Percent increase is when you find a given percent of an amount and
then add it to it. Percent decrease is when you find a given percent of
an amount and subtract it from it. One really important thing to
remember about percent increase and decrease is that percent is always
based on something, it doesn't stand alone, there is always something
that it is "of" and with percent increase and decrease problems that is
always the ORIGINAL amount. There are three kinds of problems percent
increase and decrease problems and a good first step towards solving a
problem is to decide which of them it is.
Original amount, percent known, you want to find the new amount.
This is the easiest kind. Just
change the
percent into a decimal or fraction and multiply by the original
amount to find the amount of the increase or decrease and then if it is
an increase, add it to the original amount, and if it is a decrease
subtract it from the original amount.
Example
If inflation is running at 7%, how much would you plan to set aside for
living costs next year if your current expenditure is $920 per month?
Solution
This is a percent increase because 7% inflation tells us that next year
prices will be 7% higher than they are this year so your monthly
expenditure will be 7% higher. So to get the amount of the increase in
your monthly expenditure you need to find 7% of your current
expenditure. That is you need to compute 7% of $920.
To do that you covert 7% to a decimal by
moving the decimal point two places and you get .07, so to get the
amount of the increase you multiply .07 times $920 and that gives you
$64.40. This means that your monthly expenditure next year will be
$64.40 more that this year. Then to figure out what your monthly
expenditure will be for next year you need to add $64.40 to $920, and
the result of this is $984.40.
Original amount known, new amount known, you want to find out what
percent the increase or decrease is.
This is a little bit harder, but still pretty straight forward. Just
subtract and then figure out what percent the difference is of the
ORIGINAL amount by using methods of straight percentage problems.
Example
The value of U.S exports to the Soviet Union increased from $1480
million in 1987 to $2768 million in 1988. What was the percentage
increase?
Solution
First you figure out what the increase was by subtracting. It is
convenient here to do the whole thing in millions, so the difference is
2768-1480=1288. Then we need to figure out what percent 1288 is of the
original amount 1480, so we solve this like problems like this are
normally solved (See my
Math 100 notes.)
and get the equation 1288=1480x. Dividing both sides by 1480 we get
x=1288/1480=.87 to the nearest hundredth. Then to c
onvert that to percent multiply by 100 by
moving the decimal place two places to the right and we get that
this is an 87% increase.
New amount known, percent known, you want to find the original
amount.
This is the tricky case. This is a kind of problem that you really have
to think of as an algebra problem, because if you think of it as an
arithmetic problem is seems not to be do-able. To see better what's
going on here let's first look at a problem of the
first
kind.
Problem
An item originally costing $20 is on a 10% off sale. What is the sale
price?
Solution
Figure out what 10% of 20 is to find out how much the reduction in the
price is. To find 10% of 20 change 10% to a decimal and you get .1.
Then multiply .1 times 20 to get (.1)(20)=2. So the price reduction is
$2, so to get the new price we subtract this from 20 to get 20-2=18, so
the sale price is $18.
New Problem
But now suppose we don't know the $20 dollar original price and we only
know the $18 sale price and the fact that this was the result of a 10%
discount, and we want to find out what the original price was. The
temptation as it always is in these problems is to think that if you
want to undo a 10% decrease that you should just make a 10% increase.
This sounds logical if you decrease the price by 10% and then increase
it by 10% you ought to get back to where you started. But this doesn't
work. It doesn't work because when you think this sort of thing you are
forgetting that percents don't stand alone, they are always based on
something, so it is not just like you are subtracting 10 and then
adding it back again, you are subtracting 10% of something and adding
back 10% of something. That would be fine too as long as the two
somethings were the same, but if you apply a 10% decrease and then a
10% increase they are not. You see in this example if we now applied a
10% increase, we would be finding 10% of 18 and adding that to 18. That
would not give us the original $2 that was taken away to get the 18,
but only $1.80, and if we added that to the 18 we would only get $18.80
not the $20 that we know is correct. What we would really need to do if
we were to do it this way is to find 10% of 20, the original price, and
add that to the 18 to get the 20. But that would require knowing the
original price, and if we knew that we would be done anyway, so it
looks like we are kind of stuck.
So in comes the algebraic approach to rescue. Call the unknown
original amount x, and do to the x just what you did to the 20 to get
the 18 and then set this equal to 18. Just like with the 20, to
find 10% of x we first change 10% into a decimal, and we get .1. Then
we multiply this by x to get .1x. That is the amount of the prices
reduction, so now to get the sale price we subtract this from x and get
x-.1x and since we know this is 18 we get x-.1x=18 for our equation.
Now we have an algebra equation that we can solve to get our answer. To
do that first we combine like terms and get .9x=18. Then we divide both
sides of the equation by .9 to get x=18/.9=20. So we get, as expected,
that the original price was $20.
Another Example
The population of a certain town increased by 2% to 10,000. What was
the population before the increase?
Solution
Call the population before the increase x. It increased by 2%, so the
amount it increased by can be obtained by figuring out what 2% of x is.
To do that we first change 2% to a decimal by moving the decimal point
two places and we get .02x. Then we add that to the x to get the new
population that we know is 10,000, so we get x+.02x=10,000 for the
equation that expresses what the problem has told us. Now to solve this
equation we first combine like terms to 1.02x=10,000. Then we divide
both sides by 1.02 to get our answer x=10,000/1.02=9804 to the nearest
person. So the population before the increase was 9804.