Boomtown urban planners expect the city’s population to

Look at the table below:

Percentage increase

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

= 66.66

Answer = D

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!

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%.

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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Current year population= 100
Next year population= 110
Next to next year population= 121= 2*population of last year
Last year population= 60.5
Therefore percentage increase of population over last year= {(100-60.5)/60.5}*100=~ 65%
IMO D

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This question seems to be really awkwardly worded and I thought it was fairly unclear. Is this reflective of the kind of wording we can expect on the actual exam or is this more of a MGMAT special?

Let the population of one year ago be "y".
Let population of this year be "x".
Population two years from now= 2y.
Therefore, 2y=x(1.1)^2
y/x = 2/1.21 ~= 1.65
Therefore percent increase = 65%

Let the population of the city be x this year and the population one year ago be y.

Population of city after 2 years = x*1.1*1.1 = 1.21*x

Also, it is given that 1.21*x = 2*y
Thus, y = 0.605x

Percentage increase over the last year = (x - 0.605x)/0.605x = 65.3%

Thus, the correct option is D.

I tried algebra but due to time, I opted for numbers.

121 = 2 (x)

x = 60.5

Base yr was 100; so 100-60.5/60.5 is the percentage jump being asked for. Which is roughly 65%. D

Be careful, option B is a trap when you are in a rush.

2020- 100
2021- 110 (10% increase)
2022- 121 10% increase
2019- 2022/2= 121/2= 60.5

Change in population from 2019 to 2020 is 100-60.5= 39.5
% Change= 39.5/60.5*100= 65%

Posted from my mobile device

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