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The population of a certain town increases by 50 percent every 50 year

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The population of a certain town increases by 50 percent every 50 year [#permalink]

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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?

A. 1650
B. 1700
C. 1750
D. 1800
E. 1850
[Reveal] Spoiler: OA
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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cano wrote:
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?

A. 1650
B. 1700
C. 1750
D. 1800
E. 1850


Increase by 50% = times 1.5, so if the population at a certain year were 10 then in 50 years it would be 10*1.5 and in 50*2=100 years it would be 10*1.5*1.5=10*1.5^2 --> \(160*1.5^n=810\), where \(n\) is the # of 50-year periods --> \((\frac{3}{2})^n=\frac{81}{16}=(\frac{3}{2})^4\) --> \(n=4\), so there were 4 50-year periods --> 1950-4*50=1750.

Answer: C.
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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New post 20 Sep 2010, 12:43
4 50 yrs to reach 810
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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I just worked this one backwards until I reached 160. If the population increases by 50% then you're multiplying the previous number by 3/2. So to work backwards, divide by 3/2 (which is the same as multiplying by 2/3).

So we have:

1950: 810
1900: 810*(2/3) = 540
1850: 540*(2/3) = 360
1800: 360*(2/3) = 240
1750: 240*(2/3) = 160

Answer is C
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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New post 20 Sep 2010, 17:50
redjam wrote:
I just worked this one backwards until I reached 160. If the population increases by 50% then you're multiplying the previous number by 3/2. So to work backwards, divide by 3/2 (which is the same as multiplying by 2/3).

So we have:

1950: 810
1900: 810*(2/3) = 540
1850: 540*(2/3) = 360
1800: 360*(2/3) = 240
1750: 240*(2/3) = 160

Answer is C


How long did it take you to work it like that?
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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New post 20 Sep 2010, 18:41
55 sec to work forward for C

160 -> 240 -> 360 -> 540 -> 810
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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New post 21 Sep 2010, 06:35
cano wrote:
How long did it take you to work it like that?


Didn't specifically time myself but it wasn't longer than 2 mins. I would say around 1.5 mins. They are all numbers easily divisible by 3.
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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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Re: The population of a certain town increases by 50 percent every 50 year [#permalink]

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New post 13 Feb 2017, 22:40
Hi All,

While this is an older prompt, it's still a great example of 'exponential growth', which is a concept that still appears on the GMAT occasionally. You can answer this question rather easily by TESTing THE ANSWERS.

We're told that a population increases by 50 percent every 50 years and that the population in 1950 was 810. We're asked to find the year in which the population was 160.

Let's TEST Answer B: 1700

IF....
the population was 160 in 1700 then...
the population was 240 in 1750 and...
the population was 360 in 1800 and...
the population was 540 in 1850 and...
the population was 810 in 1900

This is TOO EARLY (the population is supposed to hit 810 in 1950). Notice the pattern here though - if we just "shift" all of the values 50 years 'forward', then we'll hit 810 in 1950 (when the population in 1750 was 160).

Final Answer:
[Reveal] Spoiler:
C


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Re: The population of a certain town increases by 50 percent every 50 year   [#permalink] 13 Feb 2017, 22:40
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