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Calculating Percentage IncreaseDate: 07/09/2004 at 15:38:52 From: Sharon Subject: (no subject) If my salary increased from 45k to 65k over a period of 10 years, what percentage increase did I receive per year?
Date: 07/10/2004 at 12:24:57
From: Doctor Ian
Subject: Re: (no subject)
Hi Sharon,
It depends on how you're defining things. To show you what I mean,
suppose you're making $45K in 1990. 10% of that is $4.5K. If you
receive an increase of $4.5K per year for 10 years,
1990 $45,000
1991 $49,500 add 10% of $45,000
1992 $54,000 add 10% of $45,000
1993 $58,500 add 10% of $45,000
1994 $63,000
.
.
2000 $90,000
But suppose each year, you receive an increase amounting to 10% of
what you made that year. Now each increase is a little bigger than
the last:
1990 $45,000
1991 $49,500 add 10% of $45,000
1992 $54,450 add 10% of $49,500
1993 $59,895 add 10% of $54,450
What's going on in this case is that each year, the previous salary is
multiplied by 1.10 (100% plus 10%), so after 10 years the salary would be
10
45,000 * 1.10 = 45,000 * 2.59
= 116,550
So to answer your question, if you're talking about the first case,
you just take the total difference and divide by 10 to get the yearly
increase,
65,000 - 45,000 20,000
--------------- = ------ = 2,000
10 10
and then you figure out what percentage that is of the original
salary:
2,000 ?
------ = ---
45,000 100
2 ?
-- = ---
45 100
2*100 = 45*?
200
--- = ?
45
which comes out to about 4.4%. That is, if you increase your original
salary by 4.4% OF THE ORIGINAL SALARY for 10 years, you end up with $65K.
In the second case, the situation is this:
10
45,000 * ? = 65,000
We can start by dividing each side by 45,000:
10
? = 65,000 / 45,000
10
? = 1.44
Now what we want to know is: What number, raised to the 10th power,
gives us 1.44? It turns out that
10
1.037 = 1.438
so if you increase your original salary each year by 3.7% OF THE
PREVIOUS YEAR'S SALARY, after 10 years you end up with $65K.
Which method you prefer might depend on whom you're trying to convince
of what. :^D
Does this help?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
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