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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