situations:

1) y=kx and y=4 when x=4

2) a=kb^2 and a=6 when b=1

3) x=\frac{k}{t} and x=12 when t=2

4) a=\frac{k}{b^3} and a=3 when b=2

1) Here we need to put the two values for x and y into the equation given and rearrange to find the constant k. We then get 4=k4 so k=1. Putting this value of k back into the equation given we get y=1x.

2) Putting the values of and into the equation to find k gives 6=k1 so k=6 which can then be put into the equation for a=6b^2

3) Inserting the values into the equation gives us 12=\frac{k}{2} which when rearranged gives k=24. Putting this value into the original equation and we get x=\frac{24}{t}

4) Inserting the values gives 3=\frac{k}{2^3} and rearranging for k gives k=24 again. Putting this value into the equation gives a=\frac{24}{b^3}

We are now able to use all that we have learnt to tackle some questions that are very similar to the ones which you will face in your exam! The basic steps which you must take to complete any of the questions is to make sure that you write out an equation for the problem and then use the information given to find unknowns.

At times you may well be told what two numbers are for the two variables and then you will need to find the value of any constants; we have already tackled this on the previous page. The equation that was then found for the problem will be used to work out some different values from one given variable. An example of this is shown below.

Example

You are told that A is proportional to B. When A is equal to 4, B is equal to 20. Find the value of A when B is 30. Here we need to follow certain steps:

1) Write out the equation. First we need an equation to solve. Since A is proportional to B we can use the correct A=kB where k is a constant that should be found.

2) Find any constants. Now we need to find the value of the constant k by using the numbers that are given (A equals 4 when B is 20). By putting these values into our equation we can find k: 4=k20 so k=\frac{1}{5}.

3) Put the value of k back into the equation to give a general equation. Putting the value for k found in step 2 back into the equation found in step 1 we get A=\frac{1}{5}B.

4) Insert the known value into the general equation. We are asked to find A when B is 30. So if we insert the value for B and then rearrange the equation we can find an answer for A: A=\frac{1}{5}(30). Which tells us that A=6.

By sticking to this routine and working towards getting a general equation first you cannot go wrong with your technique.

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