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by talaangoshtari » Wed Jun 24, 2015 8:09 am
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
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by DavidG@VeritasPrep » Wed Jun 24, 2015 8:18 am
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
Not a bad question to back-solve.

If we start with B, we'll say the population was 160 in 1700. If we increase by 50% every 50 years, we'll have

1700: 160
1750: 240
1800: 360
1850: 540
1900: 810 --> this should be the population in 1950. So the starting time frame should be pushed ahead 50 years to 1750.


1750: 160
1800: 240
1850: 360
1900: 540
1950: 810

Answer is C
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by GMATGuruNY » Wed Jun 24, 2015 8:24 am
talaangoshtari 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
The population increases by 50% every 50 years.
We can PLUG IN THE ANSWERS, which represent the year in which the population was 160.
When the correct answer choice is plugged in, the population in 1950 will be 810.

Answer choice D: 1800
Population in 1850 = 160 + 50% of 160 = 160+80 = 240.
Population in 1900 = 240 + 50% of 240 = 240+120 = 360.
Population in 1950 = 360 + 50% of 360 = 360+180 = 540.
Too small.

Notice that the population must increase by 50% just ONCE MORE to reach 810.
540 + 50% of 540 = 540+270 = 810.
Since one more 50% increase is needed, the correct answer choice must be 50 years earlier than answer choice D.

The correct answer is C.
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by [email protected] » Wed Jun 24, 2015 9:31 am
Hi talaangoshtari,

This question is perfect for TESTing THE ANSWERS (as David and Mitch have shown). The concept behind the prompt is similar to an interest-rate question though, so you can set it up that way (and do a bit of estimation to get to the correct answer).

We're given a couple of facts to work with:
1) The population increases 50% every 50 years
2) The population in 1950 was 810

We're asked to find the YEAR in which the population was 160.

Using the Compound Interest Formula, we'd have....

Principal = 160
Interest rate = 50% = .5
Time = (Blocks of 50 years)

160(1.5)^T = 810

While trying to solve for the exact value of T could be a bit tedious, we can use estimation...

(1.5)^T = 810/160
(1.5)^T = about 5

1.5^1 = 1
1.5^2 = 2.25
1.5^3 = about 3.33
1.5^4 = about 5

So, 4 "blocks" of 50 years would have to have passed for the population to go from 160 to 810

1750 +50+50+50+50 = 1950

Final Answer: C

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by Ian Stewart » Wed Jun 24, 2015 3:32 pm
When you increase something by 50%, you multiply it by 1.5, or 3/2. So in this question, when we go forward 50 years, the population is multiplied by 3/2. So if instead we want to go back in time 50 years, we'd multiply by 2/3. To go from 810 to 160, we need to multiply by 160/810 = 16/81 = (4/9)^2 = (2/3)^4 so we need to multiply by 2/3 four times, which corresponds to going back in time 4*50 = 200 years. So the answer is 1750.
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by Matt@VeritasPrep » Mon Jun 29, 2015 4:03 pm
Let's say our base population is 160 in year x.

We know that every 50 years our population is multiplied by (3/2).

160 * (3/2)Ë£ = 810
(3/2)ˣ = 810/160 = 81/16 = (3/2)�

So x = 4 and our population has increased four times. It increases every 50 years, so 4 * 50 = 200 years, and the answer must be 1750.
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by nikhilgmat31 » Thu Jul 02, 2015 4:57 am
Great Great question.

plugging values may not help if it is over very big range of years.

We can use Geometric progression formula also

a(n) = a*r^(n-1)

810= 160 * (3/2)^(n-1)

81/16 = 3/2 (n-1)

n-1 =4

n=5

so 1950 is the fifth term of 50 years

1750,1800,1850,1850,1900,1950
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by Jeff@TargetTestPrep » Mon Dec 18, 2017 5:57 pm
talaangoshtari 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
We can create the following equation:

160(1.5)^n = 810

(3/2)^n = 810/160

(3/2)^n = 81/16

Since 3^4 = 81 and 2^4 = 16, we see that n must be 4 so that (3/2)^n = 81/16. However, here n represents the number of 50-year periods that it took for the population to increase from 160 to 810. Thus, the number of years was 4(50) = 200 for the population of 160 to reach a population of 810 in 1950. So, the population of 160 must have occurred in 1750.

Answer: C

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