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Boomtown urban planners expect the city’s population to

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Boomtown urban planners expect the city’s population to  [#permalink]

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New post 03 Mar 2012, 12:57
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Question Stats:

66% (02:29) correct 34% (02:27) wrong based on 295 sessions

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Boomtown urban planners expect the city’s population to increase by 10% per year over the next two years. If that projection were to come true, the population two years from now would be exactly double the population of one year ago. Which of the following is closest to the percent population increase in Boomtown over the last year?

A. 20%
B. 40%
C. 50%
D. 65%
E. 75%

This how I am trying to solve, but I am stuck after this.

If we assume that today’s population is 100, next year it would be 1.1 × 100 = 110, and the following year it would be 1.1 × 110 = 121. If this is double the population of one year ago, the population at that time must have been 0.5 × 121 = 60.5. Because the problem seeks the “closest” answer choice, we can round 60.5 to 60.

Will it be 121-60/60? But this doesn't give me the right answer. Please help.

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Re: Boomtown urban planners expect the city’s population to  [#permalink]

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New post 03 Mar 2012, 13:23
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enigma123 wrote:
Boomtown urban planners expect the city’s population to increase by 10% per year over the next two years. If that projection were to come true, the population two years from now would be exactly double the population of one year ago. Which of the following is closest to the percent population increase in Boomtown over the last year?

A. 20%
B. 40%
C. 50%
D. 65%
E. 75%

This how I am trying to solve, but I am stuck after this.

If we assume that today’s population is 100, next year it would be 1.1 × 100 = 110, and the following year it would be 1.1 × 110 = 121. If this is double the population of one year ago, the population at that time must have been 0.5 × 121 = 60.5. Because the problem seeks the “closest” answer choice, we can round 60.5 to 60.

Will it be 121-60/60? But this doesn't give me the right answer. Please help.


Population now - 100;
Population one year from now - 110;
Population two years from now - 121;

Since the population two years from now (121) is exactly double the population one year ago then the population one year ago was 121/2=60.5.

Now, the question asks about the population increase over the last year, so from 60.5 (last year) to 100 (now): percent increase=difference/original*100=(100-60.5)/60.5*100=39.5/60.5*100=~2/3*100=~65%.

Answer: D.

Hope it's clear.
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Re: Boomtown urban planners expect the city’s population to  [#permalink]

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New post 10 Apr 2014, 01:34
enigma123 wrote:
Boomtown urban planners expect the city’s population to increase by 10% per year over the next two years. If that projection were to come true, the population two years from now would be exactly double the population of one year ago. Which of the following is closest to the percent population increase in Boomtown over the last year?

A. 20%
B. 40%
C. 50%
D. 65%
E. 75%

This how I am trying to solve, but I am stuck after this.

If we assume that today’s population is 100, next year it would be 1.1 × 100 = 110, and the following year it would be 1.1 × 110 = 121. If this is double the population of one year ago, the population at that time must have been 0.5 × 121 = 60.5. Because the problem seeks the “closest” answer choice, we can round 60.5 to 60.

Will it be 121-60/60? But this doesn't give me the right answer. Please help.



You have the calculation correct till 60.5 but we need the percentage increase of 60.5 to 100.00
=> (100 - 60.5)/60.5 = 65%. even if you round 60.5 to 60 you will reach the correct answer.
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Re: Boomtown urban planners expect the city’s population to  [#permalink]

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New post 14 Apr 2014, 00:26
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Look at the table below:

Percentage increase

\(= 40 * \frac{100}{60}\)

= 66.66

Answer = D
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Re: Boomtown urban planners expect the city’s population to  [#permalink]

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New post 11 Apr 2016, 05:46
Am I correct to say that if we write \(P*(\frac{11}{10})^2=2S\) where P=present population, S= start population, so 1 year ago.
That gives \(S=\frac{121P}{200}\) but after this, how can we relate the S to P to solve the question? Does this make sense?..

Like writing \(S*\frac{x}{100}=P\) gives 160% what's wrong here?..

Thank you!
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Re: Boomtown urban planners expect the city’s population to  [#permalink]

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New post 23 May 2016, 18:54
While solving the question, I used hit and trial using some number sense.
So I know since the number doubled after two years when compared to what was year ago.
let us say year ago was 100. If I start with 50 then I get 150 in the current year, 165 the year after and 181.5 the second year after. This does not double.

If I use 75%, then I get 175 this year, 192.5 the next year and more than 200 the second year,
the answer is in between - 65%.
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Re: Boomtown urban planners expect the city’s population to  [#permalink]

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New post 27 Feb 2018, 10:30
Hi All,

This question can be solved with algebra or by TESTing THE ANSWERS. We're essentially asked to keep track of 4 years of data:

Last year
This year
Next year
2 yrs from now

We're told about the percentage increases in population over the last two years (10% each), but we're asked for the percentage increase from last year to this year (with the goal that the population has to DOUBLE over the course of all of these years).

Let's make last year's population = X

Now, let's TEST Answer B….40%

Last year = X
This year = 1.4X.... 10% of 1.4X = .14X, so....
Next year = 1.54X.... 10% of 1.54X = .154X, s....
2 years = 1.694X
This is an increase of 69.4%, which is TOO SMALL. We need a bigger percentage increase…

Let's TEST Answer D…65%

Last year = X
This year = 1.65X... 10% of 1.65X = .165X, so....
Next year = 1.815X... 10% of 1.815X = .1815X, so....
2 years = 1.9965X
This is an increase of 99.65%, which is just about DOUBLE. This is a MATCH for what we're looking for.

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Re: Boomtown urban planners expect the city’s population to &nbs [#permalink] 27 Feb 2018, 10:30
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