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The population of Linterhast was 3,600 people in 1990

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by Brent@GMATPrepNow » Thu Apr 04, 2013 7:19 am
guerrero wrote:The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?

(A) 6,000
(B) 6,400
(C) 7,200
(D) 8,000
(E) 9,600

OA B
In the 3 years between 1990 and 1993, the population increased by 1200 people.
Since the original population was 3600, we can say the population increased by 1/3 (since 1200/3600 = 1/3)

If the growth rate is constant, then in the 3 years between 1993 and 1996, the population will once again increase by 1/3

1/3 of 4800 = 1600
So, the 1996 population will be 4800 + 1600 = [spoiler]6400 = B[/spoiler]

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by Anju@Gurome » Thu Apr 04, 2013 7:27 am
guerrero wrote:The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?
Population growth rate per thousand = [(4800 - 3600)/3600]*1000 in 3 years = 1000/3 in 3 years

Hence, from 1993 to 1996, in 3 years population will increase by (4800/1000)*(1000/3) = 4800/3 = 1600

So, final population in 1996 = (4800 + 1600) = 6400

The correct answer is B.
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by srcc25anu » Thu Apr 04, 2013 11:33 am
Pop in 1990 = 3600
Pop in 1993 = 4800
Let R be the CONSTANT growth rate each year.

therefore 4800 = 3600 * (1+R)^3
=> (1+R)^3 = 4800 / 3600 = 4/3

Now we have the pop of 1993 as 4800
we have to find the population after 3 years in 1996

Pop in 1996 = 4800 * (1+R)^3
we know (1+R)^3 = 4/3

hence Pop in 1996 = 4800 * 4/3 = 6400
answer B
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