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.
