# Thread: Percentage Problem: Understanding the Solution

1. ## Percentage Problem: Understanding the Solution

Hi all,
I was wondering if you could help me with the following...

Q. The population of a certain town increases by 50 percent every 50 years. If the population in 1950 was 810, in what year was the population 160?

Ans: 1650 , 1700, 1750, 1800, 1850.

How I would have tackled this was to decrease 810 and the resulting amount by 50% and 50 years until I hit 160. But that's wrong. Here is the solution..

Sol

If the population increases by 50% every 50 years, the population in 1950 was 150% or 3/2 of the 1900 population. How; is it not just a 50% on 1900 so decreasing the value of 1950 by 50% should result in the 1900 figure?
This mean the 1900 population was 2/3 that of the 1950.... and so on. How is that? 2/3 would be 66% would it not?

A layman explanation so I can understand the principle here would be very much appreciated.

Paul

2. Ok, I'm not sure I fully understand the solution you provided, but this is how I think about it.

Say in 1900, there was a certain population. That amount is 100%, you agree with me? Then, to get the population in 1950, you add 50% f that amount, or, you multiply by 150% (50% added to 100%).

The reverse, is to divide by 150%, right?

Now, each year, you will have to divide by 150%. This is a geometric progression and you can use the general formula:

$T_n = ar^{n-1}$

and convert it to:

$P_n = pr^{n-1}$

Where P_n is the population size after going back n times (that is, n = 50 years),
p the initial population size, in this case 810.
r is the common ratio, here it's 1/150%, or 1/(150/100) = 100/150 = 2/3\

Here is where the 2/3 comes from.

Then, at n times, the population became 160:

$160 = 810\left(\dfrac23\right)^{n-1}$

Solve for n.

Then, multiply this by 50 years, subtract this from the year 1950.

3. Thanks; so if the common ratio is 2/3, why is the first calculation performed using 3/2 do you know?

Could I come to the same conculsion by reducing the amounts by 150% at a time?

4. Sure, it should bring you to the same result, because 2/3 is just a simplification of dividing by 150%

5. Ok but is 150 not 3/2?

6. Yes, 150% = 150/100 = 3/2

7. Originally Posted by dumluck
Q. The population of a certain town increases by 50 percent every 50 years. If the population in 1950 was 810, in what year was the population 160?
Ans: 1650 , 1700, 1750, 1800, 1850.
Or look at it this way:
160(1 + percentage_increase)^("n"umber of 50_year_periods) = 810
160(1 + .5)^n = 810
1.5^n = 810/160
n = log(810/160) / log(1.5)
n = 4

SO: 1950 - 4*50 = 1950 - 200 = 1750

8. Thanks. bare with me here. I'm sure we are on our way to a Eureka moment, so why are we multiplying each value by the reciprical? i.e. 2/3 if 150% = 3/2.

9. If we are going 'backwards', you do the reverse, and the reverse of multiplying by 150% is dividing by 150%.

The fact that you have to divide several times by 150% might be tiresome, and what if you have to do that operation some dozen times! This is why you use the power.

10. ahhh ok. So if the question asked for an increase in value we would be multiplying by 3/2?

Good point. Thanks.

11. Originally Posted by dumluck
ahhh ok. So if the question asked for an increase in value we would be multiplying by 3/2?
RIGHT! Same as 1.5 which I show in my example-solution ... see it?

12. Exactly right!

13. Going back to the solution I showed you, you can "re-assure" yourself that 4 periods of 50 years is correct:

160 * (1.5)^4 = 160 * 5.0625 = 810

You can look at this as a financial transaction:
if \$160 is deposited and earns interest of 50% annually (yikes!), what will be the value in 4 years?
F = A(1 + i)^n
F = 160(1 + .50)^4
F = 810 ... if no income tax!!

14. If interest was that much, banks would be ruined in no time while people would get rich extremely quickly