Boomtown urban planners expect the city’s population to incr : GMAT Problem Solving (PS)
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# Boomtown urban planners expect the city’s population to incr

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

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02 Feb 2010, 12:21
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Difficulty:

65% (hard)

Question Stats:

60% (01:45) correct 40% (00:59) wrong based on 172 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%

[Reveal] Spoiler:
Please see the above question and the explanation... In the question the city population increase by 10% per year. How then the population can be exactly double the population of one year ago? I do not get it

Explanation:
This problem can be solved most easily with the help of smart numbers. With problems involving percentages, 100 is typically the ‘smartest’ of the smart numbers.

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.

In this scenario, the population has increased from 60 to 100 over the last year, a net increase of 40 residents. To determine the percentage increase over the last year, divide the net increase by the initial population: 40/60 = 4/6 = 2/3, or roughly 67%.

For those who prefer the algebraic approach: let the current population equal p. Next year the population will equal 1.1p, and the following year it will equal 1.1 × 1.1p = 1.21p. Because the question asks for the closest answer choice, we can simplify our algebra by rounding 1.21p to 1.2p. Half of 1.2p equals 0.6p. The population increase would be equal to 0.4p/0.6p = 0.4/0.6 = 2/3, or roughly 67%.

You must select an answer to proceed.

OPEN DISCUSSION OF THIS QUESTION IS HERE: boomtown-urban-planners-expect-the-city-s-population-to-128519.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Nov 2013, 07:38, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: is something wrong with this question? or it is me mad? [#permalink]

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04 Feb 2010, 10:21
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aleyna wrote:
please see the below question and the explanation... In the question the city population increase by 10% per year. How then the population can be exactly double the population of one year ago? I do not get it

QUESTION from Manhattan test:

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?
20%
40%
50%
65%
75%

Explanation:
This problem can be solved most easily with the help of smart numbers. With problems involving percentages, 100 is typically the ‘smartest’ of the smart numbers.

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.

In this scenario, the population has increased from 60 to 100 over the last year, a net increase of 40 residents. To determine the percentage increase over the last year, divide the net increase by the initial population: 40/60 = 4/6 = 2/3, or roughly 67%.

For those who prefer the algebraic approach: let the current population equal p. Next year the population will equal 1.1p, and the following year it will equal 1.1 × 1.1p = 1.21p. Because the question asks for the closest answer choice, we can simplify our algebra by rounding 1.21p to 1.2p. Half of 1.2p equals 0.6p. The population increase would be equal to 0.4p/0.6p = 0.4/0.6 = 2/3, or roughly 67%.

You must select an answer to proceed.

Well, I think its clearly mentioned in th question that the population will increase at a rate of 10% over the next two years only & no where in the question is mentioned that at the same rate the population was increased over the last year. Actually you caught in trap set by the question.
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Re: is something wrong with this question? or it is me mad? [#permalink]

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04 Feb 2010, 16:19
we have 4 years populations :

______Y_______X______1.1X_______1.21X

we ar told that 1.21x = 2y

we are at the end of year 1 where the population is lets say x, But at the end of year 3 the population will be 1.21 x , which is double the population and the end of year zero. Which is Y

now we want to calculate the % change of year 1 the one in red . which started by number of population Yand ended with number of population X.

Y = ( 1.21x) / 2

$$\frac{ X - \frac{1.21X}{2} }{ \frac{1.21X}{ 2}$$ equate this relationship to the unknown percentage change then simplify it you will get 0.65

Last edited by Pedros on 04 Feb 2010, 17:49, edited 3 times in total.
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Re: is something wrong with this question? or it is me mad? [#permalink]

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04 Feb 2010, 16:27
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When I read the question the first thing I did was draw a timeline to make sure I had my numbers in order, which may also help you understand the progression.

x : 100 : 110 : 121

x = one year ago
100 = current day
110 = next year (10% growth)
121 = two years from now (add'l 10% growth)

121 = 2x since per the question it is double the population of x, which means x = 60.5.
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Re: is something wrong with this question? or it is me mad? [#permalink]

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07 Feb 2010, 05:40
The way I did it is similar to the official explanation, but much less efficient.

Today's population is P. in 2 years it will be 1.1x1.1xP = 1.21xP, which is the same as 2 times the population last year:
1.21xP = 2xP/1+x (1+x is the 'discount factor')
=> 1+x = 2/1.21

then I did the following:
1.21 x 1.5 ~ 1.2x1.5 = 1.2+0.6 = 1.8
1.21 x 1.6 ~ 1.2x1.6 = 1.2+0.72 = 1.92
1.21 x 1.7 ~ 1.2x1.7 = 1.2+0.84 = 2.04

so it must be between 60 and 70%
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Re: is something wrong with this question? or it is me mad? [#permalink]

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24 Nov 2013, 05:06
Pls hide explanation & OA....
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Re: is something wrong with this question? or it is me mad? [#permalink]

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24 Nov 2013, 07:40
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monirjewel wrote:
Pls hide explanation & OA....

Done.

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%

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%.

OPEN DISCUSSION OF THIS QUESTION IS HERE: boomtown-urban-planners-expect-the-city-s-population-to-128519.html
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Re: is something wrong with this question? or it is me mad?   [#permalink] 24 Nov 2013, 07:40
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