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Final population growth = S * P ^ (t/l)
S = starting population
P = progression (doubles = 2, triples = 3 etc.)
t/l = total amount of iterations
t = time
I = intervals

1,000,000 = 2,000 * 2 ^ (t years/ 1 year) = > 1,000,000 / 2000 = 2^ t = > 500 = 2^t - which from here t = 9
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jlgdr
The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?

\(2^n*2,000\geq{1,000,000}\) --> \(2^n\geq{500}\) --> \(n=9\) (since 2^9=512).

Similar questions to practice:
https://gmatclub.com/forum/a-certain-bac ... 44013.html
https://gmatclub.com/forum/it-takes-30-d ... 40269.html
https://gmatclub.com/forum/a-certain-cul ... 52258.html
https://gmatclub.com/forum/the-number-of ... 42858.html

Hope it helps.

Bunuel Why is it 2^n and not 2^(n-1)? I used the formula for the nth term of GP. The other links to similar questions provided also do not address this. Could you please help out?
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Bunuel
jlgdr
The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?

\(2^n*2,000\geq{1,000,000}\) --> \(2^n\geq{500}\) --> \(n=9\) (since 2^9=512).

Similar questions to practice:
https://gmatclub.com/forum/a-certain-bac ... 44013.html
https://gmatclub.com/forum/it-takes-30-d ... 40269.html
https://gmatclub.com/forum/a-certain-cul ... 52258.html
https://gmatclub.com/forum/the-number-of ... 42858.html

Hope it helps.

Bunuel Why is it 2^n and not 2^(n-1)? I used the formula for the nth term of GP. The other links to similar questions provided also do not address this. Could you please help out?

Hi MooneeyTunes,

The type of math notation you might choose to use will depend on the specific question that is ASKED. In this prompt, we're told that EACH YEAR the population doubles... meaning that we're going to multiply a 'starting number' by a certain number of "2s." Each year (including the first year) is another '2', so 2^N is appropriate here.

IF... that first year involved no change in the total (or a change other than a 'double') - AND you had to account for that first year while noting that it was NOT a year in which the population doubled - then 2^(N-1) would be appropriate.

GMAT assassins aren't born, they're made,
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Bunuel Why is it 2^n and not 2^(n-1)? I used the formula for the nth term of GP. The other links to similar questions provided also do not address this. Could you please help out?

Hi MooneeyTunes,

The type of math notation you might choose to use will depend on the specific question that is ASKED. In this prompt, we're told that EACH YEAR the population doubles... meaning that we're going to multiply a 'starting number' by a certain number of "2s." Each year (including the first year) is another '2', so 2^N is appropriate here.

IF... that first year involved no change in the total (or a change other than a 'double') - AND you had to account for that first year while noting that it was NOT a year in which the population doubled - then 2^(N-1) would be appropriate.

GMAT assassins aren't born, they're made,
Rich[/quote]

EMPOWERgmatRichC
I am sorry. I am not sure if I completely follow this. Because if the population doubles every year and we write the series as 2000, 2*2000, 2^2*2000 and so on and consider first term of the G.P to be 2000 then the nth term should be 2000 * 2^(n-1).
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Bunuel Why is it 2^n and not 2^(n-1)? I used the formula for the nth term of GP. The other links to similar questions provided also do not address this. Could you please help out?

Hi MooneeyTunes,

The type of math notation you might choose to use will depend on the specific question that is ASKED. In this prompt, we're told that EACH YEAR the population doubles... meaning that we're going to multiply a 'starting number' by a certain number of "2s." Each year (including the first year) is another '2', so 2^N is appropriate here.

IF... that first year involved no change in the total (or a change other than a 'double') - AND you had to account for that first year while noting that it was NOT a year in which the population doubled - then 2^(N-1) would be appropriate.

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

EMPOWERgmatRichC
I am sorry. I am not sure if I completely follow this. Because if the population doubles every year and we write the series as 2000, 2*2000, 2^2*2000 and so on and consider first term of the G.P to be 2000 then the nth term should be 2000 * 2^(n-1).[/quote]

Hi MooneeyTunes,

It might be easier if we look at this from the 'beginning' of the sequence.

Starting number: 2000
Each year, that number has DOUBLED, so at the end of the first year (re: N = 1), the total will be 4000.

Consider 2^(n-1) in this situation. When n = 1....
2^(1-1) = 2^0 = 1..... (2000)(1) = 2000.... but we need it to be 4000. Thus, 2^(n-1) is not the proper notation.

With 2^n, when n = 1...
2^(1) = 2..... (2000)(2) = 4000.... This is a match for what the total is supposed to be.

GMAT assassins aren't born, they're made,
Rich
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jlgdr
The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?
Solution:

We can just tabulate the population of grasshoppers as follows:

Year 0: 2,000

Year 1: 4,000

Year 2: 8,000

Year 3: 16,000

Year 4: 32,000

Year 5: 64,000

Year 6: 128,000

Year 7: 256,000

Year 8: 512,000

Year 9: 1,024,000

We see that it takes 9 years for the population of 2,000 grasshoppers to grow to 1,000,000 or more grasshoppers.

Answer: 9
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The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?

My Approach was:

Geometric progression where:
a0 = 2 * 1,000
an = 2^(n+1) *1,000

Then
an >= 1,000,000
2^(n+1) * 1,000 >= 1,000,000
2^(n+1) >= 1,000 = 2^3 * 5^3
2^(n-2) >= 5^3 = 125

We know that 2^7=128>125 => 2^(n-2) = 7 => n=9
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jlgdr
The population of grasshoppers doubles in a particular field every year. Approximately how many years will it take the population to grow from 2,000 grasshoppers to 1,000,000 or more?

Let initial number be x ie x = 2000 and we want to know when grasshoppers will be more than 10^6 ie more than 10^6/2000 = 500x

If we try to identify pattern, at end of each year it would be,

End of yearNumber
1x*\(2\)
2x*\(2^2\)
..
..
9x*\(2^9\) = 512x

Ans: 9 years
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