**What is a percentage?**

A percentage is the top part of a fraction whose bottom part is 100.

So 50% means 'half of' and 25% means 'a quarter of'. 100% means the
complete quantity.

Percentages are useful because they make it very easy to compare things.

For example, suppose the marks in two successive tests are 67/80 and 51/60. It is not very easy to say which of these was best. Percentages use our ordinary number system of 10's, 100's etc and, because they are out of 100 rather than 10, we avoid a lot of the decimal points which make some people twitchy.

**Changing a fraction to a %**

Taking the example of the test mark of 67 out of 80,

Multiplying both sides of this equation by 100 gives us

**Question:-** What is the second test mark of 51/60 as a %?
Try this yourself before looking.

For example, a drinks bill of £20 for a party costing £80 in total means that the drinks cost 20/80 or 1/4 or 25% of the total. The £20 on its own can't be turned into a percentage.

Then cancel down if possible.

Example:- What is 35% as a fraction?

cancelling down to the simplest form by dividing the top and bottom by 5.

*******
Remember that the value of a fraction remains unchanged when you multiply
or divide *both the top and the bottom* by the same number.
*******

**Changing a decimal to a %**

Dead easy, this one!

Suppose we want to write 0.27 as a %. Since a decimal is a kind of
fraction, all
we have to do is to multiply by 100. You just need to remember that
each time you multiply by 10 the number becomes larger by a factor
of 10 so the decimal point moves one place
to the right. Multiplying by 100 moves it 2 places to the right.
This neat rule is because decimals are fractions in our base ten number
system.

So we find that 0.27 is the same as (0.27 x 100)% = 27%.

Similarly, 0.735 is the same as (0.735 x 100)% = 73.5%

and 7.46 is the same as (7.46 x 100)% = 746%.

which just moves the decimal point 2 places to the right.

**Changing a % to a decimal**

(Again, dead easy!)

to the left.

Here are 3 examples to show you how to deal with all
possible snags.

**Example (1)** What is 37% as a decimal?

**Example (2)** What is 25.5% as a decimal?

**Example (3)** What is 50% as a decimal?

The easiest way to explain how to work these out is to look at some examples.

**Example (1)** Suppose the profits of a certain company go from £365 000
in January to £425 000 in February. What is the % increase in their profits?

value before the change took place.

Here, the actual increase in profits is £425 000 - £365 000 = £60 000.

The % profit is £60 000 as a percentage of £365 000

The actual decrease is 127 - 114 = 13.

The % decrease is 13 as a % of 127.

13 as a fraction of 127 is 13/127.

Now, just multiply by 100 so you get

There are two ways of finding this.

**Method (1)**

First find the actual increase in cost. This is 8% of £7 800 so it is

Therefore the new price is £7 800 + £624 = £8 424.

**Method (2)**

This is the all-in-one way of doing it.

The new price is 108% of the old one, so it is

At the beginning of December, the price of a certain item is increased by 5% to make a bigger Christmas profit.

At the beginning of January, there is a Sale and the unsold items are labelled

Would you now be paying the same as if you had bought the item in November? If not, would you be paying more or less?

Have a go at answering this before looking.

Suppose it cost £100 in November.

Then in December the price increased by 5% to £105.

The Sale Price is now calculated as a reduction of 5% on this current price of £105.

So, using method (1), the actual reduction in price is

Therefore you would only pay £105 - £5.25 = £99.75. You would get it cheaper in January than you would have done if you had bought it in November.

A certain computer store reckons to make 30% profit on each gizmo that it sells.

If the selling price of a particular gizmo is £1 560 what was the cost price to the dealer?

We know that the selling price of £1 560 is 130% of the cost price since the dealer is making a profit of 30% on the cost price. So we can say

Now multiply both sides of this equation by 100 and divide both sides by 130. This gives

since the two right-hand fractions cancel each other out.

Therefore the cost price to the dealer was £156 000/130 = £1 200.

In a sale, there is a rack of coats marked "All prices in this rack are reduced by 20%!".

The one I choose now has a price of £120. How much did it cost before the sale?

The working is very similar to the previous example. We know that £120 is 80% of the original selling price since 20% has been taken off this price. This time we'll save some writing in the equations by calling the original selling price P. Then we have

Now, multiplying both sides of this equation by 100 and dividing both sides by 80, we have

So the original selling price was £12 000/80 = £150.

A little bit of algebra saves a lot of writing!

People have found difficulty here when they needed to work backwards from a bill whose total includes VAT (value added tax) at 17.5% to find out what the bill would have been before the VAT was added.

The easiest way to explain how this is done is to take an actual example.

Suppose that a bill which includes VAT comes to £1602.70. We want to know what the amount was before VAT was added.

We'll save writing by calling the amount before VAT was added C.

Then, working in £, we know that

The total of the bill before VAT was added was £1364.

We can make a general rule for this if we let P stand for the total of the bill including VAT at 17.5%.

We want to find C, the amount of the bill before VAT was added.

Using the same argument as above, we get

The working here is similar to Example (5) above but VAT seems to create special difficulties.

About a year ago I added the request between the two lines of red asterisks.

I'm thinking about writing a little book on percentages particularly to help adults who need to use them but have forgotten how they work. The advantage over the web would be that there could be lots of practice questions to do mixed in with my rules and examples. Working out problems is an important part of learning but I don't think people much like doing it on the web.

I'd be very grateful to have feedback on this idea. If you can help me decide please email me at xyzjenolive@netcomuk.co.uk leaving out the xyz which is against robot viruses. Please don't send attachments as I don't risk opening them.

I'm very grateful to everyone who has emailed me about this and have responded individally to all the suggestions. The main problem as I now see it is that people want help straight away which is what I hope the web pages give. Having to buy a book (it couldn't be free because it would cost to produce) would delay the help. It might be that people who often visit the site would rather have it in a more convenient book form. I know my total numbers of visitors to each page but nothing else about them so don't know if they revisit, though my numbers suggest that quite a few people have bookmarked my "basic stuff on how percentages work" page. With additions this has now become quite long. Users are welcome to print out my pages as a reference for their own personal use.

One great advantage of the web for a writer is the possibility of changing or adding to
what is already written. It is a living medium. Also, we can both talk to each other
by email if we want to.
I'm always interested in comments or ideas and my email address is
xyzjenolive@netcomuk.co.uk leaving out the xyz which is against robot spam and viruses but
please don't send attachments as I don't open them.

You can practise all the percentage rules now by working out the answers to some problems yourself.

or back to the percentages home page