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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
     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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You mixed up formula's for depreciation & compound interest. An asset $x$ depreciates every year to $x(1-R).$ That's $x$ multiplied by a fraction $< 1.$ An asset $x$ (or population) increases every year to $x(1+R).$ That's $x$ multiplied by a constant $> 1.$ – user2468 Jul 30 '12 at 18:02
If $x$ is decreased by $y\%$ to obtain $z$, then $z$ increased by $y\%$ is not $x$ (try it with $x=100$ and $y=10\%$). So, starting in 1985, decreasing by $5\%$ each year to 1982 is not the same as starting in 1982 and increasing by $5\%$ per year to 1985. – David Mitra Jul 30 '12 at 18:03
@ David Mitra so then how do i proceed? – Nirvan Jul 30 '12 at 18:09
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. – David Mitra Jul 30 '12 at 18:13
up vote 0 down vote accepted

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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the answer is obviously wrong because the population can't be in decimal – Nirvan Jul 30 '12 at 18:20
@Nirvan Of course, silly me :) – David Mitra Jul 30 '12 at 18:25
ok so i got the answer by doing this: 185220=x(1+5/100) = 160000, thanks – Nirvan Jul 30 '12 at 18:29
@Nirvan You want $(1+5/100)^3$ there (three years of increase). So the equation is $x(1+5/100)^3=185220$. Solving this for $x$ gives $x=185220/(1.05)^3=160000$. Note, it would be wrong to write $185220=x(1+5/100)^3\color{maroon}{=160000}$. You have an equation here. Its solution needs to be written as a separate statement. – David Mitra Jul 30 '12 at 18:35
i don't understand what you're saying is missing – Nirvan Jul 30 '12 at 18:38

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.

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