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The population increases by 5% every year. What was the population in 1982, if in 1985 it was 1,85220?

My working:

     population in 1985 = 1,85,220      rate=5%      time = 3yrs    ( A=P(1-R/100)^n )      therefore, population in 1982 = 185220(1 - 5/100)^3                           = 158802.9975 

obviously that's wrong, but where's the problem?

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    As in Williams answer: take the original value (in 1982). You know that if you increase by $5\%$ per year, you get 185220. This gives you an equation whose variable is what you're looking for.2012-07-30

2 Answers 2

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You are making a basic error: If you increase a quantity $P_1$ by $y\%$ of itself to obtain the new value $P_2$, then decreasing $P_2$ by $y\%$ of itself does not get you back to $P_1$ (because you also decrease by $y\%$ of the increase previously gained). For example, increasing $100$ by $10\%$ of itself gives you $110$, but decreasing $110$ by $10\%$ of itself gives you $99$.

You need to start with the 1982 value, call this $x$, and then find an equation in $x$ and solve. William's answer explains how to do this.

(Incidentally, why do you say your answer is "obviously wrong"?)

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    oh, ok i see , well then thanks :)2012-07-30
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To increase by $5$ percents is to multiply by $1.05$. This occurred for 3 years. So you obtain the equation:

$(1.05)^{3}x = 185220$

Solve for $x$ and this is the population in 1982.