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 and get the
equation 1288=1480x. Dividing both sides
by 1480 we get x=1288/1480=.87 to the
nearest hundredth. Then to convert 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.
